\(\int (a+b \sec (c+d x))^{3/2} (a^2-b^2 \sec ^2(c+d x)) \, dx\) [733]

Optimal result

Integrand size = 32, antiderivative size = 403 \[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (4 a^2-3 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 d}+\frac {2 \sqrt {a+b} \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 d}-\frac {2 a^3 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {2 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 d}-\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-2/5*(a-b)*(a+b)^(1/2)*(4*a^2-3*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+2/5*(a+b)^(1/2)*(10*a^3-4*a^2*b-4*a*b^2+3*b 
^3)*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^3*(a+b)^(1/2)*cot(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-2/5*a*b^2*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d-2/5*b^2*( 
a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(956\) vs. \(2(403)=806\).

Time = 15.56 (sec) , antiderivative size = 956, normalized size of antiderivative = 2.37 \[ \int (a+b \sec (c+d x))^{3/2} \left (a^2-b^2 \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(-4*(a + b*Sec[c + d*x])^(3/2)*(a^2 - b^2*Sec[c + d*x]^2)*(-4*a^3*b*Tan[(c 
 + d*x)/2] - 4*a^2*b^2*Tan[(c + d*x)/2] + 3*a*b^3*Tan[(c + d*x)/2] + 3*b^4 
*Tan[(c + d*x)/2] + 8*a^3*b*Tan[(c + d*x)/2]^3 - 6*a*b^3*Tan[(c + d*x)/2]^ 
3 - 4*a^3*b*Tan[(c + d*x)/2]^5 + 4*a^2*b^2*Tan[(c + d*x)/2]^5 + 3*a*b^3*Ta 
n[(c + d*x)/2]^5 - 3*b^4*Tan[(c + d*x)/2]^5 + 10*a^4*EllipticPi[-1, ArcSin 
[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a 
+ b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 10*a^4*Ellip 
ticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sq 
rt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + 
 d*x)/2]^2)/(a + b)] + b*(-4*a^3 - 4*a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + 
 Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/ 
2]^2)/(a + b)] - (5*a^4 - 10*a^3*b - 4*a^2*b^2 + 4*a*b^3 + 3*b^4)*Elliptic 
F[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]* 
(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d 
*x)/2]^2)/(a + b)]))/(5*d*(b + a*Cos[c + d*x])^(3/2)*(a^2 - 2*b^2 + a^2*Co 
s[2*c + 2*d*x])*Sec[c + d*x]^(7/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(-1 
 + Tan[(c + d*x)/2]^2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[ 
(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (Cos[c 
 + d*x]^3*(a + b*Sec[c + d*x])^(3/2)*(a^2 - b^2*Sec[c + d*x]^2)*((-4*b*...
 

Rubi [A] (verified)

Time = 1.68 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.531, Rules used = {3042, 4530, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4530

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

\(\Big \downarrow \) 25

\(\displaystyle \int (a-b \sec (c+d x)) (a+b \sec (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4406

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{5} \left (b^2 \left (4 a^2-3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {10 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{5} \left (-\frac {10 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (4 a^2-3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {2 \sqrt {a+b} \left (10 a^3-4 a^2 b-4 a b^2+3 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\right )-\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

Input:

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

Output:

(-2*b^2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((-2*(a - b)*Sqrt 
[a + b]*(4*a^2 - 3*b^2)*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)]*S 
qrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d + (2*Sqrt[a + b]*(10*a^3 - 4*a^2 
*b - 4*a*b^2 + 3*b^3)*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)]*Sqr 
t[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (10*a^3*Sqrt[a + b]*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 - (2*a*b^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/d)/5
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4530

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_))^(m_.), x_Symbol] :> Simp[C/b^2   Int[(a + b*Csc[e + f*x])^(m + 1) 
*Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && 
 EqQ[A*b^2 + 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. \(1168\) vs. \(2(364)=728\).

Time = 53.09 (sec) , antiderivative size = 1169, normalized size of antiderivative = 2.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1169\)
parts	\(\text {Expression too large to display}\)	\(1488\)

Input:

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

Output:

2/5/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)* 
(10*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(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^4*EllipticPi(-csc(d*x+c)+cot( 
d*x+c),-1,((a-b)/(a+b))^(1/2))+4*(cos(d*x+c)^2+2*cos(d*x+c)+1)*(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^3 
*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+4*(cos(d*x+c)^2+2 
*cos(d*x+c)+1)*(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^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+ 
b))^(1/2))+3*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(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))+3*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(co 
s(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^4*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+5*(cos(d*x+ 
c)^2+2*cos(d*x+c)+1)*(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^4*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/( 
a+b))^(1/2))+10*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(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^3*b*EllipticF(-cs 
c(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+4*(-cos(d*x+c)^2-2*cos(d*x+c)-1)* 
(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^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+4...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a - b*sec(c + d*x))*(a + b*sec(c + d*x))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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