\(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [289]

Optimal result

Integrand size = 37, antiderivative size = 237 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {5 C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

-5*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+1/32*(3* 
A+115*C)*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+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/( 
a+a*sec(d*x+c))^(5/2)+1/16*(A-15*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*s 
ec(d*x+c))^(3/2)+1/16*(3*A+35*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*se 
c(d*x+c))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(488\) vs. \(2(237)=474\).

Time = 9.71 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.06 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\left (A+C \sec ^2(c+d x)\right ) \left (-4 (3 A+67 C+2 (7 A+55 C) \cos (c+d x)+(3 A+35 C) \cos (2 (c+d x))) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {1+\sec (c+d x)} \sin \left (\frac {d x}{2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-4 (3 A+67 C+2 (7 A+55 C) \cos (c+d x)+(3 A+35 C) \cos (2 (c+d x))) \sec ^2(c+d x) \sqrt {1+\sec (c+d x)} \tan \left (\frac {c}{2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\left (-640 C \log (1+\sec (c+d x))+640 C \log \left (\sqrt {\sec (c+d x)}+\sec ^{\frac {3}{2}}(c+d x)+\sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )-\sqrt {2} (3 A+115 C) \left (\log \left (1-2 \sec (c+d x)-3 \sec ^2(c+d x)-2 \sqrt {2} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )-\log \left (1-2 \sec (c+d x)-3 \sec ^2(c+d x)+2 \sqrt {2} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )\right )\right ) \sqrt {\sec (c+d x)} \sin (c+d x) \sqrt {\tan ^2(c+d x)}\right )}{64 d (A+2 C+A \cos (2 (c+d x))) (-1+\sec (c+d x)) \left (\frac {\sec (c+d x)}{1+\sec (c+d x)}\right )^{3/2} (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

-1/64*((A + C*Sec[c + d*x]^2)*(-4*(3*A + 67*C + 2*(7*A + 55*C)*Cos[c + d*x 
] + (3*A + 35*C)*Cos[2*(c + d*x)])*Sec[c/2]*Sec[(c + d*x)/2]*Sec[c + d*x]^ 
2*Sqrt[1 + Sec[c + d*x]]*Sin[(d*x)/2]*Tan[(c + d*x)/2]^2 - 4*(3*A + 67*C + 
 2*(7*A + 55*C)*Cos[c + d*x] + (3*A + 35*C)*Cos[2*(c + d*x)])*Sec[c + d*x] 
^2*Sqrt[1 + Sec[c + d*x]]*Tan[c/2]*Tan[(c + d*x)/2]^2 + (-640*C*Log[1 + Se 
c[c + d*x]] + 640*C*Log[Sqrt[Sec[c + d*x]] + Sec[c + d*x]^(3/2) + Sqrt[1 + 
 Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2]] - Sqrt[2]*(3*A + 115*C)*(Log[1 - 2*Se 
c[c + d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[ 
c + d*x]]*Sqrt[Tan[c + d*x]^2]] - Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^ 
2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^ 
2]]))*Sqrt[Sec[c + d*x]]*Sin[c + d*x]*Sqrt[Tan[c + d*x]^2]))/(d*(A + 2*C + 
 A*Cos[2*(c + d*x)])*(-1 + Sec[c + d*x])*(Sec[c + d*x]/(1 + Sec[c + d*x])) 
^(3/2)*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.62 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 4573, 27, 3042, 4507, 27, 3042, 4509, 3042, 4511, 3042, 4288, 222, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {\sec ^{\frac {5}{2}}(c+d x) (a (3 A-5 C)+2 a (A+5 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (3 A-5 C)+2 a (A+5 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (3 A-5 C)+2 a (A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (A-15 C) a^2+2 (3 A+35 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (A-15 C) a^2+2 (3 A+35 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 (A-15 C) a^2+2 (3 A+35 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^3 (3 A+35 C)-80 a^3 C \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (3 A+35 C)-80 a^3 C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4511

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-80 a^2 C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-80 a^2 C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {160 a^2 C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {160 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {\frac {-\frac {2 a^3 (3 A+115 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {160 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\frac {\sqrt {2} a^{5/2} (3 A+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {160 a^{5/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 
2)) + ((a*(A - 15*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + 
d*x])^(3/2)) + (((-160*a^(5/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a 
*Sec[c + d*x]]])/d + (Sqrt[2]*a^(5/2)*(3*A + 115*C)*ArcTanh[(Sqrt[a]*Sqrt[ 
Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (2 
*a^2*(3*A + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d 
*x]]))/(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 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 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 4507

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

rule 4509

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

rule 4511

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)/b   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b 
 Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b 
, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0]
 

rule 4573

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(433\) vs. \(2(200)=400\).

Time = 2.76 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.83

method	result	size
default	\(\frac {\cos \left (d x +c \right )^{2} \left (\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (3 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )+\sqrt {2}\, C \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (115 \cos \left (d x +c \right )^{3}+230 \cos \left (d x +c \right )^{2}+115 \cos \left (d x +c \right )\right )+C \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-80 \cos \left (d x +c \right )^{3}-160 \cos \left (d x +c \right )^{2}-80 \cos \left (d x +c \right )\right )+C \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-80 \cos \left (d x +c \right )^{3}-160 \cos \left (d x +c \right )^{2}-80 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \left (3 \cos \left (d x +c \right )+7\right ) A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (35 \cos \left (d x +c \right )^{2}+55 \cos \left (d x +c \right )+16\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(434\)
parts	\(\frac {A \left (\left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (3 \cos \left (d x +c \right )+7\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )^{3} \sqrt {2}\, \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 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 ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {C \cos \left (d x +c \right )^{4} \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (115 \cos \left (d x +c \right )^{3}+230 \cos \left (d x +c \right )^{2}+115 \cos \left (d x +c \right )\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-80 \cos \left (d x +c \right )^{3}-160 \cos \left (d x +c \right )^{2}-80 \cos \left (d x +c \right )\right )+\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (80 \cos \left (d x +c \right )^{3}+160 \cos \left (d x +c \right )^{2}+80 \cos \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (35 \cos \left (d x +c \right )^{2}+55 \cos \left (d x +c \right )+16\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sec \left (d x +c \right )^{\frac {9}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 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 ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(494\)

Input:

int(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/32/d/a^3*cos(d*x+c)^2*(2^(1/2)*A*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^ 
(1/2)*(-csc(d*x+c)+cot(d*x+c)))*(3*cos(d*x+c)^3+6*cos(d*x+c)^2+3*cos(d*x+c 
))+2^(1/2)*C*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot 
(d*x+c)))*(115*cos(d*x+c)^3+230*cos(d*x+c)^2+115*cos(d*x+c))+C*arctan(1/2/ 
(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))*(-80*cos(d*x+c)^3-160 
*cos(d*x+c)^2-80*cos(d*x+c))+C*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(c 
os(d*x+c)+1))^(1/2))*(-80*cos(d*x+c)^3-160*cos(d*x+c)^2-80*cos(d*x+c))+sin 
(d*x+c)*cos(d*x+c)*2^(1/2)*(3*cos(d*x+c)+7)*A*(-2/(cos(d*x+c)+1))^(1/2)+si 
n(d*x+c)*(35*cos(d*x+c)^2+55*cos(d*x+c)+16)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^ 
(1/2))*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(5/2)/(cos(d*x+c)^3+3*cos(d*x+c 
)^2+3*cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 752, normalized size of antiderivative = 3.17 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

[1/64*(sqrt(2)*((3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c 
)^2 + 3*(3*A + 115*C)*cos(d*x + c) + 3*A + 115*C)*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(co 
s(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)) + 80*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x 
+ c) + C)*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))*s 
in(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 
 4*((3*A + 35*C)*cos(d*x + c)^2 + (7*A + 55*C)*cos(d*x + c) + 16*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), - 
1/32*(sqrt(2)*((3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c) 
^2 + 3*(3*A + 115*C)*cos(d*x + c) + 3*A + 115*C)*sqrt(-a)*arctan(sqrt(2)*s 
qrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin( 
d*x + c))) + 80*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) 
+ C)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(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*A + 35*C)*cos(d*x + c)^2 + (7*A + 55*C)*cos(d*x + c) + 16*C)*sqrt((a* 
cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*co 
s(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. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(5/2)*(A+C*sec(d*x+c)**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. 93381 vs. \(2 (200) = 400\).

Time = 16.33 (sec) , antiderivative size = 93381, normalized size of antiderivative = 394.01 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

1/32*((512*((2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos 
(5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 2*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3* 
c) + (2*cos(2*d*x + 2*c) + cos(d*x + c))*sin(5/2*d*x + 5/2*c) + cos(4*d*x 
+ 4*c)*sin(5/2*d*x + 5/2*c) + 2*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos 
(5*d*x + 5*c)^2 + 2560*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x 
+ 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)* 
sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2*d* 
x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)*sin 
(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3*d*x 
 + 3*c)*sin(5/2*d*x + 5/2*c))*cos(8/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/ 
2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2* 
d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2 
*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos( 
2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c) 
*sin(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3 
*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(6/5*arctan2(sin(5/2*d*x + 5/2*c), co 
s(5/2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos( 
5/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 
 5/2*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10* 
cos(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Error index.cc index_gcd Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^(5/2)*(A+C*sec(d*x+c)^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)**4)/ 
(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqr 
t(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2)/(sec(c + d*x)**3 + 
 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a))/a**3