3.3.70 \(\int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\) [270]

Optimal. Leaf size=82 \[ \frac {3 F_1\left (\frac {1}{3};\frac {5}{2},1;\frac {4}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)}}{a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

3*AppellF1(1/3,5/2,1,4/3,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/2)*tan(d*x+c)^(1/3)/a/d/(a+I*a*tan(d*
x+c))^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 = {3645, 129, 441, 440} \begin {gather*} \frac {3 \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)} F_1\left (\frac {1}{3};\frac {5}{2},1;\frac {4}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{a d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[1/3, 5/2, 1, 4/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(1/3))/
(a*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 129

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 440

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

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 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 {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {1}{\left (-\frac {i x}{a}\right )^{2/3} (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 ) \text {Subst}\left (\int \frac {1}{\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 ) \text {Subst}\left (\int \frac {1}{\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 {4}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)}}{a d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [F]
time = 7.30, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 1.08, size = 0, normalized size = 0.00 \[\int \frac {1}{\tan \left (d x +c \right )^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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