3.2.52 \(\int (e x)^m \tan ^3(a+i \log (x)) \, dx\) [152]

Optimal. Leaf size=184 \[ -\frac {i (1-m) m x (e x)^m}{2 (1+m)}+\frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {i e^{-2 i a} \left (e^{2 i a} (3+m)+\frac {e^{4 i a} (1-m)}{x^2}\right ) x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}-\frac {i \left (3+2 m+m^2\right ) x (e x)^m \, _2F_1\left (1,\frac {1}{2} (-1-m);\frac {1-m}{2};-\frac {e^{2 i a}}{x^2}\right )}{1+m} \]

[Out]

-1/2*I*(1-m)*m*x*(e*x)^m/(1+m)+1/2*I*(1-exp(2*I*a)/x^2)^2*x*(e*x)^m/(1+exp(2*I*a)/x^2)^2+1/2*I*(exp(2*I*a)*(3+
m)+exp(4*I*a)*(1-m)/x^2)*x*(e*x)^m/exp(2*I*a)/(1+exp(2*I*a)/x^2)-I*(m^2+2*m+3)*x*(e*x)^m*hypergeom([1, -1/2-1/
2*m],[1/2-1/2*m],-exp(2*I*a)/x^2)/(1+m)

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Rubi [A]
time = 0.18, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {4591, 511, 479, 591, 470, 371} \begin {gather*} -\frac {i \left (m^2+2 m+3\right ) x (e x)^m \, _2F_1\left (1,\frac {1}{2} (-m-1);\frac {1-m}{2};-\frac {e^{2 i a}}{x^2}\right )}{m+1}+\frac {i e^{-2 i a} x \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right ) (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}+\frac {i x \left (1-\frac {e^{2 i a}}{x^2}\right )^2 (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}-\frac {i (1-m) m x (e x)^m}{2 (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Tan[a + I*Log[x]]^3,x]

[Out]

((-1/2*I)*(1 - m)*m*x*(e*x)^m)/(1 + m) + ((I/2)*(1 - E^((2*I)*a)/x^2)^2*x*(e*x)^m)/(1 + E^((2*I)*a)/x^2)^2 + (
(I/2)*(E^((2*I)*a)*(3 + m) + (E^((4*I)*a)*(1 - m))/x^2)*x*(e*x)^m)/(E^((2*I)*a)*(1 + E^((2*I)*a)/x^2)) - (I*(3
+ 2*m + m^2)*x*(e*x)^m*Hypergeometric2F1[1, (-1 - m)/2, (1 - m)/2, -(E^((2*I)*a)/x^2)])/(1 + m)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-(c*b -
a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Dist[1/(a*b*n*(p + 1)),
Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(
p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 511

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(-(e*x)^m
)*(x^(-1))^m, Subst[Int[(a + b/x^n)^p*((c + d/x^n)^q/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, b, c, d, e, m, p
, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0] &&  !RationalQ[m]

Rule 591

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 4591

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((I - I*E^(2*I*a*d)*
x^(2*I*b*d))/(1 + E^(2*I*a*d)*x^(2*I*b*d)))^p, x] /; FreeQ[{a, b, d, e, m, p}, x]

Rubi steps

\begin {align*} \int (e x)^m \tan ^3(a+i \log (x)) \, dx &=\int (e x)^m \tan ^3(a+i \log (x)) \, dx\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 125, normalized size = 0.68 \begin {gather*} \frac {i x (e x)^m \left (-1+6 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-x^2 (\cos (2 a)-i \sin (2 a))\right )-12 \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-x^2 (\cos (2 a)-i \sin (2 a))\right )+8 \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-x^2 (\cos (2 a)-i \sin (2 a))\right )\right )}{1+m} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Tan[a + I*Log[x]]^3,x]

[Out]

(I*x*(e*x)^m*(-1 + 6*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))] - 12*Hypergeom
etric2F1[2, (1 + m)/2, (3 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))] + 8*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/
2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]))/(1 + m)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\tan ^{3}\left (a +i \ln \left (x \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*tan(a+I*ln(x))^3,x)

[Out]

int((e*x)^m*tan(a+I*ln(x))^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((e*x)^m*tan(a+I*log(x))^3,x, algorithm="maxima")

[Out]

integrate((x*e)^m*tan(a + I*log(x))^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((e*x)^m*tan(a+I*log(x))^3,x, algorithm="fricas")

[Out]

integral((-I*x^6 + 3*I*x^4*e^(2*I*a) - 3*I*x^2*e^(4*I*a) + I*e^(6*I*a))*e^(m*log(x) + m)/(x^6 + 3*x^4*e^(2*I*a
) + 3*x^2*e^(4*I*a) + e^(6*I*a)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e x\right )^{m} \tan ^{3}{\left (a + i \log {\left (x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*tan(a+I*ln(x))**3,x)

[Out]

Integral((e*x)**m*tan(a + I*log(x))**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((e*x)^m*tan(a+I*log(x))^3,x, algorithm="giac")

[Out]

integrate((e*x)^m*tan(a + I*log(x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(a + log(x)*1i)^3*(e*x)^m,x)

[Out]

int(tan(a + log(x)*1i)^3*(e*x)^m, x)

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