3.25 \(\int (e x)^m \sinh ^2(a+b x^2) \, dx\)

Optimal. Leaf size=135 \[ -\frac {e^{2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},-2 b x^2\right )}{e}-\frac {e^{-2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \Gamma \left (\frac {m+1}{2},2 b x^2\right )}{e}-\frac {(e x)^{m+1}}{2 e (m+1)} \]

[Out]

-1/2*(e*x)^(1+m)/e/(1+m)-2^(-7/2-1/2*m)*exp(2*a)*(e*x)^(1+m)*(-b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,-2*b*x^2)/e
-2^(-7/2-1/2*m)*(e*x)^(1+m)*(b*x^2)^(-1/2-1/2*m)*GAMMA(1/2+1/2*m,2*b*x^2)/e/exp(2*a)

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Rubi [A]  time = 0.15, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5340, 5329, 2218} \[ -\frac {e^{2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (-b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},-2 b x^2\right )}{e}-\frac {e^{-2 a} 2^{-\frac {m}{2}-\frac {7}{2}} \left (b x^2\right )^{\frac {1}{2} (-m-1)} (e x)^{m+1} \text {Gamma}\left (\frac {m+1}{2},2 b x^2\right )}{e}-\frac {(e x)^{m+1}}{2 e (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^2]^2,x]

[Out]

-(e*x)^(1 + m)/(2*e*(1 + m)) - (2^(-7/2 - m/2)*E^(2*a)*(e*x)^(1 + m)*(-(b*x^2))^((-1 - m)/2)*Gamma[(1 + m)/2,
-2*b*x^2])/e - (2^(-7/2 - m/2)*(e*x)^(1 + m)*(b*x^2)^((-1 - m)/2)*Gamma[(1 + m)/2, 2*b*x^2])/(e*E^(2*a))

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 5329

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(c + d*x^n), x], x]
+ Dist[1/2, Int[(e*x)^m*E^(-c - d*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rule 5340

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

Rubi steps

\begin {align*} \int (e x)^m \sinh ^2\left (a+b x^2\right ) \, dx &=\int \left (-\frac {1}{2} (e x)^m+\frac {1}{2} (e x)^m \cosh \left (2 a+2 b x^2\right )\right ) \, dx\\ &=-\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{2} \int (e x)^m \cosh \left (2 a+2 b x^2\right ) \, dx\\ &=-\frac {(e x)^{1+m}}{2 e (1+m)}+\frac {1}{4} \int e^{-2 a-2 b x^2} (e x)^m \, dx+\frac {1}{4} \int e^{2 a+2 b x^2} (e x)^m \, dx\\ &=-\frac {(e x)^{1+m}}{2 e (1+m)}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{2 a} (e x)^{1+m} \left (-b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},-2 b x^2\right )}{e}-\frac {2^{-\frac {7}{2}-\frac {m}{2}} e^{-2 a} (e x)^{1+m} \left (b x^2\right )^{\frac {1}{2} (-1-m)} \Gamma \left (\frac {1+m}{2},2 b x^2\right )}{e}\\ \end {align*}

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Mathematica [A]  time = 0.58, size = 152, normalized size = 1.13 \[ -\frac {2^{\frac {1}{2} (-m-7)} x \left (-b^2 x^4\right )^{\frac {1}{2} (-m-1)} (e x)^m \left ((m+1) (\cosh (2 a)-\sinh (2 a)) \left (-b x^2\right )^{\frac {m+1}{2}} \Gamma \left (\frac {m+1}{2},2 b x^2\right )+(m+1) (\sinh (2 a)+\cosh (2 a)) \left (b x^2\right )^{\frac {m+1}{2}} \Gamma \left (\frac {m+1}{2},-2 b x^2\right )+2^{\frac {m+5}{2}} \left (-b^2 x^4\right )^{\frac {m+1}{2}}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^((-7 - m)/2)*x*(e*x)^m*(-(b^2*x^4))^((-1 - m)/2)*(2^((5 + m)/2)*(-(b^2*x^4))^((1 + m)/2) + (1 + m)*(-(b*x
^2))^((1 + m)/2)*Gamma[(1 + m)/2, 2*b*x^2]*(Cosh[2*a] - Sinh[2*a]) + (1 + m)*(b*x^2)^((1 + m)/2)*Gamma[(1 + m)
/2, -2*b*x^2]*(Cosh[2*a] + Sinh[2*a])))/(1 + m))

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fricas [A]  time = 0.75, size = 174, normalized size = 1.29 \[ -\frac {8 \, b x \cosh \left (m \log \left (e x\right )\right ) + {\left (e m + e\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{e^{2}}\right ) + 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) - {\left (e m + e\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{e^{2}}\right ) - 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 \, b x^{2}\right ) + 8 \, b x \sinh \left (m \log \left (e x\right )\right ) - {\left (e m + e\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{e^{2}}\right ) + 2 \, a\right ) + {\left (e m + e\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -2 \, b x^{2}\right ) \sinh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{e^{2}}\right ) - 2 \, a\right )}{16 \, {\left (b m + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(8*b*x*cosh(m*log(e*x)) + (e*m + e)*cosh(1/2*(m - 1)*log(2*b/e^2) + 2*a)*gamma(1/2*m + 1/2, 2*b*x^2) - (
e*m + e)*cosh(1/2*(m - 1)*log(-2*b/e^2) - 2*a)*gamma(1/2*m + 1/2, -2*b*x^2) + 8*b*x*sinh(m*log(e*x)) - (e*m +
e)*gamma(1/2*m + 1/2, 2*b*x^2)*sinh(1/2*(m - 1)*log(2*b/e^2) + 2*a) + (e*m + e)*gamma(1/2*m + 1/2, -2*b*x^2)*s
inh(1/2*(m - 1)*log(-2*b/e^2) - 2*a))/(b*m + b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\sinh ^{2}\left (b \,x^{2}+a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, e^{m} \int e^{\left (2 \, b x^{2} + m \log \relax (x) + 2 \, a\right )}\,{d x} + \frac {1}{4} \, e^{m} \int e^{\left (-2 \, b x^{2} + m \log \relax (x) - 2 \, a\right )}\,{d x} - \frac {\left (e x\right )^{m + 1}}{2 \, e {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^2)^2*(e*x)^m,x)

[Out]

int(sinh(a + b*x^2)^2*(e*x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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