\(\int \frac {(a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [1105]

Optimal result

Integrand size = 35, antiderivative size = 246 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {4 a^3 (27 A+17 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (21 A+11 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {8 a^3 (21 A+16 C) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (27 A+17 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 C (a+a \cos (c+d x))^3 \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)}+\frac {4 C \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{21 a d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (63 A+73 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{315 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:

-4/15*a^3*(27*A+17*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+4/21*a^3*(21 
*A+11*C)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+8/105*a^3*(21*A+16*C)*si 
n(d*x+c)/d/cos(d*x+c)^(3/2)+4/15*a^3*(27*A+17*C)*sin(d*x+c)/d/cos(d*x+c)^( 
1/2)+2/9*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/d/cos(d*x+c)^(9/2)+4/21*C*(a^2+a^ 
2*cos(d*x+c))^2*sin(d*x+c)/a/d/cos(d*x+c)^(7/2)+2/315*(63*A+73*C)*(a^3+a^3 
*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(5/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 12.54 (sec) , antiderivative size = 1135, normalized size of antiderivative = 4.61 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c + d*x]^(11/2)*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Se 
c[c + d*x]^2)*(((27*A + 17*C)*Csc[c]*Sec[c])/(15*d) + (C*Sec[c]*Sec[c + d* 
x]^5*Sin[d*x])/(18*d) + (Sec[c]*Sec[c + d*x]^4*(7*C*Sin[c] + 27*C*Sin[d*x] 
))/(126*d) + (Sec[c]*Sec[c + d*x]^3*(135*C*Sin[c] + 63*A*Sin[d*x] + 238*C* 
Sin[d*x]))/(630*d) + (Sec[c]*Sec[c + d*x]*(105*A*Sin[c] + 110*C*Sin[c] + 3 
78*A*Sin[d*x] + 238*C*Sin[d*x]))/(210*d) + (Sec[c]*Sec[c + d*x]^2*(63*A*Si 
n[c] + 238*C*Sin[c] + 315*A*Sin[d*x] + 330*C*Sin[d*x]))/(630*d)))/(A + 2*C 
 + A*Cos[2*c + 2*d*x]) - (A*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ[{1/4, 
1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + a*Sec[ 
c + d*x])^3*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[ 
d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[C 
ot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2 
*d*x])*Sqrt[1 + Cot[c]^2]) - (11*C*Cos[c + d*x]^5*Csc[c]*HypergeometricPFQ 
[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*(a + 
 a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 
 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - A 
rcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*(A + 2*C + A*C 
os[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) + (9*A*Cos[c + d*x]^5*Csc[c]*Sec[c/2 
+ (d*x)/2]^6*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2)*((Hypergeometri 
cPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan...
 

Rubi [A] (verified)

Time = 1.73 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.657, Rules used = {3042, 4602, 3042, 3523, 27, 3042, 3454, 27, 3042, 3454, 27, 3042, 3447, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3523

\(\displaystyle \frac {2 \int \frac {(\cos (c+d x) a+a)^3 (6 a C+a (9 A+C) \cos (c+d x))}{2 \cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\cos (c+d x) a+a)^3 (6 a C+a (9 A+C) \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (6 a C+a (9 A+C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {2}{7} \int \frac {(\cos (c+d x) a+a)^2 \left ((63 A+73 C) a^2+(63 A+13 C) \cos (c+d x) a^2\right )}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \int \frac {(\cos (c+d x) a+a)^2 \left ((63 A+73 C) a^2+(63 A+13 C) \cos (c+d x) a^2\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((63 A+73 C) a^2+(63 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{7} \left (\frac {2}{5} \int \frac {3 (\cos (c+d x) a+a) \left (6 (21 A+16 C) a^3+(63 A+23 C) \cos (c+d x) a^3\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \int \frac {(\cos (c+d x) a+a) \left (6 (21 A+16 C) a^3+(63 A+23 C) \cos (c+d x) a^3\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (6 (21 A+16 C) a^3+(63 A+23 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3447

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \int \frac {(63 A+23 C) \cos ^2(c+d x) a^4+6 (21 A+16 C) a^4+\left (6 (21 A+16 C) a^4+(63 A+23 C) a^4\right ) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \int \frac {(63 A+23 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^4+6 (21 A+16 C) a^4+\left (6 (21 A+16 C) a^4+(63 A+23 C) a^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (\frac {2}{3} \int \frac {3 \left (7 (27 A+17 C) a^4+5 (21 A+11 C) \cos (c+d x) a^4\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (\int \frac {7 (27 A+17 C) a^4+5 (21 A+11 C) \cos (c+d x) a^4}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (\int \frac {7 (27 A+17 C) a^4+5 (21 A+11 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (7 a^4 (27 A+17 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 a^4 (21 A+11 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (7 a^4 (27 A+17 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 a^4 (21 A+11 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (5 a^4 (21 A+11 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+7 a^4 (27 A+17 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (5 a^4 (21 A+11 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+7 a^4 (27 A+17 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {1}{7} \left (\frac {6}{5} \left (5 a^4 (21 A+11 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+7 a^4 (27 A+17 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {1}{7} \left (\frac {2 (63 A+73 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {6}{5} \left (\frac {10 a^4 (21 A+11 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {4 a^4 (21 A+16 C) \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x)}+7 a^4 (27 A+17 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )\right )+\frac {12 C \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{7 d \cos ^{\frac {7}{2}}(c+d x)}}{9 a}+\frac {2 C \sin (c+d x) (a \cos (c+d x)+a)^3}{9 d \cos ^{\frac {9}{2}}(c+d x)}\)

Input:

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

Output:

(2*C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + ((12* 
C*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((2* 
(63*A + 73*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/ 
2)) + (6*((10*a^4*(21*A + 11*C)*EllipticF[(c + d*x)/2, 2])/d + (4*a^4*(21* 
A + 16*C)*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)) + 7*a^4*(27*A + 17*C)*((-2* 
EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))))/ 
5)/7)/(9*a)
 

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 3447

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

rule 3454

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

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 3523

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

rule 4602

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (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 + A*Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] 
 && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1218\) vs. \(2(225)=450\).

Time = 9.24 (sec) , antiderivative size = 1219, normalized size of antiderivative = 4.96

method	result	size
default	\(\text {Expression too large to display}\)	\(1219\)

Input:

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

Output:

-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(1/8*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))+3/8*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/ 
2))*(sin(1/2*d*x+1/2*c)^2)^(1/2))+1/8*C*(-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*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)^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*s 
in(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))))+3/8*C*(-1/56*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)^4-5/42*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+5/21*(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+s...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.13 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (21 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (21 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (27 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (27 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (42 \, {\left (27 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (21 \, A + 22 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 7 \, {\left (9 \, A + 34 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 135 \, C a^{3} \cos \left (d x + c\right ) + 35 \, C a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{315 \, d \cos \left (d x + c\right )^{5}} \] Input:

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

Output:

-2/315*(15*I*sqrt(2)*(21*A + 11*C)*a^3*cos(d*x + c)^5*weierstrassPInverse( 
-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqrt(2)*(21*A + 11*C)*a^3*cos 
(d*x + c)^5*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21 
*I*sqrt(2)*(27*A + 17*C)*a^3*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierst 
rassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*(27*A + 
 17*C)*a^3*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 
, cos(d*x + c) - I*sin(d*x + c))) - (42*(27*A + 17*C)*a^3*cos(d*x + c)^4 + 
 15*(21*A + 22*C)*a^3*cos(d*x + c)^3 + 7*(9*A + 34*C)*a^3*cos(d*x + c)^2 + 
 135*C*a^3*cos(d*x + c) + 35*C*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*co 
s(d*x + c)^5)
 

Sympy [F]

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

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

Output:

a**3*(Integral(A/sqrt(cos(c + d*x)), x) + Integral(3*A*sec(c + d*x)/sqrt(c 
os(c + d*x)), x) + Integral(3*A*sec(c + d*x)**2/sqrt(cos(c + d*x)), x) + I 
ntegral(A*sec(c + d*x)**3/sqrt(cos(c + d*x)), x) + Integral(C*sec(c + d*x) 
**2/sqrt(cos(c + d*x)), x) + Integral(3*C*sec(c + d*x)**3/sqrt(cos(c + d*x 
)), x) + Integral(3*C*sec(c + d*x)**4/sqrt(cos(c + d*x)), x) + Integral(C* 
sec(c + d*x)**5/sqrt(cos(c + d*x)), x))
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^3/sqrt(cos(d*x + c)) 
, x)
 

Mupad [B] (verification not implemented)

Time = 16.08 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.25 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,A\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {70\,C\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {9}{4},\frac {1}{2};\ -\frac {5}{4};\ {\cos \left (c+d\,x\right )}^2\right )+270\,C\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+210\,C\,a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+378\,C\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{315\,d\,{\cos \left (c+d\,x\right )}^{9/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {6\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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