3.271 \(\int \frac{1}{\tan ^{\frac{4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=82 \[ -\frac{3 \sqrt{1+i \tan (c+d x)} F_1\left (-\frac{1}{3};\frac{5}{2},1;\frac{2}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{a d \sqrt [3]{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.146734, antiderivative size = 82, 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} \[ -\frac{3 \sqrt{1+i \tan (c+d x)} F_1\left (-\frac{1}{3};\frac{5}{2},1;\frac{2}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{a d \sqrt [3]{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [F]  time = 11.6858, size = 0, normalized size = 0. \[ \int \frac{1}{\tan ^{\frac{4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.31, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}} \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)^(4/3)/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

int(1/tan(d*x+c)^(4/3)/(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)^(4/3)/(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*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(4/3)/(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)
*(-421*I*e^(8*I*d*x + 8*I*c) + 228*I*e^(7*I*d*x + 7*I*c) - 703*I*e^(6*I*d*x + 6*I*c) + 444*I*e^(5*I*d*x + 5*I*
c) - 131*I*e^(4*I*d*x + 4*I*c) + 204*I*e^(3*I*d*x + 3*I*c) + 163*I*e^(2*I*d*x + 2*I*c) - 12*I*e^(I*d*x + I*c)
+ 12*I)*e^(I*d*x + I*c) + 36*(a^2*d*e^(8*I*d*x + 8*I*c) - 4*a^2*d*e^(7*I*d*x + 7*I*c) + 3*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*I*e^(5*I*d*x + 5*I*c) - 4530*
I*e^(4*I*d*x + 4*I*c) - 286*I*e^(3*I*d*x + 3*I*c) - 2380*I*e^(2*I*d*x + 2*I*c) - 313*I*e^(I*d*x + I*c) + 2150*
I)*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^(8*I*d*x +
8*I*c) - 4*a^2*d*e^(7*I*d*x + 7*I*c) + 3*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)**(4/3)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Timed out

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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)^(4/3)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError