3.85 \(\int x^{3+m} \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=86 \[ -\frac {e^{2 a} 2^{-m-6} x^m (-b x)^{-m} \Gamma (m+4,-2 b x)}{b^4}-\frac {e^{-2 a} 2^{-m-6} x^m (b x)^{-m} \Gamma (m+4,2 b x)}{b^4}-\frac {x^{m+4}}{2 (m+4)} \]

[Out]

-1/2*x^(4+m)/(4+m)-2^(-6-m)*exp(2*a)*x^m*GAMMA(4+m,-2*b*x)/b^4/((-b*x)^m)-2^(-6-m)*x^m*GAMMA(4+m,2*b*x)/b^4/ex
p(2*a)/((b*x)^m)

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Rubi [A]  time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3312, 3307, 2181} \[ -\frac {e^{2 a} 2^{-m-6} x^m (-b x)^{-m} \text {Gamma}(m+4,-2 b x)}{b^4}-\frac {e^{-2 a} 2^{-m-6} x^m (b x)^{-m} \text {Gamma}(m+4,2 b x)}{b^4}-\frac {x^{m+4}}{2 (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[x^(3 + m)*Sinh[a + b*x]^2,x]

[Out]

-x^(4 + m)/(2*(4 + m)) - (2^(-6 - m)*E^(2*a)*x^m*Gamma[4 + m, -2*b*x])/(b^4*(-(b*x))^m) - (2^(-6 - m)*x^m*Gamm
a[4 + m, 2*b*x])/(b^4*E^(2*a)*(b*x)^m)

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin {align*} \int x^{3+m} \sinh ^2(a+b x) \, dx &=-\int \left (\frac {x^{3+m}}{2}-\frac {1}{2} x^{3+m} \cosh (2 a+2 b x)\right ) \, dx\\ &=-\frac {x^{4+m}}{2 (4+m)}+\frac {1}{2} \int x^{3+m} \cosh (2 a+2 b x) \, dx\\ &=-\frac {x^{4+m}}{2 (4+m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} x^{3+m} \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^{3+m} \, dx\\ &=-\frac {x^{4+m}}{2 (4+m)}-\frac {2^{-6-m} e^{2 a} x^m (-b x)^{-m} \Gamma (4+m,-2 b x)}{b^4}-\frac {2^{-6-m} e^{-2 a} x^m (b x)^{-m} \Gamma (4+m,2 b x)}{b^4}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 79, normalized size = 0.92 \[ \frac {1}{64} x^m \left (-\frac {e^{2 a} 2^{-m} (-b x)^{-m} \Gamma (m+4,-2 b x)}{b^4}-\frac {e^{-2 a} 2^{-m} (b x)^{-m} \Gamma (m+4,2 b x)}{b^4}-\frac {32 x^4}{m+4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3 + m)*Sinh[a + b*x]^2,x]

[Out]

(x^m*((-32*x^4)/(4 + m) - (E^(2*a)*Gamma[4 + m, -2*b*x])/(2^m*b^4*(-(b*x))^m) - Gamma[4 + m, 2*b*x]/(2^m*b^4*E
^(2*a)*(b*x)^m)))/64

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fricas [A]  time = 0.49, size = 136, normalized size = 1.58 \[ -\frac {4 \, b x \cosh \left ({\left (m + 3\right )} \log \relax (x)\right ) + {\left (m + 4\right )} \cosh \left ({\left (m + 3\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 4, 2 \, b x\right ) - {\left (m + 4\right )} \cosh \left ({\left (m + 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 4, -2 \, b x\right ) - {\left (m + 4\right )} \Gamma \left (m + 4, 2 \, b x\right ) \sinh \left ({\left (m + 3\right )} \log \left (2 \, b\right ) + 2 \, a\right ) + {\left (m + 4\right )} \Gamma \left (m + 4, -2 \, b x\right ) \sinh \left ({\left (m + 3\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m + 3\right )} \log \relax (x)\right )}{8 \, {\left (b m + 4 \, b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3+m)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/8*(4*b*x*cosh((m + 3)*log(x)) + (m + 4)*cosh((m + 3)*log(2*b) + 2*a)*gamma(m + 4, 2*b*x) - (m + 4)*cosh((m
+ 3)*log(-2*b) - 2*a)*gamma(m + 4, -2*b*x) - (m + 4)*gamma(m + 4, 2*b*x)*sinh((m + 3)*log(2*b) + 2*a) + (m + 4
)*gamma(m + 4, -2*b*x)*sinh((m + 3)*log(-2*b) - 2*a) + 4*b*x*sinh((m + 3)*log(x)))/(b*m + 4*b)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m + 3} \sinh \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3+m)*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^(m + 3)*sinh(b*x + a)^2, x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int x^{3+m} \left (\sinh ^{2}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3+m)*sinh(b*x+a)^2,x)

[Out]

int(x^(3+m)*sinh(b*x+a)^2,x)

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maxima [A]  time = 0.74, size = 71, normalized size = 0.83 \[ -\frac {1}{4} \, \left (2 \, b x\right )^{-m - 4} x^{m + 4} e^{\left (-2 \, a\right )} \Gamma \left (m + 4, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 4} x^{m + 4} e^{\left (2 \, a\right )} \Gamma \left (m + 4, -2 \, b x\right ) - \frac {x^{m + 4}}{2 \, {\left (m + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3+m)*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/4*(2*b*x)^(-m - 4)*x^(m + 4)*e^(-2*a)*gamma(m + 4, 2*b*x) - 1/4*(-2*b*x)^(-m - 4)*x^(m + 4)*e^(2*a)*gamma(m
+ 4, -2*b*x) - 1/2*x^(m + 4)/(m + 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{m+3}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m + 3)*sinh(a + b*x)^2,x)

[Out]

int(x^(m + 3)*sinh(a + b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m + 3} \sinh ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3+m)*sinh(b*x+a)**2,x)

[Out]

Integral(x**(m + 3)*sinh(a + b*x)**2, x)

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