3.2.96 \(\int (e x)^m \coth ^3(d (a+b \log (c x^n))) \, dx\) [196]

Optimal. Leaf size=306 \[ \frac {(1+m+b d n) (1+m+2 b d n) (e x)^{1+m}}{2 b^2 d^2 e (1+m) n^2}-\frac {(e x)^{1+m} \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}+\frac {e^{-2 a d} (e x)^{1+m} \left (\frac {e^{2 a d} (1+m-2 b d n)}{n}+\frac {e^{4 a d} (1+m+2 b d n) \left (c x^n\right )^{2 b d}}{n}\right )}{2 b^2 d^2 e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {\left (1+2 m+m^2+2 b^2 d^2 n^2\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2 b d n};1+\frac {1+m}{2 b d n};e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b^2 d^2 e (1+m) n^2} \]

[Out]

1/2*(b*d*n+m+1)*(2*b*d*n+m+1)*(e*x)^(1+m)/b^2/d^2/e/(1+m)/n^2-1/2*(e*x)^(1+m)*(1+exp(2*a*d)*(c*x^n)^(2*b*d))^2
/b/d/e/n/(1-exp(2*a*d)*(c*x^n)^(2*b*d))^2+1/2*(e*x)^(1+m)*(exp(2*a*d)*(-2*b*d*n+m+1)/n+exp(4*a*d)*(2*b*d*n+m+1
)*(c*x^n)^(2*b*d)/n)/b^2/d^2/e/exp(2*a*d)/n/(1-exp(2*a*d)*(c*x^n)^(2*b*d))-(2*b^2*d^2*n^2+m^2+2*m+1)*(e*x)^(1+
m)*hypergeom([1, 1/2*(1+m)/b/d/n],[1+1/2*(1+m)/b/d/n],exp(2*a*d)*(c*x^n)^(2*b*d))/b^2/d^2/e/(1+m)/n^2

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Rubi [A]
time = 0.33, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5659, 5657, 516, 608, 470, 371} \begin {gather*} -\frac {(e x)^{m+1} \left (2 b^2 d^2 n^2+m^2+2 m+1\right ) \, _2F_1\left (1,\frac {m+1}{2 b d n};\frac {m+1}{2 b d n}+1;e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b^2 d^2 e (m+1) n^2}+\frac {e^{-2 a d} (e x)^{m+1} \left (\frac {e^{4 a d} (2 b d n+m+1) \left (c x^n\right )^{2 b d}}{n}+\frac {e^{2 a d} (-2 b d n+m+1)}{n}\right )}{2 b^2 d^2 e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {(e x)^{m+1} \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}{2 b d e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}+\frac {(e x)^{m+1} (b d n+m+1) (2 b d n+m+1)}{2 b^2 d^2 e (m+1) n^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Coth[d*(a + b*Log[c*x^n])]^3,x]

[Out]

((1 + m + b*d*n)*(1 + m + 2*b*d*n)*(e*x)^(1 + m))/(2*b^2*d^2*e*(1 + m)*n^2) - ((e*x)^(1 + m)*(1 + E^(2*a*d)*(c
*x^n)^(2*b*d))^2)/(2*b*d*e*n*(1 - E^(2*a*d)*(c*x^n)^(2*b*d))^2) + ((e*x)^(1 + m)*((E^(2*a*d)*(1 + m - 2*b*d*n)
)/n + (E^(4*a*d)*(1 + m + 2*b*d*n)*(c*x^n)^(2*b*d))/n))/(2*b^2*d^2*e*E^(2*a*d)*n*(1 - E^(2*a*d)*(c*x^n)^(2*b*d
))) - ((1 + 2*m + m^2 + 2*b^2*d^2*n^2)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/(2*b*d*n), 1 + (1 + m)/(2*b*
d*n), E^(2*a*d)*(c*x^n)^(2*b*d)])/(b^2*d^2*e*(1 + m)*n^2)

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 516

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, n}, x] && NeQ[b*c - a*d,
0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 608

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, n},
x] && 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]

Rule 5659

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

Rubi steps

\begin {align*} \int (e x)^m \coth ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int (e x)^m \coth ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}

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Mathematica [A]
time = 13.21, size = 600, normalized size = 1.96 \begin {gather*} \frac {x (e x)^m \coth \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}{1+m}-\frac {x (e x)^m \text {csch}^2\left (b d n \log (x)+d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}{2 b d n}+\frac {(1+m) x (e x)^m \text {csch}\left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \text {csch}\left (b d n \log (x)+d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \sinh (b d n \log (x))}{2 b^2 d^2 n^2}-\frac {\left (1+2 m+m^2+2 b^2 d^2 n^2\right ) x^{-m} (e x)^m \text {csch}\left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (\frac {x^{1+m} \text {csch}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \sinh (b d n \log (x))}{1+m}+\frac {e^{-\frac {(1+2 m) \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{b n}} \left (e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 b d n) \coth \left (d \left (a+b \log \left (c x^n\right )\right )\right )+e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 b d n) \, _2F_1\left (1,\frac {1+m}{2 b d n};1+\frac {1+m}{2 b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+e^{\frac {a (1+2 m+2 b d n)}{b n}+(1+m+2 b d n) \log (x)+\frac {(1+2 m+2 b d n) \left (-n \log (x)+\log \left (c x^n\right )\right )}{n}} (1+m) \, _2F_1\left (1,\frac {1+m+2 b d n}{2 b d n};\frac {1+m+4 b d n}{2 b d n};e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )\right ) \sinh \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}{(1+m) (1+m+2 b d n)}\right )}{2 b^2 d^2 n^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Coth[d*(a + b*Log[c*x^n])]^3,x]

[Out]

(x*(e*x)^m*Coth[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))])/(1 + m) - (x*(e*x)^m*Csch[b*d*n*Log[x] + d*(a + b*(-(n*
Log[x]) + Log[c*x^n]))]^2)/(2*b*d*n) + ((1 + m)*x*(e*x)^m*Csch[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*Csch[b*d*
n*Log[x] + d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*Sinh[b*d*n*Log[x]])/(2*b^2*d^2*n^2) - ((1 + 2*m + m^2 + 2*b^2
*d^2*n^2)*(e*x)^m*Csch[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*((x^(1 + m)*Csch[d*(a + b*Log[c*x^n])]*Sinh[b*d*n
*Log[x]])/(1 + m) + ((E^((a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*m)*(-(n*Log[x]) + Log[c*x^n]))/(b*n))*(1 +
m + 2*b*d*n)*Coth[d*(a + b*Log[c*x^n])] + E^((a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*m)*(-(n*Log[x]) + Log
[c*x^n]))/(b*n))*(1 + m + 2*b*d*n)*Hypergeometric2F1[1, (1 + m)/(2*b*d*n), 1 + (1 + m)/(2*b*d*n), E^(2*d*(a +
b*Log[c*x^n]))] + E^((a*(1 + 2*m + 2*b*d*n))/(b*n) + (1 + m + 2*b*d*n)*Log[x] + ((1 + 2*m + 2*b*d*n)*(-(n*Log[
x]) + Log[c*x^n]))/n)*(1 + m)*Hypergeometric2F1[1, (1 + m + 2*b*d*n)/(2*b*d*n), (1 + m + 4*b*d*n)/(2*b*d*n), E
^(2*d*(a + b*Log[c*x^n]))])*Sinh[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))])/(E^(((1 + 2*m)*(a + b*(-(n*Log[x]) + L
og[c*x^n])))/(b*n))*(1 + m)*(1 + m + 2*b*d*n))))/(2*b^2*d^2*n^2*x^m)

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Maple [F]
time = 1.40, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\coth ^{3}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*coth(d*(a+b*ln(c*x^n)))^3,x)

[Out]

int((e*x)^m*coth(d*(a+b*ln(c*x^n)))^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(d*(a+b*log(c*x^n)))^3,x, algorithm="maxima")

[Out]

-(2*b^2*d^2*n^2*e^m + m^2*e^m + 2*m*e^m + e^m)*integrate(1/2*x^m/(b^2*c^(b*d)*d^2*n^2*e^(b*d*log(x^n) + a*d) +
b^2*d^2*n^2), x) + (2*b^2*d^2*n^2*e^m + m^2*e^m + 2*m*e^m + e^m)*integrate(1/2*x^m/(b^2*c^(b*d)*d^2*n^2*e^(b*
d*log(x^n) + a*d) - b^2*d^2*n^2), x) + (b^2*c^(4*b*d)*d^2*n^2*x*e^(4*b*d*log(x^n) + 4*a*d + m*log(x) + m) + (b
^2*d^2*n^2*e^m + m^2*e^m + 2*m*e^m + e^m)*x*x^m - (2*b^2*c^(2*b*d)*d^2*n^2*e^(2*a*d + m) + 2*(m*n*e^m + n*e^m)
*b*c^(2*b*d)*d*e^(2*a*d) + (m^2*e^m + 2*m*e^m + e^m)*c^(2*b*d)*e^(2*a*d))*x*e^(2*b*d*log(x^n) + m*log(x)))/((m
*n^2 + n^2)*b^2*c^(4*b*d)*d^2*e^(4*b*d*log(x^n) + 4*a*d) - 2*(m*n^2 + n^2)*b^2*c^(2*b*d)*d^2*e^(2*b*d*log(x^n)
+ 2*a*d) + (m*n^2 + n^2)*b^2*d^2)

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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(d*(a+b*log(c*x^n)))^3,x, algorithm="fricas")

[Out]

integral((x*e)^m*coth(b*d*log(c*x^n) + a*d)^3, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*coth(d*(a+b*ln(c*x**n)))**3,x)

[Out]

Timed out

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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(d*(a+b*log(c*x^n)))^3,x, algorithm="giac")

[Out]

integrate((e*x)^m*coth((b*log(c*x^n) + a)*d)^3, x)

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

[Out]

int(coth(d*(a + b*log(c*x^n)))^3*(e*x)^m, x)

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