3.47.86 \(\int \frac {e^{\frac {100 x^2+100 x^3+25 x^4+(100 x^2+50 x^3) \log (49-14 x+x^2)+25 x^2 \log ^2(49-14 x+x^2)}{\log ^2(49-14 x+x^2)}} (-400 x^2-400 x^3-100 x^4+(-1400 x-2100 x^2-500 x^3+100 x^4) \log (49-14 x+x^2)+(-1400 x-850 x^2+150 x^3) \log ^2(49-14 x+x^2)+(-350 x+50 x^2) \log ^3(49-14 x+x^2))}{(21-3 x) \log ^3(49-14 x+x^2)+e^{\frac {100 x^2+100 x^3+25 x^4+(100 x^2+50 x^3) \log (49-14 x+x^2)+25 x^2 \log ^2(49-14 x+x^2)}{\log ^2(49-14 x+x^2)}} (-7+x) \log ^3(49-14 x+x^2)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (3-e^{x^2 \left (5+\frac {5 (2+x)}{\log \left ((7-x)^2\right )}\right )^2}\right ) \]

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Rubi [A]  time = 33.15, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, integrand size = 262, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6688, 12, 6708, 31} \begin {gather*} \log \left (3-\exp \left (\frac {25 x^2 \left (x+\log \left ((x-7)^2\right )+2\right )^2}{\log ^2\left ((x-7)^2\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((100*x^2 + 100*x^3 + 25*x^4 + (100*x^2 + 50*x^3)*Log[49 - 14*x + x^2] + 25*x^2*Log[49 - 14*x + x^2]^2)
/Log[49 - 14*x + x^2]^2)*(-400*x^2 - 400*x^3 - 100*x^4 + (-1400*x - 2100*x^2 - 500*x^3 + 100*x^4)*Log[49 - 14*
x + x^2] + (-1400*x - 850*x^2 + 150*x^3)*Log[49 - 14*x + x^2]^2 + (-350*x + 50*x^2)*Log[49 - 14*x + x^2]^3))/(
(21 - 3*x)*Log[49 - 14*x + x^2]^3 + E^((100*x^2 + 100*x^3 + 25*x^4 + (100*x^2 + 50*x^3)*Log[49 - 14*x + x^2] +
25*x^2*Log[49 - 14*x + x^2]^2)/Log[49 - 14*x + x^2]^2)*(-7 + x)*Log[49 - 14*x + x^2]^3),x]

[Out]

Log[3 - E^((25*x^2*(2 + x + Log[(-7 + x)^2])^2)/Log[(-7 + x)^2]^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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 6688

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

Rule 6708

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 \exp \left (\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}\right ) x \left (-2 x (2+x)^2+2 \left (-14-21 x-5 x^2+x^3\right ) \log \left ((-7+x)^2\right )+\left (-28-17 x+3 x^2\right ) \log ^2\left ((-7+x)^2\right )+(-7+x) \log ^3\left ((-7+x)^2\right )\right )}{\left (3-\exp \left (\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}\right )\right ) (7-x) \log ^3\left ((-7+x)^2\right )} \, dx\\ &=50 \int \frac {\exp \left (\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}\right ) x \left (-2 x (2+x)^2+2 \left (-14-21 x-5 x^2+x^3\right ) \log \left ((-7+x)^2\right )+\left (-28-17 x+3 x^2\right ) \log ^2\left ((-7+x)^2\right )+(-7+x) \log ^3\left ((-7+x)^2\right )\right )}{\left (3-\exp \left (\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}\right )\right ) (7-x) \log ^3\left ((-7+x)^2\right )} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{3+x} \, dx,x,-\exp \left (\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}\right )\right )\\ &=\log \left (3-\exp \left (\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 29, normalized size = 0.97 \begin {gather*} \log \left (-3+e^{\frac {25 x^2 \left (2+x+\log \left ((-7+x)^2\right )\right )^2}{\log ^2\left ((-7+x)^2\right )}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((100*x^2 + 100*x^3 + 25*x^4 + (100*x^2 + 50*x^3)*Log[49 - 14*x + x^2] + 25*x^2*Log[49 - 14*x + x
^2]^2)/Log[49 - 14*x + x^2]^2)*(-400*x^2 - 400*x^3 - 100*x^4 + (-1400*x - 2100*x^2 - 500*x^3 + 100*x^4)*Log[49
- 14*x + x^2] + (-1400*x - 850*x^2 + 150*x^3)*Log[49 - 14*x + x^2]^2 + (-350*x + 50*x^2)*Log[49 - 14*x + x^2]
^3))/((21 - 3*x)*Log[49 - 14*x + x^2]^3 + E^((100*x^2 + 100*x^3 + 25*x^4 + (100*x^2 + 50*x^3)*Log[49 - 14*x +
x^2] + 25*x^2*Log[49 - 14*x + x^2]^2)/Log[49 - 14*x + x^2]^2)*(-7 + x)*Log[49 - 14*x + x^2]^3),x]

[Out]

Log[-3 + E^((25*x^2*(2 + x + Log[(-7 + x)^2])^2)/Log[(-7 + x)^2]^2)]

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fricas [B]  time = 0.92, size = 66, normalized size = 2.20 \begin {gather*} \log \left (e^{\left (\frac {25 \, {\left (x^{4} + x^{2} \log \left (x^{2} - 14 \, x + 49\right )^{2} + 4 \, x^{3} + 4 \, x^{2} + 2 \, {\left (x^{3} + 2 \, x^{2}\right )} \log \left (x^{2} - 14 \, x + 49\right )\right )}}{\log \left (x^{2} - 14 \, x + 49\right )^{2}}\right )} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-350*x)*log(x^2-14*x+49)^3+(150*x^3-850*x^2-1400*x)*log(x^2-14*x+49)^2+(100*x^4-500*x^3-2100
*x^2-1400*x)*log(x^2-14*x+49)-100*x^4-400*x^3-400*x^2)*exp((25*x^2*log(x^2-14*x+49)^2+(50*x^3+100*x^2)*log(x^2
-14*x+49)+25*x^4+100*x^3+100*x^2)/log(x^2-14*x+49)^2)/((x-7)*log(x^2-14*x+49)^3*exp((25*x^2*log(x^2-14*x+49)^2
+(50*x^3+100*x^2)*log(x^2-14*x+49)+25*x^4+100*x^3+100*x^2)/log(x^2-14*x+49)^2)+(21-3*x)*log(x^2-14*x+49)^3),x,
algorithm="fricas")

[Out]

log(e^(25*(x^4 + x^2*log(x^2 - 14*x + 49)^2 + 4*x^3 + 4*x^2 + 2*(x^3 + 2*x^2)*log(x^2 - 14*x + 49))/log(x^2 -
14*x + 49)^2) - 3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-350*x)*log(x^2-14*x+49)^3+(150*x^3-850*x^2-1400*x)*log(x^2-14*x+49)^2+(100*x^4-500*x^3-2100
*x^2-1400*x)*log(x^2-14*x+49)-100*x^4-400*x^3-400*x^2)*exp((25*x^2*log(x^2-14*x+49)^2+(50*x^3+100*x^2)*log(x^2
-14*x+49)+25*x^4+100*x^3+100*x^2)/log(x^2-14*x+49)^2)/((x-7)*log(x^2-14*x+49)^3*exp((25*x^2*log(x^2-14*x+49)^2
+(50*x^3+100*x^2)*log(x^2-14*x+49)+25*x^4+100*x^3+100*x^2)/log(x^2-14*x+49)^2)+(21-3*x)*log(x^2-14*x+49)^3),x,
algorithm="giac")

[Out]

undef

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maple [B]  time = 0.08, size = 154, normalized size = 5.13

method	result	size
risch	\(25 x^{2}+\frac {25 x^{2} \left (x^{2}+2 \ln \left (x^{2}-14 x +49\right ) x +4 x +4 \ln \left (x^{2}-14 x +49\right )+4\right )}{\ln \left (x^{2}-14 x +49\right )^{2}}-\frac {25 x^{2} \ln \left (x^{2}-14 x +49\right )^{2}+\left (50 x^{3}+100 x^{2}\right ) \ln \left (x^{2}-14 x +49\right )+25 x^{4}+100 x^{3}+100 x^{2}}{\ln \left (x^{2}-14 x +49\right )^{2}}+\ln \left ({\mathrm e}^{\frac {25 x^{2} \left (\ln \left (x^{2}-14 x +49\right )+x +2\right )^{2}}{\ln \left (x^{2}-14 x +49\right )^{2}}}-3\right )\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^2-350*x)*ln(x^2-14*x+49)^3+(150*x^3-850*x^2-1400*x)*ln(x^2-14*x+49)^2+(100*x^4-500*x^3-2100*x^2-140
0*x)*ln(x^2-14*x+49)-100*x^4-400*x^3-400*x^2)*exp((25*x^2*ln(x^2-14*x+49)^2+(50*x^3+100*x^2)*ln(x^2-14*x+49)+2
5*x^4+100*x^3+100*x^2)/ln(x^2-14*x+49)^2)/((x-7)*ln(x^2-14*x+49)^3*exp((25*x^2*ln(x^2-14*x+49)^2+(50*x^3+100*x
^2)*ln(x^2-14*x+49)+25*x^4+100*x^3+100*x^2)/ln(x^2-14*x+49)^2)+(21-3*x)*ln(x^2-14*x+49)^3),x,method=_RETURNVER
BOSE)

[Out]

25*x^2+25*x^2*(x^2+2*ln(x^2-14*x+49)*x+4*x+4*ln(x^2-14*x+49)+4)/ln(x^2-14*x+49)^2-(25*x^2*ln(x^2-14*x+49)^2+(5
0*x^3+100*x^2)*ln(x^2-14*x+49)+25*x^4+100*x^3+100*x^2)/ln(x^2-14*x+49)^2+ln(exp(25*x^2*(ln(x^2-14*x+49)+x+2)^2
/ln(x^2-14*x+49)^2)-3)

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maxima [B]  time = 0.59, size = 157, normalized size = 5.23 \begin {gather*} \frac {25 \, {\left (x^{2} \log \left (x - 7\right )^{2} + x^{3} + x^{2} + {\left (x^{3} + 2 \, x^{2}\right )} \log \left (x - 7\right )\right )}}{\log \left (x - 7\right )^{2}} + \log \left ({\left (e^{\left (25 \, x^{2} + \frac {25 \, x^{4}}{4 \, \log \left (x - 7\right )^{2}} + \frac {25 \, x^{3}}{\log \left (x - 7\right )} + \frac {25 \, x^{3}}{\log \left (x - 7\right )^{2}} + \frac {50 \, x^{2}}{\log \left (x - 7\right )} + \frac {25 \, x^{2}}{\log \left (x - 7\right )^{2}}\right )} - 3\right )} e^{\left (-25 \, x^{2} - \frac {25 \, x^{3}}{\log \left (x - 7\right )} - \frac {25 \, x^{3}}{\log \left (x - 7\right )^{2}} - \frac {50 \, x^{2}}{\log \left (x - 7\right )} - \frac {25 \, x^{2}}{\log \left (x - 7\right )^{2}}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^2-350*x)*log(x^2-14*x+49)^3+(150*x^3-850*x^2-1400*x)*log(x^2-14*x+49)^2+(100*x^4-500*x^3-2100
*x^2-1400*x)*log(x^2-14*x+49)-100*x^4-400*x^3-400*x^2)*exp((25*x^2*log(x^2-14*x+49)^2+(50*x^3+100*x^2)*log(x^2
-14*x+49)+25*x^4+100*x^3+100*x^2)/log(x^2-14*x+49)^2)/((x-7)*log(x^2-14*x+49)^3*exp((25*x^2*log(x^2-14*x+49)^2
+(50*x^3+100*x^2)*log(x^2-14*x+49)+25*x^4+100*x^3+100*x^2)/log(x^2-14*x+49)^2)+(21-3*x)*log(x^2-14*x+49)^3),x,
algorithm="maxima")

[Out]

25*(x^2*log(x - 7)^2 + x^3 + x^2 + (x^3 + 2*x^2)*log(x - 7))/log(x - 7)^2 + log((e^(25*x^2 + 25/4*x^4/log(x -
7)^2 + 25*x^3/log(x - 7) + 25*x^3/log(x - 7)^2 + 50*x^2/log(x - 7) + 25*x^2/log(x - 7)^2) - 3)*e^(-25*x^2 - 25
*x^3/log(x - 7) - 25*x^3/log(x - 7)^2 - 50*x^2/log(x - 7) - 25*x^2/log(x - 7)^2))

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mupad [B]  time = 3.75, size = 95, normalized size = 3.17 \begin {gather*} \ln \left ({\mathrm {e}}^{\frac {25\,x^4}{{\ln \left (x^2-14\,x+49\right )}^2}}\,{\mathrm {e}}^{\frac {50\,x^3}{\ln \left (x^2-14\,x+49\right )}}\,{\mathrm {e}}^{\frac {100\,x^2}{\ln \left (x^2-14\,x+49\right )}}\,{\mathrm {e}}^{\frac {100\,x^2}{{\ln \left (x^2-14\,x+49\right )}^2}}\,{\mathrm {e}}^{\frac {100\,x^3}{{\ln \left (x^2-14\,x+49\right )}^2}}\,{\mathrm {e}}^{25\,x^2}-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((25*x^2*log(x^2 - 14*x + 49)^2 + log(x^2 - 14*x + 49)*(100*x^2 + 50*x^3) + 100*x^2 + 100*x^3 + 25*x^4
)/log(x^2 - 14*x + 49)^2)*(log(x^2 - 14*x + 49)^3*(350*x - 50*x^2) + log(x^2 - 14*x + 49)*(1400*x + 2100*x^2 +
500*x^3 - 100*x^4) + 400*x^2 + 400*x^3 + 100*x^4 + log(x^2 - 14*x + 49)^2*(1400*x + 850*x^2 - 150*x^3)))/(log
(x^2 - 14*x + 49)^3*(3*x - 21) - exp((25*x^2*log(x^2 - 14*x + 49)^2 + log(x^2 - 14*x + 49)*(100*x^2 + 50*x^3)
+ 100*x^2 + 100*x^3 + 25*x^4)/log(x^2 - 14*x + 49)^2)*log(x^2 - 14*x + 49)^3*(x - 7)),x)

[Out]

log(exp((25*x^4)/log(x^2 - 14*x + 49)^2)*exp((50*x^3)/log(x^2 - 14*x + 49))*exp((100*x^2)/log(x^2 - 14*x + 49)
)*exp((100*x^2)/log(x^2 - 14*x + 49)^2)*exp((100*x^3)/log(x^2 - 14*x + 49)^2)*exp(25*x^2) - 3)

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sympy [B]  time = 1.17, size = 68, normalized size = 2.27 \begin {gather*} \log {\left (e^{\frac {25 x^{4} + 100 x^{3} + 25 x^{2} \log {\left (x^{2} - 14 x + 49 \right )}^{2} + 100 x^{2} + \left (50 x^{3} + 100 x^{2}\right ) \log {\left (x^{2} - 14 x + 49 \right )}}{\log {\left (x^{2} - 14 x + 49 \right )}^{2}}} - 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**2-350*x)*ln(x**2-14*x+49)**3+(150*x**3-850*x**2-1400*x)*ln(x**2-14*x+49)**2+(100*x**4-500*x*
*3-2100*x**2-1400*x)*ln(x**2-14*x+49)-100*x**4-400*x**3-400*x**2)*exp((25*x**2*ln(x**2-14*x+49)**2+(50*x**3+10
0*x**2)*ln(x**2-14*x+49)+25*x**4+100*x**3+100*x**2)/ln(x**2-14*x+49)**2)/((x-7)*ln(x**2-14*x+49)**3*exp((25*x*
*2*ln(x**2-14*x+49)**2+(50*x**3+100*x**2)*ln(x**2-14*x+49)+25*x**4+100*x**3+100*x**2)/ln(x**2-14*x+49)**2)+(21
-3*x)*ln(x**2-14*x+49)**3),x)

[Out]

log(exp((25*x**4 + 100*x**3 + 25*x**2*log(x**2 - 14*x + 49)**2 + 100*x**2 + (50*x**3 + 100*x**2)*log(x**2 - 14
*x + 49))/log(x**2 - 14*x + 49)**2) - 3)

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