3.11.60 \(\int \frac {10 x-2 e x-2 x \log (25)}{e+\log (25)} \, dx\) [1060]

Optimal. Leaf size=18 \[ -1-x^2+\frac {5 x^2}{e+\log (25)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6, 12, 30} \begin {gather*} \frac {x^2 (5-e-\log (25))}{e+\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x - 2*E*x - 2*x*Log[25])/(E + Log[25]),x]

[Out]

(x^2*(5 - E - Log[25]))/(E + Log[25])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(10-2 e) x-2 x \log (25)}{e+\log (25)} \, dx\\ &=\int \frac {x (10-2 e-2 \log (25))}{e+\log (25)} \, dx\\ &=\frac {(2 (5-e-\log (25))) \int x \, dx}{e+\log (25)}\\ &=\frac {x^2 (5-e-\log (25))}{e+\log (25)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 0.89 \begin {gather*} -\frac {x^2 (-5+e+\log (25))}{e+\log (25)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x - 2*E*x - 2*x*Log[25])/(E + Log[25]),x]

[Out]

-((x^2*(-5 + E + Log[25]))/(E + Log[25]))

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Maple [A]
time = 0.05, size = 23, normalized size = 1.28

method	result	size
gosper	\(-\frac {x^{2} \left (2 \ln \left (5\right )+{\mathrm e}-5\right )}{2 \ln \left (5\right )+{\mathrm e}}\)	\(23\)
default	\(-\frac {x^{2} \left (2 \ln \left (5\right )+{\mathrm e}-5\right )}{2 \ln \left (5\right )+{\mathrm e}}\)	\(23\)
norman	\(-\frac {x^{2} \left (2 \ln \left (5\right )+{\mathrm e}-5\right )}{2 \ln \left (5\right )+{\mathrm e}}\)	\(23\)
risch	\(-\frac {2 x^{2} \ln \left (5\right )}{2 \ln \left (5\right )+{\mathrm e}}-\frac {x^{2} {\mathrm e}}{2 \ln \left (5\right )+{\mathrm e}}+\frac {5 x^{2}}{2 \ln \left (5\right )+{\mathrm e}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x*ln(5)-2*x*exp(1)+10*x)/(2*ln(5)+exp(1)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 30, normalized size = 1.67 \begin {gather*} -\frac {x^{2} e + 2 \, x^{2} \log \left (5\right ) - 5 \, x^{2}}{e + 2 \, \log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(5)-2*x*exp(1)+10*x)/(2*log(5)+exp(1)),x, algorithm="maxima")

[Out]

-(x^2*e + 2*x^2*log(5) - 5*x^2)/(e + 2*log(5))

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Fricas [A]
time = 0.38, size = 30, normalized size = 1.67 \begin {gather*} -\frac {x^{2} e + 2 \, x^{2} \log \left (5\right ) - 5 \, x^{2}}{e + 2 \, \log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(5)-2*x*exp(1)+10*x)/(2*log(5)+exp(1)),x, algorithm="fricas")

[Out]

-(x^2*e + 2*x^2*log(5) - 5*x^2)/(e + 2*log(5))

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Sympy [A]
time = 0.01, size = 20, normalized size = 1.11 \begin {gather*} \frac {x^{2} \left (- 2 \log {\left (5 \right )} - e + 5\right )}{e + 2 \log {\left (5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*ln(5)-2*x*exp(1)+10*x)/(2*ln(5)+exp(1)),x)

[Out]

x**2*(-2*log(5) - E + 5)/(E + 2*log(5))

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Giac [A]
time = 0.39, size = 30, normalized size = 1.67 \begin {gather*} -\frac {x^{2} e + 2 \, x^{2} \log \left (5\right ) - 5 \, x^{2}}{e + 2 \, \log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x*log(5)-2*x*exp(1)+10*x)/(2*log(5)+exp(1)),x, algorithm="giac")

[Out]

-(x^2*e + 2*x^2*log(5) - 5*x^2)/(e + 2*log(5))

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Mupad [B]
time = 0.05, size = 18, normalized size = 1.00 \begin {gather*} -\frac {x^2\,\left (\mathrm {e}+\ln \left (25\right )-5\right )}{\mathrm {e}+\ln \left (25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*exp(1) - 10*x + 4*x*log(5))/(exp(1) + 2*log(5)),x)

[Out]

-(x^2*(exp(1) + log(25) - 5))/(exp(1) + log(25))

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