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

Optimal. Leaf size=83 \[ \frac{\sqrt [3]{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{4}{3},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt [3]{a+i a \tan (c+d x)}} \]

[Out]

(AppellF1[1 + m, 4/3, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*(1 + I*Tan[c + d*x])^(1/3)*Tan[c + d*x]^(1
+ m))/(d*(1 + m)*(a + I*a*Tan[c + d*x])^(1/3))

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Rubi [A]  time = 0.103806, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3564, 135, 133} \[ \frac{\sqrt [3]{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) F_1\left (m+1;\frac{4}{3},1;m+2;-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt [3]{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[1 + m, 4/3, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*(1 + I*Tan[c + d*x])^(1/3)*Tan[c + d*x]^(1
+ m))/(d*(1 + m)*(a + I*a*Tan[c + d*x])^(1/3))

Rule 3564

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

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a}\right )^m}{(a+x)^{4/3} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=\frac{\left (i a \sqrt [3]{1+i \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{i x}{a}\right )^m}{\left (1+\frac{x}{a}\right )^{4/3} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d \sqrt [3]{a+i a \tan (c+d x)}}\\ &=\frac{F_1\left (1+m;\frac{4}{3},1;2+m;-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt [3]{1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{d (1+m) \sqrt [3]{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [F]  time = 180.002, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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Maple [F]  time = 0.18, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{m}{\frac{1}{\sqrt [3]{a+ia\tan \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{m}}{{\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)^m/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{2^{\frac{2}{3}} \left (\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \left (\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-\frac{2}{3} i \, d x - \frac{2}{3} i \, c\right )}}{2 \, a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/2*2^(2/3)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*(a/(e^(2*I*d*x + 2*I*c) + 1))^
(2/3)*(e^(2*I*d*x + 2*I*c) + 1)*e^(-2/3*I*d*x - 2/3*I*c)/a, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{m}{\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)**m/(a+I*a*tan(d*x+c))**(1/3),x)

[Out]

Integral(tan(c + d*x)**m/(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 )^{m}}{{\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)^m/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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