Integrand size = 22, antiderivative size = 290 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {e \left (A c d^2+4 a B d e-3 a A e^2\right )}{2 a \left (c d^2+a e^2\right )^2 (d+e x)}-\frac {a (B d-A e)-(A c d+a B e) x}{2 a \left (c d^2+a e^2\right ) (d+e x) \left (a+c x^2\right )}-\frac {\sqrt {c} \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \left (c d^2+a e^2\right )^3}-\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {e^2 \left (3 B c d^2-4 A c d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3} \]
1/2*e*(-3*A*a*e^2+A*c*d^2+4*B*a*d*e)/a/(a*e^2+c*d^2)^2/(e*x+d)+1/2*(-a*(-A *e+B*d)+(A*c*d+B*a*e)*x)/a/(a*e^2+c*d^2)/(e*x+d)/(c*x^2+a)-e^2*(-4*A*c*d*e -B*a*e^2+3*B*c*d^2)*ln(e*x+d)/(a*e^2+c*d^2)^3+1/2*e^2*(-4*A*c*d*e-B*a*e^2+ 3*B*c*d^2)*ln(c*x^2+a)/(a*e^2+c*d^2)^3-1/2*(2*a*B*d*e*(-3*a*e^2+c*d^2)-A*( -3*a^2*e^4+6*a*c*d^2*e^2+c^2*d^4))*arctan(x*c^(1/2)/a^(1/2))*c^(1/2)/a^(3/ 2)/(a*e^2+c*d^2)^3
Time = 0.23 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {-\frac {2 e^2 (-B d+A e) \left (c d^2+a e^2\right )}{d+e x}+\frac {\left (c d^2+a e^2\right ) \left (a^2 B e^2+A c^2 d^2 x-a c (B d (d-2 e x)+A e (-2 d+e x))\right )}{a \left (a+c x^2\right )}+\frac {\sqrt {c} \left (2 a B d e \left (-c d^2+3 a e^2\right )+A \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2}}+2 e^2 \left (-3 B c d^2+4 A c d e+a B e^2\right ) \log (d+e x)-e^2 \left (-3 B c d^2+4 A c d e+a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3} \]
((-2*e^2*(-(B*d) + A*e)*(c*d^2 + a*e^2))/(d + e*x) + ((c*d^2 + a*e^2)*(a^2 *B*e^2 + A*c^2*d^2*x - a*c*(B*d*(d - 2*e*x) + A*e*(-2*d + e*x))))/(a*(a + c*x^2)) + (Sqrt[c]*(2*a*B*d*e*(-(c*d^2) + 3*a*e^2) + A*(c^2*d^4 + 6*a*c*d^ 2*e^2 - 3*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 2*e^2*(-3*B*c*d ^2 + 4*A*c*d*e + a*B*e^2)*Log[d + e*x] - e^2*(-3*B*c*d^2 + 4*A*c*d*e + a*B *e^2)*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)
Time = 0.58 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {686, 25, 27, 657, 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.
| \(\displaystyle \int \frac {A+B x}{\left (a+c x^2\right )^2 (d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 686 |
| \(\displaystyle -\frac {\int -\frac {c \left (A c d^2-2 a B e d+3 a A e^2+2 e (A c d+a B e) x\right )}{(d+e x)^2 \left (c x^2+a\right )}dx}{2 a c \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {c \left (A c d^2-2 a B e d+3 a A e^2+2 e (A c d+a B e) x\right )}{(d+e x)^2 \left (c x^2+a\right )}dx}{2 a c \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {A c d^2-2 a B e d+3 a A e^2+2 e (A c d+a B e) x}{(d+e x)^2 \left (c x^2+a\right )}dx}{2 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {\int \left (\frac {2 a \left (-3 B c d^2+4 A c e d+a B e^2\right ) e^3}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac {\left (-A c d^2-4 a B e d+3 a A e^2\right ) e^2}{\left (c d^2+a e^2\right ) (d+e x)^2}+\frac {c \left (2 a \left (3 B c d^2-4 A c e d-a B e^2\right ) x e^2-2 a B d \left (c d^2-3 a e^2\right ) e+A \left (c^2 d^4+6 a c e^2 d^2-3 a^2 e^4\right )\right )}{\left (c d^2+a e^2\right )^2 \left (c x^2+a\right )}\right )dx}{2 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\sqrt {c} \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (2 a B d e \left (c d^2-3 a e^2\right )-A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )\right )}{\sqrt {a} \left (a e^2+c d^2\right )^2}+\frac {a e^2 \log \left (a+c x^2\right ) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac {e \left (-3 a A e^2+4 a B d e+A c d^2\right )}{(d+e x) \left (a e^2+c d^2\right )}-\frac {2 a e^2 \log (d+e x) \left (-a B e^2-4 A c d e+3 B c d^2\right )}{\left (a e^2+c d^2\right )^2}}{2 a \left (a e^2+c d^2\right )}-\frac {a (B d-A e)-x (a B e+A c d)}{2 a \left (a+c x^2\right ) (d+e x) \left (a e^2+c d^2\right )}\) |
-1/2*(a*(B*d - A*e) - (A*c*d + a*B*e)*x)/(a*(c*d^2 + a*e^2)*(d + e*x)*(a + c*x^2)) + ((e*(A*c*d^2 + 4*a*B*d*e - 3*a*A*e^2))/((c*d^2 + a*e^2)*(d + e* x)) - (Sqrt[c]*(2*a*B*d*e*(c*d^2 - 3*a*e^2) - A*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*e^2)^2) - (2 *a*e^2*(3*B*c*d^2 - 4*A*c*d*e - a*B*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^2 + (a*e^2*(3*B*c*d^2 - 4*A*c*d*e - a*B*e^2)*Log[a + c*x^2])/(c*d^2 + a*e^2)^ 2)/(2*a*(c*d^2 + a*e^2))
3.14.45.3.1 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_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.36 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(-\frac {c \left (\frac {\frac {\left (A \,a^{2} e^{4}-d^{4} A \,c^{2}-2 B \,a^{2} d \,e^{3}-2 B a c \,d^{3} e \right ) x}{2 a}-\frac {2 A a c d \,e^{3}+2 A \,c^{2} d^{3} e +B \,e^{4} a^{2}-B \,c^{2} d^{4}}{2 c}}{c \,x^{2}+a}+\frac {\frac {\left (8 A a c d \,e^{3}+2 B \,e^{4} a^{2}-6 B a c \,d^{2} e^{2}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (3 A \,a^{2} e^{4}-6 A a c \,d^{2} e^{2}-d^{4} A \,c^{2}-6 B \,a^{2} d \,e^{3}+2 B a c \,d^{3} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a}\right )}{\left (e^{2} a +c \,d^{2}\right )^{3}}+\frac {e^{2} \left (4 A c d e +B a \,e^{2}-3 B c \,d^{2}\right ) \ln \left (e x +d \right )}{\left (e^{2} a +c \,d^{2}\right )^{3}}-\frac {\left (A e -B d \right ) e^{2}}{\left (e^{2} a +c \,d^{2}\right )^{2} \left (e x +d \right )}\) | \(304\) |
| risch | \(\text {Expression too large to display}\) | \(1006\) |
-c/(a*e^2+c*d^2)^3*((1/2*(A*a^2*e^4-A*c^2*d^4-2*B*a^2*d*e^3-2*B*a*c*d^3*e) /a*x-1/2*(2*A*a*c*d*e^3+2*A*c^2*d^3*e+B*a^2*e^4-B*c^2*d^4)/c)/(c*x^2+a)+1/ 2/a*(1/2*(8*A*a*c*d*e^3+2*B*a^2*e^4-6*B*a*c*d^2*e^2)/c*ln(c*x^2+a)+(3*A*a^ 2*e^4-6*A*a*c*d^2*e^2-A*c^2*d^4-6*B*a^2*d*e^3+2*B*a*c*d^3*e)/(a*c)^(1/2)*a rctan(c*x/(a*c)^(1/2))))+e^2*(4*A*c*d*e+B*a*e^2-3*B*c*d^2)/(a*e^2+c*d^2)^3 *ln(e*x+d)-(A*e-B*d)*e^2/(a*e^2+c*d^2)^2/(e*x+d)
Leaf count of result is larger than twice the leaf count of optimal. 958 vs. \(2 (274) = 548\).
Time = 47.97 (sec) , antiderivative size = 1940, normalized size of antiderivative = 6.69 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\text {Too large to display} \]
[-1/4*(2*B*a*c^2*d^5 - 4*A*a*c^2*d^4*e - 4*B*a^2*c*d^3*e^2 - 6*B*a^3*d*e^4 + 4*A*a^3*e^5 - 2*(A*c^3*d^4*e + 4*B*a*c^2*d^3*e^2 - 2*A*a*c^2*d^2*e^3 + 4*B*a^2*c*d*e^4 - 3*A*a^2*c*e^5)*x^2 + (A*a*c^2*d^5 - 2*B*a^2*c*d^4*e + 6* A*a^2*c*d^3*e^2 + 6*B*a^3*d^2*e^3 - 3*A*a^3*d*e^4 + (A*c^3*d^4*e - 2*B*a*c ^2*d^3*e^2 + 6*A*a*c^2*d^2*e^3 + 6*B*a^2*c*d*e^4 - 3*A*a^2*c*e^5)*x^3 + (A *c^3*d^5 - 2*B*a*c^2*d^4*e + 6*A*a*c^2*d^3*e^2 + 6*B*a^2*c*d^2*e^3 - 3*A*a ^2*c*d*e^4)*x^2 + (A*a*c^2*d^4*e - 2*B*a^2*c*d^3*e^2 + 6*A*a^2*c*d^2*e^3 + 6*B*a^3*d*e^4 - 3*A*a^3*e^5)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a) - a)/(c*x^2 + a)) - 2*(A*c^3*d^5 + B*a*c^2*d^4*e + 2*A*a*c^2*d^3*e^2 + 2*B *a^2*c*d^2*e^3 + A*a^2*c*d*e^4 + B*a^3*e^5)*x - 2*(3*B*a^2*c*d^3*e^2 - 4*A *a^2*c*d^2*e^3 - B*a^3*d*e^4 + (3*B*a*c^2*d^2*e^3 - 4*A*a*c^2*d*e^4 - B*a^ 2*c*e^5)*x^3 + (3*B*a*c^2*d^3*e^2 - 4*A*a*c^2*d^2*e^3 - B*a^2*c*d*e^4)*x^2 + (3*B*a^2*c*d^2*e^3 - 4*A*a^2*c*d*e^4 - B*a^3*e^5)*x)*log(c*x^2 + a) + 4 *(3*B*a^2*c*d^3*e^2 - 4*A*a^2*c*d^2*e^3 - B*a^3*d*e^4 + (3*B*a*c^2*d^2*e^3 - 4*A*a*c^2*d*e^4 - B*a^2*c*e^5)*x^3 + (3*B*a*c^2*d^3*e^2 - 4*A*a*c^2*d^2 *e^3 - B*a^2*c*d*e^4)*x^2 + (3*B*a^2*c*d^2*e^3 - 4*A*a^2*c*d*e^4 - B*a^3*e ^5)*x)*log(e*x + d))/(a^2*c^3*d^7 + 3*a^3*c^2*d^5*e^2 + 3*a^4*c*d^3*e^4 + a^5*d*e^6 + (a*c^4*d^6*e + 3*a^2*c^3*d^4*e^3 + 3*a^3*c^2*d^2*e^5 + a^4*c*e ^7)*x^3 + (a*c^4*d^7 + 3*a^2*c^3*d^5*e^2 + 3*a^3*c^2*d^3*e^4 + a^4*c*d*e^6 )*x^2 + (a^2*c^3*d^6*e + 3*a^3*c^2*d^4*e^3 + 3*a^4*c*d^2*e^5 + a^5*e^7)...
Timed out. \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 511, normalized size of antiderivative = 1.76 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\right )} \log \left (e x + d\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {{\left (A c^{3} d^{4} - 2 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} + 6 \, B a^{2} c d e^{3} - 3 \, A a^{2} c e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c}} - \frac {B a c d^{3} - 2 \, A a c d^{2} e - 3 \, B a^{2} d e^{2} + 2 \, A a^{2} e^{3} - {\left (A c^{2} d^{2} e + 4 \, B a c d e^{2} - 3 \, A a c e^{3}\right )} x^{2} - {\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\right )} x}{2 \, {\left (a^{2} c^{2} d^{5} + 2 \, a^{3} c d^{3} e^{2} + a^{4} d e^{4} + {\left (a c^{3} d^{4} e + 2 \, a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{3} + {\left (a c^{3} d^{5} + 2 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x^{2} + {\left (a^{2} c^{2} d^{4} e + 2 \, a^{3} c d^{2} e^{3} + a^{4} e^{5}\right )} x\right )}} \]
1/2*(3*B*c*d^2*e^2 - 4*A*c*d*e^3 - B*a*e^4)*log(c*x^2 + a)/(c^3*d^6 + 3*a* c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6) - (3*B*c*d^2*e^2 - 4*A*c*d*e^3 - B*a*e^4)*log(e*x + d)/(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e ^6) + 1/2*(A*c^3*d^4 - 2*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 + 6*B*a^2*c*d*e ^3 - 3*A*a^2*c*e^4)*arctan(c*x/sqrt(a*c))/((a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3*c*d^2*e^4 + a^4*e^6)*sqrt(a*c)) - 1/2*(B*a*c*d^3 - 2*A*a*c*d^2*e - 3*B*a^2*d*e^2 + 2*A*a^2*e^3 - (A*c^2*d^2*e + 4*B*a*c*d*e^2 - 3*A*a*c*e^3) *x^2 - (A*c^2*d^3 + B*a*c*d^2*e + A*a*c*d*e^2 + B*a^2*e^3)*x)/(a^2*c^2*d^5 + 2*a^3*c*d^3*e^2 + a^4*d*e^4 + (a*c^3*d^4*e + 2*a^2*c^2*d^2*e^3 + a^3*c* e^5)*x^3 + (a*c^3*d^5 + 2*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x^2 + (a^2*c^2*d^ 4*e + 2*a^3*c*d^2*e^3 + a^4*e^5)*x)
Time = 0.26 (sec) , antiderivative size = 518, normalized size of antiderivative = 1.79 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\frac {{\left (3 \, B c d^{2} e^{2} - 4 \, A c d e^{3} - B a e^{4}\right )} \log \left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} + \frac {\frac {B d e^{6}}{e x + d} - \frac {A e^{7}}{e x + d}}{c^{2} d^{4} e^{4} + 2 \, a c d^{2} e^{6} + a^{2} e^{8}} + \frac {{\left (A c^{3} d^{4} e^{2} - 2 \, B a c^{2} d^{3} e^{3} + 6 \, A a c^{2} d^{2} e^{4} + 6 \, B a^{2} c d e^{5} - 3 \, A a^{2} c e^{6}\right )} \arctan \left (\frac {c d - \frac {c d^{2}}{e x + d} - \frac {a e^{2}}{e x + d}}{\sqrt {a c} e}\right )}{2 \, {\left (a c^{3} d^{6} + 3 \, a^{2} c^{2} d^{4} e^{2} + 3 \, a^{3} c d^{2} e^{4} + a^{4} e^{6}\right )} \sqrt {a c} e^{2}} + \frac {\frac {A c^{3} d^{3} e + 3 \, B a c^{2} d^{2} e^{2} - 3 \, A a c^{2} d e^{3} - B a^{2} c e^{4}}{c d^{2} + a e^{2}} - \frac {A c^{3} d^{4} e^{2} + 4 \, B a c^{2} d^{3} e^{3} - 6 \, A a c^{2} d^{2} e^{4} - 4 \, B a^{2} c d e^{5} + A a^{2} c e^{6}}{{\left (c d^{2} + a e^{2}\right )} {\left (e x + d\right )} e}}{2 \, {\left (c d^{2} + a e^{2}\right )}^{2} a {\left (c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {a e^{2}}{{\left (e x + d\right )}^{2}}\right )}} \]
1/2*(3*B*c*d^2*e^2 - 4*A*c*d*e^3 - B*a*e^4)*log(c - 2*c*d/(e*x + d) + c*d^ 2/(e*x + d)^2 + a*e^2/(e*x + d)^2)/(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^ 2*e^4 + a^3*e^6) + (B*d*e^6/(e*x + d) - A*e^7/(e*x + d))/(c^2*d^4*e^4 + 2* a*c*d^2*e^6 + a^2*e^8) + 1/2*(A*c^3*d^4*e^2 - 2*B*a*c^2*d^3*e^3 + 6*A*a*c^ 2*d^2*e^4 + 6*B*a^2*c*d*e^5 - 3*A*a^2*c*e^6)*arctan((c*d - c*d^2/(e*x + d) - a*e^2/(e*x + d))/(sqrt(a*c)*e))/((a*c^3*d^6 + 3*a^2*c^2*d^4*e^2 + 3*a^3 *c*d^2*e^4 + a^4*e^6)*sqrt(a*c)*e^2) + 1/2*((A*c^3*d^3*e + 3*B*a*c^2*d^2*e ^2 - 3*A*a*c^2*d*e^3 - B*a^2*c*e^4)/(c*d^2 + a*e^2) - (A*c^3*d^4*e^2 + 4*B *a*c^2*d^3*e^3 - 6*A*a*c^2*d^2*e^4 - 4*B*a^2*c*d*e^5 + A*a^2*c*e^6)/((c*d^ 2 + a*e^2)*(e*x + d)*e))/((c*d^2 + a*e^2)^2*a*(c - 2*c*d/(e*x + d) + c*d^2 /(e*x + d)^2 + a*e^2/(e*x + d)^2))
Time = 13.22 (sec) , antiderivative size = 2029, normalized size of antiderivative = 7.00 \[ \int \frac {A+B x}{(d+e x)^2 \left (a+c x^2\right )^2} \, dx=\text {Too large to display} \]
((x*(A*c*d + B*a*e))/(2*a*(a*e^2 + c*d^2)) - (2*A*a*e^3 + B*c*d^3 - 3*B*a* d*e^2 - 2*A*c*d^2*e)/(2*(a*e^2 + c*d^2)^2) + (x^2*(A*c^2*d^2*e - 3*A*a*c*e ^3 + 4*B*a*c*d*e^2))/(2*a*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)))/(a*d + a*e *x + c*d*x^2 + c*e*x^3) + (log(9*A^2*a^6*e^12*(-a^3*c)^(3/2) + A^2*c^6*d^1 2*(-a^3*c)^(3/2) - 36*B^2*a^10*e^12*(-a^3*c)^(1/2) - 558*A^2*a^2*d^2*e^10* (-a^3*c)^(5/2) + 24*B^2*a^2*d^4*e^8*(-a^3*c)^(5/2) - 108*B^2*a^6*d^2*e^10* (-a^3*c)^(3/2) - 612*A^2*c^2*d^6*e^6*(-a^3*c)^(5/2) - 308*B^2*c^2*d^8*e^4* (-a^3*c)^(5/2) + 36*B^2*a^11*c*e^12*x + A^2*a^4*c^8*d^12*x + 9*A^2*a^10*c^ 2*e^12*x + 276*A*B*a^2*d^3*e^9*(-a^3*c)^(5/2) + 808*A*B*c^2*d^7*e^5*(-a^3* c)^(5/2) - 1119*A^2*a*c*d^4*e^8*(-a^3*c)^(5/2) - 424*B^2*a*c*d^6*e^6*(-a^3 *c)^(5/2) + 14*A^2*a^5*c^7*d^10*e^2*x + 55*A^2*a^6*c^6*d^8*e^4*x + 612*A^2 *a^7*c^5*d^6*e^6*x + 1119*A^2*a^8*c^4*d^4*e^8*x + 558*A^2*a^9*c^3*d^2*e^10 *x + 4*B^2*a^6*c^6*d^10*e^2*x + 308*B^2*a^7*c^5*d^8*e^4*x + 424*B^2*a^8*c^ 4*d^6*e^6*x - 24*B^2*a^9*c^3*d^4*e^8*x - 108*B^2*a^10*c^2*d^2*e^10*x + 14* A^2*a*c^5*d^10*e^2*(-a^3*c)^(3/2) + 252*A*B*a^6*d*e^11*(-a^3*c)^(3/2) + 55 *A^2*a^2*c^4*d^8*e^4*(-a^3*c)^(3/2) + 4*B^2*a^2*c^4*d^10*e^2*(-a^3*c)^(3/2 ) - 4*A*B*a^5*c^7*d^11*e*x + 252*A*B*a^10*c^2*d*e^11*x + 1320*A*B*a*c*d^5* e^7*(-a^3*c)^(5/2) - 4*A*B*a*c^5*d^11*e*(-a^3*c)^(3/2) - 20*A*B*a^6*c^6*d^ 9*e^3*x - 808*A*B*a^7*c^5*d^7*e^5*x - 1320*A*B*a^8*c^4*d^5*e^7*x - 276*A*B *a^9*c^3*d^3*e^9*x - 20*A*B*a^2*c^4*d^9*e^3*(-a^3*c)^(3/2))*(c*(a^3*((3...