3.27.51 \(\int \frac {75 x+45 x^3+60 x^4+e^x (10-5 x+2 x^3)}{15 x^3} \, dx\)

Optimal. Leaf size=25 \[ x+\left (1+\frac {e^x}{15 x}+x\right ) \left (-\frac {5}{x}+2 x\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 14, 2199, 2194, 2177, 2178} \begin {gather*} 2 x^2-\frac {e^x}{3 x^2}+3 x+\frac {2 e^x}{15}-\frac {5}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75*x + 45*x^3 + 60*x^4 + E^x*(10 - 5*x + 2*x^3))/(15*x^3),x]

[Out]

(2*E^x)/15 - E^x/(3*x^2) - 5/x + 3*x + 2*x^2

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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{15} \int \frac {75 x+45 x^3+60 x^4+e^x \left (10-5 x+2 x^3\right )}{x^3} \, dx\\ &=\frac {1}{15} \int \left (\frac {e^x \left (10-5 x+2 x^3\right )}{x^3}+\frac {15 \left (5+3 x^2+4 x^3\right )}{x^2}\right ) \, dx\\ &=\frac {1}{15} \int \frac {e^x \left (10-5 x+2 x^3\right )}{x^3} \, dx+\int \frac {5+3 x^2+4 x^3}{x^2} \, dx\\ &=\frac {1}{15} \int \left (2 e^x+\frac {10 e^x}{x^3}-\frac {5 e^x}{x^2}\right ) \, dx+\int \left (3+\frac {5}{x^2}+4 x\right ) \, dx\\ &=-\frac {5}{x}+3 x+2 x^2+\frac {2 \int e^x \, dx}{15}-\frac {1}{3} \int \frac {e^x}{x^2} \, dx+\frac {2}{3} \int \frac {e^x}{x^3} \, dx\\ &=\frac {2 e^x}{15}-\frac {e^x}{3 x^2}-\frac {5}{x}+\frac {e^x}{3 x}+3 x+2 x^2+\frac {1}{3} \int \frac {e^x}{x^2} \, dx-\frac {1}{3} \int \frac {e^x}{x} \, dx\\ &=\frac {2 e^x}{15}-\frac {e^x}{3 x^2}-\frac {5}{x}+3 x+2 x^2-\frac {\text {Ei}(x)}{3}+\frac {1}{3} \int \frac {e^x}{x} \, dx\\ &=\frac {2 e^x}{15}-\frac {e^x}{3 x^2}-\frac {5}{x}+3 x+2 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.24 \begin {gather*} \frac {2 e^x}{15}-\frac {e^x}{3 x^2}-\frac {5}{x}+3 x+2 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75*x + 45*x^3 + 60*x^4 + E^x*(10 - 5*x + 2*x^3))/(15*x^3),x]

[Out]

(2*E^x)/15 - E^x/(3*x^2) - 5/x + 3*x + 2*x^2

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fricas [A]  time = 0.49, size = 29, normalized size = 1.16 \begin {gather*} \frac {30 \, x^{4} + 45 \, x^{3} + {\left (2 \, x^{2} - 5\right )} e^{x} - 75 \, x}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((2*x^3-5*x+10)*exp(x)+60*x^4+45*x^3+75*x)/x^3,x, algorithm="fricas")

[Out]

1/15*(30*x^4 + 45*x^3 + (2*x^2 - 5)*e^x - 75*x)/x^2

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giac [A]  time = 0.24, size = 30, normalized size = 1.20 \begin {gather*} \frac {30 \, x^{4} + 45 \, x^{3} + 2 \, x^{2} e^{x} - 75 \, x - 5 \, e^{x}}{15 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((2*x^3-5*x+10)*exp(x)+60*x^4+45*x^3+75*x)/x^3,x, algorithm="giac")

[Out]

1/15*(30*x^4 + 45*x^3 + 2*x^2*e^x - 75*x - 5*e^x)/x^2

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maple [A]  time = 0.05, size = 26, normalized size = 1.04

method	result	size
default	\(2 x^{2}+3 x -\frac {5}{x}-\frac {{\mathrm e}^{x}}{3 x^{2}}+\frac {2 \,{\mathrm e}^{x}}{15}\)	\(26\)
risch	\(2 x^{2}+3 x -\frac {5}{x}+\frac {\left (2 x^{2}-5\right ) {\mathrm e}^{x}}{15 x^{2}}\)	\(29\)
norman	\(\frac {-5 x +3 x^{3}+2 x^{4}+\frac {2 \,{\mathrm e}^{x} x^{2}}{15}-\frac {{\mathrm e}^{x}}{3}}{x^{2}}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/15*((2*x^3-5*x+10)*exp(x)+60*x^4+45*x^3+75*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

2*x^2+3*x-5/x-1/3*exp(x)/x^2+2/15*exp(x)

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maxima [C]  time = 0.72, size = 32, normalized size = 1.28 \begin {gather*} 2 \, x^{2} + 3 \, x - \frac {5}{x} + \frac {2}{15} \, e^{x} - \frac {1}{3} \, \Gamma \left (-1, -x\right ) - \frac {2}{3} \, \Gamma \left (-2, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((2*x^3-5*x+10)*exp(x)+60*x^4+45*x^3+75*x)/x^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 26, normalized size = 1.04 \begin {gather*} 3\,x+\frac {2\,{\mathrm {e}}^x}{15}-\frac {5\,x+\frac {{\mathrm {e}}^x}{3}}{x^2}+2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + (exp(x)*(2*x^3 - 5*x + 10))/15 + 3*x^3 + 4*x^4)/x^3,x)

[Out]

3*x + (2*exp(x))/15 - (5*x + exp(x)/3)/x^2 + 2*x^2

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sympy [A]  time = 0.11, size = 26, normalized size = 1.04 \begin {gather*} 2 x^{2} + 3 x - \frac {5}{x} + \frac {\left (2 x^{2} - 5\right ) e^{x}}{15 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/15*((2*x**3-5*x+10)*exp(x)+60*x**4+45*x**3+75*x)/x**3,x)

[Out]

2*x**2 + 3*x - 5/x + (2*x**2 - 5)*exp(x)/(15*x**2)

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