\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [822]

Optimal result

Integrand size = 42, antiderivative size = 475 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (18 a^3 b B-246 a b^3 B-8 a^4 C-33 a^2 b^2 C-147 b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (3 b^3 (25 B-49 C)-3 a b^2 (57 B-13 C)-6 a^2 b (3 B-C)+8 a^3 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}-\frac {2 \left (18 a^2 b B-75 b^3 B-8 a^3 C-39 a b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}-\frac {2 \left (18 a b B-8 a^2 C-49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}+\frac {2 (9 b B-4 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d} \] Output:

2/315*(a-b)*(a+b)^(1/2)*(18*B*a^3*b-246*B*a*b^3-8*C*a^4-33*C*a^2*b^2-147*C 
*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b) 
)^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^ 
4/d-2/315*(a-b)*(a+b)^(1/2)*(3*b^3*(25*B-49*C)-3*a*b^2*(57*B-13*C)-6*a^2*b 
*(3*B-C)+8*a^3*C)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2), 
((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a- 
b))^(1/2)/b^3/d-2/315*(18*B*a^2*b-75*B*b^3-8*C*a^3-39*C*a*b^2)*(a+b*sec(d* 
x+c))^(1/2)*tan(d*x+c)/b^2/d-2/315*(18*B*a*b-8*C*a^2-49*C*b^2)*(a+b*sec(d* 
x+c))^(3/2)*tan(d*x+c)/b^2/d+2/63*(9*B*b-4*C*a)*(a+b*sec(d*x+c))^(5/2)*tan 
(d*x+c)/b^2/d+2/9*C*sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 19.45 (sec) , antiderivative size = 654, normalized size of antiderivative = 1.38 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 (a+b) \left (-18 a^3 b B+246 a b^3 B+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \left (8 a^3 C-6 a^2 b (3 B+C)+3 a b^2 (57 B+13 C)+3 b^3 (25 B+49 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-18 a^3 b B+246 a b^3 B+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^3 d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {2 \left (-18 a^3 b B+246 a b^3 B+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \sin (c+d x)}{315 b^3}+\frac {2}{63} \sec ^3(c+d x) (9 b B \sin (c+d x)+10 a C \sin (c+d x))+\frac {2 \sec ^2(c+d x) \left (72 a b B \sin (c+d x)+3 a^2 C \sin (c+d x)+49 b^2 C \sin (c+d x)\right )}{315 b}+\frac {2 \sec (c+d x) \left (9 a^2 b B \sin (c+d x)+75 b^3 B \sin (c+d x)-4 a^3 C \sin (c+d x)+88 a b^2 C \sin (c+d x)\right )}{315 b^2}+\frac {2}{9} b C \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \] Input:

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

Output:

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*(a 
 + b)*(-18*a^3*b*B + 246*a*b^3*B + 8*a^4*C + 33*a^2*b^2*C + 147*b^4*C)*Sqr 
t[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + 
 Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2* 
b*(a + b)*(8*a^3*C - 6*a^2*b*(3*B + C) + 3*a*b^2*(57*B + 13*C) + 3*b^3*(25 
*B + 49*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x] 
)/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b 
)/(a + b)] + (-18*a^3*b*B + 246*a*b^3*B + 8*a^4*C + 33*a^2*b^2*C + 147*b^4 
*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]) 
)/(315*b^3*d*(b + a*Cos[c + d*x])^2*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^ 
(3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*((2*(-18*a^3*b*B + 246*a 
*b^3*B + 8*a^4*C + 33*a^2*b^2*C + 147*b^4*C)*Sin[c + d*x])/(315*b^3) + (2* 
Sec[c + d*x]^3*(9*b*B*Sin[c + d*x] + 10*a*C*Sin[c + d*x]))/63 + (2*Sec[c + 
 d*x]^2*(72*a*b*B*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d 
*x]))/(315*b) + (2*Sec[c + d*x]*(9*a^2*b*B*Sin[c + d*x] + 75*b^3*B*Sin[c + 
 d*x] - 4*a^3*C*Sin[c + d*x] + 88*a*b^2*C*Sin[c + d*x]))/(315*b^2) + (2*b* 
C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 2.27 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.03, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4560, 3042, 4521, 27, 3042, 4570, 27, 3042, 4490, 27, 3042, 4490, 27, 3042, 4493, 3042, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} (B+C \sec (c+d x))dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4521

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left ((9 b B-4 a C) \sec ^2(c+d x)+7 b C \sec (c+d x)+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left ((9 b B-4 a C) \sec ^2(c+d x)+7 b C \sec (c+d x)+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left ((9 b B-4 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+7 b C \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 b B-2 a C)-\left (-8 C a^2+18 b B a-49 b^2 C\right ) \sec (c+d x)\right )dx}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (3 b (15 b B-2 a C)-\left (-8 C a^2+18 b B a-49 b^2 C\right ) \sec (c+d x)\right )dx}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (3 b (15 b B-2 a C)+\left (8 C a^2-18 b B a+49 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {\frac {2}{5} \int \frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 C a^2+57 b B a+49 b^2 C\right )-\left (-8 C a^3+18 b B a^2-39 b^2 C a-75 b^3 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 C a^2+57 b B a+49 b^2 C\right )-\left (-8 C a^3+18 b B a^2-39 b^2 C a-75 b^3 B\right ) \sec (c+d x)\right )dx-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-2 C a^2+57 b B a+49 b^2 C\right )+\left (8 C a^3-18 b B a^2+39 b^2 C a+75 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^3+153 b B a^2+186 b^2 C a+75 b^3 B\right )-\left (-8 C a^4+18 b B a^3-33 b^2 C a^2-246 b^3 B a-147 b^4 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^3+153 b B a^2+186 b^2 C a+75 b^3 B\right )-\left (-8 C a^4+18 b B a^3-33 b^2 C a^2-246 b^3 B a-147 b^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 C a^3+153 b B a^2+186 b^2 C a+75 b^3 B\right )+\left (8 C a^4-18 b B a^3+33 b^2 C a^2+246 b^3 B a+147 b^4 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left ((a-b) \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (-\left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {3}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (-8 a^4 C+18 a^3 b B-33 a^2 b^2 C-246 a b^3 B-147 b^4 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C-6 a^2 b (3 B-C)-3 a b^2 (57 B-13 C)+3 b^3 (25 B-49 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (-8 a^3 C+18 a^2 b B-39 a b^2 C-75 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (-8 a^2 C+18 a b B-49 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 (9 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

Input:

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

Output:

(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(9*b*d) + ((2*( 
9*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((-2*(18 
*a*b*B - 8*a^2*C - 49*b^2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d 
) + (3*(((2*(a - b)*Sqrt[a + b]*(18*a^3*b*B - 246*a*b^3*B - 8*a^4*C - 33*a 
^2*b^2*C - 147*b^4*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x 
]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqr 
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(3* 
b^3*(25*B - 49*C) - 3*a*b^2*(57*B - 13*C) - 6*a^2*b*(3*B - C) + 8*a^3*C)*C 
ot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b 
)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x 
]))/(a - b))])/(b*d))/3 - (2*(18*a^2*b*B - 75*b^3*B - 8*a^3*C - 39*a*b^2*C 
)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d)))/5)/(7*b))/(9*b)
 

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 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4490

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[Csc[e + f*x]* 
(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* 
B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4493

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B)   Int[Csc[e 
 + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B   Int[Csc[e + f*x]*((1 + 
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} 
, x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
 

rule 4521

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 
2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* 
(m + n))), x] + Simp[d^2/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e 
+ f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + 
 n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B 
, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + 
n, 0] &&  !IGtQ[m, 1]
 

rule 4560

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

rule 4570

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2419\) vs. \(2(437)=874\).

Time = 97.68 (sec) , antiderivative size = 2420, normalized size of antiderivative = 5.09

method	result	size
default	\(\text {Expression too large to display}\)	\(2420\)
parts	\(\text {Expression too large to display}\)	\(2436\)

Input:

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

Output:

-2/315/d/b^3*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x 
+c)+b)*(4*sin(d*x+c)*(cos(d*x+c)-1)*a^4*b*C+147*(-cos(d*x+c)^2-2*cos(d*x+c 
)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2)*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+1 
47*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 
))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+c)+cot(d 
*x+c),((a-b)/(a+b))^(1/2))+75*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5* 
EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(-cos(d*x+c)^2-2*c 
os(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)*a^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^ 
(1/2))+18*B*a^4*b*cos(d*x+c)*sin(d*x+c)+15*(-5*cos(d*x+c)^3-5*cos(d*x+c)^2 
-3*cos(d*x+c)-3)*B*b^5*tan(d*x+c)*sec(d*x+c)^2+7*(-21*cos(d*x+c)^4-7*cos(d 
*x+c)^3-7*cos(d*x+c)^2-5*cos(d*x+c)-5)*C*b^5*tan(d*x+c)*sec(d*x+c)^3+153*( 
cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)* 
(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^3*EllipticF(-csc(d*x+c)+cot(d 
*x+c),((a-b)/(a+b))^(1/2))+246*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c) 
/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b 
^4*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+2*(cos(d*x+c)^2+2 
*cos(d*x+c)+1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*...
 

Fricas [F]

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

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

Output:

integral((C*b*sec(d*x + c)^5 + B*a*sec(d*x + c)^3 + (C*a + B*b)*sec(d*x + 
c)^4)*sqrt(b*sec(d*x + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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