3.50.14 \(\int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} (49 e^5 x+98 x^2)+e^{e^5 x+x^2} (98-7 x+2450 x^2-14 x^3+e^5 (1225 x-7 x^2))+(98-7 x+e^{e^5 x+x^2} (49 e^5 x+98 x^2)) \log (x^2)}{49 x} \, dx\)

Optimal. Leaf size=28 \[ 2+\frac {1}{2} \left (25+e^{x \left (e^5+x\right )}-\frac {x}{7}+\log \left (x^2\right )\right )^2 \]

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Rubi [C]  time = 0.99, antiderivative size = 244, normalized size of antiderivative = 8.71, number of steps used = 23, number of rules used = 11, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 14, 2244, 2236, 6686, 6742, 2234, 2204, 2240, 2241, 2554} \begin {gather*} -\frac {1}{28} e^{5-\frac {e^{10}}{4}} \left (350-e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )-\frac {1}{14} e^{-\frac {e^{10}}{4}} \left (1-175 e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )-\frac {1}{28} e^{10-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )+\frac {1}{14} e^{-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )+\frac {1}{14} e^{x^2+e^5 x+5}+\frac {1}{2} e^{2 x^2+2 e^5 x}-\frac {1}{7} e^{x^2+e^5 x} x+\frac {1}{14} \left (350-e^5\right ) e^{x^2+e^5 x}+\frac {1}{98} \left (7 \log \left (x^2\right )-x+175\right )^2+e^{x^2+e^5 x} \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2450 - 189*x + x^2 + E^(2*E^5*x + 2*x^2)*(49*E^5*x + 98*x^2) + E^(E^5*x + x^2)*(98 - 7*x + 2450*x^2 - 14*
x^3 + E^5*(1225*x - 7*x^2)) + (98 - 7*x + E^(E^5*x + x^2)*(49*E^5*x + 98*x^2))*Log[x^2])/(49*x),x]

[Out]

E^(5 + E^5*x + x^2)/14 + E^(2*E^5*x + 2*x^2)/2 + (E^(E^5*x + x^2)*(350 - E^5))/14 - (E^(E^5*x + x^2)*x)/7 + (S
qrt[Pi]*Erfi[(E^5 + 2*x)/2])/(14*E^(E^10/4)) - (E^(10 - E^10/4)*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2])/28 - ((1 - 175*E
^5)*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2])/(14*E^(E^10/4)) - (E^(5 - E^10/4)*(350 - E^5)*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2])/
28 + E^(E^5*x + x^2)*Log[x^2] + (175 - x + 7*Log[x^2])^2/98

Rule 12

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

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 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2240

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{49} \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{x} \, dx\\ &=\frac {1}{49} \int \left (49 e^{2 x \left (e^5+x\right )} \left (e^5+2 x\right )+\frac {(-14+x) \left (-175+x-7 \log \left (x^2\right )\right )}{x}+\frac {7 e^{e^5 x+x^2} \left (14-\left (1-175 e^5\right ) x+350 \left (1-\frac {e^5}{350}\right ) x^2-2 x^3+7 e^5 x \log \left (x^2\right )+14 x^2 \log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=\frac {1}{49} \int \frac {(-14+x) \left (-175+x-7 \log \left (x^2\right )\right )}{x} \, dx+\frac {1}{7} \int \frac {e^{e^5 x+x^2} \left (14-\left (1-175 e^5\right ) x+350 \left (1-\frac {e^5}{350}\right ) x^2-2 x^3+7 e^5 x \log \left (x^2\right )+14 x^2 \log \left (x^2\right )\right )}{x} \, dx+\int e^{2 x \left (e^5+x\right )} \left (e^5+2 x\right ) \, dx\\ &=\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2+\frac {1}{7} \int \left (\frac {e^{e^5 x+x^2} \left (14-\left (1-175 e^5\right ) x+\left (350-e^5\right ) x^2-2 x^3\right )}{x}+7 e^{e^5 x+x^2} \left (e^5+2 x\right ) \log \left (x^2\right )\right ) \, dx+\int e^{2 e^5 x+2 x^2} \left (e^5+2 x\right ) \, dx\\ &=\frac {1}{2} e^{2 e^5 x+2 x^2}+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2+\frac {1}{7} \int \frac {e^{e^5 x+x^2} \left (14-\left (1-175 e^5\right ) x+\left (350-e^5\right ) x^2-2 x^3\right )}{x} \, dx+\int e^{e^5 x+x^2} \left (e^5+2 x\right ) \log \left (x^2\right ) \, dx\\ &=\frac {1}{2} e^{2 e^5 x+2 x^2}+e^{e^5 x+x^2} \log \left (x^2\right )+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2+\frac {1}{7} \int \left (e^{e^5 x+x^2} \left (-1+175 e^5\right )+\frac {14 e^{e^5 x+x^2}}{x}+e^{e^5 x+x^2} \left (350-e^5\right ) x-2 e^{e^5 x+x^2} x^2\right ) \, dx-\int \frac {2 e^{e^5 x+x^2}}{x} \, dx\\ &=\frac {1}{2} e^{2 e^5 x+2 x^2}+e^{e^5 x+x^2} \log \left (x^2\right )+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2-\frac {2}{7} \int e^{e^5 x+x^2} x^2 \, dx+\frac {1}{7} \left (350-e^5\right ) \int e^{e^5 x+x^2} x \, dx+\frac {1}{7} \left (-1+175 e^5\right ) \int e^{e^5 x+x^2} \, dx\\ &=\frac {1}{2} e^{2 e^5 x+2 x^2}+\frac {1}{14} e^{e^5 x+x^2} \left (350-e^5\right )-\frac {1}{7} e^{e^5 x+x^2} x+e^{e^5 x+x^2} \log \left (x^2\right )+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2+\frac {1}{7} \int e^{e^5 x+x^2} \, dx+\frac {1}{7} e^5 \int e^{e^5 x+x^2} x \, dx-\frac {1}{7} \left (e^{-\frac {e^{10}}{4}} \left (1-175 e^5\right )\right ) \int e^{\frac {1}{4} \left (e^5+2 x\right )^2} \, dx-\frac {1}{14} \left (e^5 \left (350-e^5\right )\right ) \int e^{e^5 x+x^2} \, dx\\ &=\frac {1}{14} e^{5+e^5 x+x^2}+\frac {1}{2} e^{2 e^5 x+2 x^2}+\frac {1}{14} e^{e^5 x+x^2} \left (350-e^5\right )-\frac {1}{7} e^{e^5 x+x^2} x-\frac {1}{14} e^{-\frac {e^{10}}{4}} \left (1-175 e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )+e^{e^5 x+x^2} \log \left (x^2\right )+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2-\frac {1}{14} e^{10} \int e^{e^5 x+x^2} \, dx+\frac {1}{7} e^{-\frac {e^{10}}{4}} \int e^{\frac {1}{4} \left (e^5+2 x\right )^2} \, dx-\frac {1}{14} \left (e^{5-\frac {e^{10}}{4}} \left (350-e^5\right )\right ) \int e^{\frac {1}{4} \left (e^5+2 x\right )^2} \, dx\\ &=\frac {1}{14} e^{5+e^5 x+x^2}+\frac {1}{2} e^{2 e^5 x+2 x^2}+\frac {1}{14} e^{e^5 x+x^2} \left (350-e^5\right )-\frac {1}{7} e^{e^5 x+x^2} x+\frac {1}{14} e^{-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )-\frac {1}{14} e^{-\frac {e^{10}}{4}} \left (1-175 e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )-\frac {1}{28} e^{5-\frac {e^{10}}{4}} \left (350-e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )+e^{e^5 x+x^2} \log \left (x^2\right )+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2-\frac {1}{14} e^{10-\frac {e^{10}}{4}} \int e^{\frac {1}{4} \left (e^5+2 x\right )^2} \, dx\\ &=\frac {1}{14} e^{5+e^5 x+x^2}+\frac {1}{2} e^{2 e^5 x+2 x^2}+\frac {1}{14} e^{e^5 x+x^2} \left (350-e^5\right )-\frac {1}{7} e^{e^5 x+x^2} x+\frac {1}{14} e^{-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )-\frac {1}{28} e^{10-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )-\frac {1}{14} e^{-\frac {e^{10}}{4}} \left (1-175 e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )-\frac {1}{28} e^{5-\frac {e^{10}}{4}} \left (350-e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (e^5+2 x\right )\right )+e^{e^5 x+x^2} \log \left (x^2\right )+\frac {1}{98} \left (175-x+7 \log \left (x^2\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 26, normalized size = 0.93 \begin {gather*} \frac {1}{98} \left (-175-7 e^{x \left (e^5+x\right )}+x-7 \log \left (x^2\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2450 - 189*x + x^2 + E^(2*E^5*x + 2*x^2)*(49*E^5*x + 98*x^2) + E^(E^5*x + x^2)*(98 - 7*x + 2450*x^2
- 14*x^3 + E^5*(1225*x - 7*x^2)) + (98 - 7*x + E^(E^5*x + x^2)*(49*E^5*x + 98*x^2))*Log[x^2])/(49*x),x]

[Out]

(-175 - 7*E^(x*(E^5 + x)) + x - 7*Log[x^2])^2/98

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fricas [B]  time = 0.57, size = 65, normalized size = 2.32 \begin {gather*} \frac {1}{98} \, x^{2} - \frac {1}{7} \, {\left (x - 175\right )} e^{\left (x^{2} + x e^{5}\right )} - \frac {1}{7} \, {\left (x - 7 \, e^{\left (x^{2} + x e^{5}\right )} - 175\right )} \log \left (x^{2}\right ) + \frac {1}{2} \, \log \left (x^{2}\right )^{2} - \frac {25}{7} \, x + \frac {1}{2} \, e^{\left (2 \, x^{2} + 2 \, x e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*log(x^2)+(49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2
)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+2450*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x, algorithm="fricas")

[Out]

1/98*x^2 - 1/7*(x - 175)*e^(x^2 + x*e^5) - 1/7*(x - 7*e^(x^2 + x*e^5) - 175)*log(x^2) + 1/2*log(x^2)^2 - 25/7*
x + 1/2*e^(2*x^2 + 2*x*e^5)

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giac [B]  time = 0.19, size = 79, normalized size = 2.82 \begin {gather*} \frac {1}{98} \, x^{2} - \frac {1}{7} \, x e^{\left (x^{2} + x e^{5}\right )} - \frac {1}{7} \, x \log \left (x^{2}\right ) + e^{\left (x^{2} + x e^{5}\right )} \log \left (x^{2}\right ) + \frac {1}{2} \, \log \left (x^{2}\right )^{2} - \frac {25}{7} \, x + \frac {1}{2} \, e^{\left (2 \, x^{2} + 2 \, x e^{5}\right )} + 25 \, e^{\left (x^{2} + x e^{5}\right )} + 50 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*log(x^2)+(49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2
)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+2450*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x, algorithm="giac")

[Out]

1/98*x^2 - 1/7*x*e^(x^2 + x*e^5) - 1/7*x*log(x^2) + e^(x^2 + x*e^5)*log(x^2) + 1/2*log(x^2)^2 - 25/7*x + 1/2*e
^(2*x^2 + 2*x*e^5) + 25*e^(x^2 + x*e^5) + 50*log(x)

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maple [B]  time = 0.23, size = 103, normalized size = 3.68

method	result	size
default	\(\frac {{\mathrm e}^{2 x \,{\mathrm e}^{5}+2 x^{2}}}{2}+{\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}} \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )+25 \,{\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}}-\frac {{\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}} x}{7}+2 \ln \relax (x ) {\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}}+\frac {x^{2}}{98}-\frac {25 x}{7}+50 \ln \relax (x )+2 \ln \relax (x ) \ln \left (x^{2}\right )-2 \ln \relax (x )^{2}-\frac {x \ln \left (x^{2}\right )}{7}\)	\(103\)
risch	\(2 \ln \relax (x )^{2}+\frac {\left (-14 x +98 \,{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x}\right ) \ln \relax (x )}{49}-\frac {i \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}}{7}-i \pi \ln \relax (x ) \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\frac {x^{2}}{98}-\frac {25 x}{7}-\frac {i {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}-i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+\frac {i \pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{14}+50 \ln \relax (x )+\frac {{\mathrm e}^{2 \left ({\mathrm e}^{5}+x \right ) x}}{2}+i {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\frac {i {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )}{2}+\frac {i \pi x \mathrm {csgn}\left (i x^{2}\right )^{3}}{14}-\frac {{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} x}{7}+25 \,{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*ln(x^2)+(49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)^2+((-
7*x^2+1225*x)*exp(5)-14*x^3+2450*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x,method=_RETURNVERBOSE)

[Out]

1/2*exp(x*exp(5)+x^2)^2+exp(x*exp(5)+x^2)*(ln(x^2)-2*ln(x))+25*exp(x*exp(5)+x^2)-1/7*exp(x*exp(5)+x^2)*x+2*ln(
x)*exp(x*exp(5)+x^2)+1/98*x^2-25/7*x+50*ln(x)+2*ln(x)*ln(x^2)-2*ln(x)^2-1/7*x*ln(x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{14} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x + \frac {1}{2} i \, e^{5}\right ) e^{\left (-\frac {1}{4} \, e^{10}\right )} + \frac {1}{98} \, x^{2} - \frac {2}{7} \, x \log \relax (x) + 2 \, \log \relax (x)^{2} - \frac {25}{7} \, x + \frac {1}{2} \, e^{\left (2 \, x^{2} + 2 \, x e^{5}\right )} + \frac {1}{49} \, \int -\frac {7 \, {\left (2 \, x^{3} + x^{2} {\left (e^{5} - 350\right )} - 175 \, x e^{5} - 14 \, {\left (2 \, x^{2} + x e^{5}\right )} \log \relax (x) - 14\right )} e^{\left (x^{2} + x e^{5}\right )}}{x}\,{d x} + 50 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*log(x^2)+(49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2
)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+2450*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x, algorithm="maxima")

[Out]

1/14*I*sqrt(pi)*erf(I*x + 1/2*I*e^5)*e^(-1/4*e^10) + 1/98*x^2 - 2/7*x*log(x) + 2*log(x)^2 - 25/7*x + 1/2*e^(2*
x^2 + 2*x*e^5) + 1/49*integrate(-7*(2*x^3 + x^2*(e^5 - 350) - 175*x*e^5 - 14*(2*x^2 + x*e^5)*log(x) - 14)*e^(x
^2 + x*e^5)/x, x) + 50*log(x)

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mupad [B]  time = 4.51, size = 81, normalized size = 2.89 \begin {gather*} \frac {{\mathrm {e}}^{2\,x^2+2\,{\mathrm {e}}^5\,x}}{2}-\frac {25\,x}{7}+25\,\ln \left (x^2\right )+25\,{\mathrm {e}}^{x^2+{\mathrm {e}}^5\,x}-\frac {x\,\ln \left (x^2\right )}{7}+\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+{\mathrm {e}}^5\,x}+\frac {{\ln \left (x^2\right )}^2}{2}-\frac {x\,{\mathrm {e}}^{x^2+{\mathrm {e}}^5\,x}}{7}+\frac {x^2}{98} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x*exp(5) + x^2)*(exp(5)*(1225*x - 7*x^2) - 7*x + 2450*x^2 - 14*x^3 + 98))/49 - (27*x)/7 + (exp(2*x*e
xp(5) + 2*x^2)*(49*x*exp(5) + 98*x^2))/49 + x^2/49 + (log(x^2)*(exp(x*exp(5) + x^2)*(49*x*exp(5) + 98*x^2) - 7
*x + 98))/49 + 50)/x,x)

[Out]

exp(2*x*exp(5) + 2*x^2)/2 - (25*x)/7 + 25*log(x^2) + 25*exp(x*exp(5) + x^2) - (x*log(x^2))/7 + log(x^2)*exp(x*
exp(5) + x^2) + log(x^2)^2/2 - (x*exp(x*exp(5) + x^2))/7 + x^2/98

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sympy [B]  time = 0.61, size = 70, normalized size = 2.50 \begin {gather*} \frac {x^{2}}{98} - \frac {x \log {\left (x^{2} \right )}}{7} - \frac {25 x}{7} + \frac {\left (- 2 x + 14 \log {\left (x^{2} \right )} + 350\right ) e^{x^{2} + x e^{5}}}{14} + \frac {e^{2 x^{2} + 2 x e^{5}}}{2} + 50 \log {\relax (x )} + \frac {\log {\left (x^{2} \right )}^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/49*(((49*x*exp(5)+98*x**2)*exp(x*exp(5)+x**2)-7*x+98)*ln(x**2)+(49*x*exp(5)+98*x**2)*exp(x*exp(5)+
x**2)**2+((-7*x**2+1225*x)*exp(5)-14*x**3+2450*x**2-7*x+98)*exp(x*exp(5)+x**2)+x**2-189*x+2450)/x,x)

[Out]

x**2/98 - x*log(x**2)/7 - 25*x/7 + (-2*x + 14*log(x**2) + 350)*exp(x**2 + x*exp(5))/14 + exp(2*x**2 + 2*x*exp(
5))/2 + 50*log(x) + log(x**2)**2/2

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