Integrand size = 30, antiderivative size = 325 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=-\frac {3 \left (b^2-5 a c\right ) \left (b^2-2 a c\right )}{2 a^3 \left (b^2-4 a c\right )^2 e (d+e x)^2}+\frac {b^2-2 a c+b c (d+e x)^2}{4 a \left (b^2-4 a c\right ) e (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}+\frac {3 b^4-20 a b^2 c+20 a^2 c^2+3 b c \left (b^2-6 a c\right ) (d+e x)^2}{4 a^2 \left (b^2-4 a c\right )^2 e (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {3 \left (b^6-10 a b^4 c+30 a^2 b^2 c^2-20 a^3 c^3\right ) \text {arctanh}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{2 a^4 \left (b^2-4 a c\right )^{5/2} e}-\frac {3 b \log (d+e x)}{a^4 e}+\frac {3 b \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 a^4 e} \]
-3/2*(-5*a*c+b^2)*(-2*a*c+b^2)/a^3/(-4*a*c+b^2)^2/e/(e*x+d)^2+1/4*(b^2-2*a *c+b*c*(e*x+d)^2)/a/(-4*a*c+b^2)/e/(e*x+d)^2/(a+b*(e*x+d)^2+c*(e*x+d)^4)^2 +1/4*(3*b^4-20*a*b^2*c+20*a^2*c^2+3*b*c*(-6*a*c+b^2)*(e*x+d)^2)/a^2/(-4*a* c+b^2)^2/e/(e*x+d)^2/(a+b*(e*x+d)^2+c*(e*x+d)^4)-3/2*(-20*a^3*c^3+30*a^2*b ^2*c^2-10*a*b^4*c+b^6)*arctanh((b+2*c*(e*x+d)^2)/(-4*a*c+b^2)^(1/2))/a^4/( -4*a*c+b^2)^(5/2)/e-3*b*ln(e*x+d)/a^4/e+3/4*b*ln(a+b*(e*x+d)^2+c*(e*x+d)^4 )/a^4/e
Time = 6.16 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=-\frac {1}{2 a^3 e (d+e x)^2}+\frac {b^3-3 a b c+b^2 c (d+e x)^2-2 a c^2 (d+e x)^2}{4 a^2 \left (-b^2+4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}+\frac {-4 b^5+29 a b^3 c-46 a^2 b c^2-4 b^4 c (d+e x)^2+26 a b^2 c^2 (d+e x)^2-28 a^2 c^3 (d+e x)^2}{4 a^3 \left (-b^2+4 a c\right )^2 e \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {3 b \log (d+e x)}{a^4 e}+\frac {3 \left (b^6-10 a b^4 c+30 a^2 b^2 c^2-20 a^3 c^3+b^5 \sqrt {b^2-4 a c}-8 a b^3 c \sqrt {b^2-4 a c}+16 a^2 b c^2 \sqrt {b^2-4 a c}\right ) \log \left (b-\sqrt {b^2-4 a c}+2 c (d+e x)^2\right )}{4 a^4 \left (b^2-4 a c\right )^{5/2} e}+\frac {3 \left (-b^6+10 a b^4 c-30 a^2 b^2 c^2+20 a^3 c^3+b^5 \sqrt {b^2-4 a c}-8 a b^3 c \sqrt {b^2-4 a c}+16 a^2 b c^2 \sqrt {b^2-4 a c}\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c (d+e x)^2\right )}{4 a^4 \left (b^2-4 a c\right )^{5/2} e} \]
-1/2*1/(a^3*e*(d + e*x)^2) + (b^3 - 3*a*b*c + b^2*c*(d + e*x)^2 - 2*a*c^2* (d + e*x)^2)/(4*a^2*(-b^2 + 4*a*c)*e*(a + b*(d + e*x)^2 + c*(d + e*x)^4)^2 ) + (-4*b^5 + 29*a*b^3*c - 46*a^2*b*c^2 - 4*b^4*c*(d + e*x)^2 + 26*a*b^2*c ^2*(d + e*x)^2 - 28*a^2*c^3*(d + e*x)^2)/(4*a^3*(-b^2 + 4*a*c)^2*e*(a + b* (d + e*x)^2 + c*(d + e*x)^4)) - (3*b*Log[d + e*x])/(a^4*e) + (3*(b^6 - 10* a*b^4*c + 30*a^2*b^2*c^2 - 20*a^3*c^3 + b^5*Sqrt[b^2 - 4*a*c] - 8*a*b^3*c* Sqrt[b^2 - 4*a*c] + 16*a^2*b*c^2*Sqrt[b^2 - 4*a*c])*Log[b - Sqrt[b^2 - 4*a *c] + 2*c*(d + e*x)^2])/(4*a^4*(b^2 - 4*a*c)^(5/2)*e) + (3*(-b^6 + 10*a*b^ 4*c - 30*a^2*b^2*c^2 + 20*a^3*c^3 + b^5*Sqrt[b^2 - 4*a*c] - 8*a*b^3*c*Sqrt [b^2 - 4*a*c] + 16*a^2*b*c^2*Sqrt[b^2 - 4*a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*(d + e*x)^2])/(4*a^4*(b^2 - 4*a*c)^(5/2)*e)
Time = 0.60 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1462, 1434, 1165, 25, 1235, 27, 1200, 2009}
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.
\(\displaystyle \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx\) |
\(\Big \downarrow \) 1462 |
\(\displaystyle \frac {\int \frac {1}{(d+e x)^3 \left (c (d+e x)^4+b (d+e x)^2+a\right )^3}d(d+e x)}{e}\) |
\(\Big \downarrow \) 1434 |
\(\displaystyle \frac {\int \frac {1}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )^3}d(d+e x)^2}{2 e}\) |
\(\Big \downarrow \) 1165 |
\(\displaystyle \frac {\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}-\frac {\int -\frac {3 b^2+4 c (d+e x)^2 b-10 a c}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )^2}d(d+e x)^2}{2 a \left (b^2-4 a c\right )}}{2 e}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {3 b^2+4 c (d+e x)^2 b-10 a c}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )^2}d(d+e x)^2}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}}{2 e}\) |
\(\Big \downarrow \) 1235 |
\(\displaystyle \frac {\frac {\frac {20 a^2 c^2+3 b c \left (b^2-6 a c\right ) (d+e x)^2-20 a b^2 c+3 b^4}{a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {\int -\frac {6 \left (b c \left (b^2-6 a c\right ) (d+e x)^2+\left (b^2-5 a c\right ) \left (b^2-2 a c\right )\right )}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)^2}{a \left (b^2-4 a c\right )}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}}{2 e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {6 \int \frac {b c \left (b^2-6 a c\right ) (d+e x)^2+\left (b^2-5 a c\right ) \left (b^2-2 a c\right )}{(d+e x)^4 \left (c (d+e x)^4+b (d+e x)^2+a\right )}d(d+e x)^2}{a \left (b^2-4 a c\right )}+\frac {20 a^2 c^2+3 b c \left (b^2-6 a c\right ) (d+e x)^2-20 a b^2 c+3 b^4}{a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}}{2 e}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {\frac {\frac {6 \int \left (-\frac {b \left (4 a c-b^2\right )^2}{a^2 (d+e x)^2}+\frac {b^6-9 a c b^4+23 a^2 c^2 b^2+c \left (b^2-4 a c\right )^2 (d+e x)^2 b-10 a^3 c^3}{a^2 \left (c (d+e x)^4+b (d+e x)^2+a\right )}+\frac {\left (b^2-5 a c\right ) \left (b^2-2 a c\right )}{a (d+e x)^4}\right )d(d+e x)^2}{a \left (b^2-4 a c\right )}+\frac {20 a^2 c^2+3 b c \left (b^2-6 a c\right ) (d+e x)^2-20 a b^2 c+3 b^4}{a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\frac {20 a^2 c^2+3 b c \left (b^2-6 a c\right ) (d+e x)^2-20 a b^2 c+3 b^4}{a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac {6 \left (-\frac {b \left (b^2-4 a c\right )^2 \log \left ((d+e x)^2\right )}{a^2}+\frac {b \left (b^2-4 a c\right )^2 \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{2 a^2}-\frac {\left (-20 a^3 c^3+30 a^2 b^2 c^2-10 a b^4 c+b^6\right ) \text {arctanh}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{a^2 \sqrt {b^2-4 a c}}-\frac {\left (b^2-5 a c\right ) \left (b^2-2 a c\right )}{a (d+e x)^2}\right )}{a \left (b^2-4 a c\right )}}{2 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c (d+e x)^2}{2 a \left (b^2-4 a c\right ) (d+e x)^2 \left (a+b (d+e x)^2+c (d+e x)^4\right )^2}}{2 e}\) |
((b^2 - 2*a*c + b*c*(d + e*x)^2)/(2*a*(b^2 - 4*a*c)*(d + e*x)^2*(a + b*(d + e*x)^2 + c*(d + e*x)^4)^2) + ((3*b^4 - 20*a*b^2*c + 20*a^2*c^2 + 3*b*c*( b^2 - 6*a*c)*(d + e*x)^2)/(a*(b^2 - 4*a*c)*(d + e*x)^2*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (6*(-(((b^2 - 5*a*c)*(b^2 - 2*a*c))/(a*(d + e*x)^2)) - ((b^6 - 10*a*b^4*c + 30*a^2*b^2*c^2 - 20*a^3*c^3)*ArcTanh[(b + 2*c*(d + e *x)^2)/Sqrt[b^2 - 4*a*c]])/(a^2*Sqrt[b^2 - 4*a*c]) - (b*(b^2 - 4*a*c)^2*Lo g[(d + e*x)^2])/a^2 + (b*(b^2 - 4*a*c)^2*Log[a + b*(d + e*x)^2 + c*(d + e* x)^4])/(2*a^2)))/(a*(b^2 - 4*a*c)))/(2*a*(b^2 - 4*a*c)))/(2*e)
3.7.37.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) *x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp [1/2 Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Si mp[u^m/(Coefficient[v, x, 1]*v^m) Subst[Int[x^m*(a + b*x^2 + c*x^(2*2))^p , x], x, v], x] /; FreeQ[{a, b, c, m, p}, x] && LinearPairQ[u, v, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.94 (sec) , antiderivative size = 1141, normalized size of antiderivative = 3.51
method | result | size |
default | \(\text {Expression too large to display}\) | \(1141\) |
risch | \(\text {Expression too large to display}\) | \(2190\) |
-1/a^4*((1/2*c^2*e^5*(14*a^2*c^2-13*a*b^2*c+2*b^4)*a/(16*a^2*c^2-8*a*b^2*c +b^4)*x^6+3*(14*a^2*c^2-13*a*b^2*c+2*b^4)*a*c^2*d*e^4/(16*a^2*c^2-8*a*b^2* c+b^4)*x^5+1/4*e^3*a*c*(420*a^2*c^3*d^2-390*a*b^2*c^2*d^2+60*b^4*c*d^2+74* a^2*b*c^2-55*a*b^3*c+8*b^5)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^4+c*d*e^2*a*(140* a^2*c^3*d^2-130*a*b^2*c^2*d^2+20*b^4*c*d^2+74*a^2*b*c^2-55*a*b^3*c+8*b^5)/ (16*a^2*c^2-8*a*b^2*c+b^4)*x^3+1/2*e*a*(210*a^2*c^4*d^4-195*a*b^2*c^3*d^4+ 30*b^4*c^2*d^4+222*a^2*b*c^3*d^2-165*a*b^3*c^2*d^2+24*b^5*c*d^2+18*a^3*c^3 +7*a^2*b^2*c^2-12*a*b^4*c+2*b^6)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2+d*a*(42*a^ 2*c^4*d^4-39*a*b^2*c^3*d^4+6*b^4*c^2*d^4+74*a^2*b*c^3*d^2-55*a*b^3*c^2*d^2 +8*b^5*c*d^2+18*a^3*c^3+7*a^2*b^2*c^2-12*a*b^4*c+2*b^6)/(16*a^2*c^2-8*a*b^ 2*c+b^4)*x+1/4/e*a*(28*a^2*c^4*d^6-26*a*b^2*c^3*d^6+4*b^4*c^2*d^6+74*a^2*b *c^3*d^4-55*a*b^3*c^2*d^4+8*b^5*c*d^4+36*a^3*c^3*d^2+14*a^2*b^2*c^2*d^2-24 *a*b^4*c*d^2+4*b^6*d^2+58*a^3*b*c^2-36*a^2*b^3*c+5*a*b^5)/(16*a^2*c^2-8*a* b^2*c+b^4))/(c*e^4*x^4+4*c*d*e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2 +c*d^4+2*b*d*e*x+b*d^2+a)^2+3/2/(16*a^2*c^2-8*a*b^2*c+b^4)/e*sum((e^3*b*c* (-16*a^2*c^2+8*a*b^2*c-b^4)*_R^3+3*d*e^2*b*c*(-16*a^2*c^2+8*a*b^2*c-b^4)*_ R^2+e*(-48*a^2*b*c^3*d^2+24*a*b^3*c^2*d^2-3*b^5*c*d^2+10*a^3*c^3-23*a^2*b^ 2*c^2+9*a*b^4*c-b^6)*_R-16*a^2*b*c^3*d^3+8*a*b^3*c^2*d^3-b^5*c*d^3+10*a^3* c^3*d-23*a^2*b^2*c^2*d+9*a*b^4*c*d-b^6*d)/(2*_R^3*c*e^3+6*_R^2*c*d*e^2+6*_ R*c*d^2*e+2*c*d^3+_R*b*e+b*d)*ln(x-_R),_R=RootOf(c*e^4*_Z^4+4*c*d*e^3*_...
Leaf count of result is larger than twice the leaf count of optimal. 7517 vs. \(2 (311) = 622\).
Time = 1.61 (sec) , antiderivative size = 15165, normalized size of antiderivative = 46.66 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=\int { \frac {1}{{\left ({\left (e x + d\right )}^{4} c + {\left (e x + d\right )}^{2} b + a\right )}^{3} {\left (e x + d\right )}^{3}} \,d x } \]
-1/4*(6*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^2*c^4)*e^8*x^8 + 48*(b^4*c^2 - 7*a*b ^2*c^3 + 10*a^2*c^4)*d*e^7*x^7 + 3*(4*b^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3 + 56*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^2*c^4)*d^2)*e^6*x^6 + 6*(56*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^2*c^4)*d^3 + 3*(4*b^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3)* d)*e^5*x^5 + 6*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^2*c^4)*d^8 + (6*b^6 - 36*a*b^ 4*c + 14*a^2*b^2*c^2 + 100*a^3*c^3 + 420*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^2*c ^4)*d^4 + 45*(4*b^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3)*d^2)*e^4*x^4 + 3*(4*b ^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3)*d^6 + 4*(84*(b^4*c^2 - 7*a*b^2*c^3 + 1 0*a^2*c^4)*d^5 + 15*(4*b^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3)*d^3 + 2*(3*b^6 - 18*a*b^4*c + 7*a^2*b^2*c^2 + 50*a^3*c^3)*d)*e^3*x^3 + 2*a^2*b^4 - 16*a^ 3*b^2*c + 32*a^4*c^2 + 2*(3*b^6 - 18*a*b^4*c + 7*a^2*b^2*c^2 + 50*a^3*c^3) *d^4 + (168*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^2*c^4)*d^6 + 9*a*b^5 - 68*a^2*b^ 3*c + 122*a^3*b*c^2 + 45*(4*b^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3)*d^4 + 12* (3*b^6 - 18*a*b^4*c + 7*a^2*b^2*c^2 + 50*a^3*c^3)*d^2)*e^2*x^2 + (9*a*b^5 - 68*a^2*b^3*c + 122*a^3*b*c^2)*d^2 + 2*(24*(b^4*c^2 - 7*a*b^2*c^3 + 10*a^ 2*c^4)*d^7 + 9*(4*b^5*c - 29*a*b^3*c^2 + 46*a^2*b*c^3)*d^5 + 4*(3*b^6 - 18 *a*b^4*c + 7*a^2*b^2*c^2 + 50*a^3*c^3)*d^3 + (9*a*b^5 - 68*a^2*b^3*c + 122 *a^3*b*c^2)*d)*e*x)/((a^3*b^4*c^2 - 8*a^4*b^2*c^3 + 16*a^5*c^4)*e^11*x^10 + 10*(a^3*b^4*c^2 - 8*a^4*b^2*c^3 + 16*a^5*c^4)*d*e^10*x^9 + (2*a^3*b^5*c - 16*a^4*b^3*c^2 + 32*a^5*b*c^3 + 45*(a^3*b^4*c^2 - 8*a^4*b^2*c^3 + 16*...
Time = 0.32 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=\frac {3 \, {\left (b^{6} - 10 \, a b^{4} c + 30 \, a^{2} b^{2} c^{2} - 20 \, a^{3} c^{3}\right )} \arctan \left (-\frac {b + \frac {2 \, a}{{\left (e x + d\right )}^{2}}}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (a^{4} b^{4} - 8 \, a^{5} b^{2} c + 16 \, a^{6} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} e} + \frac {3 \, b \log \left (c + \frac {b}{{\left (e x + d\right )}^{2}} + \frac {a}{{\left (e x + d\right )}^{4}}\right )}{4 \, a^{4} e} - \frac {1}{2 \, {\left (e x + d\right )}^{2} a^{3} e} + \frac {5 \, b^{5} c^{2} - 36 \, a b^{3} c^{3} + 58 \, a^{2} b c^{4} + \frac {2 \, {\left (5 \, b^{6} c e - 38 \, a b^{4} c^{2} e + 71 \, a^{2} b^{2} c^{3} e - 14 \, a^{3} c^{4} e\right )}}{{\left (e x + d\right )}^{2} e} + \frac {5 \, b^{7} e^{2} - 34 \, a b^{5} c e^{2} + 41 \, a^{2} b^{3} c^{2} e^{2} + 42 \, a^{3} b c^{3} e^{2}}{{\left (e x + d\right )}^{4} e^{2}} + \frac {6 \, {\left (a b^{6} e^{3} - 8 \, a^{2} b^{4} c e^{3} + 17 \, a^{3} b^{2} c^{2} e^{3} - 6 \, a^{4} c^{3} e^{3}\right )}}{{\left (e x + d\right )}^{6} e^{3}}}{4 \, {\left (b^{2} - 4 \, a c\right )}^{2} a^{4} {\left (c + \frac {b}{{\left (e x + d\right )}^{2}} + \frac {a}{{\left (e x + d\right )}^{4}}\right )}^{2} e} \]
3/2*(b^6 - 10*a*b^4*c + 30*a^2*b^2*c^2 - 20*a^3*c^3)*arctan(-(b + 2*a/(e*x + d)^2)/sqrt(-b^2 + 4*a*c))/((a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2)*sqrt(-b ^2 + 4*a*c)*e) + 3/4*b*log(c + b/(e*x + d)^2 + a/(e*x + d)^4)/(a^4*e) - 1/ 2/((e*x + d)^2*a^3*e) + 1/4*(5*b^5*c^2 - 36*a*b^3*c^3 + 58*a^2*b*c^4 + 2*( 5*b^6*c*e - 38*a*b^4*c^2*e + 71*a^2*b^2*c^3*e - 14*a^3*c^4*e)/((e*x + d)^2 *e) + (5*b^7*e^2 - 34*a*b^5*c*e^2 + 41*a^2*b^3*c^2*e^2 + 42*a^3*b*c^3*e^2) /((e*x + d)^4*e^2) + 6*(a*b^6*e^3 - 8*a^2*b^4*c*e^3 + 17*a^3*b^2*c^2*e^3 - 6*a^4*c^3*e^3)/((e*x + d)^6*e^3))/((b^2 - 4*a*c)^2*a^4*(c + b/(e*x + d)^2 + a/(e*x + d)^4)^2*e)
Time = 22.78 (sec) , antiderivative size = 21465, normalized size of antiderivative = 66.05 \[ \int \frac {1}{(d+e x)^3 \left (a+b (d+e x)^2+c (d+e x)^4\right )^3} \, dx=\text {Too large to display} \]
(log(((27*c^4*e^14*(b^4 + 10*a^2*c^2 - 7*a*b^2*c)^2*(b^5 + 16*a^2*b*c^2 + b^4*c*d^2 + 10*a^2*c^3*d^2 - 8*a*b^3*c - 7*a*b^2*c^2*d^2))/(a^9*(4*a*c - b ^2)^6) - ((3*b - 3*a^4*e*(-(b^6 - 20*a^3*c^3 + 30*a^2*b^2*c^2 - 10*a*b^4*c )^2/(a^8*e^2*(4*a*c - b^2)^5))^(1/2))*((9*c^3*e^15*(b^4 + 10*a^2*c^2 - 7*a *b^2*c)*(4*b^6 - 10*a^3*c^3 + 6*b^5*c*d^2 + 71*a^2*b^2*c^2 - 33*a*b^4*c - 47*a*b^3*c^2*d^2 + 90*a^2*b*c^3*d^2))/(a^6*(4*a*c - b^2)^4) - ((3*b - 3*a^ 4*e*(-(b^6 - 20*a^3*c^3 + 30*a^2*b^2*c^2 - 10*a*b^4*c)^2/(a^8*e^2*(4*a*c - b^2)^5))^(1/2))*((6*c^2*e^16*(2*b^7 - 20*a^3*b*c^3 + b^6*c*d^2 + 46*a^2*b ^3*c^2 + 100*a^3*c^4*d^2 - 18*a*b^5*c - 2*a*b^4*c^2*d^2 - 30*a^2*b^2*c^3*d ^2))/(a^3*(4*a*c - b^2)^2) + (6*c^3*e^18*x^2*(b^6 + 100*a^3*c^3 - 30*a^2*b ^2*c^2 - 2*a*b^4*c))/(a^3*(4*a*c - b^2)^2) + (b*c^2*e^16*(3*b - 3*a^4*e*(- (b^6 - 20*a^3*c^3 + 30*a^2*b^2*c^2 - 10*a*b^4*c)^2/(a^8*e^2*(4*a*c - b^2)^ 5))^(1/2))*(a*b + 3*b^2*d^2 + 3*b^2*e^2*x^2 - 10*a*c*d^2 + 6*b^2*d*e*x - 1 0*a*c*e^2*x^2 - 20*a*c*d*e*x))/a^4 + (12*c^3*d*e^17*x*(b^6 + 100*a^3*c^3 - 30*a^2*b^2*c^2 - 2*a*b^4*c))/(a^3*(4*a*c - b^2)^2)))/(4*a^4*e) + (9*b*c^4 *e^17*x^2*(6*b^8 + 900*a^4*c^4 + 479*a^2*b^4*c^2 - 1100*a^3*b^2*c^3 - 89*a *b^6*c))/(a^6*(4*a*c - b^2)^4) + (18*b*c^4*d*e^16*x*(6*b^8 + 900*a^4*c^4 + 479*a^2*b^4*c^2 - 1100*a^3*b^2*c^3 - 89*a*b^6*c))/(a^6*(4*a*c - b^2)^4))) /(4*a^4*e) + (27*c^5*e^16*x^2*(b^4 + 10*a^2*c^2 - 7*a*b^2*c)^3)/(a^9*(4*a* c - b^2)^6) + (54*c^5*d*e^15*x*(b^4 + 10*a^2*c^2 - 7*a*b^2*c)^3)/(a^9*(...