\(\int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [94]

Optimal result

Integrand size = 24, antiderivative size = 92 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 92, 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 = {3608, 3559, 3561, 212} \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 (a+i a \tan (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

(-2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (2*a*Sqrt[a + I*a*Tan[c + d*x]]
)/d + (2*(a + I*a*Tan[c + d*x])^(3/2))/(3*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 3559

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))
), x] + Dist[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] && G
tQ[n, 1]

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 3608

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

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.88 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {-6 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a (4+i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[In]

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

[Out]

(-6*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + 2*a*(4 + I*Tan[c + d*x])*Sqrt[a +
I*a*Tan[c + d*x]])/(3*d)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{d}\)	\(70\)
default	\(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{d}\)	\(70\)

[In]

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

[Out]

1/d*(2/3*(a+I*a*tan(d*x+c))^(3/2)+2*a*(a+I*a*tan(d*x+c))^(1/2)-2*a^(3/2)*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. 258 vs. \(2 (71) = 142\).

Time = 0.24 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.80 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {3 \, \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 3 \, \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 2 \, \sqrt {2} {\left (5 \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 3 \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.11 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {3 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} + 6 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}}{3 \, a^{2} d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.80 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {2\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {2\,\sqrt {2}\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{d} \]

[In]

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

[Out]

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