3.5.41 \(\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx\) [441]

Optimal. Leaf size=83 \[ -\frac {3 i \, _2F_1\left (-\frac {1}{6},\frac {5}{3};\frac {5}{6};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{2/3}}{2^{2/3} d \sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-3/2*I*hypergeom([-1/6, 5/3],[5/6],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x+c))^(2/3)*2^(1/3)/d/(e*sec(d*x+c))^(1/3)
/(a+I*a*tan(d*x+c))^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 72, 71} \begin {gather*} -\frac {3 i (1+i \tan (c+d x))^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {5}{3};\frac {5}{6};\frac {1}{2} (1-i \tan (c+d x))\right )}{2^{2/3} d \sqrt {a+i a \tan (c+d x)} \sqrt [3]{e \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*Hypergeometric2F1[-1/6, 5/3, 5/6, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(2/3))/(2^(2/3)*d*(e*Se
c[c + d*x])^(1/3)*Sqrt[a + I*a*Tan[c + d*x]])

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 3586

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/2)*(a - b*Tan[e + f*x])^(m/2)), Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a
- b*Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {\left (\sqrt [6]{a-i a \tan (c+d x)} \sqrt [6]{a+i a \tan (c+d x)}\right ) \int \frac {1}{\sqrt [6]{a-i a \tan (c+d x)} (a+i a \tan (c+d x))^{2/3}} \, dx}{\sqrt [3]{e \sec (c+d x)}}\\ &=\frac {\left (a^2 \sqrt [6]{a-i a \tan (c+d x)} \sqrt [6]{a+i a \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{(a-i a x)^{7/6} (a+i a x)^{5/3}} \, dx,x,\tan (c+d x)\right )}{d \sqrt [3]{e \sec (c+d x)}}\\ &=\frac {\left (a \sqrt [6]{a-i a \tan (c+d x)} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {1}{2}+\frac {i x}{2}\right )^{5/3} (a-i a x)^{7/6}} \, dx,x,\tan (c+d x)\right )}{2\ 2^{2/3} d \sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ &=-\frac {3 i \, _2F_1\left (-\frac {1}{6},\frac {5}{3};\frac {5}{6};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{2/3}}{2^{2/3} d \sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.86, size = 95, normalized size = 1.14 \begin {gather*} \frac {12 i-\frac {30 i e^{2 i (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {4}{3};-e^{2 i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{5/6}}}{16 d \sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*I - ((30*I)*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/6, 1/3, 4/3, -E^((2*I)*(c + d*x))])/(1 + E^((2*I)*(c +
d*x)))^(5/6))/(16*d*(e*Sec[c + d*x])^(1/3)*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [F]
time = 0.89, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}} \sqrt {a +i a \tan \left (d x +c \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(e*sec(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-1/8*(3*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(4*I*e^(6*I*d*x + 6*I*c) + 9*I*e^(4*I*d*x + 4*I*c) + 6*I*e^(
2*I*d*x + 2*I*c) + I)*e^(2/3*I*d*x + 2/3*I*c)/(e^(2*I*d*x + 2*I*c) + 1)^(2/3) - 8*(a*d*e^(4*I*d*x + 4*I*c + 1/
3) - a*d*e^(2*I*d*x + 2*I*c + 1/3))*integral(-15/16*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(3*I*e^(4*I*d*x
+ 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(2/3*I*d*x + 2/3*I*c)/((a*d*e^(6*I*d*x + 6*I*c + 1/3) - 2*a*d*e^(4*I
*d*x + 4*I*c + 1/3) + a*d*e^(2*I*d*x + 2*I*c + 1/3))*(e^(2*I*d*x + 2*I*c) + 1)^(2/3)), x))/(a*d*e^(4*I*d*x + 4
*I*c + 1/3) - a*d*e^(2*I*d*x + 2*I*c + 1/3))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{e \sec {\left (c + d x \right )}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(e*sec(d*x+c))^(1/3)/(a+I*a*tan(d*x+c))^(1/2),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}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{1/3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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