3.84.6 \(\int \frac {e^x (-140+8 x-4 x^2+5 \log (3)+16 \log (5))}{19600+1120 x^2+16 x^4+(-1400-40 x^2) \log (3)+25 \log ^2(3)+(-4480-128 x^2+160 \log (3)) \log (5)+256 \log ^2(5)} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^x}{8+5 \log (3)-4 \left (1+x^2-4 (-9+\log (5))\right )} \]

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Rubi [B]  time = 0.40, antiderivative size = 73, normalized size of antiderivative = 2.92, number of steps used = 1, number of rules used = 1, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2288} \begin {gather*} -\frac {e^x \left (4 x^2+140-\log (243)-16 \log (5)\right )}{16 x^4+1120 x^2-32 \log (5) \left (4 x^2+5 (28-\log (3))\right )-40 \left (x^2+35\right ) \log (3)+19600+256 \log ^2(5)+25 \log ^2(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-140 + 8*x - 4*x^2 + 5*Log[3] + 16*Log[5]))/(19600 + 1120*x^2 + 16*x^4 + (-1400 - 40*x^2)*Log[3] + 2
5*Log[3]^2 + (-4480 - 128*x^2 + 160*Log[3])*Log[5] + 256*Log[5]^2),x]

[Out]

-((E^x*(140 + 4*x^2 - 16*Log[5] - Log[243]))/(19600 + 1120*x^2 + 16*x^4 - 40*(35 + x^2)*Log[3] + 25*Log[3]^2 -
32*(4*x^2 + 5*(28 - Log[3]))*Log[5] + 256*Log[5]^2))

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 {e^x \left (140+4 x^2-16 \log (5)-\log (243)\right )}{19600+1120 x^2+16 x^4-40 \left (35+x^2\right ) \log (3)+25 \log ^2(3)-32 \left (4 x^2+5 (28-\log (3))\right ) \log (5)+256 \log ^2(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 34, normalized size = 1.36 \begin {gather*} \frac {e^x \left (-140-4 x^2+16 \log (5)+\log (243)\right )}{\left (140+4 x^2-5 \log (3)-16 \log (5)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-140 + 8*x - 4*x^2 + 5*Log[3] + 16*Log[5]))/(19600 + 1120*x^2 + 16*x^4 + (-1400 - 40*x^2)*Log[
3] + 25*Log[3]^2 + (-4480 - 128*x^2 + 160*Log[3])*Log[5] + 256*Log[5]^2),x]

[Out]

(E^x*(-140 - 4*x^2 + 16*Log[5] + Log[243]))/(140 + 4*x^2 - 5*Log[3] - 16*Log[5])^2

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fricas [A]  time = 1.04, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{x}}{4 \, x^{2} - 16 \, \log \relax (5) - 5 \, \log \relax (3) + 140} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(5)+5*log(3)-4*x^2+8*x-140)*exp(x)/(256*log(5)^2+(160*log(3)-128*x^2-4480)*log(5)+25*log(3)^2
+(-40*x^2-1400)*log(3)+16*x^4+1120*x^2+19600),x, algorithm="fricas")

[Out]

-e^x/(4*x^2 - 16*log(5) - 5*log(3) + 140)

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giac [A]  time = 0.20, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{x}}{4 \, x^{2} - 16 \, \log \relax (5) - 5 \, \log \relax (3) + 140} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(5)+5*log(3)-4*x^2+8*x-140)*exp(x)/(256*log(5)^2+(160*log(3)-128*x^2-4480)*log(5)+25*log(3)^2
+(-40*x^2-1400)*log(3)+16*x^4+1120*x^2+19600),x, algorithm="giac")

[Out]

-e^x/(4*x^2 - 16*log(5) - 5*log(3) + 140)

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maple [A]  time = 0.16, size = 21, normalized size = 0.84

method	result	size
gosper	\(\frac {{\mathrm e}^{x}}{-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)}\)	\(21\)
norman	\(\frac {{\mathrm e}^{x}}{-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)}\)	\(21\)
default	\(-\frac {70 \,{\mathrm e}^{x} x}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}+\frac {35 \,{\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{4 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}+\frac {35 \,{\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{4 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}+\frac {5 i \ln \relax (3) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {2 i \ln \relax (5) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}+\frac {{\mathrm e}^{x}}{-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)}-\frac {5 i \ln \relax (3) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {i {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {{\mathrm e}^{x} x}{2 \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}+\frac {{\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16}+\frac {{\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16}+\frac {5 \ln \relax (3) {\mathrm e}^{x} x}{2 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}-\frac {5 \ln \relax (3) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}-\frac {5 \ln \relax (3) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}+\frac {2 i \ln \relax (5) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {35 i {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{2 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}+\frac {8 \ln \relax (5) {\mathrm e}^{x} x}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}-\frac {\ln \relax (5) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \ln \relax (5)+5 \ln \relax (3)-140}-\frac {\ln \relax (5) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \ln \relax (5)+5 \ln \relax (3)-140}+\frac {35 i {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{2 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}+\frac {i {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}\)	\(1038\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*ln(5)+5*ln(3)-4*x^2+8*x-140)*exp(x)/(256*ln(5)^2+(160*ln(3)-128*x^2-4480)*ln(5)+25*ln(3)^2+(-40*x^2-14
00)*ln(3)+16*x^4+1120*x^2+19600),x,method=_RETURNVERBOSE)

[Out]

exp(x)/(-140-4*x^2+16*ln(5)+5*ln(3))

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maxima [A]  time = 0.49, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{x}}{4 \, x^{2} - 16 \, \log \relax (5) - 5 \, \log \relax (3) + 140} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(5)+5*log(3)-4*x^2+8*x-140)*exp(x)/(256*log(5)^2+(160*log(3)-128*x^2-4480)*log(5)+25*log(3)^2
+(-40*x^2-1400)*log(3)+16*x^4+1120*x^2+19600),x, algorithm="maxima")

[Out]

-e^x/(4*x^2 - 16*log(5) - 5*log(3) + 140)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (-4\,x^2+8\,x+5\,\ln \relax (3)+16\,\ln \relax (5)-140\right )}{25\,{\ln \relax (3)}^2-\ln \relax (3)\,\left (40\,x^2+1400\right )-\ln \relax (5)\,\left (128\,x^2-160\,\ln \relax (3)+4480\right )+256\,{\ln \relax (5)}^2+1120\,x^2+16\,x^4+19600} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(8*x + 5*log(3) + 16*log(5) - 4*x^2 - 140))/(25*log(3)^2 - log(3)*(40*x^2 + 1400) - log(5)*(128*x^
2 - 160*log(3) + 4480) + 256*log(5)^2 + 1120*x^2 + 16*x^4 + 19600),x)

[Out]

int((exp(x)*(8*x + 5*log(3) + 16*log(5) - 4*x^2 - 140))/(25*log(3)^2 - log(3)*(40*x^2 + 1400) - log(5)*(128*x^
2 - 160*log(3) + 4480) + 256*log(5)^2 + 1120*x^2 + 16*x^4 + 19600), x)

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sympy [A]  time = 0.17, size = 20, normalized size = 0.80 \begin {gather*} - \frac {e^{x}}{4 x^{2} - 16 \log {\relax (5 )} - 5 \log {\relax (3 )} + 140} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*ln(5)+5*ln(3)-4*x**2+8*x-140)*exp(x)/(256*ln(5)**2+(160*ln(3)-128*x**2-4480)*ln(5)+25*ln(3)**2+(
-40*x**2-1400)*ln(3)+16*x**4+1120*x**2+19600),x)

[Out]

-exp(x)/(4*x**2 - 16*log(5) - 5*log(3) + 140)

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