3.2.95 \(\int (e x)^m \coth ^2(d (a+b \log (c x^n))) \, dx\) [195]

Optimal. Leaf size=168 \[ \frac {(1+m+b d n) (e x)^{1+m}}{b d e (1+m) n}+\frac {(e x)^{1+m} \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2 b d n};1+\frac {1+m}{2 b d n};e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d e n} \]

[Out]

(b*d*n+m+1)*(e*x)^(1+m)/b/d/e/(1+m)/n+(e*x)^(1+m)*(1+exp(2*a*d)*(c*x^n)^(2*b*d))/b/d/e/n/(1-exp(2*a*d)*(c*x^n)
^(2*b*d))-2*(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/d/e/n

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Rubi [A]
time = 0.15, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5659, 5657, 516, 470, 371} \begin {gather*} -\frac {2 (e x)^{m+1} \, _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 d e n}+\frac {(e x)^{m+1} \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}{b d e n \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}+\frac {(e x)^{m+1} (b d n+m+1)}{b d e (m+1) n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Coth[d*(a + b*Log[c*x^n])]^2,x]

[Out]

((1 + m + b*d*n)*(e*x)^(1 + m))/(b*d*e*(1 + m)*n) + ((e*x)^(1 + m)*(1 + E^(2*a*d)*(c*x^n)^(2*b*d)))/(b*d*e*n*(
1 - E^(2*a*d)*(c*x^n)^(2*b*d))) - (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*d*e*n)

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 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 ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=\int (e x)^m \coth ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\\ \end {align*}

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Mathematica [A]
time = 9.90, size = 312, normalized size = 1.86 \begin {gather*} (e x)^m \left (\frac {x}{1+m}-\frac {e^{-\frac {(1+2 m) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} x^{-2 m} \left (e^{\frac {(1+2 m) \left (a+b \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 {(1+2 m) \left (a+b \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 {(1+2 m+2 b d n) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} (1+m) x^{1+2 m+2 b d n} \, _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 )}{b d n (1+m+2 b d n)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Coth[d*(a + b*Log[c*x^n])]^2,x]

[Out]

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

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Maple [F]
time = 0.43, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\coth ^{2}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*coth(d*(a+b*ln(c*x^n)))^2,x)

[Out]

int((e*x)^m*coth(d*(a+b*ln(c*x^n)))^2,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)))^2,x, algorithm="maxima")

[Out]

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

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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)))^2,x, algorithm="fricas")

[Out]

integral((x*e)^m*coth(b*d*log(c*x^n) + a*d)^2, 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 ^{2}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \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)))**2,x)

[Out]

Integral((e*x)**m*coth(a*d + b*d*log(c*x**n))**2, 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(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")

[Out]

integrate((e*x)^m*coth((b*log(c*x^n) + a)*d)^2, x)

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

[Out]

int(coth(d*(a + b*log(c*x^n)))^2*(e*x)^m, x)

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