3.3.78 \(\int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx\) [278]

Optimal. Leaf size=71 \[ -\frac {3 i \, _2F_1\left (-\frac {5}{6},\frac {23}{6};\frac {1}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{20\ 2^{5/6} a^2 f (d \sec (e+f x))^{5/3}} \]

[Out]

-3/40*I*hypergeom([-5/6, 23/6],[1/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e))^(5/6)*2^(1/6)/a^2/f/(d*sec(f*x+e))
^(5/3)

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Rubi [A]
time = 0.15, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 = {3586, 3604, 72, 71} \begin {gather*} -\frac {3 i (1+i \tan (e+f x))^{5/6} \, _2F_1\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}} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 3586

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 3604

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 {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 ) \text {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 ) \text {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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(71)=142\).
time = 0.97, size = 143, normalized size = 2.01 \begin {gather*} \frac {3 i \sec ^4(e+f x) \left (-46-40 \cos (2 (e+f x))+6 \cos (4 (e+f x))+128 e^{2 i (e+f x)} \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 )-10 i \sin (2 (e+f x))+11 i \sin (4 (e+f x))\right )}{680 a^2 f (d \sec (e+f x))^{5/3} (-i+\tan (e+f x))^2} \end {gather*}

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 = 0.81, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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) - 3*2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(17*I*e^(8*I*f*x + 8*I*e)
- 50*I*e^(6*I*f*x + 6*I*e) - 92*I*e^(4*I*f*x + 4*I*e) - 30*I*e^(2*I*f*x + 2*I*e) - 5*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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {1}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}} \tan ^{2}{\left (e + f x \right )} - 2 i \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}} \tan {\left (e + f x \right )} - \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx}{a^{2}} \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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