3.1.25 \(\int (e x)^m \sinh ^2(a+b x^2) \, dx\) [25]

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

[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 = 0.11, 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 = {5448, 5437, 2250} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sinh[a + b*x^2]^2,x]

[Out]

-1/2*(e*x)^(1 + m)/(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 2250

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

Rule 5437

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 5448

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 = 0.43, size = 152, normalized size = 1.13 \begin {gather*} -\frac {2^{\frac {1}{2} (-7-m)} x (e x)^m \left (-b^2 x^4\right )^{\frac {1}{2} (-1-m)} \left (2^{\frac {5+m}{2}} \left (-b^2 x^4\right )^{\frac {1+m}{2}}+(1+m) \left (-b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},2 b x^2\right ) (\cosh (2 a)-\sinh (2 a))+(1+m) \left (b x^2\right )^{\frac {1+m}{2}} \Gamma \left (\frac {1+m}{2},-2 b x^2\right ) (\cosh (2 a)+\sinh (2 a))\right )}{1+m} \end {gather*}

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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Maple [F]
time = 1.44, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{m} \left (\sinh ^{2}\left (x^{2} b +a \right )\right )\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (111) = 222\).
time = 0.09, size = 274, normalized size = 2.03 \begin {gather*} -\frac {8 \, b x \cosh \left (m \log \left (x \cosh \left (1\right ) + x \sinh \left (1\right )\right )\right ) + {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (\frac {2 \, b}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) + 2 \, a\right ) \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, 2 \, b x^{2}\right ) - {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\right )} \cosh \left (\frac {1}{2} \, {\left (m - 1\right )} \log \left (-\frac {2 \, b}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{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 (x \cosh \left (1\right ) + x \sinh \left (1\right )\right )\right ) - {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\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}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) + 2 \, a\right ) + {\left ({\left (m + 1\right )} \cosh \left (1\right ) + {\left (m + 1\right )} \sinh \left (1\right )\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}{\cosh \left (1\right )^{2} + 2 \, \cosh \left (1\right ) \sinh \left (1\right ) + \sinh \left (1\right )^{2}}\right ) - 2 \, a\right )}{16 \, {\left (b m + b\right )}} \end {gather*}

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

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

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

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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