\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [378]

Optimal result

Integrand size = 33, antiderivative size = 329 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} d}-\frac {2 (2 a+b) (3 A b-(4 a+b) B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} d}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 B \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 d} \] Output:

-2/3*(6*A*a^2*b-3*A*b^3-8*B*a^3+5*B*a*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^4/(a+b)^(1/2)/d-2/3*(2*a+b)*(3*A*b-(4 
*a+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^3/(a+b)^(1/2)/d-2*a^2*(A*b-B*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x 
+c))^(1/2)+2/3*B*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d
 

Mathematica [A] (warning: unable to verify)

Time = 17.07 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.60 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-6 a^2 A b+3 A b^3+8 a^3 B-5 a b^2 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (-2 a^2-a b+b^2\right ) (3 A b+(-4 a+b) B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-6 a^2 A b+3 A b^3+8 a^3 B-5 a b^2 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^3 \left (-a^2+b^2\right ) d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \left (\frac {2 \left (-6 a^2 A b+3 A b^3+8 a^3 B-5 a b^2 B\right ) \sin (c+d x)}{3 b^3 \left (-a^2+b^2\right )}+\frac {2 \left (a^2 A b \sin (c+d x)-a^3 B \sin (c+d x)\right )}{b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {2 B \tan (c+d x)}{3 b^2}\right )}{d (a+b \sec (c+d x))^{3/2}} \] Input:

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

Output:

(-2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c 
+ d*x]]*(2*(a + b)*(-6*a^2*A*b + 3*A*b^3 + 8*a^3*B - 5*a*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*(-2*a 
^2 - a*b + b^2)*(3*A*b + (-4*a + b)*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[ArcSi 
n[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-6*a^2*A*b + 3*A*b^3 + 8*a^3*B - 
5*a*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d 
*x)/2]))/(3*b^3*(-a^2 + b^2)*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x 
])^(3/2)) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*(-6*a^2*A*b + 3*A*b 
^3 + 8*a^3*B - 5*a*b^2*B)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)) + (2*(a^2*A*b 
*Sin[c + d*x] - a^3*B*Sin[c + d*x]))/(b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x] 
)) + (2*B*Tan[c + d*x])/(3*b^2)))/(d*(a + b*Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4516, 27, 3042, 4570, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4516

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \sec (c+d x) b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \sec (c+d x) b\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {b \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) (2 a+b) (-4 a B+3 A b-b B) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) (2 a+b) (-4 a B+3 A b-b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {b \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (2 a+b) (-4 a B+3 A b-b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{3 b}+\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {2 B \left (a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} (2 a+b) (-4 a B+3 A b-b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{3 b}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

Input:

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

Output:

(-2*a^2*(A*b - a*B)*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d* 
x]]) + (((-2*(a - b)*Sqrt[a + b]*(6*a^2*A*b - 3*A*b^3 - 8*a^3*B + 5*a*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*d) - (2*(a - b)*Sqrt[a + b]*(2*a + b)*(3*A*b - 4*a*B 
 - 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 + Se 
c[c + d*x]))/(a - b))])/d)/(3*b) + (2*(a^2 - b^2)*B*Sqrt[a + b*Sec[c + d*x 
]]*Tan[c + d*x])/(3*d))/(b^2*(a^2 - b^2))
 

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 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]
 

rule 4516

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

rule 4570

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-C)*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*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1785\) vs. \(2(301)=602\).

Time = 50.80 (sec) , antiderivative size = 1786, normalized size of antiderivative = 5.43

method	result	size
default	\(\text {Expression too large to display}\)	\(1786\)
parts	\(\text {Expression too large to display}\)	\(1793\)

Input:

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

Output:

2/3/d/(a-b)/(a+b)/b^3*(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)*(6*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(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*b*EllipticE 
(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+6*(cos(d*x+c)^2+2*cos(d*x+c)+ 
1)*A*(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^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)*A*(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*b^3*EllipticE(-csc(d*x+c)+co 
t(d*x+c),((a-b)/(a+b))^(1/2))+3*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(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)*b 
^4*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(-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^4*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b 
))^(1/2))+8*(-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*b*EllipticE(-csc( 
d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+5*(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^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+5*(co 
s(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*b^3*EllipticE(-csc(d*x+c)+cot(d*...
 

Fricas [F]

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

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

Output:

integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b 
^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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