3.271 \(\int x^m \sinh ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=120 \[ \frac{(m+1) x^{m+1} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac{2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}+\frac{2 b^2 n^2 x^{m+1}}{(m+1) \left ((m+1)^2-4 b^2 n^2\right )} \]

[Out]

(2*b^2*n^2*x^(1 + m))/((1 + m)*((1 + m)^2 - 4*b^2*n^2)) - (2*b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*L
og[c*x^n]])/((1 + m)^2 - 4*b^2*n^2) + ((1 + m)*x^(1 + m)*Sinh[a + b*Log[c*x^n]]^2)/((1 + m)^2 - 4*b^2*n^2)

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Rubi [A]  time = 0.0480274, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {5529, 30} \[ \frac{(m+1) x^{m+1} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}-\frac{2 b n x^{m+1} \sinh \left (a+b \log \left (c x^n\right )\right ) \cosh \left (a+b \log \left (c x^n\right )\right )}{(m+1)^2-4 b^2 n^2}+\frac{2 b^2 n^2 x^{m+1}}{(m+1) \left ((m+1)^2-4 b^2 n^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Sinh[a + b*Log[c*x^n]]^2,x]

[Out]

(2*b^2*n^2*x^(1 + m))/((1 + m)*((1 + m)^2 - 4*b^2*n^2)) - (2*b*n*x^(1 + m)*Cosh[a + b*Log[c*x^n]]*Sinh[a + b*L
og[c*x^n]])/((1 + m)^2 - 4*b^2*n^2) + ((1 + m)*x^(1 + m)*Sinh[a + b*Log[c*x^n]]^2)/((1 + m)^2 - 4*b^2*n^2)

Rule 5529

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^m \sinh ^2\left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{2 b n x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-4 b^2 n^2}+\frac{(1+m) x^{1+m} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-4 b^2 n^2}+\frac{\left (2 b^2 n^2\right ) \int x^m \, dx}{(1+m)^2-4 b^2 n^2}\\ &=\frac{2 b^2 n^2 x^{1+m}}{(1+m) \left ((1+m)^2-4 b^2 n^2\right )}-\frac{2 b n x^{1+m} \cosh \left (a+b \log \left (c x^n\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-4 b^2 n^2}+\frac{(1+m) x^{1+m} \sinh ^2\left (a+b \log \left (c x^n\right )\right )}{(1+m)^2-4 b^2 n^2}\\ \end{align*}

Mathematica [A]  time = 0.280334, size = 89, normalized size = 0.74 \[ \frac{x^{m+1} \left (-2 b (m+1) n \sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+(m+1)^2 \cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+4 b^2 n^2-m^2-2 m-1\right )}{2 (m+1) (-2 b n+m+1) (2 b n+m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*Sinh[a + b*Log[c*x^n]]^2,x]

[Out]

(x^(1 + m)*(-1 - 2*m - m^2 + 4*b^2*n^2 + (1 + m)^2*Cosh[2*(a + b*Log[c*x^n])] - 2*b*(1 + m)*n*Sinh[2*(a + b*Lo
g[c*x^n])]))/(2*(1 + m)*(1 + m - 2*b*n)*(1 + m + 2*b*n))

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \sinh \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*sinh(a+b*ln(c*x^n))^2,x)

[Out]

int(x^m*sinh(a+b*ln(c*x^n))^2,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sinh(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.18709, size = 726, normalized size = 6.05 \begin{align*} \frac{{\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} \cosh \left (m \log \left (x\right )\right ) +{\left (4 \, b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} x \cosh \left (m \log \left (x\right )\right ) +{\left ({\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (m \log \left (x\right )\right ) +{\left (m^{2} + 2 \, m + 1\right )} x \sinh \left (m \log \left (x\right )\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - 4 \,{\left ({\left (b m + b\right )} n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \cosh \left (m \log \left (x\right )\right ) +{\left (b m + b\right )} n x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (m \log \left (x\right )\right )\right )} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) +{\left ({\left (m^{2} + 2 \, m + 1\right )} x \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} +{\left (4 \, b^{2} n^{2} - m^{2} - 2 \, m - 1\right )} x\right )} \sinh \left (m \log \left (x\right )\right )}{2 \,{\left (m^{3} - 4 \,{\left (b^{2} m + b^{2}\right )} n^{2} + 3 \, m^{2} + 3 \, m + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sinh(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

1/2*((m^2 + 2*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^2*cosh(m*log(x)) + (4*b^2*n^2 - m^2 - 2*m - 1)*x*cosh(m
*log(x)) + ((m^2 + 2*m + 1)*x*cosh(m*log(x)) + (m^2 + 2*m + 1)*x*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) +
a)^2 - 4*((b*m + b)*n*x*cosh(b*n*log(x) + b*log(c) + a)*cosh(m*log(x)) + (b*m + b)*n*x*cosh(b*n*log(x) + b*log
(c) + a)*sinh(m*log(x)))*sinh(b*n*log(x) + b*log(c) + a) + ((m^2 + 2*m + 1)*x*cosh(b*n*log(x) + b*log(c) + a)^
2 + (4*b^2*n^2 - m^2 - 2*m - 1)*x)*sinh(m*log(x)))/(m^3 - 4*(b^2*m + b^2)*n^2 + 3*m^2 + 3*m + 1)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*sinh(a+b*ln(c*x**n))**2,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.27568, size = 1023, normalized size = 8.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*sinh(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

1/2*b*c^(2*b)*m*n*x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/4*c^(2*b)*m^2*
x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) + 1/2*b*c^(2*b)*n*x*x^(2*b*n)*x^m*e^
(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 2*b^2*n^2*x*x^m/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m
^2 - 3*m - 1) - 1/2*c^(2*b)*m*x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/4*
c^(2*b)*x*x^(2*b*n)*x^m*e^(2*a)/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) + 1/2*m^2*x*x^m/(4*b^2*m*n^2
+ 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/2*b*m*n*x*x^m*e^(-2*a)/((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m
- 1)*c^(2*b)*x^(2*b*n)) + m*x*x^m/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1) - 1/4*m^2*x*x^m*e^(-2*a)/
((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n)) - 1/2*b*n*x*x^m*e^(-2*a)/((4*b^2*m*n^2 +
4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n)) + 1/2*x*x^m/(4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*
m - 1) - 1/2*m*x*x^m*e^(-2*a)/((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n)) - 1/4*x*x^
m*e^(-2*a)/((4*b^2*m*n^2 + 4*b^2*n^2 - m^3 - 3*m^2 - 3*m - 1)*c^(2*b)*x^(2*b*n))