3.6.38 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx\) [538]

3.6.38.1 Optimal result
3.6.38.2 Mathematica [A] (warning: unable to verify)
3.6.38.3 Rubi [A] (verified)
3.6.38.4 Maple [B] (verified)
3.6.38.5 Fricas [F]
3.6.38.6 Sympy [F]
3.6.38.7 Maxima [F]
3.6.38.8 Giac [F]
3.6.38.9 Mupad [F(-1)]

3.6.38.1 Optimal result

Integrand size = 23, antiderivative size = 342 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}-\frac {2 \left (6 a^2-25 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \]

output
4/105*a*(a-b)*(3*a^2-41*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))*(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^3/d+2/105*(a-b)*(6*a^2+57*a*b-25*b^2)*c 
ot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2) 
)*(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-4/35*a*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/7*(a+b*sec(d*x+c)) 
^(5/2)*tan(d*x+c)/b/d-2/105*(6*a^2-25*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+ 
c)/b/d
 
3.6.38.2 Mathematica [A] (warning: unable to verify)

Time = 11.79 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x))^{3/2} \, dx=\frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 a \left (3 a^3+3 a^2 b-41 a b^2-41 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b \left (-6 a^3+51 a^2 b+82 a b^2+25 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \left (3 a^2-41 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 b^2 d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {4 a \left (3 a^2-41 b^2\right ) \sin (c+d x)}{105 b^2}+\frac {2 \sec (c+d x) \left (3 a^2 \sin (c+d x)+25 b^2 \sin (c+d x)\right )}{105 b}+\frac {16}{35} a \sec (c+d x) \tan (c+d x)+\frac {2}{7} b \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \]

input
Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2),x]
 
output
(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*a*( 
3*a^3 + 3*a^2*b - 41*a*b^2 - 41*b^3)*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)] + b*(-6*a^3 + 51*a^2*b + 82*a*b^2 + 25 
*b^3)*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)] + a*(3*a^2 - 41*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/ 
2]^2*Tan[(c + d*x)/2]))/(105*b^2*d*(b + a*Cos[c + d*x])^2*Sqrt[Sec[(c + d* 
x)/2]^2]*Sec[c + d*x]^(3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(( 
-4*a*(3*a^2 - 41*b^2)*Sin[c + d*x])/(105*b^2) + (2*Sec[c + d*x]*(3*a^2*Sin 
[c + d*x] + 25*b^2*Sin[c + d*x]))/(105*b) + (16*a*Sec[c + d*x]*Tan[c + d*x 
])/35 + (2*b*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*x]))
 
3.6.38.3 Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4327, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4327

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (5 b-2 a \sec (c+d x)) (a+b \sec (c+d x))^{3/2}dx}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (5 b-2 a \sec (c+d x)) (a+b \sec (c+d x))^{3/2}dx}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (5 b-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (19 a b-\left (6 a^2-25 b^2\right ) \sec (c+d x)\right )dx-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (19 a b-\left (6 a^2-25 b^2\right ) \sec (c+d x)\right )dx-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (19 a b+\left (25 b^2-6 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (51 a^2+25 b^2\right )-2 a \left (3 a^2-41 b^2\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (51 a^2+25 b^2\right )-2 a \left (3 a^2-41 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (51 a^2+25 b^2\right )-2 a \left (3 a^2-41 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2+57 a b-25 b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-2 a \left (3 a^2-41 b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2+57 a b-25 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-2 a \left (3 a^2-41 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-2 a \left (3 a^2-41 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (6 a^2+57 a b-25 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {4 a (a-b) \sqrt {a+b} \left (3 a^2-41 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )-\frac {2 \left (6 a^2-25 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

input
Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2),x]
 
output
(2*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((-4*a*(a + b*Sec[c 
+ d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((4*a*(a - b)*Sqrt[a + b]*(3*a^2 - 41 
*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)]*Sqrt[-((b*(1 + Sec[ 
c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(6*a^2 + 57*a*b - 2 
5*b^2)*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*(6*a^2 - 25*b^2)*Sqrt[a + b*Sec[c + d 
*x]]*Tan[c + d*x])/(3*d))/5)/(7*b)
 

3.6.38.3.1 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 4327
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 
))), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - 
b^2, 0] &&  !LtQ[m, -1]
 

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]
 
3.6.38.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2342\) vs. \(2(308)=616\).

Time = 15.60 (sec) , antiderivative size = 2343, normalized size of antiderivative = 6.85

method result size
default \(\text {Expression too large to display}\) \(2343\)

input
int(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 
output
2/105/d/b^2*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(-3*a^3 
*b*sin(d*x+c)+107*a*b^3*sin(d*x+c)+25*b^4*sin(d*x+c)+27*a^2*b^2*sin(d*x+c) 
-6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-cs 
c(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b+82*( 
1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d* 
x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2+82*(1/ 
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+ 
c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^3-25*Ellipti 
cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(c 
os(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*cos(d*x+c)^2-6*( 
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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*cos(d*x+c) 
^2-50*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*cos 
(d*x+c)-12*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^ 
4*cos(d*x+c)+6*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(c 
ot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)*a^3*b-51*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d 
*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)...
 
3.6.38.5 Fricas [F]

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

input
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
 
output
integral((b*sec(d*x + c)^4 + a*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), x 
)
 
3.6.38.6 Sympy [F]

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

input
integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**(3/2),x)
 
output
Integral((a + b*sec(c + d*x))**(3/2)*sec(c + d*x)**3, x)
 
3.6.38.7 Maxima [F]

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

input
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
 
output
integrate((b*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^3, x)
 
3.6.38.8 Giac [F]

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

input
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
 
output
integrate((b*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^3, x)
 
3.6.38.9 Mupad [F(-1)]

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

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