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

Optimal. Leaf size=81 \[ \frac{3 \sqrt{1+i \tan (c+d x)} \tan ^{\frac{5}{3}}(c+d x) F_1\left (\frac{5}{3};\frac{3}{2},1;\frac{8}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{5 d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

(3*AppellF1[5/3, 3/2, 1, 8/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(5/3))/
(5*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.142085, antiderivative size = 81, 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} \[ \frac{3 \sqrt{1+i \tan (c+d x)} \tan ^{\frac{5}{3}}(c+d x) F_1\left (\frac{5}{3};\frac{3}{2},1;\frac{8}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{5 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[5/3, 3/2, 1, 8/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(5/3))/
(5*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{\tan ^{\frac{2}{3}}(c+d x)}{\sqrt{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 )^{2/3}}{(a+x)^{3/2} \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^4}{\left (a+i a x^3\right )^{3/2} \left (-a^2+i a^2 x^3\right )} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{d}\\ &=-\frac{\left (3 a^2 \sqrt{1+i \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+i x^3\right )^{3/2} \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{5}{3};\frac{3}{2},1;\frac{8}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt{1+i \tan (c+d x)} \tan ^{\frac{5}{3}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [F]  time = 2.47798, size = 0, normalized size = 0. \[ \int \frac{\tan ^{\frac{2}{3}}(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.402, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}{\frac{1}{\sqrt{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)^(2/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 (2 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (i \, d x + i \, c\right )} +{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + 4 \, a d e^{\left (2 i \, d x + 2 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 (-3 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 66 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 32 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 64 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 29 i \, e^{\left (i \, d x + i \, c\right )} - 2 i\right )} e^{\left (i \, d x + i \, c\right )}}{6 \,{\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 11 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - 12 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a d e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{a d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + 4 \, a d e^{\left (2 i \, d x + 2 i \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)*(2*I
*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(I*d*x + I*c) + (a*d*e^(4*I*d*x + 4*I*c) - 4*a*d*e^(3*
I*d*x + 3*I*c) + 4*a*d*e^(2*I*d*x + 2*I*c))*integral(1/6*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)*(-3*I*e^(5*I*d*x + 5*I*c) + 66*I*e^(4*I*d*x + 4*I*c) - 32*
I*e^(3*I*d*x + 3*I*c) + 64*I*e^(2*I*d*x + 2*I*c) - 29*I*e^(I*d*x + I*c) - 2*I)*e^(I*d*x + I*c)/(a*d*e^(6*I*d*x
+ 6*I*c) - 6*a*d*e^(5*I*d*x + 5*I*c) + 11*a*d*e^(4*I*d*x + 4*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) - 12*a*d*e^(2*I
*d*x + 2*I*c) + 8*a*d*e^(I*d*x + I*c)), x))/(a*d*e^(4*I*d*x + 4*I*c) - 4*a*d*e^(3*I*d*x + 3*I*c) + 4*a*d*e^(2*
I*d*x + 2*I*c))

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

[Out]

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

[Out]

Exception raised: TypeError