3.442 \(\int \frac{1}{(e \sec (c+d x))^{4/3} \sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=88 \[ -\frac{3 i \sqrt [6]{1+i \tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{13}{6},\frac{1}{3},\frac{1}{2} (1-i \tan (c+d x))\right )}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}} \]

[Out]

(((-3*I)/8)*Hypergeometric2F1[-2/3, 13/6, 1/3, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(1/6)*Sqrt[a + I*a
*Tan[c + d*x]])/(2^(1/6)*a*d*(e*Sec[c + d*x])^(4/3))

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Rubi [A]  time = 0.209879, antiderivative size = 88, 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} \[ -\frac{3 i \sqrt [6]{1+i \tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{13}{6},\frac{1}{3},\frac{1}{2} (1-i \tan (c+d x))\right )}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(((-3*I)/8)*Hypergeometric2F1[-2/3, 13/6, 1/3, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[c + d*x])^(1/6)*Sqrt[a + I*a
*Tan[c + d*x]])/(2^(1/6)*a*d*(e*Sec[c + d*x])^(4/3))

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}{(e \sec (c+d x))^{4/3} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{\left ((a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3}\right ) \int \frac{1}{(a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{7/6}} \, dx}{(e \sec (c+d x))^{4/3}}\\ &=\frac{\left (a^2 (a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{(a-i a x)^{5/3} (a+i a x)^{13/6}} \, dx,x,\tan (c+d x)\right )}{d (e \sec (c+d x))^{4/3}}\\ &=\frac{\left ((a-i a \tan (c+d x))^{2/3} \sqrt{a+i a \tan (c+d x)} \sqrt [6]{\frac{a+i a \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{1}{2}+\frac{i x}{2}\right )^{13/6} (a-i a x)^{5/3}} \, dx,x,\tan (c+d x)\right )}{4 \sqrt [6]{2} d (e \sec (c+d x))^{4/3}}\\ &=-\frac{3 i \, _2F_1\left (-\frac{2}{3},\frac{13}{6};\frac{1}{3};\frac{1}{2} (1-i \tan (c+d x))\right ) \sqrt [6]{1+i \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.655092, size = 112, normalized size = 1.27 \[ -\frac{3 i \sec ^2(c+d x) \left (-55 \sqrt [6]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{6},\frac{5}{6},-e^{2 i (c+d x)}\right )+11 i \sin (2 (c+d x))+3 \cos (2 (c+d x))+3\right )}{112 d \sqrt{a+i a \tan (c+d x)} (e \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(((-3*I)/112)*Sec[c + d*x]^2*(3 + 3*Cos[2*(c + d*x)] - 55*(1 + E^((2*I)*(c + d*x)))^(1/6)*Hypergeometric2F1[-1
/6, 1/6, 5/6, -E^((2*I)*(c + d*x))] + (11*I)*Sin[2*(c + d*x)]))/(d*(e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c +
d*x]])

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\sec \left ( dx+c \right ) \right ) ^{-{\frac{4}{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(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

int(1/(e*sec(d*x+c))^(4/3)/(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{4}{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))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((e*sec(d*x + c))^(4/3)*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 (-21 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 42 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 114 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 60 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 303 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 180 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 24 i \, e^{\left (i \, d x + i \, c\right )} + 12 i\right )} e^{\left (\frac{5}{3} i \, d x + \frac{5}{3} i \, c\right )} + 112 \,{\left (a d e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + a d e^{2} e^{\left (4 i \, d x + 4 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 (55 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 385 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 275 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 385 i \, e^{\left (i \, d x + i \, c\right )} + 220 i\right )} e^{\left (\frac{5}{3} i \, d x + \frac{5}{3} i \, c\right )}}{112 \,{\left (a d e^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 3 \, a d e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a d e^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )}\right )}}, x\right )}{112 \,{\left (a d e^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + a d e^{2} e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/112*(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)*(-21*I*e^(8*I*d*x + 8*I*c
) + 42*I*e^(7*I*d*x + 7*I*c) + 114*I*e^(6*I*d*x + 6*I*c) + 60*I*e^(5*I*d*x + 5*I*c) + 303*I*e^(4*I*d*x + 4*I*c
) - 6*I*e^(3*I*d*x + 3*I*c) + 180*I*e^(2*I*d*x + 2*I*c) - 24*I*e^(I*d*x + I*c) + 12*I)*e^(5/3*I*d*x + 5/3*I*c)
+ 112*(a*d*e^2*e^(6*I*d*x + 6*I*c) - 2*a*d*e^2*e^(5*I*d*x + 5*I*c) + a*d*e^2*e^(4*I*d*x + 4*I*c))*integral(1/
112*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)*(55*I*e^(4*I*d*x + 4*I*c) +
385*I*e^(3*I*d*x + 3*I*c) + 275*I*e^(2*I*d*x + 2*I*c) + 385*I*e^(I*d*x + I*c) + 220*I)*e^(5/3*I*d*x + 5/3*I*c)
/(a*d*e^2*e^(5*I*d*x + 5*I*c) - 3*a*d*e^2*e^(4*I*d*x + 4*I*c) + 3*a*d*e^2*e^(3*I*d*x + 3*I*c) - a*d*e^2*e^(2*I
*d*x + 2*I*c)), x))/(a*d*e^2*e^(6*I*d*x + 6*I*c) - 2*a*d*e^2*e^(5*I*d*x + 5*I*c) + a*d*e^2*e^(4*I*d*x + 4*I*c)
)

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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(1/(e*sec(d*x+c))**(4/3)/(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{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{4}{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))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(4/3)*sqrt(I*a*tan(d*x + c) + a)), x)