\(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [119]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {11}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d} \]

[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)-11/6/a/d/(a+I*a*tan(d*x+c))^(1/2)-
7/3*(a+I*a*tan(d*x+c))^(1/2)/a^2/d-1/3*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3639, 3673, 3607, 3561, 212} \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}-\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {11}{6 a d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]/(2*Sqrt[2]*a^(3/2)*d) - Tan[c + d*x]^2/(3*d*(a + I*a*Tan
[c + d*x])^(3/2)) - 11/(6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - (7*Sqrt[a + I*a*Tan[c + d*x]])/(3*a^2*d)

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 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]

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[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))*Tan[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] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3673

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan (c+d x) \left (-2 a+\frac {7}{2} i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2} \\ & = -\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}-\frac {\int \frac {-\frac {7 i a}{2}-2 a \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2} \\ & = -\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {11}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = -\frac {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {11}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\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 {\tan ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {11}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \sqrt {a+i a \tan (c+d x)}}{3 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (25+39 i \tan (c+d x)-12 \tan ^2(c+d x)\right )}{(-i+\tan (c+d x))^2}}{12 a^2 d} \]

[In]

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

[Out]

(3*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + (2*Sqrt[a + I*a*Tan[c + d*x]]*(25 +
(39*I)*Tan[c + d*x] - 12*Tan[c + d*x]^2))/(-I + Tan[c + d*x])^2)/(12*a^2*d)

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70

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

[In]

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

[Out]

2/d/a^2*(-(a+I*a*tan(d*x+c))^(1/2)-5/4*a/(a+I*a*tan(d*x+c))^(1/2)+1/6*a^2/(a+I*a*tan(d*x+c))^(3/2)+1/8*a^(1/2)
*2^(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. 273 vs. \(2 (102) = 204\).

Time = 0.24 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.05 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (38 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 13 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

[In]

integrate(tan(d*x+c)^3/(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*sq
rt(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))*(38*e^(4*I*d*x + 4*I*c) + 13*e^(2*I*d*x + 2*I*c) - 1))*e^(-3*I*d*x - 3*I*c)/(a^2*
d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 48 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - 2 \, a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{4} d} \]

[In]

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

[Out]

-1/24*(3*sqrt(2)*a^(5/2)*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))) + 48*sqrt(I*a*tan(d*x + c) + a)*a^2 + 4*(15*(I*a*tan(d*x + c) + a)*a^3 - 2*a^4)/(I*a*tan(d*x +
c) + a)^(3/2))/(a^4*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d}-\frac {\frac {13\,a}{6}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \]

[In]

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

[Out]

(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) - (2*(a + a*tan(c + d*x)*1i
)^(1/2))/(a^2*d) - ((13*a)/6 + (a*tan(c + d*x)*5i)/2)/(a*d*(a + a*tan(c + d*x)*1i)^(3/2))