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

Optimal. Leaf size=213 \[ -\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

[Out]

x/(4*2^(1/3)*a^(1/3)) - ((I/2)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3
))])/(2^(1/3)*a^(1/3)*d) - ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) - (((3*I)/4)*Log[2^(1/3)*a^(1/3) - (a
+ I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(1/3)*d) - ((3*I)/2)/(d*(a + I*a*Tan[c + d*x])^(1/3)) - (((3*I)/2)*(a
+ I*a*Tan[c + d*x])^(2/3))/(a*d)

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Rubi [A]  time = 0.155633, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3543, 3479, 3481, 55, 617, 204, 31} \[ -\frac{i \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt{3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(4*2^(1/3)*a^(1/3)) - ((I/2)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqrt[3]*a^(1/3
))])/(2^(1/3)*a^(1/3)*d) - ((I/4)*Log[Cos[c + d*x]])/(2^(1/3)*a^(1/3)*d) - (((3*I)/4)*Log[2^(1/3)*a^(1/3) - (a
+ I*a*Tan[c + d*x])^(1/3)])/(2^(1/3)*a^(1/3)*d) - ((3*I)/2)/(d*(a + I*a*Tan[c + d*x])^(1/3)) - (((3*I)/2)*(a
+ I*a*Tan[c + d*x])^(2/3))/(a*d)

Rule 3543

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

Rule 3479

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 3481

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\tan ^2(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\int \frac{1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\\ &=-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac{\int (a+i a \tan (c+d x))^{2/3} \, dx}{2 a}\\ &=-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}+\frac{i \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{2 d}\\ &=\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}\\ &=\frac{x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac{i \sqrt{3} \tan ^{-1}\left (\frac{1+\frac{2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac{3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac{3 i (a+i a \tan (c+d x))^{2/3}}{2 a d}\\ \end{align*}

Mathematica [C]  time = 0.615205, size = 82, normalized size = 0.38 \[ -\frac{3 \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\tan (c+d x)-i)-4 \tan (c+d x)+8 i\right )}{8 d \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(8*I - 4*Tan[c + d*x] + Hypergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(-I
+ Tan[c + d*x])))/(8*d*(a + I*a*Tan[c + d*x])^(1/3))

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Maple [A]  time = 0.017, size = 181, normalized size = 0.9 \begin{align*}{\frac{-{\frac{3\,i}{2}}}{ad} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}}-{\frac{{\frac{3\,i}{2}}}{d}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}-{\frac{{\frac{i}{4}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \sqrt [3]{a+ia\tan \left ( dx+c \right ) }-\sqrt [3]{2}\sqrt [3]{a} \right ){\frac{1}{\sqrt [3]{a}}}}+{\frac{{\frac{i}{8}}{2}^{{\frac{2}{3}}}}{d}\ln \left ( \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{2}\sqrt [3]{a}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }+{2}^{{\frac{2}{3}}}{a}^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{a}}}}-{\frac{{\frac{i}{4}}\sqrt{3}{2}^{{\frac{2}{3}}}}{d}\arctan \left ({\frac{\sqrt{3}}{3} \left ({{2}^{{\frac{2}{3}}}\sqrt [3]{a+ia\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [3]{a}}}}+1 \right ) } \right ){\frac{1}{\sqrt [3]{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*I*(a+I*a*tan(d*x+c))^(2/3)/a/d-3/2*I/d/(a+I*a*tan(d*x+c))^(1/3)-1/4*I/d/a^(1/3)*2^(2/3)*ln((a+I*a*tan(d*x
+c))^(1/3)-2^(1/3)*a^(1/3))+1/8*I/d/a^(1/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan(d*x
+c))^(1/3)+2^(2/3)*a^(2/3))-1/4*I/d/a^(1/3)*3^(1/2)*2^(2/3)*arctan(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x
+c))^(1/3)+1))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.71951, size = 934, normalized size = 4.38 \begin{align*} \frac{{\left (4 \, a d \left (\frac{i}{16 \, a d^{3}}\right )^{\frac{1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (8 \, a d^{2} \left (\frac{i}{16 \, a d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) + 2^{\frac{2}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (-9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )} e^{\left (\frac{4}{3} i \, d x + \frac{4}{3} i \, c\right )} +{\left (-2 i \, \sqrt{3} a d - 2 \, a d\right )} \left (\frac{i}{16 \, a d^{3}}\right )^{\frac{1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left (4 i \, \sqrt{3} a d^{2} - 4 \, a d^{2}\right )} \left (\frac{i}{16 \, a d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right ) +{\left (2 i \, \sqrt{3} a d - 2 \, a d\right )} \left (\frac{i}{16 \, a d^{3}}\right )^{\frac{1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left (-4 i \, \sqrt{3} a d^{2} - 4 \, a d^{2}\right )} \left (\frac{i}{16 \, a d^{3}}\right )^{\frac{2}{3}} + 2^{\frac{1}{3}} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{2}{3} i \, d x + \frac{2}{3} i \, c\right )}\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

1/4*(4*a*d*(1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(8*a*d^2*(1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*
d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)) + 2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-9*I*e^(2*I*d
*x + 2*I*c) - 3*I)*e^(4/3*I*d*x + 4/3*I*c) + (-2*I*sqrt(3)*a*d - 2*a*d)*(1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*
I*c)*log((4*I*sqrt(3)*a*d^2 - 4*a*d^2)*(1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^
(2/3*I*d*x + 2/3*I*c)) + (2*I*sqrt(3)*a*d - 2*a*d)*(1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log((-4*I*sqrt(3
)*a*d^2 - 4*a*d^2)*(1/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c
)))*e^(-2*I*d*x - 2*I*c)/(a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (c + d x \right )}}{\sqrt [3]{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2/(a+I*a*tan(d*x+c))**(1/3),x)

[Out]

Integral(tan(c + d*x)**2/(a*(I*tan(c + d*x) + 1))**(1/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^2/(I*a*tan(d*x + c) + a)^(1/3), x)