3.89 \(\int x \sinh ((a+b x)^2) \, dx\)

Optimal. Leaf size=54 \[ \frac{\sqrt{\pi } a \text{Erf}(a+b x)}{4 b^2}-\frac{\sqrt{\pi } a \text{Erfi}(a+b x)}{4 b^2}+\frac{\cosh \left ((a+b x)^2\right )}{2 b^2} \]

[Out]

Cosh[(a + b*x)^2]/(2*b^2) + (a*Sqrt[Pi]*Erf[a + b*x])/(4*b^2) - (a*Sqrt[Pi]*Erfi[a + b*x])/(4*b^2)

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Rubi [A]  time = 0.0532086, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5364, 6742, 5298, 2204, 2205, 5320, 2638} \[ \frac{\sqrt{\pi } a \text{Erf}(a+b x)}{4 b^2}-\frac{\sqrt{\pi } a \text{Erfi}(a+b x)}{4 b^2}+\frac{\cosh \left ((a+b x)^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[(a + b*x)^2],x]

[Out]

Cosh[(a + b*x)^2]/(2*b^2) + (a*Sqrt[Pi]*Erf[a + b*x])/(4*b^2) - (a*Sqrt[Pi]*Erfi[a + b*x])/(4*b^2)

Rule 5364

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(u_)^(n_)])^(p_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1]^(
m + 1), Subst[Int[(x - Coefficient[u, x, 0])^m*(a + b*Sinh[c + d*x^n])^p, x], x, u], x] /; FreeQ[{a, b, c, d,
n, p}, x] && LinearQ[u, x] && NeQ[u, x] && IntegerQ[m]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 5298

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

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 5320

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x \sinh \left ((a+b x)^2\right ) \, dx &=\frac{\operatorname{Subst}\left (\int (-a+x) \sinh \left (x^2\right ) \, dx,x,a+b x\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a \sinh \left (x^2\right )+x \sinh \left (x^2\right )\right ) \, dx,x,a+b x\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int x \sinh \left (x^2\right ) \, dx,x,a+b x\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \sinh \left (x^2\right ) \, dx,x,a+b x\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \sinh (x) \, dx,x,(a+b x)^2\right )}{2 b^2}+\frac{a \operatorname{Subst}\left (\int e^{-x^2} \, dx,x,a+b x\right )}{2 b^2}-\frac{a \operatorname{Subst}\left (\int e^{x^2} \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac{\cosh \left ((a+b x)^2\right )}{2 b^2}+\frac{a \sqrt{\pi } \text{erf}(a+b x)}{4 b^2}-\frac{a \sqrt{\pi } \text{erfi}(a+b x)}{4 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0269452, size = 44, normalized size = 0.81 \[ \frac{\cosh \left ((a+b x)^2\right )}{2 b^2}-\frac{\sqrt{\pi } a (\text{Erfi}(a+b x)-\text{Erf}(a+b x))}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[(a + b*x)^2],x]

[Out]

Cosh[(a + b*x)^2]/(2*b^2) - (a*Sqrt[Pi]*(-Erf[a + b*x] + Erfi[a + b*x]))/(4*b^2)

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Maple [C]  time = 0.029, size = 66, normalized size = 1.2 \begin{align*}{\frac{{{\rm e}^{- \left ( bx+a \right ) ^{2}}}}{4\,{b}^{2}}}+{\frac{a{\it Erf} \left ( bx+a \right ) \sqrt{\pi }}{4\,{b}^{2}}}+{\frac{{{\rm e}^{ \left ( bx+a \right ) ^{2}}}}{4\,{b}^{2}}}+{\frac{{\frac{i}{4}}a\sqrt{\pi }{\it Erf} \left ( ibx+ia \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b^2*exp(-(b*x+a)^2)+1/4*a*erf(b*x+a)*Pi^(1/2)/b^2+1/4/b^2*exp((b*x+a)^2)+1/4*I*a/b^2*Pi^(1/2)*erf(I*b*x+I*
a)

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Maxima [B]  time = 1.67452, size = 948, normalized size = 17.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*sinh((b*x + a)^2) - 1/4*((sqrt(pi)*(b^2*x + a*b)*a^2*b^2*(erf(sqrt(-(b^2*x + a*b)^2/b^2)) - 1)/((b^2)^
(5/2)*sqrt(-(b^2*x + a*b)^2/b^2)) - 2*a*b^3*e^((b^2*x + a*b)^2/b^2)/(b^2)^(5/2) - (b^2*x + a*b)^3*gamma(3/2, -
(b^2*x + a*b)^2/b^2)/((b^2)^(5/2)*(-(b^2*x + a*b)^2/b^2)^(3/2)))*a/sqrt(b^2) - (sqrt(pi)*(b^2*x + a*b)*a^2*b^2
*(erf(sqrt((b^2*x + a*b)^2/b^2)) - 1)/((-b^2)^(5/2)*sqrt((b^2*x + a*b)^2/b^2)) + 2*a*b^3*e^(-(b^2*x + a*b)^2/b
^2)/(-b^2)^(5/2) - (b^2*x + a*b)^3*gamma(3/2, (b^2*x + a*b)^2/b^2)/((-b^2)^(5/2)*((b^2*x + a*b)^2/b^2)^(3/2)))
*a/sqrt(-b^2) - (sqrt(pi)*(b^2*x + a*b)*a^3*b^3*(erf(sqrt((b^2*x + a*b)^2/b^2)) - 1)/((-b^2)^(7/2)*sqrt((b^2*x
+ a*b)^2/b^2)) + 3*a^2*b^4*e^(-(b^2*x + a*b)^2/b^2)/(-b^2)^(7/2) + b^4*gamma(2, (b^2*x + a*b)^2/b^2)/(-b^2)^(
7/2) - 3*(b^2*x + a*b)^3*a*b*gamma(3/2, (b^2*x + a*b)^2/b^2)/((-b^2)^(7/2)*((b^2*x + a*b)^2/b^2)^(3/2)))*b/sqr
t(-b^2) - (sqrt(pi)*(b^2*x + a*b)*a^3*b^3*(erf(sqrt(-(b^2*x + a*b)^2/b^2)) - 1)/((b^2)^(7/2)*sqrt(-(b^2*x + a*
b)^2/b^2)) - 3*a^2*b^4*e^((b^2*x + a*b)^2/b^2)/(b^2)^(7/2) + b^4*gamma(2, -(b^2*x + a*b)^2/b^2)/(b^2)^(7/2) -
3*(b^2*x + a*b)^3*a*b*gamma(3/2, -(b^2*x + a*b)^2/b^2)/((b^2)^(7/2)*(-(b^2*x + a*b)^2/b^2)^(3/2)))*b/sqrt(b^2)
)*b

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Fricas [B]  time = 1.8869, size = 317, normalized size = 5.87 \begin{align*} \frac{{\left (\sqrt{\pi } a \sqrt{b^{2}} \operatorname{erf}\left (\frac{\sqrt{b^{2}}{\left (b x + a\right )}}{b}\right ) e^{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} - \sqrt{\pi } a \sqrt{b^{2}} \operatorname{erfi}\left (\frac{\sqrt{b^{2}}{\left (b x + a\right )}}{b}\right ) e^{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} + b e^{\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2}\right )} + b\right )} e^{\left (-b^{2} x^{2} - 2 \, a b x - a^{2}\right )}}{4 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(pi)*a*sqrt(b^2)*erf(sqrt(b^2)*(b*x + a)/b)*e^(b^2*x^2 + 2*a*b*x + a^2) - sqrt(pi)*a*sqrt(b^2)*erfi(s
qrt(b^2)*(b*x + a)/b)*e^(b^2*x^2 + 2*a*b*x + a^2) + b*e^(2*b^2*x^2 + 4*a*b*x + 2*a^2) + b)*e^(-b^2*x^2 - 2*a*b
*x - a^2)/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.15458, size = 134, normalized size = 2.48 \begin{align*} -\frac{-\frac{i \, \sqrt{\pi } a \operatorname{erf}\left (i \, b{\left (x + \frac{a}{b}\right )}\right )}{b} - \frac{e^{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}}{b}}{4 \, b} - \frac{\frac{\sqrt{\pi } a \operatorname{erf}\left (-b{\left (x + \frac{a}{b}\right )}\right )}{b} - \frac{e^{\left (-b^{2} x^{2} - 2 \, a b x - a^{2}\right )}}{b}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-I*sqrt(pi)*a*erf(I*b*(x + a/b))/b - e^(b^2*x^2 + 2*a*b*x + a^2)/b)/b - 1/4*(sqrt(pi)*a*erf(-b*(x + a/b)
)/b - e^(-b^2*x^2 - 2*a*b*x - a^2)/b)/b