\(\int (a+b \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [720]

Optimal result

Integrand size = 27, antiderivative size = 415 \[ \int (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (a^2 C+b^2 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^2 d}-\frac {2 \sqrt {a+b} \left (a^2 C-2 a b (5 A+2 C)+b^2 (5 A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b d}-\frac {2 a A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 a C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 d}+\frac {2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-2/5*(a-b)*(a+b)^(1/2)*(C*a^2+b^2*(5*A+3*C))*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^2/d-2/5*(a+b)^(1/2)*(C*a^2-2*a*b*(5 
*A+2*C)+b^2*(5*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/d-2*a*A*(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*C*(a+b*sec(d*x+c))^(1/2)*tan 
(d*x+c)/d+2/5*C*(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. \(6117\) vs. \(2(415)=830\).

Time = 22.37 (sec) , antiderivative size = 6117, normalized size of antiderivative = 14.74 \[ \int (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.60 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 4545, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4545

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

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

Output:

(2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + ((-2*(a - b)*Sqrt[a 
+ b]*(a^2*C + b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Se 
c[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^2*d) + (2*Sqrt[a + b]*(1 
0*a*A*b - 5*A*b^2 - a^2*C + 4*a*b*C - 3*b^2*C)*Cot[c + d*x]*EllipticF[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))])/(b*d) - (10 
*a*A*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*C*Sqrt[a + b*Sec[c + 
 d*x]]*Tan[c + d*x])/d)/5
 

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 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 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 4545

Int[((A_.) + 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*(m + 1) + b*C*m)*Csc[e + f*x] + a*C*m*Csc[e + f*x]^2, x 
], x], x] /; FreeQ[{a, b, e, f, A, 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. \(1488\) vs. \(2(376)=752\).

Time = 51.64 (sec) , antiderivative size = 1489, normalized size of antiderivative = 3.59

method	result	size
parts	\(\text {Expression too large to display}\)	\(1489\)
default	\(\text {Expression too large to display}\)	\(1502\)

Input:

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

Output:

1/2*A/d/(b+a*cos(d*x+c))*(cos(d*x+c)+1)^2*((-(1-cos(d*x+c))^3*csc(d*x+c)^3 
-csc(d*x+c)+cot(d*x+c))*b^2+((1-cos(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot( 
d*x+c))*a*b-2*(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)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2) 
)*a^2+4*(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)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a*b+ 
2*(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)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*b^2-2*(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)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a*b-2*(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)*El 
lipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*b^2+4*(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)*Elliptic 
Pi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^2)*((1-cos(d*x+c))^2*c 
sc(d*x+c)^2-1)*(a+b*sec(d*x+c))^(1/2)*sec(d*x+c)+2/5*C/d/b*(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)*((cos(d*x+c)^2+2*cos( 
d*x+c)+1)*(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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2) 
)+(cos(d*x+c)^2+2*cos(d*x+c)+1)*(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*EllipticE(-csc(d*x+c)+cot...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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