\(\int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx\) [5178]

Optimal result

Integrand size = 32, antiderivative size = 29 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=3 \left (5-x+\frac {\left (-4+(-5+x)^2-x\right ) x}{e^9+\log ^2(16)}\right ) \]

[Out]

3*x*((-5+x)^2-4-x)/(16*ln(2)^2+exp(9))-3*x+15

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12} \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=\frac {3 x^3}{e^9+\log ^2(16)}-\frac {33 x^2}{e^9+\log ^2(16)}+\frac {3 x \left (21-e^9-\log ^2(16)\right )}{e^9+\log ^2(16)} \]

[In]

Int[(63 - 3*E^9 - 66*x + 9*x^2 - 3*Log[16]^2)/(E^9 + Log[16]^2),x]

[Out]

(-33*x^2)/(E^9 + Log[16]^2) + (3*x^3)/(E^9 + Log[16]^2) + (3*x*(21 - E^9 - Log[16]^2))/(E^9 + Log[16]^2)

Rule 12

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (63-3 e^9-66 x+9 x^2-3 \log ^2(16)\right ) \, dx}{e^9+\log ^2(16)} \\ & = -\frac {33 x^2}{e^9+\log ^2(16)}+\frac {3 x^3}{e^9+\log ^2(16)}+\frac {3 x \left (21-e^9-\log ^2(16)\right )}{e^9+\log ^2(16)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=-\frac {3 \left (-21 x+e^9 x+11 x^2-x^3+x \log ^2(16)\right )}{e^9+\log ^2(16)} \]

[In]

Integrate[(63 - 3*E^9 - 66*x + 9*x^2 - 3*Log[16]^2)/(E^9 + Log[16]^2),x]

[Out]

(-3*(-21*x + E^9*x + 11*x^2 - x^3 + x*Log[16]^2))/(E^9 + Log[16]^2)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14

method	result	size
gosper	\(-\frac {3 x \left (16 \ln \left (2\right )^{2}-x^{2}+{\mathrm e}^{9}+11 x -21\right )}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}\)	\(33\)
default	\(\frac {-48 x \ln \left (2\right )^{2}+3 x^{3}-3 x \,{\mathrm e}^{9}-33 x^{2}+63 x}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}\)	\(38\)
parallelrisch	\(\frac {3 x^{3}-33 x^{2}+\left (-48 \ln \left (2\right )^{2}-3 \,{\mathrm e}^{9}+63\right ) x}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}\)	\(38\)
norman	\(-\frac {33 x^{2}}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}+\frac {3 x^{3}}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}-\frac {3 \left (16 \ln \left (2\right )^{2}+{\mathrm e}^{9}-21\right ) x}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}\)	\(58\)
risch	\(-\frac {48 x \ln \left (2\right )^{2}}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}+\frac {3 x^{3}}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}-\frac {3 x \,{\mathrm e}^{9}}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}-\frac {33 x^{2}}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}+\frac {63 x}{16 \ln \left (2\right )^{2}+{\mathrm e}^{9}}\)	\(82\)

[In]

int((-48*ln(2)^2-3*exp(9)+9*x^2-66*x+63)/(16*ln(2)^2+exp(9)),x,method=_RETURNVERBOSE)

[Out]

-3*x*(16*ln(2)^2-x^2+exp(9)+11*x-21)/(16*ln(2)^2+exp(9))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=\frac {3 \, {\left (x^{3} - 16 \, x \log \left (2\right )^{2} - 11 \, x^{2} - x e^{9} + 21 \, x\right )}}{16 \, \log \left (2\right )^{2} + e^{9}} \]

[In]

integrate((-48*log(2)^2-3*exp(9)+9*x^2-66*x+63)/(16*log(2)^2+exp(9)),x, algorithm="fricas")

[Out]

3*(x^3 - 16*x*log(2)^2 - 11*x^2 - x*e^9 + 21*x)/(16*log(2)^2 + e^9)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=\frac {3 x^{3}}{16 \log {\left (2 \right )}^{2} + e^{9}} - \frac {33 x^{2}}{16 \log {\left (2 \right )}^{2} + e^{9}} + \frac {x \left (- 3 e^{9} - 48 \log {\left (2 \right )}^{2} + 63\right )}{16 \log {\left (2 \right )}^{2} + e^{9}} \]

[In]

integrate((-48*ln(2)**2-3*exp(9)+9*x**2-66*x+63)/(16*ln(2)**2+exp(9)),x)

[Out]

3*x**3/(16*log(2)**2 + exp(9)) - 33*x**2/(16*log(2)**2 + exp(9)) + x*(-3*exp(9) - 48*log(2)**2 + 63)/(16*log(2
)**2 + exp(9))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=\frac {3 \, {\left (x^{3} - 16 \, x \log \left (2\right )^{2} - 11 \, x^{2} - x e^{9} + 21 \, x\right )}}{16 \, \log \left (2\right )^{2} + e^{9}} \]

[In]

integrate((-48*log(2)^2-3*exp(9)+9*x^2-66*x+63)/(16*log(2)^2+exp(9)),x, algorithm="maxima")

[Out]

3*(x^3 - 16*x*log(2)^2 - 11*x^2 - x*e^9 + 21*x)/(16*log(2)^2 + e^9)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=\frac {3 \, {\left (x^{3} - 16 \, x \log \left (2\right )^{2} - 11 \, x^{2} - x e^{9} + 21 \, x\right )}}{16 \, \log \left (2\right )^{2} + e^{9}} \]

[In]

integrate((-48*log(2)^2-3*exp(9)+9*x^2-66*x+63)/(16*log(2)^2+exp(9)),x, algorithm="giac")

[Out]

3*(x^3 - 16*x*log(2)^2 - 11*x^2 - x*e^9 + 21*x)/(16*log(2)^2 + e^9)

Mupad [B] (verification not implemented)

Time = 11.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {63-3 e^9-66 x+9 x^2-3 \log ^2(16)}{e^9+\log ^2(16)} \, dx=-\frac {3\,x\,\left (-x^2+11\,x+{\mathrm {e}}^9+16\,{\ln \left (2\right )}^2-21\right )}{{\mathrm {e}}^9+16\,{\ln \left (2\right )}^2} \]

[In]

int(-(66*x + 3*exp(9) + 48*log(2)^2 - 9*x^2 - 63)/(exp(9) + 16*log(2)^2),x)

[Out]

-(3*x*(11*x + exp(9) + 16*log(2)^2 - x^2 - 21))/(exp(9) + 16*log(2)^2)