3.3.75 \(\int \frac {(d \sec (e+f x))^{5/3}}{(a+i a \tan (e+f x))^2} \, dx\) [275]

Optimal. Leaf size=87 \[ \frac {3 i \, _2F_1\left (\frac {5}{6},\frac {13}{6};\frac {11}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3} \sqrt [6]{1+i \tan (e+f x)}}{10 \sqrt [6]{2} f \left (a^2+i a^2 \tan (e+f x)\right )} \]

[Out]

3/20*I*hypergeom([5/6, 13/6],[11/6],1/2-1/2*I*tan(f*x+e))*(d*sec(f*x+e))^(5/3)*(1+I*tan(f*x+e))^(1/6)*2^(5/6)/
f/(a^2+I*a^2*tan(f*x+e))

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Rubi [A]
time = 0.14, antiderivative size = 87, 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 \sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \, _2F_1\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 )} \end {gather*}

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 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 {(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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.82, size = 128, normalized size = 1.47 \begin {gather*} -\frac {3 e^{-i (4 e+5 f x)} \left (1+e^{2 i (e+f x)}\right ) \left (1+e^{2 i (e+f x)}+2 e^{2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {2}{3};\frac {5}{6};-e^{2 i (e+f x)}\right )\right ) (d \sec (e+f x))^{5/3} (-i \cos (f x)+\sin (f x))}{28 a^2 f} \end {gather*}

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 = 0.52, size = 0, normalized size = 0.00 \[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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]

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

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

[Out]

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

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