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

Optimal. Leaf size=87 \[ \frac{3 i \sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \text{Hypergeometric2F1}\left (\frac{5}{6},\frac{13}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{10 \sqrt [6]{2} f \left (a^2+i a^2 \tan (e+f x)\right )} \]

[Out]

(((3*I)/10)*Hypergeometric2F1[5/6, 13/6, 11/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(5/3)*(1 + I*Tan[e + f
*x])^(1/6))/(2^(1/6)*f*(a^2 + I*a^2*Tan[e + f*x]))

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Rubi [A]  time = 0.185338, antiderivative size = 87, 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} \[ \frac{3 i \sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \text{Hypergeometric2F1}\left (\frac{5}{6},\frac{13}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{10 \sqrt [6]{2} f \left (a^2+i a^2 \tan (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/10)*Hypergeometric2F1[5/6, 13/6, 11/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(5/3)*(1 + I*Tan[e + f
*x])^(1/6))/(2^(1/6)*f*(a^2 + I*a^2*Tan[e + f*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{(d \sec (e+f x))^{5/3}}{(a+i a \tan (e+f x))^2} \, dx &=\frac{(d \sec (e+f x))^{5/3} \int \frac{(a-i a \tan (e+f x))^{5/6}}{(a+i a \tan (e+f x))^{7/6}} \, dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\\ &=\frac{\left (a^2 (d \sec (e+f x))^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [6]{a-i a x} (a+i a x)^{13/6}} \, dx,x,\tan (e+f x)\right )}{f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\\ &=\frac{\left ((d \sec (e+f x))^{5/3} \sqrt [6]{\frac{a+i a \tan (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{1}{2}+\frac{i x}{2}\right )^{13/6} \sqrt [6]{a-i a x}} \, dx,x,\tan (e+f x)\right )}{4 \sqrt [6]{2} f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))}\\ &=\frac{3 i \, _2F_1\left (\frac{5}{6},\frac{13}{6};\frac{11}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3} \sqrt [6]{1+i \tan (e+f x)}}{10 \sqrt [6]{2} f \left (a^2+i a^2 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.704386, size = 128, normalized size = 1.47 \[ -\frac{3 e^{-i (4 e+5 f x)} \left (1+e^{2 i (e+f x)}\right ) (\sin (f x)-i \cos (f x)) \left (2 e^{2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{2}{3},\frac{5}{6},-e^{2 i (e+f x)}\right )+e^{2 i (e+f x)}+1\right ) (d \sec (e+f x))^{5/3}}{28 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 + E^((2*I)*(e + f*x)))*(1 + E^((2*I)*(e + f*x)) + 2*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(2/3)
*Hypergeometric2F1[-1/6, 2/3, 5/6, -E^((2*I)*(e + f*x))])*(d*Sec[e + f*x])^(5/3)*((-I)*Cos[f*x] + Sin[f*x]))/(
28*a^2*E^(I*(4*e + 5*f*x))*f)

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Maple [F]  time = 0.174, 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((d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e))^2,x)

[Out]

int((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((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 (14 \, a^{2} f e^{\left (3 i \, f x + 3 i \, e\right )}{\rm integral}\left (-\frac{i \cdot 2^{\frac{2}{3}} d \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )}}{7 \, a^{2} f}, x\right ) + 2^{\frac{2}{3}}{\left (6 i \, d e^{\left (4 i \, f x + 4 i \, e\right )} + 9 i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, d\right )} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{14 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*(14*a^2*f*e^(3*I*f*x + 3*I*e)*integral(-1/7*I*2^(2/3)*d*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x
+ 2/3*I*e)/(a^2*f), x) + 2^(2/3)*(6*I*d*e^(4*I*f*x + 4*I*e) + 9*I*d*e^(2*I*f*x + 2*I*e) + 3*I*d)*(d/(e^(2*I*f*
x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e))*e^(-3*I*f*x - 3*I*e)/(a^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((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{\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((d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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