3.95.95 \(\int \frac {-256+16 x^4+15 e^{\frac {5 x^3}{2}} x^5+e^{\frac {5 x^3}{4}} (32 x+8 x^3-120 x^4+30 x^6)}{2 x^3} \, dx\)

Optimal. Leaf size=20 \[ \left (e^{\frac {5 x^3}{4}}-\frac {8}{x}+2 x\right )^2 \]

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Rubi [B]  time = 0.10, antiderivative size = 45, normalized size of antiderivative = 2.25, number of steps used = 7, number of rules used = 4, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2209, 2288} \begin {gather*} e^{\frac {5 x^3}{2}}+4 x^2+\frac {64}{x^2}-\frac {4 e^{\frac {5 x^3}{4}} \left (4 x^3-x^5\right )}{x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-256 + 16*x^4 + 15*E^((5*x^3)/2)*x^5 + E^((5*x^3)/4)*(32*x + 8*x^3 - 120*x^4 + 30*x^6))/(2*x^3),x]

[Out]

E^((5*x^3)/2) + 64/x^2 + 4*x^2 - (4*E^((5*x^3)/4)*(4*x^3 - x^5))/x^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {-256+16 x^4+15 e^{\frac {5 x^3}{2}} x^5+e^{\frac {5 x^3}{4}} \left (32 x+8 x^3-120 x^4+30 x^6\right )}{x^3} \, dx\\ &=\frac {1}{2} \int \left (15 e^{\frac {5 x^3}{2}} x^2+\frac {16 \left (-16+x^4\right )}{x^3}+\frac {2 e^{\frac {5 x^3}{4}} \left (16+4 x^2-60 x^3+15 x^5\right )}{x^2}\right ) \, dx\\ &=\frac {15}{2} \int e^{\frac {5 x^3}{2}} x^2 \, dx+8 \int \frac {-16+x^4}{x^3} \, dx+\int \frac {e^{\frac {5 x^3}{4}} \left (16+4 x^2-60 x^3+15 x^5\right )}{x^2} \, dx\\ &=e^{\frac {5 x^3}{2}}-\frac {4 e^{\frac {5 x^3}{4}} \left (4 x^3-x^5\right )}{x^4}+8 \int \left (-\frac {16}{x^3}+x\right ) \, dx\\ &=e^{\frac {5 x^3}{2}}+\frac {64}{x^2}+4 x^2-\frac {4 e^{\frac {5 x^3}{4}} \left (4 x^3-x^5\right )}{x^4}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 46, normalized size = 2.30 \begin {gather*} e^{\frac {5 x^3}{2}}+\frac {64}{x^2}-\frac {16 e^{\frac {5 x^3}{4}}}{x}+4 e^{\frac {5 x^3}{4}} x+4 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-256 + 16*x^4 + 15*E^((5*x^3)/2)*x^5 + E^((5*x^3)/4)*(32*x + 8*x^3 - 120*x^4 + 30*x^6))/(2*x^3),x]

[Out]

E^((5*x^3)/2) + 64/x^2 - (16*E^((5*x^3)/4))/x + 4*E^((5*x^3)/4)*x + 4*x^2

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fricas [B]  time = 0.63, size = 36, normalized size = 1.80 \begin {gather*} \frac {4 \, x^{4} + x^{2} e^{\left (\frac {5}{2} \, x^{3}\right )} + 4 \, {\left (x^{3} - 4 \, x\right )} e^{\left (\frac {5}{4} \, x^{3}\right )} + 64}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x^5*exp(5/4*x^3)^2+(30*x^6-120*x^4+8*x^3+32*x)*exp(5/4*x^3)+16*x^4-256)/x^3,x, algorithm="fr
icas")

[Out]

(4*x^4 + x^2*e^(5/2*x^3) + 4*(x^3 - 4*x)*e^(5/4*x^3) + 64)/x^2

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giac [B]  time = 2.68, size = 41, normalized size = 2.05 \begin {gather*} \frac {4 \, x^{4} + 4 \, x^{3} e^{\left (\frac {5}{4} \, x^{3}\right )} + x^{2} e^{\left (\frac {5}{2} \, x^{3}\right )} - 16 \, x e^{\left (\frac {5}{4} \, x^{3}\right )} + 64}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x^5*exp(5/4*x^3)^2+(30*x^6-120*x^4+8*x^3+32*x)*exp(5/4*x^3)+16*x^4-256)/x^3,x, algorithm="gi
ac")

[Out]

(4*x^4 + 4*x^3*e^(5/4*x^3) + x^2*e^(5/2*x^3) - 16*x*e^(5/4*x^3) + 64)/x^2

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maple [A]  time = 0.06, size = 34, normalized size = 1.70

method	result	size
risch	\(4 x^{2}+\frac {64}{x^{2}}+{\mathrm e}^{\frac {5 x^{3}}{2}}+\frac {4 \left (x^{2}-4\right ) {\mathrm e}^{\frac {5 x^{3}}{4}}}{x}\)	\(34\)
norman	\(\frac {64+x^{2} {\mathrm e}^{\frac {5 x^{3}}{2}}+4 x^{4}-16 \,{\mathrm e}^{\frac {5 x^{3}}{4}} x +4 \,{\mathrm e}^{\frac {5 x^{3}}{4}} x^{3}}{x^{2}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(15*x^5*exp(5/4*x^3)^2+(30*x^6-120*x^4+8*x^3+32*x)*exp(5/4*x^3)+16*x^4-256)/x^3,x,method=_RETURNVERBOS
E)

[Out]

4*x^2+64/x^2+exp(5/2*x^3)+4*(x^2-4)/x*exp(5/4*x^3)

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maxima [C]  time = 0.38, size = 103, normalized size = 5.15 \begin {gather*} -\frac {16 \, \left (\frac {5}{4}\right )^{\frac {2}{3}} x^{4} \Gamma \left (\frac {4}{3}, -\frac {5}{4} \, x^{3}\right )}{5 \, \left (-x^{3}\right )^{\frac {4}{3}}} + \frac {16 \, \left (\frac {5}{4}\right )^{\frac {1}{3}} x^{2} \Gamma \left (\frac {2}{3}, -\frac {5}{4} \, x^{3}\right )}{\left (-x^{3}\right )^{\frac {2}{3}}} - \frac {16 \, \left (\frac {5}{4}\right )^{\frac {2}{3}} x \Gamma \left (\frac {1}{3}, -\frac {5}{4} \, x^{3}\right )}{15 \, \left (-x^{3}\right )^{\frac {1}{3}}} + 4 \, x^{2} - \frac {16 \, \left (\frac {5}{4}\right )^{\frac {1}{3}} \left (-x^{3}\right )^{\frac {1}{3}} \Gamma \left (-\frac {1}{3}, -\frac {5}{4} \, x^{3}\right )}{3 \, x} + \frac {64}{x^{2}} + e^{\left (\frac {5}{2} \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x^5*exp(5/4*x^3)^2+(30*x^6-120*x^4+8*x^3+32*x)*exp(5/4*x^3)+16*x^4-256)/x^3,x, algorithm="ma
xima")

[Out]

-16/5*(5/4)^(2/3)*x^4*gamma(4/3, -5/4*x^3)/(-x^3)^(4/3) + 16*(5/4)^(1/3)*x^2*gamma(2/3, -5/4*x^3)/(-x^3)^(2/3)
- 16/15*(5/4)^(2/3)*x*gamma(1/3, -5/4*x^3)/(-x^3)^(1/3) + 4*x^2 - 16/3*(5/4)^(1/3)*(-x^3)^(1/3)*gamma(-1/3, -
5/4*x^3)/x + 64/x^2 + e^(5/2*x^3)

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mupad [B]  time = 5.77, size = 37, normalized size = 1.85 \begin {gather*} {\mathrm {e}}^{\frac {5\,x^3}{2}}-\frac {16\,x\,{\mathrm {e}}^{\frac {5\,x^3}{4}}-64}{x^2}+4\,x\,{\mathrm {e}}^{\frac {5\,x^3}{4}}+4\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15*x^5*exp((5*x^3)/2))/2 + (exp((5*x^3)/4)*(32*x + 8*x^3 - 120*x^4 + 30*x^6))/2 + 8*x^4 - 128)/x^3,x)

[Out]

exp((5*x^3)/2) - (16*x*exp((5*x^3)/4) - 64)/x^2 + 4*x*exp((5*x^3)/4) + 4*x^2

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sympy [B]  time = 0.14, size = 36, normalized size = 1.80 \begin {gather*} 4 x^{2} + \frac {x e^{\frac {5 x^{3}}{2}} + \left (4 x^{2} - 16\right ) e^{\frac {5 x^{3}}{4}}}{x} + \frac {64}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(15*x**5*exp(5/4*x**3)**2+(30*x**6-120*x**4+8*x**3+32*x)*exp(5/4*x**3)+16*x**4-256)/x**3,x)

[Out]

4*x**2 + (x*exp(5*x**3/2) + (4*x**2 - 16)*exp(5*x**3/4))/x + 64/x**2

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