3.6 \(\int \frac {\sinh (a+b x^2)}{x^2} \, dx\)

Optimal. Leaf size=66 \[ \frac {1}{2} \sqrt {\pi } e^{-a} \sqrt {b} \text {erf}\left (\sqrt {b} x\right )+\frac {1}{2} \sqrt {\pi } e^a \sqrt {b} \text {erfi}\left (\sqrt {b} x\right )-\frac {\sinh \left (a+b x^2\right )}{x} \]

[Out]

-sinh(b*x^2+a)/x+1/2*erf(x*b^(1/2))*b^(1/2)*Pi^(1/2)/exp(a)+1/2*exp(a)*erfi(x*b^(1/2))*b^(1/2)*Pi^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5326, 5299, 2204, 2205} \[ \frac {1}{2} \sqrt {\pi } e^{-a} \sqrt {b} \text {Erf}\left (\sqrt {b} x\right )+\frac {1}{2} \sqrt {\pi } e^a \sqrt {b} \text {Erfi}\left (\sqrt {b} x\right )-\frac {\sinh \left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]/x^2,x]

[Out]

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

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 5299

Int[Cosh[(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 5326

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

Rubi steps

\begin {align*} \int \frac {\sinh \left (a+b x^2\right )}{x^2} \, dx &=-\frac {\sinh \left (a+b x^2\right )}{x}+(2 b) \int \cosh \left (a+b x^2\right ) \, dx\\ &=-\frac {\sinh \left (a+b x^2\right )}{x}+b \int e^{-a-b x^2} \, dx+b \int e^{a+b x^2} \, dx\\ &=\frac {1}{2} \sqrt {b} e^{-a} \sqrt {\pi } \text {erf}\left (\sqrt {b} x\right )+\frac {1}{2} \sqrt {b} e^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x\right )-\frac {\sinh \left (a+b x^2\right )}{x}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 70, normalized size = 1.06 \[ \frac {\sqrt {\pi } \sqrt {b} x (\cosh (a)-\sinh (a)) \text {erf}\left (\sqrt {b} x\right )+\sqrt {\pi } \sqrt {b} x (\sinh (a)+\cosh (a)) \text {erfi}\left (\sqrt {b} x\right )-2 \sinh \left (a+b x^2\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]/x^2,x]

[Out]

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

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fricas [B]  time = 0.48, size = 184, normalized size = 2.79 \[ -\frac {\sqrt {\pi } {\left (x \cosh \left (b x^{2} + a\right ) \cosh \relax (a) + x \cosh \left (b x^{2} + a\right ) \sinh \relax (a) + {\left (x \cosh \relax (a) + x \sinh \relax (a)\right )} \sinh \left (b x^{2} + a\right )\right )} \sqrt {-b} \operatorname {erf}\left (\sqrt {-b} x\right ) - \sqrt {\pi } {\left (x \cosh \left (b x^{2} + a\right ) \cosh \relax (a) - x \cosh \left (b x^{2} + a\right ) \sinh \relax (a) + {\left (x \cosh \relax (a) - x \sinh \relax (a)\right )} \sinh \left (b x^{2} + a\right )\right )} \sqrt {b} \operatorname {erf}\left (\sqrt {b} x\right ) + \cosh \left (b x^{2} + a\right )^{2} + 2 \, \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right ) + \sinh \left (b x^{2} + a\right )^{2} - 1}{2 \, {\left (x \cosh \left (b x^{2} + a\right ) + x \sinh \left (b x^{2} + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(pi)*(x*cosh(b*x^2 + a)*cosh(a) + x*cosh(b*x^2 + a)*sinh(a) + (x*cosh(a) + x*sinh(a))*sinh(b*x^2 + a
))*sqrt(-b)*erf(sqrt(-b)*x) - sqrt(pi)*(x*cosh(b*x^2 + a)*cosh(a) - x*cosh(b*x^2 + a)*sinh(a) + (x*cosh(a) - x
*sinh(a))*sinh(b*x^2 + a))*sqrt(b)*erf(sqrt(b)*x) + cosh(b*x^2 + a)^2 + 2*cosh(b*x^2 + a)*sinh(b*x^2 + a) + si
nh(b*x^2 + a)^2 - 1)/(x*cosh(b*x^2 + a) + x*sinh(b*x^2 + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x^2 + a)/x^2, x)

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maple [A]  time = 0.03, size = 70, normalized size = 1.06 \[ \frac {{\mathrm e}^{-a} {\mathrm e}^{-b \,x^{2}}}{2 x}+\frac {{\mathrm e}^{-a} \sqrt {b}\, \sqrt {\pi }\, \erf \left (x \sqrt {b}\right )}{2}-\frac {{\mathrm e}^{a} {\mathrm e}^{b \,x^{2}}}{2 x}+\frac {{\mathrm e}^{a} b \sqrt {\pi }\, \erf \left (\sqrt {-b}\, x \right )}{2 \sqrt {-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 54, normalized size = 0.82 \[ \frac {1}{2} \, {\left (\frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {b} x\right ) e^{\left (-a\right )}}{\sqrt {b}} + \frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {-b} x\right ) e^{a}}{\sqrt {-b}}\right )} b - \frac {\sinh \left (b x^{2} + a\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {sinh}\left (b\,x^2+a\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^2)/x^2,x)

[Out]

int(sinh(a + b*x^2)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**2)/x**2, x)

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