\(\int (a+b \sec (c+d x))^{3/2} (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)) \, dx\) [975]

Optimal result

Integrand size = 50, antiderivative size = 448 \[ \int (a+b \sec (c+d x))^{3/2} \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (35 a b B-12 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 \sqrt {a+b} \left (a b^2 (35 B-12 C)-b^3 (5 B-9 C)+30 a^3 C-3 a^2 b (15 B+4 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 d}-\frac {2 a^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 b^2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-2/15*(a-b)*(a+b)^(1/2)*(35*B*a*b-12*C*a^2+9*C*b^2)*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)/d-2/15*(a+b)^(1/2)*(a*b^2*(35* 
B-12*C)-b^3*(5*B-9*C)+30*a^3*C-3*a^2*b*(15*B+4*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)/d-2*a^2*(a+b)^(1/2)*(B*b-C*a)*c 
ot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((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)/ 
d+2/15*b^2*(5*B*b+3*C*a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/5*b^2*C*(a+ 
b*sec(d*x+c))^(3/2)*tan(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 18.76 (sec) , antiderivative size = 689, normalized size of antiderivative = 1.54 \[ \int (a+b \sec (c+d x))^{3/2} \left (a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} (b B-a C+b C \sec (c+d x)) \left (-b (a+b) \left (35 a b B-12 a^2 C+9 b^2 C\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) \left (-15 a^2 C+3 a b (10 B+C)+b^2 (5 B+9 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-15 a^2 (b B-a C) \left ((a-b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-2 a \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-b \left (35 a b B-12 a^2 C+9 b^2 C\right ) (b+a \cos (c+d x)) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 d (b+a \cos (c+d x))^3 (b C+b B \cos (c+d x)-a C \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sec ^{\frac {7}{2}}(c+d x)}+\frac {\cos ^3(c+d x) (a+b \sec (c+d x))^{5/2} (b B-a C+b C \sec (c+d x)) \left (\frac {2}{15} b \left (35 a b B-12 a^2 C+9 b^2 C\right ) \sin (c+d x)+\frac {2}{15} \sec (c+d x) \left (5 b^3 B \sin (c+d x)+6 a b^2 C \sin (c+d x)\right )+\frac {2}{5} b^3 C \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2 (b C+b B \cos (c+d x)-a C \cos (c+d x))} \] Input:

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

Output:

(2*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d 
*x]]*(a + b*Sec[c + d*x])^(5/2)*(b*B - a*C + b*C*Sec[c + d*x])*(-(b*(a + b 
)*(35*a*b*B - 12*a^2*C + 9*b^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - 
 b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2 
]^2)/(a + b)]) + b*(a + b)*(-15*a^2*C + 3*a*b*(10*B + C) + b^2*(5*B + 9*C) 
)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sec[(c + d*x)/2]^2* 
Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - 15*a^2*(b*B - a* 
C)*((a - b)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*a*Ell 
ipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*Sec[(c + d*x)/2]^2 
*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - b*(35*a*b*B - 1 
2*a^2*C + 9*b^2*C)*(b + a*Cos[c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^ 
(3/2)*Sec[c + d*x]*Tan[(c + d*x)/2]))/(15*d*(b + a*Cos[c + d*x])^3*(b*C + 
b*B*Cos[c + d*x] - a*C*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d* 
x]^(7/2)) + (Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(5/2)*(b*B - a*C + b*C*Se 
c[c + d*x])*((2*b*(35*a*b*B - 12*a^2*C + 9*b^2*C)*Sin[c + d*x])/15 + (2*Se 
c[c + d*x]*(5*b^3*B*Sin[c + d*x] + 6*a*b^2*C*Sin[c + d*x]))/15 + (2*b^3*C* 
Sec[c + d*x]*Tan[c + d*x])/5))/(d*(b + a*Cos[c + d*x])^2*(b*C + b*B*Cos[c 
+ d*x] - a*C*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 2.02 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4529, 3042, 4406, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4406

\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left ((5 b B+3 a C) \sec ^2(c+d x) b^4+\left (-5 C a^2+10 b B a+3 b^2 C\right ) \sec (c+d x) b^3+5 a^2 (b B-a C) b^2\right )dx+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left ((5 b B+3 a C) \sec ^2(c+d x) b^4+\left (-5 C a^2+10 b B a+3 b^2 C\right ) \sec (c+d x) b^3+5 a^2 (b B-a C) b^2\right )dx+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left ((5 b B+3 a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+\left (-5 C a^2+10 b B a+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+5 a^2 (b B-a C) b^2\right )dx+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 4544

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\left (-12 C a^2+35 b B a+9 b^2 C\right ) \sec ^2(c+d x) b^4+\left (-30 C a^3+45 b B a^2+12 b^2 C a+5 b^3 B\right ) \sec (c+d x) b^3+15 a^3 (b B-a C) b^2}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\left (-12 C a^2+35 b B a+9 b^2 C\right ) \sec ^2(c+d x) b^4+\left (-30 C a^3+45 b B a^2+12 b^2 C a+5 b^3 B\right ) \sec (c+d x) b^3+15 a^3 (b B-a C) b^2}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\left (-12 C a^2+35 b B a+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 b^4+\left (-30 C a^3+45 b B a^2+12 b^2 C a+5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+15 a^3 (b B-a C) b^2}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (b^4 \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {15 b^2 (b B-a C) a^3+\left (b^3 \left (-30 C a^3+45 b B a^2+12 b^2 C a+5 b^3 B\right )-b^4 \left (-12 C a^2+35 b B a+9 b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (b^4 \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {15 b^2 (b B-a C) a^3+\left (b^3 \left (-30 C a^3+45 b B a^2+12 b^2 C a+5 b^3 B\right )-b^4 \left (-12 C a^2+35 b B a+9 b^2 C\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 a^3 b^2 (b B-a C) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b^4 \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^3 \left (30 a^3 C-3 a^2 b (15 B+4 C)+a b^2 (35 B-12 C)-b^3 (5 B-9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (15 a^3 b^2 (b B-a C) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b^4 \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^3 \left (30 a^3 C-3 a^2 b (15 B+4 C)+a b^2 (35 B-12 C)-b^3 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (b^4 \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b^3 \left (30 a^3 C-3 a^2 b (15 B+4 C)+a b^2 (35 B-12 C)-b^3 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a^2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (b^4 \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a^2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 b^2 \sqrt {a+b} \left (30 a^3 C-3 a^2 b (15 B+4 C)+a b^2 (35 B-12 C)-b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 b^2 (a-b) \sqrt {a+b} \left (-12 a^2 C+35 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {30 a^2 b^2 \sqrt {a+b} (b B-a C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 b^2 \sqrt {a+b} \left (30 a^3 C-3 a^2 b (15 B+4 C)+a b^2 (35 B-12 C)-b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 b^4 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 b^4 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{b^2}\)

Input:

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

Output:

((2*b^4*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*b 
^2*Sqrt[a + b]*(35*a*b*B - 12*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[ArcS 
in[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Se 
c[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*b^2* 
Sqrt[a + b]*(a*b^2*(35*B - 12*C) - b^3*(5*B - 9*C) + 30*a^3*C - 3*a^2*b*(1 
5*B + 4*C))*Cot[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))])/d - (30*a^2*b^2*Sqrt[a + b]*(b*B - a*C)*Cot[c 
+ d*x]*EllipticPi[(a + b)/a, 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))])/d)/3 + (2*b^4*(5*b*B + 3*a*C)*Sqrt[a + b*Sec[c + d*x 
]]*Tan[c + d*x])/(3*d))/5)/b^2
 

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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4406

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

rule 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 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 4529

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

rule 4544

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

rule 4546

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1954\) vs. \(2(409)=818\).

Time = 69.45 (sec) , antiderivative size = 1955, normalized size of antiderivative = 4.36

method	result	size
default	\(\text {Expression too large to display}\)	\(1955\)
parts	\(\text {Expression too large to display}\)	\(2100\)

Input:

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

Output:

2/15/d*(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) 
*(30*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(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^3*b*EllipticPi(-csc(d*x+c) 
+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+30*(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)*a^4*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+35*(cos 
(d*x+c)^2+2*cos(d*x+c)+1)*B*(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^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+ 
c),((a-b)/(a+b))^(1/2))+35*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^3*E 
llipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+12*(-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^3*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b) 
)^(1/2))+12*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(co 
s(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*EllipticE(-cs 
c(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(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*x+c))/(cos(d*x+c)+1 
))^(1/2)*a*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(co 
s(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^4*EllipticE(-csc(d*x+c)+cot(d*x+...
 

Fricas [F]

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

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

Output:

integral((C*b^3*sec(d*x + c)^3 - C*a^3 + B*a^2*b + (C*a*b^2 + B*b^3)*sec(d 
*x + c)^2 - (C*a^2*b - 2*B*a*b^2)*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), 
x)
 

Sympy [F]

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

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

Output:

-Integral(C*a**3*sqrt(a + b*sec(c + d*x)), x) - Integral(-B*a**2*b*sqrt(a 
+ b*sec(c + d*x)), x) - Integral(-B*b**3*sqrt(a + b*sec(c + d*x))*sec(c + 
d*x)**2, x) - Integral(-C*b**3*sqrt(a + b*sec(c + d*x))*sec(c + d*x)**3, x 
) - Integral(-2*B*a*b**2*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x) - Integ 
ral(-C*a*b**2*sqrt(a + b*sec(c + d*x))*sec(c + d*x)**2, x) - Integral(C*a* 
*2*b*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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