Integrand size = 105, antiderivative size = 33 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=x+x^2-\frac {9 x^2 \left (-e^4+x\right )^2}{25 \left (x+\frac {1}{x \log (4)}\right )^2} \] Output:
x+x^2-9/25*x^2*(x-exp(4))^2/(x+1/2/x/ln(2))^2
Leaf count is larger than twice the leaf count of optimal. \(84\) vs. \(2(33)=66\).
Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.55 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=\frac {1}{25} \left (\left (25+18 e^4\right ) x+16 x^2-\frac {9 \left (-1+e^8 \log (4)-2 e^4 x \log (4)\right )}{\log (4) \left (1+x^2 \log (4)\right )^2}+\frac {9 \left (-3+2 e^8 \log (4)-4 e^4 x \log (4)\right )}{\log (4) \left (1+x^2 \log (4)\right )}\right ) \] Input:
Integrate[(25 + 50*x + (75*x^2 + 150*x^3)*Log[4] + (-36*E^8*x^3 + 75*x^4 + 90*E^4*x^4 + 96*x^5)*Log[4]^2 + (25*x^6 + 18*E^4*x^6 + 32*x^7)*Log[4]^3)/ (25 + 75*x^2*Log[4] + 75*x^4*Log[4]^2 + 25*x^6*Log[4]^3),x]
Output:
((25 + 18*E^4)*x + 16*x^2 - (9*(-1 + E^8*Log[4] - 2*E^4*x*Log[4]))/(Log[4] *(1 + x^2*Log[4])^2) + (9*(-3 + 2*E^8*Log[4] - 4*E^4*x*Log[4]))/(Log[4]*(1 + x^2*Log[4])))/25
Leaf count is larger than twice the leaf count of optimal. \(90\) vs. \(2(33)=66\).
Time = 0.62 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.73, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2070, 2345, 27, 2345, 27, 2019, 17}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (32 x^7+18 e^4 x^6+25 x^6\right ) \log ^3(4)+\left (150 x^3+75 x^2\right ) \log (4)+\left (96 x^5+90 e^4 x^4+75 x^4-36 e^8 x^3\right ) \log ^2(4)+50 x+25}{25 x^6 \log ^3(4)+75 x^4 \log ^2(4)+75 x^2 \log (4)+25} \, dx\) |
\(\Big \downarrow \) 2070 |
\(\displaystyle \int \frac {\left (32 x^7+18 e^4 x^6+25 x^6\right ) \log ^3(4)+\left (150 x^3+75 x^2\right ) \log (4)+\left (96 x^5+90 e^4 x^4+75 x^4-36 e^8 x^3\right ) \log ^2(4)+50 x+25}{\left (5^{2/3} x^2 \log (4)+5^{2/3}\right )^3}dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}-\frac {\int -\frac {4 \left (32 \log ^2(4) x^5+\left (25+18 e^4\right ) \log ^2(4) x^4+64 \log (4) x^3+2 \left (25+36 e^4\right ) \log (4) x^2+2 \left (43-18 e^8 \log (4)\right ) x-18 e^4+25\right )}{5 \sqrt [3]{5} \left (\log (4) x^2+1\right )^2}dx}{4\ 5^{2/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \frac {32 \log ^2(4) x^5+\left (25+18 e^4\right ) \log ^2(4) x^4+64 \log (4) x^3+2 \left (25+36 e^4\right ) \log (4) x^2+2 \left (43-18 e^8 \log (4)\right ) x-18 e^4+25}{\left (\log (4) x^2+1\right )^2}dx+\frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {1}{25} \left (-\frac {1}{2} \int -\frac {2 \left (32 \log (4) x^3+\left (25+18 e^4\right ) \log (4) x^2+32 x+18 e^4+25\right )}{\log (4) x^2+1}dx-\frac {9 \left (4 e^4 x \log (4)+3-2 e^8 \log (4)\right )}{\log (4) \left (x^2 \log (4)+1\right )}\right )+\frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \left (\int \frac {32 \log (4) x^3+\left (25+18 e^4\right ) \log (4) x^2+32 x+18 e^4+25}{\log (4) x^2+1}dx-\frac {9 \left (4 e^4 x \log (4)+3-2 e^8 \log (4)\right )}{\log (4) \left (x^2 \log (4)+1\right )}\right )+\frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle \frac {1}{25} \left (\int \left (32 x+18 e^4+25\right )dx-\frac {9 \left (4 e^4 x \log (4)+3-2 e^8 \log (4)\right )}{\log (4) \left (x^2 \log (4)+1\right )}\right )+\frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {9 \left (2 e^4 x \log (4)+1-e^8 \log (4)\right )}{25 \log (4) \left (x^2 \log (4)+1\right )^2}+\frac {1}{25} \left (\frac {1}{64} \left (32 x+18 e^4+25\right )^2-\frac {9 \left (4 e^4 x \log (4)+3-2 e^8 \log (4)\right )}{\log (4) \left (x^2 \log (4)+1\right )}\right )\) |
Input:
Int[(25 + 50*x + (75*x^2 + 150*x^3)*Log[4] + (-36*E^8*x^3 + 75*x^4 + 90*E^ 4*x^4 + 96*x^5)*Log[4]^2 + (25*x^6 + 18*E^4*x^6 + 32*x^7)*Log[4]^3)/(25 + 75*x^2*Log[4] + 75*x^4*Log[4]^2 + 25*x^6*Log[4]^3),x]
Output:
(9*(1 - E^8*Log[4] + 2*E^4*x*Log[4]))/(25*Log[4]*(1 + x^2*Log[4])^2) + ((2 5 + 18*E^4 + 32*x)^2/64 - (9*(3 - 2*E^8*Log[4] + 4*E^4*x*Log[4]))/(Log[4]* (1 + x^2*Log[4])))/25
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x^2, 0], Expon[Px , x^2]], b = Rt[Coeff[Px, x^2, Expon[Px, x^2]], Expon[Px, x^2]]}, Int[u*(a + b*x^2)^(Expon[Px, x^2]*p), x] /; EqQ[Px, (a + b*x^2)^Expon[Px, x^2]]] /; IntegerQ[p] && PolyQ[Px, x^2] && GtQ[Expon[Px, x^2], 1] && NeQ[Coeff[Px, x^ 2, 0], 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(31)=62\).
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.97
method | result | size |
default | \(\frac {16 x^{2}}{25}+\frac {18 x \,{\mathrm e}^{4}}{25}+x -\frac {72 \left ({\mathrm e}^{4} \ln \left (2\right ) x^{3}+\left (-\frac {{\mathrm e}^{8} \ln \left (2\right )}{2}+\frac {3}{8}\right ) x^{2}+\frac {x \,{\mathrm e}^{4}}{4}-\frac {{\mathrm e}^{8} \ln \left (2\right )-1}{8 \ln \left (2\right )}\right )}{25 \left (2 x^{2} \ln \left (2\right )+1\right )^{2}}\) | \(65\) |
norman | \(\frac {x +\left (\frac {72 \,{\mathrm e}^{4} \ln \left (2\right )^{2}}{25}+4 \ln \left (2\right )^{2}\right ) x^{5}+x^{2}+\left (-\frac {36 \,{\mathrm e}^{8} \ln \left (2\right )^{2}}{25}+4 \ln \left (2\right )\right ) x^{4}+4 x^{3} \ln \left (2\right )+\frac {64 x^{6} \ln \left (2\right )^{2}}{25}}{\left (2 x^{2} \ln \left (2\right )+1\right )^{2}}\) | \(72\) |
risch | \(\frac {18 x \,{\mathrm e}^{4}}{25}+\frac {16 x^{2}}{25}+x +\frac {-\frac {18 \,{\mathrm e}^{4} \ln \left (2\right ) x^{3}}{25}+\frac {\left (9 \,{\mathrm e}^{8} \ln \left (2\right )-\frac {27}{4}\right ) x^{2}}{25}-\frac {9 x \,{\mathrm e}^{4}}{50}+\frac {\frac {9 \,{\mathrm e}^{8} \ln \left (2\right )}{100}-\frac {9}{100}}{\ln \left (2\right )}}{x^{4} \ln \left (2\right )^{2}+x^{2} \ln \left (2\right )+\frac {1}{4}}\) | \(73\) |
gosper | \(-\frac {x \left (36 \,{\mathrm e}^{8} \ln \left (2\right )^{2} x^{3}-72 \,{\mathrm e}^{4} \ln \left (2\right )^{2} x^{4}-64 \ln \left (2\right )^{2} x^{5}-100 x^{4} \ln \left (2\right )^{2}-100 x^{3} \ln \left (2\right )-100 x^{2} \ln \left (2\right )-25 x -25\right )}{25 \left (4 x^{4} \ln \left (2\right )^{2}+4 x^{2} \ln \left (2\right )+1\right )}\) | \(85\) |
parallelrisch | \(-\frac {36 \,{\mathrm e}^{8} \ln \left (2\right )^{2} x^{4}-72 \,{\mathrm e}^{4} \ln \left (2\right )^{2} x^{5}-64 x^{6} \ln \left (2\right )^{2}-100 \ln \left (2\right )^{2} x^{5}-100 x^{4} \ln \left (2\right )-100 x^{3} \ln \left (2\right )-25 x^{2}-25 x}{25 \left (4 x^{4} \ln \left (2\right )^{2}+4 x^{2} \ln \left (2\right )+1\right )}\) | \(88\) |
Input:
int((8*(18*x^6*exp(4)+32*x^7+25*x^6)*ln(2)^3+4*(-36*x^3*exp(4)^2+90*x^4*ex p(4)+96*x^5+75*x^4)*ln(2)^2+2*(150*x^3+75*x^2)*ln(2)+50*x+25)/(200*x^6*ln( 2)^3+300*x^4*ln(2)^2+150*x^2*ln(2)+25),x,method=_RETURNVERBOSE)
Output:
16/25*x^2+18/25*x*exp(4)+x-72/25*(exp(4)*ln(2)*x^3+(-1/2*exp(8)*ln(2)+3/8) *x^2+1/4*x*exp(4)-1/8/ln(2)*(exp(8)*ln(2)-1))/(2*x^2*ln(2)+1)^2
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (32) = 64\).
Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.79 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=\frac {4 \, {\left (16 \, x^{6} + 18 \, x^{5} e^{4} + 25 \, x^{5}\right )} \log \left (2\right )^{3} + 4 \, {\left (16 \, x^{4} + 25 \, x^{3} + 9 \, x^{2} e^{8}\right )} \log \left (2\right )^{2} - {\left (11 \, x^{2} - 25 \, x - 9 \, e^{8}\right )} \log \left (2\right ) - 9}{25 \, {\left (4 \, x^{4} \log \left (2\right )^{3} + 4 \, x^{2} \log \left (2\right )^{2} + \log \left (2\right )\right )}} \] Input:
integrate((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90 *x^4*exp(4)+96*x^5+75*x^4)*log(2)^2+2*(150*x^3+75*x^2)*log(2)+50*x+25)/(20 0*x^6*log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x, algorithm="fricas")
Output:
1/25*(4*(16*x^6 + 18*x^5*e^4 + 25*x^5)*log(2)^3 + 4*(16*x^4 + 25*x^3 + 9*x ^2*e^8)*log(2)^2 - (11*x^2 - 25*x - 9*e^8)*log(2) - 9)/(4*x^4*log(2)^3 + 4 *x^2*log(2)^2 + log(2))
Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (29) = 58\).
Time = 0.89 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.85 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=\frac {16 x^{2}}{25} + x \left (1 + \frac {18 e^{4}}{25}\right ) + \frac {- 72 x^{3} e^{4} \log {\left (2 \right )}^{2} + x^{2} \left (- 27 \log {\left (2 \right )} + 36 e^{8} \log {\left (2 \right )}^{2}\right ) - 18 x e^{4} \log {\left (2 \right )} - 9 + 9 e^{8} \log {\left (2 \right )}}{100 x^{4} \log {\left (2 \right )}^{3} + 100 x^{2} \log {\left (2 \right )}^{2} + 25 \log {\left (2 \right )}} \] Input:
integrate((8*(18*x**6*exp(4)+32*x**7+25*x**6)*ln(2)**3+4*(-36*x**3*exp(4)* *2+90*x**4*exp(4)+96*x**5+75*x**4)*ln(2)**2+2*(150*x**3+75*x**2)*ln(2)+50* x+25)/(200*x**6*ln(2)**3+300*x**4*ln(2)**2+150*x**2*ln(2)+25),x)
Output:
16*x**2/25 + x*(1 + 18*exp(4)/25) + (-72*x**3*exp(4)*log(2)**2 + x**2*(-27 *log(2) + 36*exp(8)*log(2)**2) - 18*x*exp(4)*log(2) - 9 + 9*exp(8)*log(2)) /(100*x**4*log(2)**3 + 100*x**2*log(2)**2 + 25*log(2))
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (32) = 64\).
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.55 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=\frac {16}{25} \, x^{2} + \frac {1}{25} \, x {\left (18 \, e^{4} + 25\right )} - \frac {9 \, {\left (8 \, x^{3} e^{4} \log \left (2\right )^{2} - {\left (4 \, e^{8} \log \left (2\right )^{2} - 3 \, \log \left (2\right )\right )} x^{2} + 2 \, x e^{4} \log \left (2\right ) - e^{8} \log \left (2\right ) + 1\right )}}{25 \, {\left (4 \, x^{4} \log \left (2\right )^{3} + 4 \, x^{2} \log \left (2\right )^{2} + \log \left (2\right )\right )}} \] Input:
integrate((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90 *x^4*exp(4)+96*x^5+75*x^4)*log(2)^2+2*(150*x^3+75*x^2)*log(2)+50*x+25)/(20 0*x^6*log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x, algorithm="maxima")
Output:
16/25*x^2 + 1/25*x*(18*e^4 + 25) - 9/25*(8*x^3*e^4*log(2)^2 - (4*e^8*log(2 )^2 - 3*log(2))*x^2 + 2*x*e^4*log(2) - e^8*log(2) + 1)/(4*x^4*log(2)^3 + 4 *x^2*log(2)^2 + log(2))
Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (32) = 64\).
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.85 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=-\frac {9 \, {\left (8 \, x^{3} e^{4} \log \left (2\right )^{2} - 4 \, x^{2} e^{8} \log \left (2\right )^{2} + 3 \, x^{2} \log \left (2\right ) + 2 \, x e^{4} \log \left (2\right ) - e^{8} \log \left (2\right ) + 1\right )}}{25 \, {\left (2 \, x^{2} \log \left (2\right ) + 1\right )}^{2} \log \left (2\right )} + \frac {16 \, x^{2} \log \left (2\right )^{6} + 18 \, x e^{4} \log \left (2\right )^{6} + 25 \, x \log \left (2\right )^{6}}{25 \, \log \left (2\right )^{6}} \] Input:
integrate((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90 *x^4*exp(4)+96*x^5+75*x^4)*log(2)^2+2*(150*x^3+75*x^2)*log(2)+50*x+25)/(20 0*x^6*log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x, algorithm="giac")
Output:
-9/25*(8*x^3*e^4*log(2)^2 - 4*x^2*e^8*log(2)^2 + 3*x^2*log(2) + 2*x*e^4*lo g(2) - e^8*log(2) + 1)/((2*x^2*log(2) + 1)^2*log(2)) + 1/25*(16*x^2*log(2) ^6 + 18*x*e^4*log(2)^6 + 25*x*log(2)^6)/log(2)^6
Time = 3.84 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.33 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=\frac {16\,x^2}{25}-\frac {72\,{\mathrm {e}}^4\,\ln \left (2\right )\,x^3+\left (27-36\,{\mathrm {e}}^8\,\ln \left (2\right )\right )\,x^2+18\,{\mathrm {e}}^4\,x-\frac {9\,\left ({\mathrm {e}}^8\,\ln \left (2\right )-1\right )}{\ln \left (2\right )}}{100\,{\ln \left (2\right )}^2\,x^4+100\,\ln \left (2\right )\,x^2+25}+x\,\left (\frac {18\,{\mathrm {e}}^4}{25}+1\right ) \] Input:
int((50*x + 4*log(2)^2*(90*x^4*exp(4) - 36*x^3*exp(8) + 75*x^4 + 96*x^5) + 8*log(2)^3*(18*x^6*exp(4) + 25*x^6 + 32*x^7) + 2*log(2)*(75*x^2 + 150*x^3 ) + 25)/(300*x^4*log(2)^2 + 200*x^6*log(2)^3 + 150*x^2*log(2) + 25),x)
Output:
(16*x^2)/25 - (18*x*exp(4) - (9*(exp(8)*log(2) - 1))/log(2) - x^2*(36*exp( 8)*log(2) - 27) + 72*x^3*exp(4)*log(2))/(100*x^4*log(2)^2 + 100*x^2*log(2) + 25) + x*((18*exp(4))/25 + 1)
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.82 \[ \int \frac {25+50 x+\left (75 x^2+150 x^3\right ) \log (4)+\left (-36 e^8 x^3+75 x^4+90 e^4 x^4+96 x^5\right ) \log ^2(4)+\left (25 x^6+18 e^4 x^6+32 x^7\right ) \log ^3(4)}{25+75 x^2 \log (4)+75 x^4 \log ^2(4)+25 x^6 \log ^3(4)} \, dx=\frac {-144 \mathrm {log}\left (2\right )^{3} e^{8} x^{4}+288 \mathrm {log}\left (2\right )^{3} e^{4} x^{5}+256 \mathrm {log}\left (2\right )^{3} x^{6}+400 \mathrm {log}\left (2\right )^{3} x^{5}+300 \mathrm {log}\left (2\right )^{2} x^{4}+400 \mathrm {log}\left (2\right )^{2} x^{3}+100 \,\mathrm {log}\left (2\right ) x -25}{100 \,\mathrm {log}\left (2\right ) \left (4 \mathrm {log}\left (2\right )^{2} x^{4}+4 \,\mathrm {log}\left (2\right ) x^{2}+1\right )} \] Input:
int((8*(18*x^6*exp(4)+32*x^7+25*x^6)*log(2)^3+4*(-36*x^3*exp(4)^2+90*x^4*e xp(4)+96*x^5+75*x^4)*log(2)^2+2*(150*x^3+75*x^2)*log(2)+50*x+25)/(200*x^6* log(2)^3+300*x^4*log(2)^2+150*x^2*log(2)+25),x)
Output:
( - 144*log(2)**3*e**8*x**4 + 288*log(2)**3*e**4*x**5 + 256*log(2)**3*x**6 + 400*log(2)**3*x**5 + 300*log(2)**2*x**4 + 400*log(2)**2*x**3 + 100*log( 2)*x - 25)/(100*log(2)*(4*log(2)**2*x**4 + 4*log(2)*x**2 + 1))