\(\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [358]

Optimal result

Integrand size = 31, antiderivative size = 312 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (20 a A b+3 a^2 B+9 b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (15 a A-5 A b-3 a B+9 b B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {2 (5 A b+3 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-2/15*(a-b)*(a+b)^(1/2)*(20*A*a*b+3*B*a^2+9*B*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)/b^2/d+2/15*(a-b)*(a+b)^(1/2)*(1 
5*A*a-5*A*b-3*B*a+9*B*b)*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+2/15*(5*A*b+3*B*a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/ 
d+2/5*B*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 14.73 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.61 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \left (2 (a+b) \left (20 a A b+3 a^2 B+9 b^2 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) (3 a (5 A+B)+b (5 A+9 B)) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (20 a A b+3 a^2 B+9 b^2 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b d (b+a \cos (c+d x))^2 (B+A \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \left (\frac {2 \left (20 a A b+3 a^2 B+9 b^2 B\right ) \sin (c+d x)}{15 b}+\frac {2}{15} \sec (c+d x) (5 A b \sin (c+d x)+6 a B \sin (c+d x))+\frac {2}{5} b B \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x)) (B+A \cos (c+d x))} \] Input:

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

Output:

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(A + 
B*Sec[c + d*x])*(2*(a + b)*(20*a*A*b + 3*a^2*B + 9*b^2*B)*Sqrt[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)*(3* 
a*(5*A + B) + b*(5*A + 9*B))*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)] + (20*a*A*b + 3*a^2*B + 9*b^2*B)*Cos[c + d*x]* 
(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15*b*d*(b + a* 
Cos[c + d*x])^2*(B + A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c + d*x] 
^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x])* 
((2*(20*a*A*b + 3*a^2*B + 9*b^2*B)*Sin[c + d*x])/(15*b) + (2*Sec[c + d*x]* 
(5*A*b*Sin[c + d*x] + 6*a*B*Sin[c + d*x]))/15 + (2*b*B*Sec[c + d*x]*Tan[c 
+ d*x])/5))/(d*(b + a*Cos[c + d*x])*(B + A*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (15 A a^2+12 b B a+5 A b^2+\left (3 B a^2+20 A b a+9 b^2 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (15 A a^2+12 b B a+5 A b^2+\left (3 B a^2+20 A b a+9 b^2 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (15 A a^2+12 b B a+5 A b^2+\left (3 B a^2+20 A b a+9 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 B+20 a A b+9 b^2 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+(a-b) (15 a A-3 a B-5 A b+9 b B) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 B+20 a A b+9 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) (15 a A-3 a B-5 A b+9 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 B+20 a A b+9 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} (15 a A-3 a B-5 A b+9 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} (15 a A-3 a B-5 A b+9 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^2 B+20 a A b+9 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {2 (3 a B+5 A b) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

Input:

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

Output:

(2*B*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a 
 + b]*(20*a*A*b + 3*a^2*B + 9*b^2*B)*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)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)* 
Sqrt[a + b]*(15*a*A - 5*A*b - 3*a*B + 9*b*B)*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))])/(b*d))/3 + (2 
*(5*A*b + 3*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1434\) vs. \(2(282)=564\).

Time = 34.06 (sec) , antiderivative size = 1435, normalized size of antiderivative = 4.60

method	result	size
default	\(\text {Expression too large to display}\)	\(1435\)
parts	\(\text {Expression too large to display}\)	\(1452\)

Input:

int(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 

Output:

2/15/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)*(20*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*( 
1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE(-csc(d*x+c) 
+cot(d*x+c),((a-b)/(a+b))^(1/2))+20*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(cos(d 
*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2 
)*a*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)*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos 
(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^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)*B*(1/(a+b)*(b+a*cos(d*x+c))/ 
(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE(-c 
sc(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)* 
B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c 
)))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(c 
os(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*( 
b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c) 
,((a-b)/(a+b))^(1/2))+15*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(cos(d*x+c)/(1+c 
os(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*El 
lipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+20*(-cos(d*x+c)^2-2*co 
s(d*x+c)-1)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/ 
(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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