3.3.65 \(\int \frac {a+i a \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [265]

Optimal. Leaf size=67 \[ -\frac {3 i 2^{5/6} a \, _2F_1\left (-\frac {1}{6},\frac {1}{6};\frac {5}{6};\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [6]{1+i \tan (e+f x)}}{f \sqrt [3]{d \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 67, 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 = {3586, 3604, 72, 71} \begin {gather*} -\frac {3 i 2^{5/6} a \sqrt [6]{1+i \tan (e+f x)} \, _2F_1\left (-\frac {1}{6},\frac {1}{6};\frac {5}{6};\frac {1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [3]{d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.45, size = 98, normalized size = 1.46 \begin {gather*} -\frac {3 i 2^{2/3} a e^{2 i (e+f x)} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i (e+f x)}\right )}{5 \sqrt [3]{\frac {d e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt [3]{1+e^{2 i (e+f x)}} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.26, size = 0, normalized size = 0.00 \[\int \frac {a +i a \tan \left (f x +e \right )}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i a \left (\int \left (- \frac {i}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\right )\, dx + \int \frac {\tan {\left (e + f x \right )}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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