\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\) [267]

Optimal result

Integrand size = 35, antiderivative size = 250 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {(163 A-75 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {(17 A-9 B) \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(299 A-147 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/32*(163*A-75*B)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/ 
(a+a*sec(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A-B)*sin(d*x+c)/d/sec(d*x+c 
)^(1/2)/(a+a*sec(d*x+c))^(5/2)-1/16*(17*A-9*B)*sin(d*x+c)/a/d/sec(d*x+c)^( 
1/2)/(a+a*sec(d*x+c))^(3/2)+1/48*(95*A-39*B)*sin(d*x+c)/a^2/d/sec(d*x+c)^( 
1/2)/(a+a*sec(d*x+c))^(1/2)-1/48*(299*A-147*B)*sec(d*x+c)^(1/2)*sin(d*x+c) 
/a^2/d/(a+a*sec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 1.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {-12 \sqrt {2} (163 A-75 B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)+2 (-379 A+195 B+(-479 A+255 B) \cos (c+d x)+(-80 A+48 B) \cos (2 (c+d x))+8 A \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{96 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (a (1+\sec (c+d x)))^{5/2}} \] Input:

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5 
/2)),x]
 

Output:

(-12*Sqrt[2]*(163*A - 75*B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - S 
ec[c + d*x]]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(7/2)*Sin[c + d*x] + 2*(-379 
*A + 195*B + (-479*A + 255*B)*Cos[c + d*x] + (-80*A + 48*B)*Cos[2*(c + d*x 
)] + 8*A*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^2*Tan[c + d 
*x])/(96*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*(a*(1 + Sec[c + d*x]) 
)^(5/2))
 

Rubi [A] (verified)

Time = 1.49 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4508, 27, 3042, 4508, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\int \frac {a (11 A-3 B)-6 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (11 A-3 B)-6 a (A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (11 A-3 B)-6 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {a^2 (95 A-39 B)-4 a^2 (17 A-9 B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a^2 (95 A-39 B)-4 a^2 (17 A-9 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (95 A-39 B)-4 a^2 (17 A-9 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {a^3 (299 A-147 B)-2 a^3 (95 A-39 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (299 A-147 B)-2 a^3 (95 A-39 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (299 A-147 B)-2 a^3 (95 A-39 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (299 A-147 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^3 (163 A-75 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (299 A-147 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^3 (163 A-75 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a^3 (163 A-75 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (299 A-147 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 a^2 (95 A-39 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (299 A-147 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} a^{5/2} (163 A-75 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {a (17 A-9 B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\)

Input:

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)),x 
]
 

Output:

-1/4*((A - B)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/ 
2)) + (-1/2*(a*(17*A - 9*B)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec 
[c + d*x])^(3/2)) + ((2*a^2*(95*A - 39*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + 
d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-3*Sqrt[2]*a^(5/2)*(163*A - 75*B)*ArcT 
anh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + 
d*x]])])/d + (2*a^3*(299*A - 147*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sq 
rt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2))/(8*a^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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 4295

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

rule 4501

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

rule 4508

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

rule 4510

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

Maple [A] (warning: unable to verify)

Time = 2.40 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.14

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (489 \cos \left (d x +c \right )^{2}+1467 \cos \left (d x +c \right )+1467+489 \sec \left (d x +c \right )\right )+B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \left (-225 \cos \left (d x +c \right )^{2}-675 \cos \left (d x +c \right )-675-225 \sec \left (d x +c \right )\right )+\left (-64 \cos \left (d x +c \right )^{3}+320 \cos \left (d x +c \right )^{2}+1006 \cos \left (d x +c \right )+598\right ) A \tan \left (d x +c \right )+\left (-192 \cos \left (d x +c \right )^{2}-510 \cos \left (d x +c \right )-294\right ) B \tan \left (d x +c \right )\right )}{96 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(284\)
parts	\(\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (64 \cos \left (d x +c \right )^{3}-320 \cos \left (d x +c \right )^{2}-1006 \cos \left (d x +c \right )-598\right ) \tan \left (d x +c \right )+\arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-489 \cos \left (d x +c \right )^{2}-1467 \cos \left (d x +c \right )-1467-489 \sec \left (d x +c \right )\right )\right )}{96 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {B \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}+\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}+\frac {75 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{32}-\frac {83 \cot \left (d x +c \right )}{32}+\frac {83 \csc \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3} \sqrt {\sec \left (d x +c \right )}}\)	\(317\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/96/d/a^3*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d* 
x+c)+1)/sec(d*x+c)^(3/2)*((-2/(1+cos(d*x+c)))^(1/2)*A*arctan(1/2*2^(1/2)*( 
-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(489*cos(d*x+c)^2+1467* 
cos(d*x+c)+1467+489*sec(d*x+c))+B*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^( 
1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-225*cos(d*x+c)^ 
2-675*cos(d*x+c)-675-225*sec(d*x+c))+(-64*cos(d*x+c)^3+320*cos(d*x+c)^2+10 
06*cos(d*x+c)+598)*A*tan(d*x+c)+(-192*cos(d*x+c)^2-510*cos(d*x+c)-294)*B*t 
an(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 564, normalized size of antiderivative = 2.26 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="fricas")
 

Output:

[-1/192*(3*sqrt(2)*((163*A - 75*B)*cos(d*x + c)^3 + 3*(163*A - 75*B)*cos(d 
*x + c)^2 + 3*(163*A - 75*B)*cos(d*x + c) + 163*A - 75*B)*sqrt(a)*log(-(a* 
cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) 
*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 
 + 2*cos(d*x + c) + 1)) - 4*(32*A*cos(d*x + c)^4 - 32*(5*A - 3*B)*cos(d*x 
+ c)^3 - (503*A - 255*B)*cos(d*x + c)^2 - (299*A - 147*B)*cos(d*x + c))*sq 
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^ 
3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d 
), -1/96*(3*sqrt(2)*((163*A - 75*B)*cos(d*x + c)^3 + 3*(163*A - 75*B)*cos( 
d*x + c)^2 + 3*(163*A - 75*B)*cos(d*x + c) + 163*A - 75*B)*sqrt(-a)*arctan 
(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c 
))/(a*sin(d*x + c))) - 2*(32*A*cos(d*x + c)^4 - 32*(5*A - 3*B)*cos(d*x + c 
)^3 - (503*A - 255*B)*cos(d*x + c)^2 - (299*A - 147*B)*cos(d*x + c))*sqrt( 
(a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d 
*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 407281 vs. \(2 (213) = 426\).

Time = 5.11 (sec) , antiderivative size = 407281, normalized size of antiderivative = 1629.12 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="maxima")
 

Output:

-1/96*(3*(576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2* 
sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 
1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*cos(5/2* 
d*x + 5/2*c)^6 + 14400*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2* 
c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1 
/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c)) 
*cos(3/2*d*x + 3/2*c)^6 + 187500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x 
 + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + s 
in(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^ 
6 + 576*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/ 
2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c) 
^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(5/2*d*x + 
5/2*c)^6 + 5184*(75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 
2*sin(1/2*d*x + 1/2*c) + 1) - 75*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x 
+ 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 64*sin(1/2*d*x + 1/2*c))*sin(3/ 
2*d*x + 3/2*c)^6 + 262500*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2* 
c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2* 
d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(1/2*d*x + 1/2*c)^4*sin(1 
/2*d*x + 1/2*c)^2 + 77700*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2* 
c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1...
 

Giac [A] (verification not implemented)

Time = 3.33 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.85 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {{\left ({\left (3 \, {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{6}} - \frac {\sqrt {2} {\left (23 \, A a^{5} - 15 \, B a^{5}\right )}}{a^{6}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {4 \, \sqrt {2} {\left (167 \, A a^{5} - 75 \, B a^{5}\right )}}{a^{6}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, \sqrt {2} {\left (155 \, A a^{5} - 83 \, B a^{5}\right )}}{a^{6}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {3 \, \sqrt {2} {\left (163 \, A - 75 \, B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{96 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="giac")
 

Output:

1/96*(((3*(2*sqrt(2)*(A*a^5 - B*a^5)*tan(1/2*d*x + 1/2*c)^2/a^6 - sqrt(2)* 
(23*A*a^5 - 15*B*a^5)/a^6)*tan(1/2*d*x + 1/2*c)^2 - 4*sqrt(2)*(167*A*a^5 - 
 75*B*a^5)/a^6)*tan(1/2*d*x + 1/2*c)^2 - 3*sqrt(2)*(155*A*a^5 - 83*B*a^5)/ 
a^6)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 3*sqrt(2) 
*(163*A - 75*B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x 
 + 1/2*c)^2 + a)))/a^(5/2))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2 
)),x)
 

Output:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2 
)), x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}+3 \sec \left (d x +c \right )^{4}+3 \sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}+3 \sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right ) b \right )}{a^{3}} \] Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/(sec(c + d*x)**5 
 + 3*sec(c + d*x)**4 + 3*sec(c + d*x)**3 + sec(c + d*x)**2),x)*a + int((sq 
rt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/(sec(c + d*x)**4 + 3*sec(c + d*x) 
**3 + 3*sec(c + d*x)**2 + sec(c + d*x)),x)*b))/a**3