\(\int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [375]

Optimal result

Integrand size = 24, antiderivative size = 122 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i \sec (c+d x)}{16 a d (a+i a \tan (c+d x))^{3/2}} \] Output:

3/32*I*arctanh(1/2*a^(1/2)*sec(d*x+c)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*2^ 
(1/2)/a^(5/2)/d+1/4*I*sec(d*x+c)/d/(a+I*a*tan(d*x+c))^(5/2)+3/16*I*sec(d*x 
+c)/a/d/(a+I*a*tan(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \sec ^3(c+d x) \left (7+3 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )+7 \cos (2 (c+d x))+3 i \sin (2 (c+d x))\right )}{32 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

((-1/32*I)*Sec[c + d*x]^3*(7 + 3*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c 
+ d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] + 7*Cos[2*(c + d*x)] + (3* 
I)*Sin[2*(c + d*x)]))/(a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d* 
x]])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3983, 3042, 3983, 3042, 3970, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

\(\displaystyle \frac {3 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {\sec (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{4 a}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3970

\(\displaystyle \frac {3 \left (\frac {i \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{2 a d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3 \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \sec (c+d x)}{2 d (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {i \sec (c+d x)}{4 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

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

Output:

((I/4)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (3*(((I/2)*ArcTanh 
[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^ 
(3/2)*d) + ((I/2)*Sec[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(3/2))))/(8*a)
 

Defintions of rubi rules used

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 3970

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S 
ymbol] :> Simp[-2*(a/(b*f))   Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ 
Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 0 
]
 

rule 3983

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ 
(b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n))   Int[(d*Sec[e + 
f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x 
] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* 
n]
 

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (97 ) = 194\).

Time = 7.97 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.39

method	result	size
default	\(-\frac {i \operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \left (12 \sin \left (d x +c \right )+6 \tan \left (d x +c \right )-3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+\operatorname {arctanh}\left (\frac {\left (i-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {2}}{2 \sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \left (12 \cos \left (d x +c \right )+6-9 \sec \left (d x +c \right )-3 \sec \left (d x +c \right )^{2}\right )+3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-4 \sin \left (d x +c \right )+\sec \left (d x +c \right ) \tan \left (d x +c \right )\right )+3 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (4 \cos \left (d x +c \right )-3 \sec \left (d x +c \right )\right )+6 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (4 \sin \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (7+12 \cos \left (d x +c \right )-2 \sec \left (d x +c \right )\right )}{32 d \left (1+i \tan \left (d x +c \right )\right )^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{2} \left (\cos \left (d x +c \right )+1\right )}\)	\(413\)

Input:

int(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32/d/(1+I*tan(d*x+c))^2/(a*(1+I*tan(d*x+c)))^(1/2)/(-cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2)/a^2/(cos(d*x+c)+1)*(I*arctanh(1/2/(cot(d*x+c)^2-2*cot(d*x+c 
)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(I-cot(d*x+c)+csc(d*x+c))*2^(1/2))*(12* 
sin(d*x+c)+6*tan(d*x+c)-3*sec(d*x+c)*tan(d*x+c))+arctanh(1/2/(cot(d*x+c)^2 
-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(I-cot(d*x+c)+csc(d*x+c))*2 
^(1/2))*(12*cos(d*x+c)+6-9*sec(d*x+c)-3*sec(d*x+c)^2)+3*2^(1/2)*(-2*cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*(-4*sin(d*x+c)+sec(d*x+c)*tan(d*x+c))+3*I*2^(1/ 
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(4*cos(d*x+c)-3*sec(d*x+c))+6*(-co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(4*sin(d*x+c)+tan(d*x+c))-2*I*(-cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)*(7+12*cos(d*x+c)-2*sec(d*x+c)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (91) = 182\).

Time = 0.08 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.19 \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-3 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, a^{2} d}\right ) + 3 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, a^{2} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{3} d} \] Input:

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/32*(-3*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(4*I*d*x + 4*I*c)*log(-3/8* 
(sqrt(2)*sqrt(1/2)*(I*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(a/(e^(2*I* 
d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) + 3*I* 
sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(4*I*d*x + 4*I*c)*log(-3/8*(sqrt(2)*sq 
rt(1/2)*(-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I* 
c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) + sqrt(2)*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1))*(5*I*e^(4*I*d*x + 4*I*c) + 7*I*e^(2*I*d*x + 2* 
I*c) + 2*I))*e^(-4*I*d*x - 4*I*c)/(a^3*d)
 

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral(sec(c + d*x)/(I*a*(tan(c + d*x) - I))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(sec(d*x + c)/(I*a*tan(d*x + c) + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(sec(c + d*x)/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt(ta 
n(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt(a)* 
a**2)