\(\int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx\) [67]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [B] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 114 \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=-\frac {2 \cosh (f x)}{3 d (d x)^{3/2}}+\frac {2 f^{3/2} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 f^{3/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {4 f \sinh (f x)}{3 d^2 \sqrt {d x}} \]

[Out]

-2/3*cosh(f*x)/d/(d*x)^(3/2)+2/3*f^(3/2)*erf(f^(1/2)*(d*x)^(1/2)/d^(1/2))*Pi^(1/2)/d^(5/2)+2/3*f^(3/2)*erfi(f^
(1/2)*(d*x)^(1/2)/d^(1/2))*Pi^(1/2)/d^(5/2)-4/3*f*sinh(f*x)/d^2/(d*x)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3378, 3388, 2211, 2235, 2236} \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=\frac {2 \sqrt {\pi } f^{3/2} \text {erf}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 \sqrt {\pi } f^{3/2} \text {erfi}\left (\frac {\sqrt {f} \sqrt {d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {4 f \sinh (f x)}{3 d^2 \sqrt {d x}}-\frac {2 \cosh (f x)}{3 d (d x)^{3/2}} \]

[In]

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

[Out]

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

Rule 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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 2236

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 3378

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 3388

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=\frac {x \left (-2 e^{f x} (1+2 f x)-4 (-f x)^{3/2} \Gamma \left (\frac {1}{2},-f x\right )+e^{-f x} \left (-2+4 f x-4 e^{f x} (f x)^{3/2} \Gamma \left (\frac {1}{2},f x\right )\right )\right )}{6 (d x)^{5/2}} \]

[In]

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

[Out]

(x*(-2*E^(f*x)*(1 + 2*f*x) - 4*(-(f*x))^(3/2)*Gamma[1/2, -(f*x)] + (-2 + 4*f*x - 4*E^(f*x)*(f*x)^(3/2)*Gamma[1
/2, f*x])/E^(f*x)))/(6*(d*x)^(5/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.11

method result size
meijerg \(-\frac {i \sqrt {\pi }\, x^{\frac {5}{2}} \sqrt {2}\, \left (i f \right )^{\frac {5}{2}} \left (-\frac {8 \sqrt {2}\, \left (-f x +\frac {1}{2}\right ) {\mathrm e}^{-f x}}{3 \sqrt {\pi }\, x^{\frac {3}{2}} \left (i f \right )^{\frac {3}{2}}}-\frac {8 \sqrt {2}\, \left (f x +\frac {1}{2}\right ) {\mathrm e}^{f x}}{3 \sqrt {\pi }\, x^{\frac {3}{2}} \left (i f \right )^{\frac {3}{2}}}+\frac {8 \sqrt {2}\, f^{\frac {3}{2}} \operatorname {erf}\left (\sqrt {x}\, \sqrt {f}\right )}{3 \left (i f \right )^{\frac {3}{2}}}+\frac {8 \sqrt {2}\, f^{\frac {3}{2}} \operatorname {erfi}\left (\sqrt {x}\, \sqrt {f}\right )}{3 \left (i f \right )^{\frac {3}{2}}}\right )}{8 \left (d x \right )^{\frac {5}{2}} f}\) \(126\)

[In]

int(cosh(f*x)/(d*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (78) = 156\).

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.57 \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=\frac {2 \, \sqrt {\pi } {\left (d f x^{2} \cosh \left (f x\right ) + d f x^{2} \sinh \left (f x\right )\right )} \sqrt {\frac {f}{d}} \operatorname {erf}\left (\sqrt {d x} \sqrt {\frac {f}{d}}\right ) - 2 \, \sqrt {\pi } {\left (d f x^{2} \cosh \left (f x\right ) + d f x^{2} \sinh \left (f x\right )\right )} \sqrt {-\frac {f}{d}} \operatorname {erf}\left (\sqrt {d x} \sqrt {-\frac {f}{d}}\right ) - {\left ({\left (2 \, f x + 1\right )} \cosh \left (f x\right )^{2} + 2 \, {\left (2 \, f x + 1\right )} \cosh \left (f x\right ) \sinh \left (f x\right ) + {\left (2 \, f x + 1\right )} \sinh \left (f x\right )^{2} - 2 \, f x + 1\right )} \sqrt {d x}}{3 \, {\left (d^{3} x^{2} \cosh \left (f x\right ) + d^{3} x^{2} \sinh \left (f x\right )\right )}} \]

[In]

integrate(cosh(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))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.48 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.09 \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=- \frac {\sqrt {2} \sqrt {\pi } f^{\frac {3}{2}} e^{- \frac {i \pi }{4}} C\left (\frac {\sqrt {2} \sqrt {f} \sqrt {x} e^{\frac {i \pi }{4}}}{\sqrt {\pi }}\right ) \Gamma \left (- \frac {3}{4}\right )}{d^{\frac {5}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {f \sinh {\left (f x \right )} \Gamma \left (- \frac {3}{4}\right )}{d^{\frac {5}{2}} \sqrt {x} \Gamma \left (\frac {1}{4}\right )} + \frac {\cosh {\left (f x \right )} \Gamma \left (- \frac {3}{4}\right )}{2 d^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.51 \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=\frac {\frac {f {\left (\frac {\sqrt {f x} \Gamma \left (-\frac {1}{2}, f x\right )}{\sqrt {d x}} - \frac {\sqrt {-f x} \Gamma \left (-\frac {1}{2}, -f x\right )}{\sqrt {d x}}\right )}}{d} - \frac {2 \, \cosh \left (f x\right )}{\left (d x\right )^{\frac {3}{2}}}}{3 \, d} \]

[In]

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

Giac [F]

\[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=\int { \frac {\cosh \left (f x\right )}{\left (d x\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate(cosh(f*x)/(d*x)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh (f x)}{(d x)^{5/2}} \, dx=\int \frac {\mathrm {cosh}\left (f\,x\right )}{{\left (d\,x\right )}^{5/2}} \,d x \]

[In]

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

[Out]

int(cosh(f*x)/(d*x)^(5/2), x)