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

Optimal. Leaf size=69 \[ -\frac {3 i \sqrt [6]{2} a (1+i \tan (e+f x))^{5/6} \, _2F_1\left (-\frac {5}{6},\frac {5}{6};\frac {1}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (d \sec (e+f x))^{5/3}} \]

[Out]

-3/5*I*2^(1/6)*a*hypergeom([-5/6, 5/6],[1/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e))^(5/6)/f/(d*sec(f*x+e))^(5/
3)

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Rubi [A]  time = 0.15, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3505, 3523, 70, 69} \[ -\frac {3 i \sqrt [6]{2} a (1+i \tan (e+f x))^{5/6} \text {Hypergeometric2F1}\left (-\frac {5}{6},\frac {5}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (d \sec (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 106, normalized size = 1.54 \[ -\frac {3 i a e^{i (e+f x)} \left (4 \sqrt [3]{1+e^{2 i (e+f x)}} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-e^{2 i (e+f x)}\right )+e^{2 i (e+f x)}+1\right )}{5 d f \left (1+e^{2 i (e+f x)}\right ) (d \sec (e+f x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/5)*a*E^(I*(e + f*x))*(1 + E^((2*I)*(e + f*x)) + 4*(1 + E^((2*I)*(e + f*x)))^(1/3)*Hypergeometric2F1[1
/6, 1/3, 7/6, -E^((2*I)*(e + f*x))]))/(d*(1 + E^((2*I)*(e + f*x)))*f*(d*Sec[e + f*x])^(2/3))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \frac {10 \, d^{2} f {\rm integral}\left (-\frac {2 i \cdot 2^{\frac {1}{3}} a \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}} e^{\left (-\frac {2}{3} i \, f x - \frac {2}{3} i \, e\right )}}{5 \, d^{2} f}, x\right ) + 2^{\frac {1}{3}} {\left (-3 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a\right )} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {1}{3} i \, f x + \frac {1}{3} i \, e\right )}}{10 \, d^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(10*d^2*f*integral(-2/5*I*2^(1/3)*a*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(-2/3*I*f*x - 2/3*I*e)/(d^2*f),
x) + 2^(1/3)*(-3*I*a*e^(2*I*f*x + 2*I*e) - 3*I*a)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(1/3*I*f*x + 1/3*I*e)
)/(d^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \frac {a +i a \tan \left (f x +e \right )}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)/(d/cos(e + f*x))^(5/3),x)

[Out]

int((a + a*tan(e + f*x)*1i)/(d/cos(e + f*x))^(5/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- \frac {i}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\right )\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*(Integral(-I/(d*sec(e + f*x))**(5/3), x) + Integral(tan(e + f*x)/(d*sec(e + f*x))**(5/3), x))

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