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

Optimal. Leaf size=81 \[ \frac {3 \sqrt {1+i \tan (c+d x)} \tan ^{\frac {4}{3}}(c+d x) F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{4 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

3/4*AppellF1(4/3,3/2,1,7/3,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/2)*tan(d*x+c)^(4/3)/d/(a+I*a*tan(d*
x+c))^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 81, 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 = {3564, 130, 511, 510} \[ \frac {3 \sqrt {1+i \tan (c+d x)} \tan ^{\frac {4}{3}}(c+d x) F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{4 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[4/3, 3/2, 1, 7/3, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^(4/3))/
(4*d*Sqrt[a + I*a*Tan[c + d*x]])

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 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])

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{-\frac {i x}{a}}}{(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^3}{\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^3}{\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 {4}{3};\frac {3}{2},1;\frac {7}{3};-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{\frac {4}{3}}(c+d x)}{4 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 5.61, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.39, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{\frac {1}{3}}\left (d x +c \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt [3]{\tan {\left (c + d x \right )}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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