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

Optimal. Leaf size=83 \[ -\frac{3 i (1+i \tan (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{5}{3},\frac{5}{6},\frac{1}{2} (1-i \tan (c+d x))\right )}{2^{2/3} d \sqrt{a+i a \tan (c+d x)} \sqrt [3]{e \sec (c+d x)}} \]

[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]])

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Rubi [A]  time = 0.196309, antiderivative size = 83, normalized size of antiderivative = 1., 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 = {3505, 3523, 70, 69} \[ -\frac{3 i (1+i \tan (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{5}{3},\frac{5}{6},\frac{1}{2} (1-i \tan (c+d x))\right )}{2^{2/3} d \sqrt{a+i a \tan (c+d x)} \sqrt [3]{e \sec (c+d x)}} \]

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 3505

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 3523

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]

Rule 70

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.738996, size = 95, normalized size = 1.14 \[ \frac{12 i-\frac{30 i e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{4}{3},-e^{2 i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{5/6}}}{16 d \sqrt{a+i a \tan (c+d x)} \sqrt [3]{e \sec (c+d x)}} \]

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 = 0.359, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt [3]{e\sec \left ( dx+c \right ) }}}{\frac{1}{\sqrt{a+ia\tan \left ( dx+c \right ) }}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{1}{3}} \sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

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]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2^{\frac{1}{6}} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (-12 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 27 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )} e^{\left (\frac{5}{3} i \, d x + \frac{5}{3} i \, c\right )} + 8 \,{\left (a d e e^{\left (5 i \, d x + 5 i \, c\right )} - a d e e^{\left (3 i \, d x + 3 i \, c\right )}\right )}{\rm integral}\left (\frac{2^{\frac{1}{6}} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac{2}{3}}{\left (-45 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 60 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 15 i\right )} e^{\left (\frac{5}{3} i \, d x + \frac{5}{3} i \, c\right )}}{16 \,{\left (a d e e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, a d e e^{\left (5 i \, d x + 5 i \, c\right )} + a d e e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}, x\right )}{8 \,{\left (a d e e^{\left (5 i \, d x + 5 i \, c\right )} - a d e e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \end{align*}

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*(2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-12*I*e^(6*I*d*x + 6*I*c)
- 27*I*e^(4*I*d*x + 4*I*c) - 18*I*e^(2*I*d*x + 2*I*c) - 3*I)*e^(5/3*I*d*x + 5/3*I*c) + 8*(a*d*e*e^(5*I*d*x + 5
*I*c) - a*d*e*e^(3*I*d*x + 3*I*c))*integral(1/16*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*
I*c) + 1))^(2/3)*(-45*I*e^(4*I*d*x + 4*I*c) - 60*I*e^(2*I*d*x + 2*I*c) - 15*I)*e^(5/3*I*d*x + 5/3*I*c)/(a*d*e*
e^(7*I*d*x + 7*I*c) - 2*a*d*e*e^(5*I*d*x + 5*I*c) + a*d*e*e^(3*I*d*x + 3*I*c)), x))/(a*d*e*e^(5*I*d*x + 5*I*c)
- a*d*e*e^(3*I*d*x + 3*I*c))

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

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{1}{3}} \sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

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]

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