\(\int \cos (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [360]

Optimal result

Integrand size = 31, antiderivative size = 361 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {(a-b) \sqrt {a+b} (a A-2 b B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}+\frac {\sqrt {a+b} (2 b (A-B)+a (A+4 B)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}-\frac {\sqrt {a+b} (3 A b+2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {a A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{d} \] Output:

(a-b)*(a+b)^(1/2)*(A*a-2*B*b)*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/d+(a+b)^(1/2)*(2*b*(A-B)+a*(A+4*B))*cot(d*x+c)*Ell 
ipticF((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-(a+b)^(1/2)*(3*A*b+2 
*B*a)*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+a*A*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(971\) vs. \(2(361)=722\).

Time = 15.66 (sec) , antiderivative size = 971, normalized size of antiderivative = 2.69 \[ \int \cos (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

(2*b*B*Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(d*(b + a*Cos 
[c + d*x])) + ((a + b*Sec[c + d*x])^(3/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(- 
1)]*(a^2*A*Tan[(c + d*x)/2] + a*A*b*Tan[(c + d*x)/2] - 2*a*b*B*Tan[(c + d* 
x)/2] - 2*b^2*B*Tan[(c + d*x)/2] - 2*a^2*A*Tan[(c + d*x)/2]^3 + 4*a*b*B*Ta 
n[(c + d*x)/2]^3 + a^2*A*Tan[(c + d*x)/2]^5 - a*A*b*Tan[(c + d*x)/2]^5 - 2 
*a*b*B*Tan[(c + d*x)/2]^5 + 2*b^2*B*Tan[(c + d*x)/2]^5 + 6*a*A*b*EllipticP 
i[-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)] + 
4*a^2*B*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)] + 6*a*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/ 
(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*T 
an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a^2*B*EllipticPi[-1 
, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - T 
an[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2] 
^2)/(a + b)] + (a + b)*(a*A - 2*b*B)*EllipticE[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)] - 2*(2*a*b 
*(A - B) + a^2*B - b^2*(A + B))*EllipticF[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[(...
 

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 368, 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, 4513, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4513

\(\displaystyle \frac {a A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}-\int -\frac {-b (a A-2 b B) \sec ^2(c+d x)+2 b (A b+2 a B) \sec (c+d x)+a (3 A b+2 a B)}{2 \sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{2} \left (\int \frac {a (3 A b+2 a B)+(2 b (A b+2 a B)+b (a A-2 b B)) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b (a A-2 b B) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {a A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{2} \left (-b (a A-2 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a (2 a B+3 A b) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+b (a (A+4 B)+2 b (A-B)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {a A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (a (2 a B+3 A b) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b (a (A+4 B)+2 b (A-B)) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a A-2 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {a A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\)

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{2} \left (\frac {2 \sqrt {a+b} (a (A+4 B)+2 b (A-B)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} (a A-2 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} (2 a B+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {a A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{d}\)

Input:

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

Output:

((2*(a - b)*Sqrt[a + b]*(a*A - 2*b*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*d) + (2*Sqrt[a + 
b]*(2*b*(A - B) + a*(A + 4*B))*Cot[c + d*x]*EllipticF[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))])/d - (2*Sqrt[a + b]*(3*A*b + 
 2*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*A*Sqrt[a + b*Sec[c + d*x]]*S 
in[c + d*x])/d
 

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 4513

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

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. \(1220\) vs. \(2(332)=664\).

Time = 12.12 (sec) , antiderivative size = 1221, normalized size of antiderivative = 3.38

method	result	size
default	\(\text {Expression too large to display}\)	\(1221\)

Input:

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

Output:

1/d*(6*(-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*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)*B*(co 
s(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*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b))^(1/2))+(-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*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)*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*EllipticE(- 
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) 
*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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+2*(co 
s(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^2*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)*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*Ellipti 
cF(-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)*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)*b^2*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)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(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)*cos(d*x + c), x)
 

Giac [F]

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

integrate(cos(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)*cos(d*x + c), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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