3.269 \(\int \frac{1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=84 \[ \frac{3 \sqrt{1+i \tan (c+d x)} \tan ^{\frac{2}{3}}(c+d x) F_1\left (\frac{2}{3};\frac{5}{2},1;\frac{5}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(3*AppellF1[2/3, 5/2, 1, 5/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(2/3))/
(2*a*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.132579, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3564, 130, 511, 510} \[ \frac{3 \sqrt{1+i \tan (c+d x)} \tan ^{\frac{2}{3}}(c+d x) F_1\left (\frac{2}{3};\frac{5}{2},1;\frac{5}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{2 a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[2/3, 5/2, 1, 5/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(2/3))/
(2*a*d*Sqrt[a + I*a*Tan[c + d*x]])

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 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{-\frac{i x}{a}} (a+x)^{5/2} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a+i a x^3\right )^{5/2} \left (-a^2+i a^2 x^3\right )} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d}\\ &=-\frac{\left (3 a \sqrt{1+i \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+i x^3\right )^{5/2} \left (-a^2+i a^2 x^3\right )} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d \sqrt{a+i a \tan (c+d x)}}\\ &=\frac{3 F_1\left (\frac{2}{3};\frac{5}{2},1;\frac{5}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt{1+i \tan (c+d x)} \tan ^{\frac{2}{3}}(c+d x)}{2 a d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [F]  time = 10.9793, size = 0, normalized size = 0. \[ \int \frac{1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.356, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt [3]{\tan \left ( dx+c \right ) }}} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 )^{\frac{2}{3}}{\left (79 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 12 \, e^{\left (5 i \, d x + 5 i \, c\right )} + 170 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 24 \, e^{\left (3 i \, d x + 3 i \, c\right )} + 103 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 12 \, e^{\left (i \, d x + i \, c\right )} + 12\right )} e^{\left (i \, d x + i \, c\right )} + 36 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}{\rm integral}\left (\frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 )^{\frac{2}{3}}{\left (27 \, e^{\left (5 i \, d x + 5 i \, c\right )} + 750 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 484 \, e^{\left (3 i \, d x + 3 i \, c\right )} + 40 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 457 \, e^{\left (i \, d x + i \, c\right )} - 710\right )} e^{\left (i \, d x + i \, c\right )}}{108 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 11 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} - 12 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{2} d e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{36 \,{\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} - 4 \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)
*(79*e^(6*I*d*x + 6*I*c) - 12*e^(5*I*d*x + 5*I*c) + 170*e^(4*I*d*x + 4*I*c) - 24*e^(3*I*d*x + 3*I*c) + 103*e^(
2*I*d*x + 2*I*c) - 12*e^(I*d*x + I*c) + 12)*e^(I*d*x + I*c) + 36*(a^2*d*e^(6*I*d*x + 6*I*c) - 4*a^2*d*e^(5*I*d
*x + 5*I*c) + 4*a^2*d*e^(4*I*d*x + 4*I*c))*integral(1/108*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*((-I*e^(2*
I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(27*e^(5*I*d*x + 5*I*c) + 750*e^(4*I*d*x + 4*I*c) + 484*e
^(3*I*d*x + 3*I*c) + 40*e^(2*I*d*x + 2*I*c) + 457*e^(I*d*x + I*c) - 710)*e^(I*d*x + I*c)/(a^2*d*e^(6*I*d*x + 6
*I*c) - 6*a^2*d*e^(5*I*d*x + 5*I*c) + 11*a^2*d*e^(4*I*d*x + 4*I*c) - 2*a^2*d*e^(3*I*d*x + 3*I*c) - 12*a^2*d*e^
(2*I*d*x + 2*I*c) + 8*a^2*d*e^(I*d*x + I*c)), x))/(a^2*d*e^(6*I*d*x + 6*I*c) - 4*a^2*d*e^(5*I*d*x + 5*I*c) + 4
*a^2*d*e^(4*I*d*x + 4*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError