\(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [128]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {13}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {31}{20 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

1/8*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/2)/ 
d-1/5*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(5/2)-13/30/a/d/(a+I*a*tan(d*x+c)) 
^(3/2)+31/20/a^2/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\sec ^2(c+d x) \left (2 \sqrt {a} (-19+86 \cos (2 (c+d x))+80 i \sin (2 (c+d x)))+15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{120 a^{5/2} d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

-1/120*(Sec[c + d*x]^2*(2*Sqrt[a]*(-19 + 86*Cos[2*(c + d*x)] + (80*I)*Sin[ 
2*(c + d*x)]) + 15*Sqrt[2]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqr 
t[a])]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]]) 
)/(a^(5/2)*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4041, 27, 3042, 4073, 3042, 4009, 3042, 3961, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\tan (c+d x) (4 a-9 i a \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan (c+d x) (4 a-9 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x) (4 a-9 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4073

\(\displaystyle \frac {-\frac {i \int \frac {13 a^2-18 i a^2 \tan (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {13 a}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {i \int \frac {13 a^2-18 i a^2 \tan (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {13 a}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {-\frac {i \left (\frac {31 i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {5}{2} a \int \sqrt {i \tan (c+d x) a+a}dx\right )}{2 a^2}-\frac {13 a}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {i \left (\frac {31 i a^2}{d \sqrt {a+i a \tan (c+d x)}}-\frac {5}{2} a \int \sqrt {i \tan (c+d x) a+a}dx\right )}{2 a^2}-\frac {13 a}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {-\frac {i \left (\frac {5 i a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {31 i a^2}{d \sqrt {a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {13 a}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {i \left (\frac {5 i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} d}+\frac {31 i a^2}{d \sqrt {a+i a \tan (c+d x)}}\right )}{2 a^2}-\frac {13 a}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}-\frac {\tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

Int[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

-1/5*Tan[c + d*x]^2/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((-13*a)/(3*d*(a + 
I*a*Tan[c + d*x])^(3/2)) - ((I/2)*(((5*I)*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan 
[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + ((31*I)*a^2)/(d*Sqrt[a + I*a* 
Tan[c + d*x]])))/a^2)/(10*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 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

rule 4009

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

rule 4041

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

rule 4073

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

Maple [A] (verified)

Time = 1.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {\frac {7}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 a}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{2}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2} d}\)	\(93\)
default	\(\frac {\frac {7}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 a}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{2}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2} d}\)	\(93\)

Input:

int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/d/a^2*(7/8/(a+I*a*tan(d*x+c))^(1/2)-5/12*a/(a+I*a*tan(d*x+c))^(3/2)+1/10 
*a^2/(a+I*a*tan(d*x+c))^(5/2)+1/16*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan( 
d*x+c))^(1/2)*2^(1/2)/a^(1/2)))
 

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. 283 vs. \(2 (102) = 204\).

Time = 0.09 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.13 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (83 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 64 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 16 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/120*(15*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqr 
t(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2* 
I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*s 
qrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt( 
1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) 
*sqrt(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1))*(83*e^(6*I*d*x + 6*I*c) + 64*e^(4*I*d*x + 4*I* 
c) - 16*e^(2*I*d*x + 2*I*c) + 3))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

\[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral(tan(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {15 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - \frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 50 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} + 12 \, a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{4} d} \] Input:

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

-1/240*(15*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + 
 a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) - 4*(105*(I*a*tan(d*x 
 + c) + a)^2*a^2 - 50*(I*a*tan(d*x + c) + a)*a^3 + 12*a^4)/(I*a*tan(d*x + 
c) + a)^(5/2))/(a^4*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^3/(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 [B] (verification not implemented)

Time = 1.00 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {5\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}+\frac {a^2}{5}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \] Input:

int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

((7*(a + a*tan(c + d*x)*1i)^2)/4 - (5*a*(a + a*tan(c + d*x)*1i))/6 + a^2/5 
)/(a^2*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (2^(1/2)*atanh((2^(1/2)*(a + a*t 
an(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)
 

Reduce [F]

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

int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(tan(c + d*x)**3/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt 
(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt( 
a)*a**2)