3.40.100 \(\int \frac {8 e^x \log (-7+4 x)+e^x (-7+4 x) \log ^2(-7+4 x)}{-175+100 x} \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{25} e^x \log ^2(-7+4 x) \]

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Rubi [B]  time = 0.14, antiderivative size = 38, normalized size of antiderivative = 2.53, number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6741, 2288} \begin {gather*} \frac {e^x \log (4 x-7) (7 \log (4 x-7)-4 x \log (4 x-7))}{25 (7-4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^x*Log[-7 + 4*x] + E^x*(-7 + 4*x)*Log[-7 + 4*x]^2)/(-175 + 100*x),x]

[Out]

(E^x*Log[-7 + 4*x]*(7*Log[-7 + 4*x] - 4*x*Log[-7 + 4*x]))/(25*(7 - 4*x))

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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \log (-7+4 x) (-8+7 \log (-7+4 x)-4 x \log (-7+4 x))}{175-100 x} \, dx\\ &=\frac {e^x \log (-7+4 x) (7 \log (-7+4 x)-4 x \log (-7+4 x))}{25 (7-4 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{25} e^x \log ^2(-7+4 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^x*Log[-7 + 4*x] + E^x*(-7 + 4*x)*Log[-7 + 4*x]^2)/(-175 + 100*x),x]

[Out]

(E^x*Log[-7 + 4*x]^2)/25

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fricas [A]  time = 0.84, size = 12, normalized size = 0.80 \begin {gather*} \frac {1}{25} \, e^{x} \log \left (4 \, x - 7\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-7)*exp(x)*log(4*x-7)^2+8*exp(x)*log(4*x-7))/(100*x-175),x, algorithm="fricas")

[Out]

1/25*e^x*log(4*x - 7)^2

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giac [A]  time = 0.15, size = 12, normalized size = 0.80 \begin {gather*} \frac {1}{25} \, e^{x} \log \left (4 \, x - 7\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-7)*exp(x)*log(4*x-7)^2+8*exp(x)*log(4*x-7))/(100*x-175),x, algorithm="giac")

[Out]

1/25*e^x*log(4*x - 7)^2

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maple [A]  time = 0.06, size = 13, normalized size = 0.87

method	result	size
norman	\(\frac {{\mathrm e}^{x} \ln \left (4 x -7\right )^{2}}{25}\)	\(13\)
risch	\(\frac {{\mathrm e}^{x} \ln \left (4 x -7\right )^{2}}{25}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x-7)*exp(x)*ln(4*x-7)^2+8*exp(x)*ln(4*x-7))/(100*x-175),x,method=_RETURNVERBOSE)

[Out]

1/25*exp(x)*ln(4*x-7)^2

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maxima [A]  time = 0.42, size = 12, normalized size = 0.80 \begin {gather*} \frac {1}{25} \, e^{x} \log \left (4 \, x - 7\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-7)*exp(x)*log(4*x-7)^2+8*exp(x)*log(4*x-7))/(100*x-175),x, algorithm="maxima")

[Out]

1/25*e^x*log(4*x - 7)^2

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mupad [B]  time = 2.76, size = 12, normalized size = 0.80 \begin {gather*} \frac {{\mathrm {e}}^x\,{\ln \left (4\,x-7\right )}^2}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(x)*log(4*x - 7) + exp(x)*log(4*x - 7)^2*(4*x - 7))/(100*x - 175),x)

[Out]

(exp(x)*log(4*x - 7)^2)/25

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sympy [A]  time = 0.33, size = 12, normalized size = 0.80 \begin {gather*} \frac {e^{x} \log {\left (4 x - 7 \right )}^{2}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-7)*exp(x)*ln(4*x-7)**2+8*exp(x)*ln(4*x-7))/(100*x-175),x)

[Out]

exp(x)*log(4*x - 7)**2/25

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