3.63 \(\int \frac {\sinh (f x)}{(d x)^{3/2}} \, dx\)

Optimal. Leaf size=87 \[ \frac {\sqrt {\pi } \sqrt {f} \text {erf}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {\pi } \sqrt {f} \text {erfi}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sinh (f x)}{d \sqrt {d x}} \]

[Out]

erf(f^(1/2)*(d*x)^(1/2)/d^(1/2))*f^(1/2)*Pi^(1/2)/d^(3/2)+erfi(f^(1/2)*(d*x)^(1/2)/d^(1/2))*f^(1/2)*Pi^(1/2)/d
^(3/2)-2*sinh(f*x)/d/(d*x)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3297, 3307, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } \sqrt {f} \text {Erf}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {\pi } \sqrt {f} \text {Erfi}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sinh (f x)}{d \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[f*x]/(d*x)^(3/2),x]

[Out]

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

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 3307

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.56 \[ \frac {x \left (-2 \sinh (f x)+\sqrt {-f x} \Gamma \left (\frac {1}{2},-f x\right )-\sqrt {f x} \Gamma \left (\frac {1}{2},f x\right )\right )}{(d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[f*x]/(d*x)^(3/2),x]

[Out]

(x*(Sqrt[-(f*x)]*Gamma[1/2, -(f*x)] - Sqrt[f*x]*Gamma[1/2, f*x] - 2*Sinh[f*x]))/(d*x)^(3/2)

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fricas [B]  time = 0.57, size = 137, normalized size = 1.57 \[ \frac {\sqrt {\pi } {\left (d x \cosh \left (f x\right ) + d x \sinh \left (f x\right )\right )} \sqrt {\frac {f}{d}} \operatorname {erf}\left (\sqrt {d x} \sqrt {\frac {f}{d}}\right ) - \sqrt {\pi } {\left (d x \cosh \left (f x\right ) + d x \sinh \left (f x\right )\right )} \sqrt {-\frac {f}{d}} \operatorname {erf}\left (\sqrt {d x} \sqrt {-\frac {f}{d}}\right ) - \sqrt {d x} {\left (\cosh \left (f x\right )^{2} + 2 \, \cosh \left (f x\right ) \sinh \left (f x\right ) + \sinh \left (f x\right )^{2} - 1\right )}}{d^{2} x \cosh \left (f x\right ) + d^{2} x \sinh \left (f x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (f x\right )}{\left (d x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)^(3/2),x, algorithm="giac")

[Out]

integrate(sinh(f*x)/(d*x)^(3/2), x)

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maple [C]  time = 0.03, size = 120, normalized size = 1.38 \[ -\frac {\sqrt {\pi }\, x^{\frac {3}{2}} \sqrt {2}\, \left (i f \right )^{\frac {3}{2}} \left (\frac {2 \sqrt {2}\, \sqrt {i f}\, {\mathrm e}^{-f x}}{\sqrt {\pi }\, \sqrt {x}\, f}-\frac {2 \sqrt {2}\, \sqrt {i f}\, {\mathrm e}^{f x}}{\sqrt {\pi }\, \sqrt {x}\, f}+\frac {2 \sqrt {i f}\, \sqrt {2}\, \erf \left (\sqrt {x}\, \sqrt {f}\right )}{\sqrt {f}}+\frac {2 \sqrt {i f}\, \sqrt {2}\, \erfi \left (\sqrt {x}\, \sqrt {f}\right )}{\sqrt {f}}\right )}{4 \left (d x \right )^{\frac {3}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x)/(d*x)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.64, size = 74, normalized size = 0.85 \[ \frac {\frac {f {\left (\frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {d x} \sqrt {\frac {f}{d}}\right )}{\sqrt {\frac {f}{d}}} + \frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {d x} \sqrt {-\frac {f}{d}}\right )}{\sqrt {-\frac {f}{d}}}\right )}}{d} - \frac {2 \, \sinh \left (f x\right )}{\sqrt {d x}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)^(3/2),x, algorithm="maxima")

[Out]

(f*(sqrt(pi)*erf(sqrt(d*x)*sqrt(f/d))/sqrt(f/d) + sqrt(pi)*erf(sqrt(d*x)*sqrt(-f/d))/sqrt(-f/d))/d - 2*sinh(f*
x)/sqrt(d*x))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (f\,x\right )}{{\left (d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x)/(d*x)^(3/2),x)

[Out]

int(sinh(f*x)/(d*x)^(3/2), x)

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sympy [C]  time = 2.79, size = 94, normalized size = 1.08 \[ \frac {\sqrt {2} \sqrt {\pi } \sqrt {f} e^{- \frac {i \pi }{4}} C\left (\frac {\sqrt {2} \sqrt {f} \sqrt {x} e^{\frac {i \pi }{4}}}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{2 d^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} - \frac {\sinh {\left (f x \right )} \Gamma \left (\frac {1}{4}\right )}{2 d^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x)/(d*x)**(3/2),x)

[Out]

sqrt(2)*sqrt(pi)*sqrt(f)*exp(-I*pi/4)*fresnelc(sqrt(2)*sqrt(f)*sqrt(x)*exp(I*pi/4)/sqrt(pi))*gamma(1/4)/(2*d**
(3/2)*gamma(5/4)) - sinh(f*x)*gamma(1/4)/(2*d**(3/2)*sqrt(x)*gamma(5/4))

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