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

Optimal. Leaf size=111 \[ \frac{3 \sqrt{\pi } d^{3/2} \text{Erf}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{8 f^{5/2}}+\frac{3 \sqrt{\pi } d^{3/2} \text{Erfi}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{8 f^{5/2}}-\frac{3 d \sqrt{d x} \cosh (f x)}{2 f^2}+\frac{(d x)^{3/2} \sinh (f x)}{f} \]

[Out]

(-3*d*Sqrt[d*x]*Cosh[f*x])/(2*f^2) + (3*d^(3/2)*Sqrt[Pi]*Erf[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(8*f^(5/2)) + (3*d^
(3/2)*Sqrt[Pi]*Erfi[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(8*f^(5/2)) + ((d*x)^(3/2)*Sinh[f*x])/f

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Rubi [A]  time = 0.15545, antiderivative size = 111, 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 = {3296, 3307, 2180, 2204, 2205} \[ \frac{3 \sqrt{\pi } d^{3/2} \text{Erf}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{8 f^{5/2}}+\frac{3 \sqrt{\pi } d^{3/2} \text{Erfi}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{8 f^{5/2}}-\frac{3 d \sqrt{d x} \cosh (f x)}{2 f^2}+\frac{(d x)^{3/2} \sinh (f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*Sqrt[d*x]*Cosh[f*x])/(2*f^2) + (3*d^(3/2)*Sqrt[Pi]*Erf[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(8*f^(5/2)) + (3*d^
(3/2)*Sqrt[Pi]*Erfi[(Sqrt[f]*Sqrt[d*x])/Sqrt[d]])/(8*f^(5/2)) + ((d*x)^(3/2)*Sinh[f*x])/f

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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]

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 (d x)^{3/2} \cosh (f x) \, dx &=\frac{(d x)^{3/2} \sinh (f x)}{f}-\frac{(3 d) \int \sqrt{d x} \sinh (f x) \, dx}{2 f}\\ &=-\frac{3 d \sqrt{d x} \cosh (f x)}{2 f^2}+\frac{(d x)^{3/2} \sinh (f x)}{f}+\frac{\left (3 d^2\right ) \int \frac{\cosh (f x)}{\sqrt{d x}} \, dx}{4 f^2}\\ &=-\frac{3 d \sqrt{d x} \cosh (f x)}{2 f^2}+\frac{(d x)^{3/2} \sinh (f x)}{f}+\frac{\left (3 d^2\right ) \int \frac{e^{-f x}}{\sqrt{d x}} \, dx}{8 f^2}+\frac{\left (3 d^2\right ) \int \frac{e^{f x}}{\sqrt{d x}} \, dx}{8 f^2}\\ &=-\frac{3 d \sqrt{d x} \cosh (f x)}{2 f^2}+\frac{(d x)^{3/2} \sinh (f x)}{f}+\frac{(3 d) \operatorname{Subst}\left (\int e^{-\frac{f x^2}{d}} \, dx,x,\sqrt{d x}\right )}{4 f^2}+\frac{(3 d) \operatorname{Subst}\left (\int e^{\frac{f x^2}{d}} \, dx,x,\sqrt{d x}\right )}{4 f^2}\\ &=-\frac{3 d \sqrt{d x} \cosh (f x)}{2 f^2}+\frac{3 d^{3/2} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{8 f^{5/2}}+\frac{3 d^{3/2} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{f} \sqrt{d x}}{\sqrt{d}}\right )}{8 f^{5/2}}+\frac{(d x)^{3/2} \sinh (f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0133719, size = 51, normalized size = 0.46 \[ \frac{d^2 \left (\sqrt{-f x} \text{Gamma}\left (\frac{5}{2},-f x\right )-\sqrt{f x} \text{Gamma}\left (\frac{5}{2},f x\right )\right )}{2 f^3 \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.027, size = 133, normalized size = 1.2 \begin{align*}{\frac{-2\,i\sqrt{2}\sqrt{\pi }}{f} \left ( dx \right ) ^{{\frac{3}{2}}} \left ( -{\frac{\sqrt{2} \left ( 10\,fx+15 \right ){{\rm e}^{-fx}}}{80\,\sqrt{\pi }{f}^{2}}\sqrt{x} \left ( if \right ) ^{{\frac{5}{2}}}}-{\frac{\sqrt{2} \left ( -10\,fx+15 \right ){{\rm e}^{fx}}}{80\,\sqrt{\pi }{f}^{2}}\sqrt{x} \left ( if \right ) ^{{\frac{5}{2}}}}+{\frac{3\,\sqrt{2}}{32} \left ( if \right ) ^{{\frac{5}{2}}}{\it Erf} \left ( \sqrt{x}\sqrt{f} \right ){f}^{-{\frac{5}{2}}}}+{\frac{3\,\sqrt{2}}{32} \left ( if \right ) ^{{\frac{5}{2}}}{\it erfi} \left ( \sqrt{x}\sqrt{f} \right ){f}^{-{\frac{5}{2}}}} \right ){x}^{-{\frac{3}{2}}} \left ( if \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I*(d*x)^(3/2)/x^(3/2)*2^(1/2)/(I*f)^(3/2)*Pi^(1/2)/f*(-1/80/Pi^(1/2)*x^(1/2)*2^(1/2)*(I*f)^(5/2)*(10*f*x+15
)/f^2*exp(-f*x)-1/80/Pi^(1/2)*x^(1/2)*2^(1/2)*(I*f)^(5/2)*(-10*f*x+15)/f^2*exp(f*x)+3/32*(I*f)^(5/2)*2^(1/2)/f
^(5/2)*erf(x^(1/2)*f^(1/2))+3/32*(I*f)^(5/2)*2^(1/2)/f^(5/2)*erfi(x^(1/2)*f^(1/2)))

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Maxima [B]  time = 1.06195, size = 235, normalized size = 2.12 \begin{align*} \frac{16 \, \left (d x\right )^{\frac{5}{2}} \cosh \left (f x\right ) + \frac{f{\left (\frac{15 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (\sqrt{d x} \sqrt{\frac{f}{d}}\right )}{f^{3} \sqrt{\frac{f}{d}}} + \frac{15 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (\sqrt{d x} \sqrt{-\frac{f}{d}}\right )}{f^{3} \sqrt{-\frac{f}{d}}} - \frac{2 \,{\left (4 \, \left (d x\right )^{\frac{5}{2}} d f^{2} - 10 \, \left (d x\right )^{\frac{3}{2}} d^{2} f + 15 \, \sqrt{d x} d^{3}\right )} e^{\left (f x\right )}}{f^{3}} - \frac{2 \,{\left (4 \, \left (d x\right )^{\frac{5}{2}} d f^{2} + 10 \, \left (d x\right )^{\frac{3}{2}} d^{2} f + 15 \, \sqrt{d x} d^{3}\right )} e^{\left (-f x\right )}}{f^{3}}\right )}}{d}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(16*(d*x)^(5/2)*cosh(f*x) + f*(15*sqrt(pi)*d^3*erf(sqrt(d*x)*sqrt(f/d))/(f^3*sqrt(f/d)) + 15*sqrt(pi)*d^3
*erf(sqrt(d*x)*sqrt(-f/d))/(f^3*sqrt(-f/d)) - 2*(4*(d*x)^(5/2)*d*f^2 - 10*(d*x)^(3/2)*d^2*f + 15*sqrt(d*x)*d^3
)*e^(f*x)/f^3 - 2*(4*(d*x)^(5/2)*d*f^2 + 10*(d*x)^(3/2)*d^2*f + 15*sqrt(d*x)*d^3)*e^(-f*x)/f^3)/d)/d

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Fricas [B]  time = 1.83915, size = 466, normalized size = 4.2 \begin{align*} \frac{3 \, \sqrt{\pi }{\left (d^{2} \cosh \left (f x\right ) + d^{2} \sinh \left (f x\right )\right )} \sqrt{\frac{f}{d}} \operatorname{erf}\left (\sqrt{d x} \sqrt{\frac{f}{d}}\right ) - 3 \, \sqrt{\pi }{\left (d^{2} \cosh \left (f x\right ) + d^{2} \sinh \left (f x\right )\right )} \sqrt{-\frac{f}{d}} \operatorname{erf}\left (\sqrt{d x} \sqrt{-\frac{f}{d}}\right ) - 2 \,{\left (2 \, d f^{2} x -{\left (2 \, d f^{2} x - 3 \, d f\right )} \cosh \left (f x\right )^{2} - 2 \,{\left (2 \, d f^{2} x - 3 \, d f\right )} \cosh \left (f x\right ) \sinh \left (f x\right ) -{\left (2 \, d f^{2} x - 3 \, d f\right )} \sinh \left (f x\right )^{2} + 3 \, d f\right )} \sqrt{d x}}{8 \,{\left (f^{3} \cosh \left (f x\right ) + f^{3} \sinh \left (f x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*sqrt(pi)*(d^2*cosh(f*x) + d^2*sinh(f*x))*sqrt(f/d)*erf(sqrt(d*x)*sqrt(f/d)) - 3*sqrt(pi)*(d^2*cosh(f*x)
+ d^2*sinh(f*x))*sqrt(-f/d)*erf(sqrt(d*x)*sqrt(-f/d)) - 2*(2*d*f^2*x - (2*d*f^2*x - 3*d*f)*cosh(f*x)^2 - 2*(2
*d*f^2*x - 3*d*f)*cosh(f*x)*sinh(f*x) - (2*d*f^2*x - 3*d*f)*sinh(f*x)^2 + 3*d*f)*sqrt(d*x))/(f^3*cosh(f*x) + f
^3*sinh(f*x))

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Sympy [C]  time = 142.406, size = 131, normalized size = 1.18 \begin{align*} \frac{5 d^{\frac{3}{2}} x^{\frac{3}{2}} \sinh{\left (f x \right )} \Gamma \left (\frac{5}{4}\right )}{4 f \Gamma \left (\frac{9}{4}\right )} - \frac{15 d^{\frac{3}{2}} \sqrt{x} \cosh{\left (f x \right )} \Gamma \left (\frac{5}{4}\right )}{8 f^{2} \Gamma \left (\frac{9}{4}\right )} + \frac{15 \sqrt{2} \sqrt{\pi } d^{\frac{3}{2}} 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{5}{4}\right )}{16 f^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*d**(3/2)*x**(3/2)*sinh(f*x)*gamma(5/4)/(4*f*gamma(9/4)) - 15*d**(3/2)*sqrt(x)*cosh(f*x)*gamma(5/4)/(8*f**2*g
amma(9/4)) + 15*sqrt(2)*sqrt(pi)*d**(3/2)*exp(-I*pi/4)*fresnelc(sqrt(2)*sqrt(f)*sqrt(x)*exp(I*pi/4)/sqrt(pi))*
gamma(5/4)/(16*f**(5/2)*gamma(9/4))

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Giac [A]  time = 1.23246, size = 194, normalized size = 1.75 \begin{align*} -\frac{\frac{3 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (-\frac{\sqrt{d f} \sqrt{d x}}{d}\right )}{\sqrt{d f} f^{2}} + \frac{2 \,{\left (2 \, \sqrt{d x} d^{2} f x + 3 \, \sqrt{d x} d^{2}\right )} e^{\left (-f x\right )}}{f^{2}}}{8 \, d} - \frac{\frac{3 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (-\frac{\sqrt{-d f} \sqrt{d x}}{d}\right )}{\sqrt{-d f} f^{2}} - \frac{2 \,{\left (2 \, \sqrt{d x} d^{2} f x - 3 \, \sqrt{d x} d^{2}\right )} e^{\left (f x\right )}}{f^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*sqrt(pi)*d^3*erf(-sqrt(d*f)*sqrt(d*x)/d)/(sqrt(d*f)*f^2) + 2*(2*sqrt(d*x)*d^2*f*x + 3*sqrt(d*x)*d^2)*e
^(-f*x)/f^2)/d - 1/8*(3*sqrt(pi)*d^3*erf(-sqrt(-d*f)*sqrt(d*x)/d)/(sqrt(-d*f)*f^2) - 2*(2*sqrt(d*x)*d^2*f*x -
3*sqrt(d*x)*d^2)*e^(f*x)/f^2)/d