\(\int \frac {(a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1311]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F(-1)]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 43, antiderivative size = 357 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (21 a^2 b B+5 b^3 B+7 a^3 (3 A+C)+3 a b^2 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b \left (63 A b^2+99 a b B+24 a^2 C+49 b^2 C\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (54 a^2 b B+15 b^3 B+8 a^3 C+9 a b^2 (7 A+5 C)\right ) \sin (c+d x)}{63 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^3 B+27 a b^2 B+9 a^2 b (5 A+3 C)+b^3 (9 A+7 C)\right ) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 (3 b B+2 a C) (b+a \cos (c+d x))^2 \sin (c+d x)}{21 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 C (b+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)} \] Output:

-2/15*(15*B*a^3+27*B*a*b^2+9*a^2*b*(5*A+3*C)+b^3*(9*A+7*C))*EllipticE(sin( 
1/2*d*x+1/2*c),2^(1/2))/d+2/21*(21*B*a^2*b+5*B*b^3+7*a^3*(3*A+C)+3*a*b^2*( 
7*A+5*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/315*b*(63*A*b^2+99*B* 
a*b+24*C*a^2+49*C*b^2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/63*(54*B*a^2*b+15*B 
*b^3+8*a^3*C+9*a*b^2*(7*A+5*C))*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/15*(15*B*a 
^3+27*B*a*b^2+9*a^2*b*(5*A+3*C)+b^3*(9*A+7*C))*sin(d*x+c)/d/cos(d*x+c)^(1/ 
2)+2/21*(3*B*b+2*C*a)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/9 
*C*(b+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(9/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 11.98 (sec) , antiderivative size = 3345, normalized size of antiderivative = 9.37 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Result too large to show} \] Input:

Integrate[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)) 
/Sqrt[Cos[c + d*x]],x]
 

Output:

(Cos[c + d*x]^(11/2)*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c 
+ d*x]^2)*((4*(45*a^2*A*b + 9*A*b^3 + 15*a^3*B + 27*a*b^2*B + 27*a^2*b*C + 
 7*b^3*C)*Csc[c]*Sec[c])/(15*d) + (4*b^3*C*Sec[c]*Sec[c + d*x]^5*Sin[d*x]) 
/(9*d) + (4*Sec[c]*Sec[c + d*x]^4*(7*b^3*C*Sin[c] + 9*b^3*B*Sin[d*x] + 27* 
a*b^2*C*Sin[d*x]))/(63*d) + (4*Sec[c]*Sec[c + d*x]^2*(63*A*b^3*Sin[c] + 18 
9*a*b^2*B*Sin[c] + 189*a^2*b*C*Sin[c] + 49*b^3*C*Sin[c] + 315*a*A*b^2*Sin[ 
d*x] + 315*a^2*b*B*Sin[d*x] + 75*b^3*B*Sin[d*x] + 105*a^3*C*Sin[d*x] + 225 
*a*b^2*C*Sin[d*x]))/(315*d) + (4*Sec[c]*Sec[c + d*x]^3*(45*b^3*B*Sin[c] + 
135*a*b^2*C*Sin[c] + 63*A*b^3*Sin[d*x] + 189*a*b^2*B*Sin[d*x] + 189*a^2*b* 
C*Sin[d*x] + 49*b^3*C*Sin[d*x]))/(315*d) + (4*Sec[c]*Sec[c + d*x]*(105*a*A 
*b^2*Sin[c] + 105*a^2*b*B*Sin[c] + 25*b^3*B*Sin[c] + 35*a^3*C*Sin[c] + 75* 
a*b^2*C*Sin[c] + 315*a^2*A*b*Sin[d*x] + 63*A*b^3*Sin[d*x] + 105*a^3*B*Sin[ 
d*x] + 189*a*b^2*B*Sin[d*x] + 189*a^2*b*C*Sin[d*x] + 49*b^3*C*Sin[d*x]))/( 
105*d)))/((b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 
 2*d*x])) - (4*a^3*A*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, { 
5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d 
*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTa 
n[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*S 
qrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2 
*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*a*A*b^2*...
 

Rubi [A] (verified)

Time = 2.16 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.93, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.488, Rules used = {3042, 4600, 3042, 3526, 27, 3042, 3526, 27, 3042, 3510, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\sqrt {\cos (c+d x)}}dx\)

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {(a \cos (c+d x)+b)^3 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )}{\cos ^{\frac {11}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^3 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\)

\(\Big \downarrow \) 3526

\(\displaystyle \frac {2}{9} \int \frac {(b+a \cos (c+d x))^2 \left (a (9 A+C) \cos ^2(c+d x)+(9 A b+7 C b+9 a B) \cos (c+d x)+3 (3 b B+2 a C)\right )}{2 \cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \int \frac {(b+a \cos (c+d x))^2 \left (a (9 A+C) \cos ^2(c+d x)+(9 A b+7 C b+9 a B) \cos (c+d x)+3 (3 b B+2 a C)\right )}{\cos ^{\frac {9}{2}}(c+d x)}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (a (9 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(9 A b+7 C b+9 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 (3 b B+2 a C)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3526

\(\displaystyle \frac {1}{9} \left (\frac {2}{7} \int \frac {(b+a \cos (c+d x)) \left (24 C a^2+(63 a A+9 b B+13 a C) \cos ^2(c+d x) a+99 b B a+63 A b^2+49 b^2 C+\left (63 B a^2+126 A b a+86 b C a+45 b^2 B\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \int \frac {(b+a \cos (c+d x)) \left (24 C a^2+(63 a A+9 b B+13 a C) \cos ^2(c+d x) a+99 b B a+63 A b^2+49 b^2 C+\left (63 B a^2+126 A b a+86 b C a+45 b^2 B\right ) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \int \frac {\left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (24 C a^2+(63 a A+9 b B+13 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+99 b B a+63 A b^2+49 b^2 C+\left (63 B a^2+126 A b a+86 b C a+45 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3510

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int -\frac {5 a^2 (63 a A+9 b B+13 a C) \cos ^2(c+d x)+21 \left (15 B a^3+9 b (5 A+3 C) a^2+27 b^2 B a+b^3 (9 A+7 C)\right ) \cos (c+d x)+15 \left (8 C a^3+54 b B a^2+9 b^2 (7 A+5 C) a+15 b^3 B\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \int \frac {5 a^2 (63 a A+9 b B+13 a C) \cos ^2(c+d x)+21 \left (15 B a^3+9 b (5 A+3 C) a^2+27 b^2 B a+b^3 (9 A+7 C)\right ) \cos (c+d x)+15 \left (8 C a^3+54 b B a^2+9 b^2 (7 A+5 C) a+15 b^3 B\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \int \frac {5 a^2 (63 a A+9 b B+13 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+21 \left (15 B a^3+9 b (5 A+3 C) a^2+27 b^2 B a+b^3 (9 A+7 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+15 \left (8 C a^3+54 b B a^2+9 b^2 (7 A+5 C) a+15 b^3 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {9 \left (7 \left (15 B a^3+9 b (5 A+3 C) a^2+27 b^2 B a+b^3 (9 A+7 C)\right )+5 \left (7 (3 A+C) a^3+21 b B a^2+3 b^2 (7 A+5 C) a+5 b^3 B\right ) \cos (c+d x)\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \int \frac {7 \left (15 B a^3+9 b (5 A+3 C) a^2+27 b^2 B a+b^3 (9 A+7 C)\right )+5 \left (7 (3 A+C) a^3+21 b B a^2+3 b^2 (7 A+5 C) a+5 b^3 B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \int \frac {7 \left (15 B a^3+9 b (5 A+3 C) a^2+27 b^2 B a+b^3 (9 A+7 C)\right )+5 \left (7 (3 A+C) a^3+21 b B a^2+3 b^2 (7 A+5 C) a+5 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \left (7 \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \left (7 \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \left (5 \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+7 \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \left (5 \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+7 \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {1}{5} \left (3 \left (5 \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+7 \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {2 b \sin (c+d x) \left (24 a^2 C+99 a b B+63 A b^2+49 b^2 C\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {10 \sin (c+d x) \left (8 a^3 C+54 a^2 b B+9 a b^2 (7 A+5 C)+15 b^3 B\right )}{d \cos ^{\frac {3}{2}}(c+d x)}+3 \left (\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (7 a^3 (3 A+C)+21 a^2 b B+3 a b^2 (7 A+5 C)+5 b^3 B\right )}{d}+7 \left (15 a^3 B+9 a^2 b (5 A+3 C)+27 a b^2 B+b^3 (9 A+7 C)\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )\right )+\frac {6 (2 a C+3 b B) \sin (c+d x) (a \cos (c+d x)+b)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )+\frac {2 C \sin (c+d x) (a \cos (c+d x)+b)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

Input:

Int[((a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[ 
Cos[c + d*x]],x]
 

Output:

(2*C*(b + a*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + ((6*( 
3*b*B + 2*a*C)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2 
)) + ((2*b*(63*A*b^2 + 99*a*b*B + 24*a^2*C + 49*b^2*C)*Sin[c + d*x])/(5*d* 
Cos[c + d*x]^(5/2)) + ((10*(54*a^2*b*B + 15*b^3*B + 8*a^3*C + 9*a*b^2*(7*A 
 + 5*C))*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)) + 3*((10*(21*a^2*b*B + 5*b^3 
*B + 7*a^3*(3*A + C) + 3*a*b^2*(7*A + 5*C))*EllipticF[(c + d*x)/2, 2])/d + 
 7*(15*a^3*B + 27*a*b^2*B + 9*a^2*b*(5*A + 3*C) + b^3*(9*A + 7*C))*((-2*El 
lipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))))/5) 
/7)/9
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3500
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3510
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 3526
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d 
, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4600
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1264\) vs. \(2(336)=672\).

Time = 10.67 (sec) , antiderivative size = 1265, normalized size of antiderivative = 3.54

method result size
default \(\text {Expression too large to display}\) \(1265\)

Input:

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x, 
method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^3*A*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 
2*C*b^3*(-1/144*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/ 
2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)*(-2*si 
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^ 
3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2 
+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos( 
1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 
1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d 
*x+1/2*c),2^(1/2))))+2*a*(3*A*b^2+3*B*a*b+C*a^2)*(-1/6*cos(1/2*d*x+1/2*c)* 
(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2 
-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2) 
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d* 
x+1/2*c),2^(1/2)))+2*a^2*(3*A*b+B*a)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1 
/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2 
*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic 
E(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2))+2/5*b*(A*b^...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (7 i \, {\left (3 \, A + C\right )} a^{3} + 21 i \, B a^{2} b + 3 i \, {\left (7 \, A + 5 \, C\right )} a b^{2} + 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, \sqrt {2} {\left (-7 i \, {\left (3 \, A + C\right )} a^{3} - 21 i \, B a^{2} b - 3 i \, {\left (7 \, A + 5 \, C\right )} a b^{2} - 5 i \, B b^{3}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (15 i \, B a^{3} + 9 i \, {\left (5 \, A + 3 \, C\right )} a^{2} b + 27 i \, B a b^{2} + i \, {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-15 i \, B a^{3} - 9 i \, {\left (5 \, A + 3 \, C\right )} a^{2} b - 27 i \, B a b^{2} - i \, {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (21 \, {\left (15 \, B a^{3} + 9 \, {\left (5 \, A + 3 \, C\right )} a^{2} b + 27 \, B a b^{2} + {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 35 \, C b^{3} + 15 \, {\left (7 \, C a^{3} + 21 \, B a^{2} b + 3 \, {\left (7 \, A + 5 \, C\right )} a b^{2} + 5 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 7 \, {\left (27 \, C a^{2} b + 27 \, B a b^{2} + {\left (9 \, A + 7 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} + 45 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, d \cos \left (d x + c\right )^{5}} \] Input:

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1 
/2),x, algorithm="fricas")
 

Output:

-1/315*(15*sqrt(2)*(7*I*(3*A + C)*a^3 + 21*I*B*a^2*b + 3*I*(7*A + 5*C)*a*b 
^2 + 5*I*B*b^3)*cos(d*x + c)^5*weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c)) + 15*sqrt(2)*(-7*I*(3*A + C)*a^3 - 21*I*B*a^2*b - 3*I*(7*A 
+ 5*C)*a*b^2 - 5*I*B*b^3)*cos(d*x + c)^5*weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(15*I*B*a^3 + 9*I*(5*A + 3*C)*a^2*b 
+ 27*I*B*a*b^2 + I*(9*A + 7*C)*b^3)*cos(d*x + c)^5*weierstrassZeta(-4, 0, 
weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(- 
15*I*B*a^3 - 9*I*(5*A + 3*C)*a^2*b - 27*I*B*a*b^2 - I*(9*A + 7*C)*b^3)*cos 
(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) 
 - I*sin(d*x + c))) - 2*(21*(15*B*a^3 + 9*(5*A + 3*C)*a^2*b + 27*B*a*b^2 + 
 (9*A + 7*C)*b^3)*cos(d*x + c)^4 + 35*C*b^3 + 15*(7*C*a^3 + 21*B*a^2*b + 3 
*(7*A + 5*C)*a*b^2 + 5*B*b^3)*cos(d*x + c)^3 + 7*(27*C*a^2*b + 27*B*a*b^2 
+ (9*A + 7*C)*b^3)*cos(d*x + c)^2 + 45*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*s 
qrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
 

Sympy [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{3} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sqrt {\cos {\left (c + d x \right )}}}\, dx \] Input:

integrate((a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)* 
*(1/2),x)
 

Output:

Integral((a + b*sec(c + d*x))**3*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/ 
sqrt(cos(c + d*x)), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1 
/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1 
/2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^3/s 
qrt(cos(d*x + c)), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 18.22 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

int(((a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c 
 + d*x)^(1/2),x)
 

Output:

((2*B*b^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/7 + 2 
*B*a^3*cos(c + d*x)^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x 
)^2) + 2*B*a^2*b*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, c 
os(c + d*x)^2) + (6*B*a*b^2*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2 
], -1/4, cos(c + d*x)^2))/5)/(d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1 
/2)) + (70*C*b^3*sin(c + d*x)*hypergeom([-9/4, 1/2], -5/4, cos(c + d*x)^2) 
 + 210*C*a^3*cos(c + d*x)^3*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c 
 + d*x)^2) + 378*C*a^2*b*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-5/4, 1/2] 
, -1/4, cos(c + d*x)^2) + 270*C*a*b^2*cos(c + d*x)*sin(c + d*x)*hypergeom( 
[-7/4, 1/2], -3/4, cos(c + d*x)^2))/(315*d*cos(c + d*x)^(9/2)*(1 - cos(c + 
 d*x)^2)^(1/2)) + (2*A*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*b^3*sin(c 
 + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5 
/2)*(sin(c + d*x)^2)^(1/2)) + (6*A*a^2*b*sin(c + d*x)*hypergeom([-1/4, 1/2 
], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + ( 
2*A*a*b^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos 
(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \frac {(a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) a^{4}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}d x \right ) b^{3} c +3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}d x \right ) a \,b^{2} c +\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}d x \right ) b^{4}+3 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )}d x \right ) a^{2} b c +4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}}{\cos \left (d x +c \right )}d x \right ) a \,b^{3}+\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )}d x \right ) a^{3} c +6 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )}d x \right ) a^{2} b^{2}+4 \left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )}d x \right ) a^{3} b \] Input:

int((a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2),x)
 

Output:

int(sqrt(cos(c + d*x))/cos(c + d*x),x)*a**4 + int((sqrt(cos(c + d*x))*sec( 
c + d*x)**5)/cos(c + d*x),x)*b**3*c + 3*int((sqrt(cos(c + d*x))*sec(c + d* 
x)**4)/cos(c + d*x),x)*a*b**2*c + int((sqrt(cos(c + d*x))*sec(c + d*x)**4) 
/cos(c + d*x),x)*b**4 + 3*int((sqrt(cos(c + d*x))*sec(c + d*x)**3)/cos(c + 
 d*x),x)*a**2*b*c + 4*int((sqrt(cos(c + d*x))*sec(c + d*x)**3)/cos(c + d*x 
),x)*a*b**3 + int((sqrt(cos(c + d*x))*sec(c + d*x)**2)/cos(c + d*x),x)*a** 
3*c + 6*int((sqrt(cos(c + d*x))*sec(c + d*x)**2)/cos(c + d*x),x)*a**2*b**2 
 + 4*int((sqrt(cos(c + d*x))*sec(c + d*x))/cos(c + d*x),x)*a**3*b