3.278 \(\int \frac{1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=71 \[ -\frac{3 i (1+i \tan (e+f x))^{5/6} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{23}{6},\frac{1}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{20\ 2^{5/6} a^2 f (d \sec (e+f x))^{5/3}} \]

[Out]

(((-3*I)/20)*Hypergeometric2F1[-5/6, 23/6, 1/6, (1 - I*Tan[e + f*x])/2]*(1 + I*Tan[e + f*x])^(5/6))/(2^(5/6)*a
^2*f*(d*Sec[e + f*x])^(5/3))

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Rubi [A]  time = 0.190183, antiderivative size = 71, normalized size of antiderivative = 1., 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 = {3505, 3523, 70, 69} \[ -\frac{3 i (1+i \tan (e+f x))^{5/6} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{23}{6},\frac{1}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{20\ 2^{5/6} a^2 f (d \sec (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Sec[e + f*x])^(5/3)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

(((-3*I)/20)*Hypergeometric2F1[-5/6, 23/6, 1/6, (1 - I*Tan[e + f*x])/2]*(1 + I*Tan[e + f*x])^(5/6))/(2^(5/6)*a
^2*f*(d*Sec[e + f*x])^(5/3))

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}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx &=\frac{\left ((a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \int \frac{1}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{17/6}} \, dx}{(d \sec (e+f x))^{5/3}}\\ &=\frac{\left (a^2 (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{(a-i a x)^{11/6} (a+i a x)^{23/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3}}\\ &=\frac{\left ((a-i a \tan (e+f x))^{5/6} \left (\frac{a+i a \tan (e+f x)}{a}\right )^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{1}{2}+\frac{i x}{2}\right )^{23/6} (a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{8\ 2^{5/6} a f (d \sec (e+f x))^{5/3}}\\ &=-\frac{3 i \, _2F_1\left (-\frac{5}{6},\frac{23}{6};\frac{1}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{20\ 2^{5/6} a^2 f (d \sec (e+f x))^{5/3}}\\ \end{align*}

Mathematica [B]  time = 0.809679, size = 143, normalized size = 2.01 \[ \frac{3 i \sec ^4(e+f x) \left (128 e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )-10 i \sin (2 (e+f x))+11 i \sin (4 (e+f x))-40 \cos (2 (e+f x))+6 \cos (4 (e+f x))-46\right )}{680 a^2 f (\tan (e+f x)-i)^2 (d \sec (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Sec[e + f*x])^(5/3)*(a + I*a*Tan[e + f*x])^2),x]

[Out]

(((3*I)/680)*Sec[e + f*x]^4*(-46 - 40*Cos[2*(e + f*x)] + 6*Cos[4*(e + f*x)] + 128*E^((2*I)*(e + f*x))*(1 + E^(
(2*I)*(e + f*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(e + f*x))] - (10*I)*Sin[2*(e + f*x)] + (11
*I)*Sin[4*(e + f*x)]))/(a^2*f*(d*Sec[e + f*x])^(5/3)*(-I + Tan[e + f*x])^2)

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Maple [F]  time = 0.151, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}} \left ( d\sec \left ( fx+e \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e))^2,x)

[Out]

int(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e))^2,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (1360 \, a^{2} d^{2} f e^{\left (6 i \, f x + 6 i \, e\right )}{\rm integral}\left (-\frac{16 i \cdot 2^{\frac{1}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}} e^{\left (-\frac{2}{3} i \, f x - \frac{2}{3} i \, e\right )}}{85 \, a^{2} d^{2} f}, x\right ) + 2^{\frac{1}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}}{\left (-51 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 150 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 276 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 90 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i\right )} e^{\left (\frac{1}{3} i \, f x + \frac{1}{3} i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1360 \, a^{2} d^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1360*(1360*a^2*d^2*f*e^(6*I*f*x + 6*I*e)*integral(-16/85*I*2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(-2
/3*I*f*x - 2/3*I*e)/(a^2*d^2*f), x) + 2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(-51*I*e^(8*I*f*x + 8*I*e) +
150*I*e^(6*I*f*x + 6*I*e) + 276*I*e^(4*I*f*x + 4*I*e) + 90*I*e^(2*I*f*x + 2*I*e) + 15*I)*e^(1/3*I*f*x + 1/3*I
*e))*e^(-6*I*f*x - 6*I*e)/(a^2*d^2*f)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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