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

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

[Out]

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

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Rubi [A]  time = 0.13, 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} \[ \frac {3 \sqrt {1+i \tan (c+d x)} \tan ^{\frac {2}{3}}(c+d x) F_1\left (\frac {2}{3};\frac {3}{2},1;\frac {5}{3};-i \tan (c+d x),i \tan (c+d x)\right )}{2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[2/3, 3/2, 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*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 {1}{\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 {1}{\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}{\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}{\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 {2}{3};\frac {3}{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 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [F]  time = 3.50, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \frac {2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - 4 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, a d e^{\left (i \, d x + i \, c\right )}\right )} {\rm integral}\left (\frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (3 \, e^{\left (5 i \, d x + 5 i \, c\right )} + 30 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 46 \, e^{\left (3 i \, d x + 3 i \, c\right )} - 20 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 43 \, e^{\left (i \, d x + i \, c\right )} - 50\right )}}{6 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} - 6 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 11 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 12 \, a d e^{\left (i \, d x + i \, c\right )} + 8 \, a d\right )}}, x\right )}{a d e^{\left (3 i \, d x + 3 i \, c\right )} - 4 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, a d e^{\left (i \, d x + i \, c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*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)*(e
^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1) + (a*d*e^(3*I*d*x + 3*I*c) - 4*a*d*e^(2*I*d*x + 2*I*c) + 4*a*d
*e^(I*d*x + 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*e^(5*I*d*x + 5*I*c) + 30*e^(4*I*d*x + 4*I*c) + 46*e^(3*I*d*x + 3*I*c) - 20*e^(2*I
*d*x + 2*I*c) + 43*e^(I*d*x + I*c) - 50)/(a*d*e^(5*I*d*x + 5*I*c) - 6*a*d*e^(4*I*d*x + 4*I*c) + 11*a*d*e^(3*I*
d*x + 3*I*c) - 2*a*d*e^(2*I*d*x + 2*I*c) - 12*a*d*e^(I*d*x + I*c) + 8*a*d), x))/(a*d*e^(3*I*d*x + 3*I*c) - 4*a
*d*e^(2*I*d*x + 2*I*c) + 4*a*d*e^(I*d*x + I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.52, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a +i a \tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \[ \int \frac {1}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sqrt [3]{\tan {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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