\(\int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [290]

Optimal result

Integrand size = 37, antiderivative size = 192 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(5 A-43 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(5 A-11 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \] Output:

2*C*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+1/32*(5*A 
-43*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)^(5/2)*sin(d*x+c)/d/(a+ 
a*sec(d*x+c))^(5/2)+1/16*(5*A-11*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*s 
ec(d*x+c))^(3/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(445\) vs. \(2(192)=384\).

Time = 5.52 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.32 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left (A+C \sec ^2(c+d x)\right ) \left (8 (A-15 C+(5 A-11 C) \cos (c+d x)) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {1+\sec (c+d x)} \sin \left (\frac {d x}{2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+8 (A-15 C+(5 A-11 C) \cos (c+d x)) \sec (c+d x) \sqrt {1+\sec (c+d x)} \tan \left (\frac {c}{2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\left (-256 C \log (1+\sec (c+d x))+256 C \log \left (\sqrt {\sec (c+d x)}+\sec ^{\frac {3}{2}}(c+d x)+\sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )+\sqrt {2} (5 A-43 C) \left (\log \left (1-2 \sec (c+d x)-3 \sec ^2(c+d x)-2 \sqrt {2} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )-\log \left (1-2 \sec (c+d x)-3 \sec ^2(c+d x)+2 \sqrt {2} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )\right )\right ) \sqrt {\sec (c+d x)} \sin (c+d x) \sqrt {\tan ^2(c+d x)}\right )}{64 d (A+2 C+A \cos (2 (c+d x))) (-1+\sec (c+d x)) \left (\frac {\sec (c+d x)}{1+\sec (c+d x)}\right )^{3/2} (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

((A + C*Sec[c + d*x]^2)*(8*(A - 15*C + (5*A - 11*C)*Cos[c + d*x])*Sec[c/2] 
*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[1 + Sec[c + d*x]]*Sin[(d*x)/2]*Tan[(c 
+ d*x)/2]^2 + 8*(A - 15*C + (5*A - 11*C)*Cos[c + d*x])*Sec[c + d*x]*Sqrt[1 
 + Sec[c + d*x]]*Tan[c/2]*Tan[(c + d*x)/2]^2 + (-256*C*Log[1 + Sec[c + d*x 
]] + 256*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]*(5*A - 43*C)*(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]] - 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]))/(64*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 = 1.20 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 4573, 27, 3042, 4507, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((5 A-11 C) a^2+32 C \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((5 A-11 C) a^2+32 C \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((5 A-11 C) a^2+32 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4511

\(\displaystyle \frac {\frac {a^2 (5 A-43 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx+32 a C \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (5 A-43 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+32 a C \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {a^2 (5 A-43 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {64 a C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {a^2 (5 A-43 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {64 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^2 (5 A-43 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\sqrt {2} a^{3/2} (5 A-43 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {64 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {a (5 A-11 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

Int[(Sec[c + d*x]^(3/2)*(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]^(5/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 
2)) + (((64*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d* 
x]]])/d + (Sqrt[2]*a^(3/2)*(5*A - 43*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x] 
]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/(4*a^2) + (a*(5*A 
- 11*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2))) 
/(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 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. \(381\) vs. \(2(161)=322\).

Time = 2.70 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.99

method	result	size
default	\(\frac {\cos \left (d x +c \right )^{2} \left (\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (-43 \cos \left (d x +c \right )^{2}-86 \cos \left (d x +c \right )-43\right ) \sqrt {2}\, C \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (32 \cos \left (d x +c \right )^{2}+64 \cos \left (d x +c \right )+32\right ) C \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (32 \cos \left (d x +c \right )^{2}+64 \cos \left (d x +c \right )+32\right ) C \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \left (5 \cos \left (d x +c \right )+1\right ) \sqrt {2}\, A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (-11 \cos \left (d x +c \right )-15\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(382\)
parts	\(\frac {A \left (\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (5 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {C \sec \left (d x +c \right )^{\frac {7}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+16 \sqrt {2}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-16 \sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+13 \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+43 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {2}}{64 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{3} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(451\)

Input:

int(sec(d*x+c)^(3/2)*(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*((5*cos(d*x+c)^2+10*cos(d*x+c)+5)*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)))+(-43*cos( 
d*x+c)^2-86*cos(d*x+c)-43)*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)))+(32*cos(d*x+c)^2+64*cos(d*x+c)+32)*C*arc 
tan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))+(32*cos(d*x+c 
)^2+64*cos(d*x+c)+32)*C*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+ 
c)+1))^(1/2))+sin(d*x+c)*(5*cos(d*x+c)+1)*2^(1/2)*A*(-2/(cos(d*x+c)+1))^(1 
/2)+sin(d*x+c)*(-11*cos(d*x+c)-15)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2))*se 
c(d*x+c)^(3/2)*(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)/(-1/(cos(d*x+c)+1))^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (161) = 322\).

Time = 0.14 (sec) , antiderivative size = 742, normalized size of antiderivative = 3.86 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

[-1/64*(sqrt(2)*((5*A - 43*C)*cos(d*x + c)^3 + 3*(5*A - 43*C)*cos(d*x + c) 
^2 + 3*(5*A - 43*C)*cos(d*x + c) + 5*A - 43*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(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)) - 32*(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))*sin( 
d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4* 
((5*A - 11*C)*cos(d*x + c)^2 + (A - 15*C)*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), -1/32*(sqrt(2 
)*((5*A - 43*C)*cos(d*x + c)^3 + 3*(5*A - 43*C)*cos(d*x + c)^2 + 3*(5*A - 
43*C)*cos(d*x + c) + 5*A - 43*C)*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))) - 32* 
(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*sqrt(-a)*ar 
ctan(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*((5*A - 11*C)*c 
os(d*x + c)^2 + (A - 15*C)*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*co 
s(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

Sympy [F(-1)]

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

integrate(sec(d*x+c)**(3/2)*(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. 7863 vs. \(2 (161) = 322\).

Time = 0.68 (sec) , antiderivative size = 7863, normalized size of antiderivative = 40.95 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

1/32*((4*(3*sin(3/2*d*x + 3/2*c) + 5*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), 
 cos(3/2*d*x + 3/2*c))) - 3*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* 
d*x + 3/2*c))) - 5*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))) - 40*( 
2*sin(3*d*x + 3*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
 3/2*c))) + 2*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) 
)*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 24*(2*sin 
(3*d*x + 3*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* 
c))) + 2*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos 
(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 24*(3*sin(3/2* 
d*x + 3/2*c) - 5*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))) + 16*(3* 
sin(3/2*d*x + 3/2*c) - 5*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x 
 + 3/2*c))))*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) 
+ 5*(16*cos(3*d*x + 3*c)^2 + 2*(4*cos(3*d*x + 3*c) + 6*cos(4/3*arctan2(sin 
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*cos(2/3*arctan2(sin(3/2*d*x 
+ 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(8/3*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)))^2 + 12*(4*cos(3*d*x + 3*c) + 4*cos(2/3*arctan2(sin(3/2*d*x + 
 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(sec(d*x+c)^(3/2)*(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:Warning, need to choose a branch for the root of a poly 
nomial with parameters. This might be wrong.Non regular value [0,0] was di 
scarded and
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^(3/2)*(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)**3)/ 
(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))/(sec(c + d*x)**3 + 3* 
sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a))/a**3