\(\int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [121]

Optimal result

Integrand size = 24, antiderivative size = 98 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-1/4*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)+1/2/a/d/(a+I*a*tan(d*x+c))^(1/2)-
1/3/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3607, 3560, 3561, 212} \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {1}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3560

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

Rule 3561

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

Rule 3607

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {i \int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{2 a} \\ & = -\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = -\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{2 a d} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{2 a d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.35 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.64 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-2+\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (3+3 i \tan (c+d x))}{6 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

(-2 + Hypergeometric2F1[-1/2, 1, 1/2, (1 + I*Tan[c + d*x])/2]*(3 + (3*I)*Tan[c + d*x]))/(6*d*(a + I*a*Tan[c +
d*x])^(3/2))

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {3}{2}}}-\frac {1}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{2 a \sqrt {a +i a \tan \left (d x +c \right )}}}{d}\)	\(72\)
default	\(\frac {-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {3}{2}}}-\frac {1}{3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{2 a \sqrt {a +i a \tan \left (d x +c \right )}}}{d}\)	\(72\)

[In]

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

[Out]

1/d*(-1/4/a^(3/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1/3/(a+I*a*tan(d*x+c))^(3/2)+1
/2/a/(a+I*a*tan(d*x+c))^(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. 271 vs. \(2 (73) = 146\).

Time = 0.24 (sec) , antiderivative size = 271, normalized size of antiderivative = 2.77 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2 \, e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

[In]

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

[Out]

-1/12*(3*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*
I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*s
qrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) + a^
2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*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))*(2*e^(4*I*d*x + 4*I*c) + e^(2*I*d*x + 2*I*c) - 1))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{2} d} \]

[In]

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

[Out]

1/24*(3*sqrt(2)*sqrt(a)*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*(3*(I*a*tan(d*x + c) + a)*a - 2*a^2)/(I*a*tan(d*x + c) + a)^(3/2))/(a^2*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.14 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.73 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{2\,a}-\frac {1}{3}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \]

[In]

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

[Out]

((a + a*tan(c + d*x)*1i)/(2*a) - 1/3)/(d*(a + a*tan(c + d*x)*1i)^(3/2)) - (2^(1/2)*atanh((2^(1/2)*(a + a*tan(c
+ d*x)*1i)^(1/2))/(2*a^(1/2))))/(4*a^(3/2)*d)