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

Optimal. Leaf size=83 \[ \frac {3 i \sqrt [3]{1+i \tan (c+d x)} \sqrt [3]{e \sec (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {4}{3};\frac {7}{6};\frac {1}{2} (1-i \tan (c+d x))\right )}{\sqrt [3]{2} d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

3/2*I*hypergeom([1/6, 4/3],[7/6],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(1/3)*(1+I*tan(d*x+c))^(1/3)*2^(2/3)/d/(
a+I*a*tan(d*x+c))^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3505, 3523, 70, 69} \[ \frac {3 i \sqrt [3]{1+i \tan (c+d x)} \sqrt [3]{e \sec (c+d x)} \text {Hypergeometric2F1}\left (\frac {1}{6},\frac {4}{3},\frac {7}{6},\frac {1}{2} (1-i \tan (c+d x))\right )}{\sqrt [3]{2} d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*I)*Hypergeometric2F1[1/6, 4/3, 7/6, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(1/3)*(1 + I*Tan[c + d*x])^(1
/3))/(2^(1/3)*d*Sqrt[a + I*a*Tan[c + d*x]])

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

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Mathematica [A]  time = 0.57, size = 95, normalized size = 1.14 \[ \frac {3 \left (8 i-\frac {2 i e^{2 i (c+d x)} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {5}{3};-e^{2 i (c+d x)}\right )}{\sqrt [6]{1+e^{2 i (c+d x)}}}\right ) \sqrt [3]{e \sec (c+d x)}}{16 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(8*I - ((2*I)*E^((2*I)*(c + d*x))*Hypergeometric2F1[2/3, 5/6, 5/3, -E^((2*I)*(c + d*x))])/(1 + E^((2*I)*(c
+ d*x)))^(1/6))*(e*Sec[c + d*x])^(1/3))/(16*d*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ \frac {{\left (4 \, a d e^{\left (i \, d x + i \, c\right )} {\rm integral}\left (-\frac {i \cdot 2^{\frac {5}{6}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {1}{3} i \, d x + \frac {1}{3} i \, c\right )}}{4 \, a d}, x\right ) + 2^{\frac {5}{6}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \left (\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} {\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac {1}{3} i \, d x + \frac {1}{3} i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/4*(4*a*d*e^(I*d*x + I*c)*integral(-1/4*I*2^(5/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) +
1))^(1/3)*e^(1/3*I*d*x + 1/3*I*c)/(a*d), x) + 2^(5/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*
c) + 1))^(1/3)*(3*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(1/3*I*d*x + 1/3*I*c))*e^(-I*d*x - I*c)/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}}}{\sqrt {a +i a \tan \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/3)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(1/3)/(a+I*a*tan(d*x+c))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/3)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

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

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