3.95.74 \(\int \frac {-45+50 x^3-45 e^{1+x} x^3+90 \log (x)}{x^3} \, dx\) [9474]

Optimal. Leaf size=27 \[ 5 \left (x+9 \left (e^{e^2}-e^{1+x}+x-\frac {\log (x)}{x^2}\right )\right ) \]

[Out]

45*exp(exp(2))-45*ln(x)/x^2+50*x-45*exp(1+x)

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Rubi [A]
time = 0.03, antiderivative size = 18, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {14, 2225, 2341} \begin {gather*} -\frac {45 \log (x)}{x^2}+50 x-45 e^{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-45 + 50*x^3 - 45*E^(1 + x)*x^3 + 90*Log[x])/x^3,x]

[Out]

-45*E^(1 + x) + 50*x - (45*Log[x])/x^2

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 2225

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 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-45 e^{1+x}+\frac {5 \left (-9+10 x^3+18 \log (x)\right )}{x^3}\right ) \, dx\\ &=5 \int \frac {-9+10 x^3+18 \log (x)}{x^3} \, dx-45 \int e^{1+x} \, dx\\ &=-45 e^{1+x}+5 \int \left (\frac {-9+10 x^3}{x^3}+\frac {18 \log (x)}{x^3}\right ) \, dx\\ &=-45 e^{1+x}+5 \int \frac {-9+10 x^3}{x^3} \, dx+90 \int \frac {\log (x)}{x^3} \, dx\\ &=-45 e^{1+x}-\frac {45}{2 x^2}-\frac {45 \log (x)}{x^2}+5 \int \left (10-\frac {9}{x^3}\right ) \, dx\\ &=-45 e^{1+x}+50 x-\frac {45 \log (x)}{x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.04, size = 18, normalized size = 0.67 \begin {gather*} -45 e^{1+x}+50 x-\frac {45 \log (x)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 + 50*x^3 - 45*E^(1 + x)*x^3 + 90*Log[x])/x^3,x]

[Out]

-45*E^(1 + x) + 50*x - (45*Log[x])/x^2

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Maple [A]
time = 0.25, size = 18, normalized size = 0.67

method	result	size
default	\(50 x -\frac {45 \ln \left (x \right )}{x^{2}}-45 \,{\mathrm e}^{x +1}\)	\(18\)
risch	\(50 x -\frac {45 \ln \left (x \right )}{x^{2}}-45 \,{\mathrm e}^{x +1}\)	\(18\)
norman	\(\frac {50 x^{3}-45 x^{2} {\mathrm e}^{x +1}-45 \ln \left (x \right )}{x^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*ln(x)-45*x^3*exp(x+1)+50*x^3-45)/x^3,x,method=_RETURNVERBOSE)

[Out]

50*x-45*ln(x)/x^2-45*exp(x+1)

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Maxima [A]
time = 0.27, size = 17, normalized size = 0.63 \begin {gather*} 50 \, x - \frac {45 \, \log \left (x\right )}{x^{2}} - 45 \, e^{\left (x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*log(x)-45*x^3*exp(1+x)+50*x^3-45)/x^3,x, algorithm="maxima")

[Out]

50*x - 45*log(x)/x^2 - 45*e^(x + 1)

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Fricas [A]
time = 0.40, size = 24, normalized size = 0.89 \begin {gather*} \frac {5 \, {\left (10 \, x^{3} - 9 \, x^{2} e^{\left (x + 1\right )} - 9 \, \log \left (x\right )\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*log(x)-45*x^3*exp(1+x)+50*x^3-45)/x^3,x, algorithm="fricas")

[Out]

5*(10*x^3 - 9*x^2*e^(x + 1) - 9*log(x))/x^2

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Sympy [A]
time = 0.07, size = 17, normalized size = 0.63 \begin {gather*} 50 x - 45 e^{x + 1} - \frac {45 \log {\left (x \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*ln(x)-45*x**3*exp(1+x)+50*x**3-45)/x**3,x)

[Out]

50*x - 45*exp(x + 1) - 45*log(x)/x**2

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Giac [A]
time = 0.41, size = 24, normalized size = 0.89 \begin {gather*} \frac {5 \, {\left (10 \, x^{3} - 9 \, x^{2} e^{\left (x + 1\right )} - 9 \, \log \left (x\right )\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((90*log(x)-45*x^3*exp(1+x)+50*x^3-45)/x^3,x, algorithm="giac")

[Out]

5*(10*x^3 - 9*x^2*e^(x + 1) - 9*log(x))/x^2

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Mupad [B]
time = 5.64, size = 17, normalized size = 0.63 \begin {gather*} 50\,x-\frac {45\,\ln \left (x\right )}{x^2}-45\,\mathrm {e}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((90*log(x) - 45*x^3*exp(x + 1) + 50*x^3 - 45)/x^3,x)

[Out]

50*x - (45*log(x))/x^2 - 45*exp(1)*exp(x)

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