3.64 \(\int \frac{\sinh (f x)}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=114 \[ -\frac{2 \sqrt{\pi } f^{3/2} \text{Erf}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 \sqrt{\pi } f^{3/2} \text{Erfi}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 f \cosh (f x)}{3 d^2 \sqrt{d x}}-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}} \]

[Out]

(-4*f*Cosh[f*x])/(3*d^2*Sqrt[d*x]) - (2*f^(3/2)*Sqrt[Pi]*Erf[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2)) + (2*f^
(3/2)*Sqrt[Pi]*Erfi[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2)) - (2*Sinh[f*x])/(3*d*(d*x)^(3/2))

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Rubi [A]  time = 0.149009, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3297, 3308, 2180, 2204, 2205} \[ -\frac{2 \sqrt{\pi } f^{3/2} \text{Erf}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 \sqrt{\pi } f^{3/2} \text{Erfi}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 f \cosh (f x)}{3 d^2 \sqrt{d x}}-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[f*x]/(d*x)^(5/2),x]

[Out]

(-4*f*Cosh[f*x])/(3*d^2*Sqrt[d*x]) - (2*f^(3/2)*Sqrt[Pi]*Erf[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2)) + (2*f^
(3/2)*Sqrt[Pi]*Erfi[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(3*d^(5/2)) - (2*Sinh[f*x])/(3*d*(d*x)^(3/2))

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3308

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

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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]

Rubi steps

\begin{align*} \int \frac{\sinh (f x)}{(d x)^{5/2}} \, dx &=-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}}+\frac{(2 f) \int \frac{\cosh (f x)}{(d x)^{3/2}} \, dx}{3 d}\\ &=-\frac{4 f \cosh (f x)}{3 d^2 \sqrt{d x}}-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}}+\frac{\left (4 f^2\right ) \int \frac{\sinh (f x)}{\sqrt{d x}} \, dx}{3 d^2}\\ &=-\frac{4 f \cosh (f x)}{3 d^2 \sqrt{d x}}-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}}-\frac{\left (2 f^2\right ) \int \frac{e^{-f x}}{\sqrt{d x}} \, dx}{3 d^2}+\frac{\left (2 f^2\right ) \int \frac{e^{f x}}{\sqrt{d x}} \, dx}{3 d^2}\\ &=-\frac{4 f \cosh (f x)}{3 d^2 \sqrt{d x}}-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}}-\frac{\left (4 f^2\right ) \operatorname{Subst}\left (\int e^{-\frac{f x^2}{d}} \, dx,x,\sqrt{d x}\right )}{3 d^3}+\frac{\left (4 f^2\right ) \operatorname{Subst}\left (\int e^{\frac{f x^2}{d}} \, dx,x,\sqrt{d x}\right )}{3 d^3}\\ &=-\frac{4 f \cosh (f x)}{3 d^2 \sqrt{d x}}-\frac{2 f^{3/2} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 f^{3/2} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{2 \sinh (f x)}{3 d (d x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0830307, size = 84, normalized size = 0.74 \[ -\frac{x e^{-f x} \left (2 e^{f x} (-f x)^{3/2} \text{Gamma}\left (\frac{1}{2},-f x\right )-2 e^{f x} (f x)^{3/2} \text{Gamma}\left (\frac{1}{2},f x\right )+e^{2 f x}+2 f x e^{2 f x}+2 f x-1\right )}{3 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[f*x]/(d*x)^(5/2),x]

[Out]

-(x*(-1 + E^(2*f*x) + 2*f*x + 2*E^(2*f*x)*f*x + 2*E^(f*x)*(-(f*x))^(3/2)*Gamma[1/2, -(f*x)] - 2*E^(f*x)*(f*x)^
(3/2)*Gamma[1/2, f*x]))/(3*E^(f*x)*(d*x)^(5/2))

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Maple [C]  time = 0.024, size = 132, normalized size = 1.2 \begin{align*} -{\frac{\sqrt{\pi }\sqrt{2}}{8\,f}{x}^{{\frac{5}{2}}} \left ( if \right ) ^{{\frac{5}{2}}} \left ( -{\frac{4\,\sqrt{2} \left ( 2\,fx+1 \right ){{\rm e}^{fx}}}{3\,\sqrt{\pi }f}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{if}}}}+{\frac{4\,\sqrt{2} \left ( -2\,fx+1 \right ){{\rm e}^{-fx}}}{3\,\sqrt{\pi }f}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{if}}}}-{\frac{8\,\sqrt{2}}{3}\sqrt{f}{\it Erf} \left ( \sqrt{x}\sqrt{f} \right ){\frac{1}{\sqrt{if}}}}+{\frac{8\,\sqrt{2}}{3}\sqrt{f}{\it erfi} \left ( \sqrt{x}\sqrt{f} \right ){\frac{1}{\sqrt{if}}}} \right ) \left ( dx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x)/(d*x)^(5/2),x)

[Out]

-1/8*Pi^(1/2)/(d*x)^(5/2)*x^(5/2)*2^(1/2)*(I*f)^(5/2)/f*(-4/3/Pi^(1/2)/x^(3/2)*2^(1/2)/(I*f)^(1/2)*(2*f*x+1)/f
*exp(f*x)+4/3/Pi^(1/2)/x^(3/2)*2^(1/2)/(I*f)^(1/2)*(-2*f*x+1)/f*exp(-f*x)-8/3/(I*f)^(1/2)*2^(1/2)*f^(1/2)*erf(
x^(1/2)*f^(1/2))+8/3/(I*f)^(1/2)*2^(1/2)*f^(1/2)*erfi(x^(1/2)*f^(1/2)))

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Maxima [A]  time = 1.18749, size = 77, normalized size = 0.68 \begin{align*} -\frac{\frac{f{\left (\frac{\sqrt{f x} \Gamma \left (-\frac{1}{2}, f x\right )}{\sqrt{d x}} + \frac{\sqrt{-f x} \Gamma \left (-\frac{1}{2}, -f x\right )}{\sqrt{d x}}\right )}}{d} + \frac{2 \, \sinh \left (f x\right )}{\left (d x\right )^{\frac{3}{2}}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(f*(sqrt(f*x)*gamma(-1/2, f*x)/sqrt(d*x) + sqrt(-f*x)*gamma(-1/2, -f*x)/sqrt(d*x))/d + 2*sinh(f*x)/(d*x)^
(3/2))/d

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Fricas [B]  time = 2.64832, size = 454, normalized size = 3.98 \begin{align*} -\frac{2 \, \sqrt{\pi }{\left (d f x^{2} \cosh \left (f x\right ) + d f x^{2} \sinh \left (f x\right )\right )} \sqrt{\frac{f}{d}} \operatorname{erf}\left (\sqrt{d x} \sqrt{\frac{f}{d}}\right ) + 2 \, \sqrt{\pi }{\left (d f x^{2} \cosh \left (f x\right ) + d f x^{2} \sinh \left (f x\right )\right )} \sqrt{-\frac{f}{d}} \operatorname{erf}\left (\sqrt{d x} \sqrt{-\frac{f}{d}}\right ) +{\left ({\left (2 \, f x + 1\right )} \cosh \left (f x\right )^{2} + 2 \,{\left (2 \, f x + 1\right )} \cosh \left (f x\right ) \sinh \left (f x\right ) +{\left (2 \, f x + 1\right )} \sinh \left (f x\right )^{2} + 2 \, f x - 1\right )} \sqrt{d x}}{3 \,{\left (d^{3} x^{2} \cosh \left (f x\right ) + d^{3} x^{2} \sinh \left (f x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*sqrt(pi)*(d*f*x^2*cosh(f*x) + d*f*x^2*sinh(f*x))*sqrt(f/d)*erf(sqrt(d*x)*sqrt(f/d)) + 2*sqrt(pi)*(d*f*
x^2*cosh(f*x) + d*f*x^2*sinh(f*x))*sqrt(-f/d)*erf(sqrt(d*x)*sqrt(-f/d)) + ((2*f*x + 1)*cosh(f*x)^2 + 2*(2*f*x
+ 1)*cosh(f*x)*sinh(f*x) + (2*f*x + 1)*sinh(f*x)^2 + 2*f*x - 1)*sqrt(d*x))/(d^3*x^2*cosh(f*x) + d^3*x^2*sinh(f
*x))

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Sympy [C]  time = 175.807, size = 129, normalized size = 1.13 \begin{align*} - \frac{\sqrt{2} \sqrt{\pi } f^{\frac{3}{2}} e^{- \frac{3 i \pi }{4}} S\left (\frac{\sqrt{2} \sqrt{f} \sqrt{x} e^{\frac{i \pi }{4}}}{\sqrt{\pi }}\right ) \Gamma \left (- \frac{1}{4}\right )}{3 d^{\frac{5}{2}} \Gamma \left (\frac{3}{4}\right )} + \frac{f \cosh{\left (f x \right )} \Gamma \left (- \frac{1}{4}\right )}{3 d^{\frac{5}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{\sinh{\left (f x \right )} \Gamma \left (- \frac{1}{4}\right )}{6 d^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)**(5/2),x)

[Out]

-sqrt(2)*sqrt(pi)*f**(3/2)*exp(-3*I*pi/4)*fresnels(sqrt(2)*sqrt(f)*sqrt(x)*exp(I*pi/4)/sqrt(pi))*gamma(-1/4)/(
3*d**(5/2)*gamma(3/4)) + f*cosh(f*x)*gamma(-1/4)/(3*d**(5/2)*sqrt(x)*gamma(3/4)) + sinh(f*x)*gamma(-1/4)/(6*d*
*(5/2)*x**(3/2)*gamma(3/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (f x\right )}{\left (d x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sinh(f*x)/(d*x)^(5/2), x)