3.3.88 \(\int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [288]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 = {3645, 140, 138} \begin {gather*} \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)}} \end {gather*}

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 138

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

Rule 140

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 3645

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/b)*x)^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]

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 ) \text {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 ) \text {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*}

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Mathematica [F]
time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

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 = 0.40, size = 0, normalized size = 0.00 \[\int \frac {\tan ^{m}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{m}{\left (c + d x \right )}}{\sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

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/(I*a*(tan(c + d*x) - I))**(1/3), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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