3.25.16 \(\int \frac {25 e^2-30 e^4+e^{x^2} (50 e^2 x+e^4 (15-30 x^2))}{200-240 e^2 x+72 e^4 x^2} \, dx\)

Optimal. Leaf size=27 \[ 2+\frac {-2+e^{x^2}+x}{4 \left (\frac {2}{e^2}-\frac {6 x}{5}\right )} \]

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Rubi [B]  time = 0.48, antiderivative size = 58, normalized size of antiderivative = 2.15, number of steps used = 8, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {27, 12, 6741, 6742, 2288} \begin {gather*} \frac {5 e^{x^2+2} \left (5 x-3 e^2 x^2\right )}{8 x \left (5-3 e^2 x\right )^2}+\frac {5 \left (5-6 e^2\right )}{24 \left (5-3 e^2 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*E^2 - 30*E^4 + E^x^2*(50*E^2*x + E^4*(15 - 30*x^2)))/(200 - 240*E^2*x + 72*E^4*x^2),x]

[Out]

(5*(5 - 6*E^2))/(24*(5 - 3*E^2*x)) + (5*E^(2 + x^2)*(5*x - 3*E^2*x^2))/(8*x*(5 - 3*E^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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 e^2-30 e^4+e^{x^2} \left (50 e^2 x+e^4 \left (15-30 x^2\right )\right )}{8 \left (-5+3 e^2 x\right )^2} \, dx\\ &=\frac {1}{8} \int \frac {25 e^2-30 e^4+e^{x^2} \left (50 e^2 x+e^4 \left (15-30 x^2\right )\right )}{\left (-5+3 e^2 x\right )^2} \, dx\\ &=\frac {1}{8} \int \frac {25 e^2 \left (1-\frac {6 e^2}{5}\right )+e^{x^2} \left (50 e^2 x+e^4 \left (15-30 x^2\right )\right )}{\left (5-3 e^2 x\right )^2} \, dx\\ &=\frac {1}{8} \int \frac {5 e^2 \left (3 e^{2+x^2}+5 \left (1-\frac {6 e^2}{5}\right )+10 e^{x^2} x-6 e^{2+x^2} x^2\right )}{\left (5-3 e^2 x\right )^2} \, dx\\ &=\frac {1}{8} \left (5 e^2\right ) \int \frac {3 e^{2+x^2}+5 \left (1-\frac {6 e^2}{5}\right )+10 e^{x^2} x-6 e^{2+x^2} x^2}{\left (5-3 e^2 x\right )^2} \, dx\\ &=\frac {1}{8} \left (5 e^2\right ) \int \left (\frac {5-6 e^2}{\left (-5+3 e^2 x\right )^2}-\frac {e^{x^2} \left (-3 e^2-10 x+6 e^2 x^2\right )}{\left (-5+3 e^2 x\right )^2}\right ) \, dx\\ &=\frac {5 \left (5-6 e^2\right )}{24 \left (5-3 e^2 x\right )}-\frac {1}{8} \left (5 e^2\right ) \int \frac {e^{x^2} \left (-3 e^2-10 x+6 e^2 x^2\right )}{\left (-5+3 e^2 x\right )^2} \, dx\\ &=\frac {5 \left (5-6 e^2\right )}{24 \left (5-3 e^2 x\right )}+\frac {5 e^{2+x^2} \left (5 x-3 e^2 x^2\right )}{8 x \left (5-3 e^2 x\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 37, normalized size = 1.37 \begin {gather*} \frac {5 e^2 \left (5-6 e^2+3 e^{2+x^2}\right )}{8 \left (15 e^2-9 e^4 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*E^2 - 30*E^4 + E^x^2*(50*E^2*x + E^4*(15 - 30*x^2)))/(200 - 240*E^2*x + 72*E^4*x^2),x]

[Out]

(5*E^2*(5 - 6*E^2 + 3*E^(2 + x^2)))/(8*(15*E^2 - 9*E^4*x))

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fricas [A]  time = 0.47, size = 25, normalized size = 0.93 \begin {gather*} \frac {5 \, {\left (6 \, e^{2} - 3 \, e^{\left (x^{2} + 2\right )} - 5\right )}}{24 \, {\left (3 \, x e^{2} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x^2+15)*exp(2)^2+50*exp(2)*x)*exp(x^2)-30*exp(2)^2+25*exp(2))/(72*x^2*exp(2)^2-240*exp(2)*x+2
00),x, algorithm="fricas")

[Out]

5/24*(6*e^2 - 3*e^(x^2 + 2) - 5)/(3*x*e^2 - 5)

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giac [A]  time = 0.24, size = 25, normalized size = 0.93 \begin {gather*} \frac {5 \, {\left (6 \, e^{2} - 3 \, e^{\left (x^{2} + 2\right )} - 5\right )}}{24 \, {\left (3 \, x e^{2} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x^2+15)*exp(2)^2+50*exp(2)*x)*exp(x^2)-30*exp(2)^2+25*exp(2))/(72*x^2*exp(2)^2-240*exp(2)*x+2
00),x, algorithm="giac")

[Out]

5/24*(6*e^2 - 3*e^(x^2 + 2) - 5)/(3*x*e^2 - 5)

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maple [A]  time = 2.31, size = 33, normalized size = 1.22

method	result	size
norman	\(\frac {\left (\frac {3 \,{\mathrm e}^{4}}{4}-\frac {5 \,{\mathrm e}^{2}}{8}\right ) x -\frac {5 \,{\mathrm e}^{x^{2}} {\mathrm e}^{2}}{8}}{3 \,{\mathrm e}^{2} x -5}\)	\(33\)
risch	\(\frac {5 \,{\mathrm e}^{2}}{12 \left ({\mathrm e}^{2} x -\frac {5}{3}\right )}-\frac {25}{72 \left ({\mathrm e}^{2} x -\frac {5}{3}\right )}-\frac {5 \,{\mathrm e}^{x^{2}+2}}{8 \left (3 \,{\mathrm e}^{2} x -5\right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-30*x^2+15)*exp(2)^2+50*exp(2)*x)*exp(x^2)-30*exp(2)^2+25*exp(2))/(72*x^2*exp(2)^2-240*exp(2)*x+200),x,
method=_RETURNVERBOSE)

[Out]

((3/4*exp(2)^2-5/8*exp(2))*x-5/8*exp(x^2)*exp(2))/(3*exp(2)*x-5)

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maxima [B]  time = 0.45, size = 50, normalized size = 1.85 \begin {gather*} \frac {5 \, e^{4}}{4 \, {\left (3 \, x e^{4} - 5 \, e^{2}\right )}} - \frac {25 \, e^{2}}{24 \, {\left (3 \, x e^{4} - 5 \, e^{2}\right )}} - \frac {5 \, e^{\left (x^{2} + 2\right )}}{8 \, {\left (3 \, x e^{2} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x^2+15)*exp(2)^2+50*exp(2)*x)*exp(x^2)-30*exp(2)^2+25*exp(2))/(72*x^2*exp(2)^2-240*exp(2)*x+2
00),x, algorithm="maxima")

[Out]

5/4*e^4/(3*x*e^4 - 5*e^2) - 25/24*e^2/(3*x*e^4 - 5*e^2) - 5/8*e^(x^2 + 2)/(3*x*e^2 - 5)

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mupad [B]  time = 0.25, size = 31, normalized size = 1.15 \begin {gather*} -\frac {5\,{\mathrm {e}}^{x^2+2}-\frac {x\,{\mathrm {e}}^2\,\left (30\,{\mathrm {e}}^2-25\right )}{5}}{24\,x\,{\mathrm {e}}^2-40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(2) - 30*exp(4) + exp(x^2)*(50*x*exp(2) - exp(4)*(30*x^2 - 15)))/(72*x^2*exp(4) - 240*x*exp(2) + 20
0),x)

[Out]

-(5*exp(x^2 + 2) - (x*exp(2)*(30*exp(2) - 25))/5)/(24*x*exp(2) - 40)

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sympy [A]  time = 0.22, size = 41, normalized size = 1.52 \begin {gather*} - \frac {- 30 e^{4} + 25 e^{2}}{72 x e^{4} - 120 e^{2}} - \frac {5 e^{2} e^{x^{2}}}{24 x e^{2} - 40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-30*x**2+15)*exp(2)**2+50*exp(2)*x)*exp(x**2)-30*exp(2)**2+25*exp(2))/(72*x**2*exp(2)**2-240*exp(
2)*x+200),x)

[Out]

-(-30*exp(4) + 25*exp(2))/(72*x*exp(4) - 120*exp(2)) - 5*exp(2)*exp(x**2)/(24*x*exp(2) - 40)

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