\(\int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\) [232]

Optimal result

Integrand size = 35, antiderivative size = 180 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^{3/2} (14 A+11 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^2 (14 A+11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (6 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \] Output:

1/8*a^(3/2)*(14*A+11*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2)) 
/d+1/8*a^2*(14*A+11*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2 
)+1/12*a^2*(6*A+7*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+ 
1/3*a*B*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [F(-1)]

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

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

Output:

$Aborted
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4506, 27, 3042, 4504, 3042, 4290, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{3} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+B)+a (6 A+7 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+B)+a (6 A+7 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (2 A+B)+a (6 A+7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 4290

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {1}{6} \left (\frac {3}{4} a (14 A+11 B) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{6} \left (\frac {a^2 (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a (14 A+11 B) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\)

Input:

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

Output:

(a*B*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a 
^2*(6*A + 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d* 
x]]) + (3*a*(14*A + 11*B)*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a 
+ a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a* 
Sec[c + d*x]])))/4)/6
 

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288

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

rule 4290

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( 
f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 
1)))   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre 
eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
 

rule 4504

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C 
ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] 
 + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1))   Int[Sqrt[a + b*Csc[e + f* 
x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ 
[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && 
!LtQ[n, 0]
 

rule 4506

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

Maple [A] (verified)

Time = 4.29 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.71

method	result	size
default	\(\frac {a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (42 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+33 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+42 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+33 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \left (42 \cos \left (d x +c \right )+12\right ) \sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\left (33 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+8\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(308\)
parts	\(\frac {A a \left (7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-7 \cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )+2\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B a \left (33 \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{3}-33 \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sin \left (d x +c \right ) \left (33 \cos \left (d x +c \right )^{2}+22 \cos \left (d x +c \right )+8\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}}}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(356\)

Input:

int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x,method=_RET 
URNVERBOSE)
 

Output:

1/48/d*a*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(3/2)/(1+cos(d*x+c))/(-1/(1+c 
os(d*x+c)))^(1/2)*(42*A*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d*x+c)+1)/ 
(-1/(1+cos(d*x+c)))^(1/2))+33*B*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d* 
x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+42*A*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c 
)-csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))+33*B*cos(d*x+c)^2*arctan(1/2*(c 
ot(d*x+c)-csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))+sin(d*x+c)*(42*cos(d*x+ 
c)+12)*2^(1/2)*A*(-2/(1+cos(d*x+c)))^(1/2)+(33*cos(d*x+c)^2+22*cos(d*x+c)+ 
8)*2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.53 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (3 \, {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + 8 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {3 \, {\left ({\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (3 \, {\left (14 \, A + 11 \, B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 11 \, B\right )} a \cos \left (d x + c\right ) + 8 \, B a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/96*(3*((14*A + 11*B)*a*cos(d*x + c)^3 + (14*A + 11*B)*a*cos(d*x + c)^2) 
*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 
2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + 
c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(14 
*A + 11*B)*a*cos(d*x + c)^2 + 2*(6*A + 11*B)*a*cos(d*x + c) + 8*B*a)*sqrt( 
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos 
(d*x + c)^3 + d*cos(d*x + c)^2), 1/48*(3*((14*A + 11*B)*a*cos(d*x + c)^3 + 
 (14*A + 11*B)*a*cos(d*x + c)^2)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*c 
os(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos( 
d*x + c))*sin(d*x + c))) + 2*(3*(14*A + 11*B)*a*cos(d*x + c)^2 + 2*(6*A + 
11*B)*a*cos(d*x + c) + 8*B*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin( 
d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4606 vs. \(2 (154) = 308\).

Time = 0.46 (sec) , antiderivative size = 4606, normalized size of antiderivative = 25.59 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/96*(6*(56*sqrt(2)*a*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
 3/2*c)))*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2 
4*sqrt(2)*a*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*s 
in(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*sqrt(2)*a 
*sin(3/2*d*x + 3/2*c) + 28*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), 
 cos(3/2*d*x + 3/2*c))) - 4*(3*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 7*sqrt(2)* 
a*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sqrt(2) 
*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 7*sqrt(2 
)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(8/3* 
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*(3*sqrt(2)*a*sin( 
3/2*d*x + 3/2*c) - 7*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3 
/2*d*x + 3/2*c))))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 
*c))) - 7*(a*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^ 
2 + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + a 
*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 4*a*sin( 
8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(4/3*arctan2(s 
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*a*sin(4/3*arctan2(sin(3/2* 
d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*(2*a*cos(4/3*arctan2(sin(3/2*d* 
x + 3/2*c), cos(3/2*d*x + 3/2*c))) + a)*cos(8/3*arctan2(sin(3/2*d*x + 3/2* 
c), cos(3/2*d*x + 3/2*c))) + 4*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (154) = 308\).

Time = 2.44 (sec) , antiderivative size = 783, normalized size of antiderivative = 4.35 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(3*(14*A*a^(3/2)*sgn(cos(d*x + c)) + 11*B*a^(3/2)*sgn(cos(d*x + c)))* 
log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) 
)^2 - a*(2*sqrt(2) + 3))) - 3*(14*A*a^(3/2)*sgn(cos(d*x + c)) + 11*B*a^(3/ 
2)*sgn(cos(d*x + c)))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1 
/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*(42*sqrt(2)*(sqrt(a)*t 
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(5/2)*sgn 
(cos(d*x + c)) + 33*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2 
*d*x + 1/2*c)^2 + a))^10*B*a^(5/2)*sgn(cos(d*x + c)) - 822*sqrt(2)*(sqrt(a 
)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*a^(7/2)*s 
gn(cos(d*x + c)) - 303*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan( 
1/2*d*x + 1/2*c)^2 + a))^8*B*a^(7/2)*sgn(cos(d*x + c)) + 3780*sqrt(2)*(sqr 
t(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(9/2 
)*sgn(cos(d*x + c)) + 2394*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a* 
tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(9/2)*sgn(cos(d*x + c)) - 2508*sqrt(2)* 
(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^ 
(11/2)*sgn(cos(d*x + c)) - 1806*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sq 
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(11/2)*sgn(cos(d*x + c)) + 498*sqr 
t(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 
*A*a^(13/2)*sgn(cos(d*x + c)) + 309*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) 
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(13/2)*sgn(cos(d*x + c)) - ...
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2) 
,x)
                                                                                    
                                                                                    
 

Output:

int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(3/2) 
, x)
 

Reduce [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:

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

Output:

sqrt(a)*a*(int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x 
)*b + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*a + 
 int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*b + int( 
sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*a)