3.2.67 \(\int (e x)^m \coth ^3(a+2 \log (x)) \, dx\) [167]

Optimal. Leaf size=177 \[ \frac {(3+m) (5+m) (e x)^{1+m}}{8 e (1+m)}-\frac {(e x)^{1+m} \left (1+e^{2 a} x^4\right )^2}{4 e \left (1-e^{2 a} x^4\right )^2}-\frac {e^{-2 a} (e x)^{1+m} \left (e^{2 a} (3-m)-e^{4 a} (5+m) x^4\right )}{8 e \left (1-e^{2 a} x^4\right )}-\frac {\left (9+2 m+m^2\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{4};\frac {5+m}{4};e^{2 a} x^4\right )}{4 e (1+m)} \]

[Out]

1/8*(3+m)*(5+m)*(e*x)^(1+m)/e/(1+m)-1/4*(e*x)^(1+m)*(1+exp(2*a)*x^4)^2/e/(1-exp(2*a)*x^4)^2-1/8*(e*x)^(1+m)*(e
xp(2*a)*(3-m)-exp(4*a)*(5+m)*x^4)/e/exp(2*a)/(1-exp(2*a)*x^4)-1/4*(m^2+2*m+9)*(e*x)^(1+m)*hypergeom([1, 1/4+1/
4*m],[5/4+1/4*m],exp(2*a)*x^4)/e/(1+m)

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Rubi [A]
time = 0.13, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5657, 479, 591, 470, 371} \begin {gather*} -\frac {\left (m^2+2 m+9\right ) (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{4};\frac {m+5}{4};e^{2 a} x^4\right )}{4 e (m+1)}-\frac {\left (e^{2 a} x^4+1\right )^2 (e x)^{m+1}}{4 e \left (1-e^{2 a} x^4\right )^2}-\frac {e^{-2 a} \left (e^{2 a} (3-m)-e^{4 a} (m+5) x^4\right ) (e x)^{m+1}}{8 e \left (1-e^{2 a} x^4\right )}+\frac {(m+3) (m+5) (e x)^{m+1}}{8 e (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Coth[a + 2*Log[x]]^3,x]

[Out]

((3 + m)*(5 + m)*(e*x)^(1 + m))/(8*e*(1 + m)) - ((e*x)^(1 + m)*(1 + E^(2*a)*x^4)^2)/(4*e*(1 - E^(2*a)*x^4)^2)
- ((e*x)^(1 + m)*(E^(2*a)*(3 - m) - E^(4*a)*(5 + m)*x^4))/(8*e*E^(2*a)*(1 - E^(2*a)*x^4)) - ((9 + 2*m + m^2)*(
e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, E^(2*a)*x^4])/(4*e*(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 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 5657

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

Rubi steps

\begin {align*} \int (e x)^m \coth ^3(a+2 \log (x)) \, dx &=\int (e x)^m \coth ^3(a+2 \log (x)) \, dx\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 108, normalized size = 0.61 \begin {gather*} -\frac {x (e x)^m \left (-1+6 \, _2F_1\left (1,\frac {1+m}{4};\frac {5+m}{4};x^4 (\cosh (2 a)+\sinh (2 a))\right )-12 \, _2F_1\left (2,\frac {1+m}{4};\frac {5+m}{4};x^4 (\cosh (2 a)+\sinh (2 a))\right )+8 \, _2F_1\left (3,\frac {1+m}{4};\frac {5+m}{4};x^4 (\cosh (2 a)+\sinh (2 a))\right )\right )}{1+m} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Coth[a + 2*Log[x]]^3,x]

[Out]

-((x*(e*x)^m*(-1 + 6*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, x^4*(Cosh[2*a] + Sinh[2*a])] - 12*Hypergeometr
ic2F1[2, (1 + m)/4, (5 + m)/4, x^4*(Cosh[2*a] + Sinh[2*a])] + 8*Hypergeometric2F1[3, (1 + m)/4, (5 + m)/4, x^4
*(Cosh[2*a] + Sinh[2*a])]))/(1 + m))

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Maple [F]
time = 0.60, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\coth ^{3}\left (a +2 \ln \left (x \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*coth(a+2*ln(x))^3,x)

[Out]

int((e*x)^m*coth(a+2*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*coth(a+2*log(x))^3,x, algorithm="maxima")

[Out]

integrate((x*e)^m*coth(a + 2*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*coth(a+2*log(x))^3,x, algorithm="fricas")

[Out]

integral((x*e)^m*coth(a + 2*log(x))^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e x\right )^{m} \coth ^{3}{\left (a + 2 \log {\left (x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*coth(a+2*ln(x))**3,x)

[Out]

Integral((e*x)**m*coth(a + 2*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*coth(a+2*log(x))^3,x, algorithm="giac")

[Out]

integrate((e*x)^m*coth(a + 2*log(x))^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {coth}\left (a+2\,\ln \left (x\right )\right )}^3\,{\left (e\,x\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(a + 2*log(x))^3*(e*x)^m,x)

[Out]

int(coth(a + 2*log(x))^3*(e*x)^m, x)

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