3.438 \(\int \frac{(e \sec (c+d x))^{5/3}}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=86 \[ \frac{3 i \sqrt [3]{2} a (1+i \tan (c+d x))^{2/3} (e \sec (c+d x))^{5/3} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(((3*I)/5)*2^(1/3)*a*Hypergeometric2F1[2/3, 5/6, 11/6, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(5/3)*(1 + I*T
an[c + d*x])^(2/3))/(d*(a + I*a*Tan[c + d*x])^(3/2))

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Rubi [A]  time = 0.198889, antiderivative size = 86, 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} \[ \frac{3 i \sqrt [3]{2} a (1+i \tan (c+d x))^{2/3} (e \sec (c+d x))^{5/3} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(5/3)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((3*I)/5)*2^(1/3)*a*Hypergeometric2F1[2/3, 5/6, 11/6, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(5/3)*(1 + I*T
an[c + d*x])^(2/3))/(d*(a + I*a*Tan[c + d*x])^(3/2))

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{(e \sec (c+d x))^{5/3}}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{(e \sec (c+d x))^{5/3} \int (a-i a \tan (c+d x))^{5/6} \sqrt [3]{a+i a \tan (c+d x)} \, dx}{(a-i a \tan (c+d x))^{5/6} (a+i a \tan (c+d x))^{5/6}}\\ &=\frac{\left (a^2 (e \sec (c+d x))^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [6]{a-i a x} (a+i a x)^{2/3}} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/6} (a+i a \tan (c+d x))^{5/6}}\\ &=\frac{\left (a^2 (e \sec (c+d x))^{5/3} \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 )^{2/3} \sqrt [6]{a-i a x}} \, dx,x,\tan (c+d x)\right )}{2^{2/3} d (a-i a \tan (c+d x))^{5/6} (a+i a \tan (c+d x))^{3/2}}\\ &=\frac{3 i \sqrt [3]{2} a \, _2F_1\left (\frac{2}{3},\frac{5}{6};\frac{11}{6};\frac{1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{5/3} (1+i \tan (c+d x))^{2/3}}{5 d (a+i a \tan (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.550094, size = 116, normalized size = 1.35 \[ \frac{3 i 2^{2/3} e e^{i (c+d x)} \left (\frac{e e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left (-2+\sqrt [6]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{4}{3},-e^{2 i (c+d x)}\right )\right )}{d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(5/3)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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Maple [F]  time = 0.33, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\sec \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}}{\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((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(5/3)/(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{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{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((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(5/3)/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{{\left (a d e^{\left (i \, d x + i \, c\right )}{\rm integral}\left (\frac{2^{\frac{1}{6}}{\left (i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e\right )} \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}} e^{\left (-\frac{4}{3} i \, d x - \frac{4}{3} i \, c\right )}}{a d}, x\right ) + 2 \cdot 2^{\frac{1}{6}}{\left (-3 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, e\right )} \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}} e^{\left (\frac{5}{3} i \, d x + \frac{5}{3} i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

(a*d*e^(I*d*x + I*c)*integral(2^(1/6)*(I*e*e^(2*I*d*x + 2*I*c) + I*e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^
(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(-4/3*I*d*x - 4/3*I*c)/(a*d), x) + 2*2^(1/6)*(-3*I*e*e^(2*I*d*x + 2*I*c) - 3*I
*e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*e^(5/3*I*d*x + 5/3*I*c))*e^(-I*d*x -
I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(5/3)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{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((e*sec(d*x+c))^(5/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(5/3)/sqrt(I*a*tan(d*x + c) + a), x)