Integrand size = 29, antiderivative size = 220 \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=-\frac {f}{2 (b e-a f) (d e-c f) (e+f x)^2}-\frac {f (2 b d e-b c f-a d f)}{(b e-a f)^2 (d e-c f)^2 (e+f x)}+\frac {b^3 \log (a+b x)}{(b c-a d) (b e-a f)^3}-\frac {d^3 \log (c+d x)}{(b c-a d) (d e-c f)^3}+\frac {f \left (a^2 d^2 f^2-a b d f (3 d e-c f)+b^2 \left (3 d^2 e^2-3 c d e f+c^2 f^2\right )\right ) \log (e+f x)}{(b e-a f)^3 (d e-c f)^3} \] Output:
-1/2*f/(-a*f+b*e)/(-c*f+d*e)/(f*x+e)^2-f*(-a*d*f-b*c*f+2*b*d*e)/(-a*f+b*e) ^2/(-c*f+d*e)^2/(f*x+e)+b^3*ln(b*x+a)/(-a*d+b*c)/(-a*f+b*e)^3-d^3*ln(d*x+c )/(-a*d+b*c)/(-c*f+d*e)^3+f*(a^2*d^2*f^2-a*b*d*f*(-c*f+3*d*e)+b^2*(c^2*f^2 -3*c*d*e*f+3*d^2*e^2))*ln(f*x+e)/(-a*f+b*e)^3/(-c*f+d*e)^3
Time = 0.18 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=-\frac {f}{2 (b e-a f) (d e-c f) (e+f x)^2}+\frac {f (-2 b d e+b c f+a d f)}{(b e-a f)^2 (d e-c f)^2 (e+f x)}+\frac {b^3 \log (a+b x)}{(b c-a d) (b e-a f)^3}+\frac {d^3 \log (c+d x)}{(b c-a d) (-d e+c f)^3}+\frac {f \left (a^2 d^2 f^2+a b d f (-3 d e+c f)+b^2 \left (3 d^2 e^2-3 c d e f+c^2 f^2\right )\right ) \log (e+f x)}{(b e-a f)^3 (d e-c f)^3} \] Input:
Integrate[1/((e + f*x)^3*(a*c + (b*c + a*d)*x + b*d*x^2)),x]
Output:
-1/2*f/((b*e - a*f)*(d*e - c*f)*(e + f*x)^2) + (f*(-2*b*d*e + b*c*f + a*d* f))/((b*e - a*f)^2*(d*e - c*f)^2*(e + f*x)) + (b^3*Log[a + b*x])/((b*c - a *d)*(b*e - a*f)^3) + (d^3*Log[c + d*x])/((b*c - a*d)*(-(d*e) + c*f)^3) + ( f*(a^2*d^2*f^2 + a*b*d*f*(-3*d*e + c*f) + b^2*(3*d^2*e^2 - 3*c*d*e*f + c^2 *f^2))*Log[e + f*x])/((b*e - a*f)^3*(d*e - c*f)^3)
Time = 0.63 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1141, 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.
| \(\displaystyle \int \frac {1}{(e+f x)^3 \left (x (a d+b c)+a c+b d x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1141 |
| \(\displaystyle b d \int \left (\frac {b^3}{d (b c-a d) (b e-a f)^3 (a+b x)}-\frac {d^3}{(b c-a d) (d e-c f)^3 (c+d x) b}+\frac {f^2 \left (\left (3 d^2 e^2-3 c d f e+c^2 f^2\right ) b^2-a d f (3 d e-c f) b+a^2 d^2 f^2\right )}{d (b e-a f)^3 (d e-c f)^3 (e+f x) b}+\frac {f^2 (2 b d e-b c f-a d f)}{d (b e-a f)^2 (d e-c f)^2 (e+f x)^2 b}+\frac {f^2}{d (b e-a f) (d e-c f) (e+f x)^3 b}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b d \left (\frac {f \log (e+f x) \left (a^2 d^2 f^2-a b d f (3 d e-c f)+b^2 \left (c^2 f^2-3 c d e f+3 d^2 e^2\right )\right )}{b d (b e-a f)^3 (d e-c f)^3}+\frac {b^2 \log (a+b x)}{d (b c-a d) (b e-a f)^3}-\frac {d^2 \log (c+d x)}{b (b c-a d) (d e-c f)^3}-\frac {f (-a d f-b c f+2 b d e)}{b d (e+f x) (b e-a f)^2 (d e-c f)^2}-\frac {f}{2 b d (e+f x)^2 (b e-a f) (d e-c f)}\right )\) |
Input:
Int[1/((e + f*x)^3*(a*c + (b*c + a*d)*x + b*d*x^2)),x]
Output:
b*d*(-1/2*f/(b*d*(b*e - a*f)*(d*e - c*f)*(e + f*x)^2) - (f*(2*b*d*e - b*c* f - a*d*f))/(b*d*(b*e - a*f)^2*(d*e - c*f)^2*(e + f*x)) + (b^2*Log[a + b*x ])/(d*(b*c - a*d)*(b*e - a*f)^3) - (d^2*Log[c + d*x])/(b*(b*c - a*d)*(d*e - c*f)^3) + (f*(a^2*d^2*f^2 - a*b*d*f*(3*d*e - c*f) + b^2*(3*d^2*e^2 - 3*c *d*e*f + c^2*f^2))*Log[e + f*x])/(b*d*(b*e - a*f)^3*(d*e - c*f)^3))
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
Time = 1.28 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(\frac {b^{3} \ln \left (b x +a \right )}{\left (a f -b e \right )^{3} \left (a d -b c \right )}-\frac {f}{2 \left (a f -b e \right ) \left (c f -d e \right ) \left (f x +e \right )^{2}}+\frac {f \left (a^{2} d^{2} f^{2}+a b c d \,f^{2}-3 a b \,d^{2} e f +b^{2} c^{2} f^{2}-3 b^{2} c d e f +3 d^{2} e^{2} b^{2}\right ) \ln \left (f x +e \right )}{\left (a f -b e \right )^{3} \left (c f -d e \right )^{3}}+\frac {f \left (a d f +b c f -2 b d e \right )}{\left (a f -b e \right )^{2} \left (c f -d e \right )^{2} \left (f x +e \right )}-\frac {d^{3} \ln \left (d x +c \right )}{\left (c f -d e \right )^{3} \left (a d -b c \right )}\) | \(222\) |
| norman | \(\frac {\frac {\left (a d \,f^{4}+b c \,f^{4}-2 b d e \,f^{3}\right ) x}{f \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \left (a^{2} f^{2}-2 a b e f +b^{2} e^{2}\right )}+\frac {-a c \,f^{5}+3 a d e \,f^{4}+3 b c e \,f^{4}-5 b d \,e^{2} f^{3}}{2 f^{2} \left (c^{2} f^{2}-2 c d e f +d^{2} e^{2}\right ) \left (a^{2} f^{2}-2 a b e f +b^{2} e^{2}\right )}}{\left (f x +e \right )^{2}}+\frac {b^{3} \ln \left (b x +a \right )}{\left (a^{3} f^{3}-3 a^{2} b e \,f^{2}+3 a \,b^{2} e^{2} f -b^{3} e^{3}\right ) \left (a d -b c \right )}+\frac {\left (a^{2} d^{2} f^{2}+a b c d \,f^{2}-3 a b \,d^{2} e f +b^{2} c^{2} f^{2}-3 b^{2} c d e f +3 d^{2} e^{2} b^{2}\right ) f \ln \left (f x +e \right )}{\left (a^{3} f^{3}-3 a^{2} b e \,f^{2}+3 a \,b^{2} e^{2} f -b^{3} e^{3}\right ) \left (c^{3} f^{3}-3 c^{2} d e \,f^{2}+3 c \,d^{2} e^{2} f -d^{3} e^{3}\right )}-\frac {d^{3} \ln \left (d x +c \right )}{a \,c^{3} d \,f^{3}-3 a \,c^{2} d^{2} e \,f^{2}+3 a c \,d^{3} e^{2} f -a \,d^{4} e^{3}-b \,c^{4} f^{3}+3 b \,c^{3} d e \,f^{2}-3 b \,c^{2} d^{2} e^{2} f +b c \,d^{3} e^{3}}\) | \(466\) |
| parallelrisch | \(\text {Expression too large to display}\) | \(1307\) |
| risch | \(\text {Expression too large to display}\) | \(1945\) |
Input:
int(1/(f*x+e)^3/(a*c+(a*d+b*c)*x+b*d*x^2),x,method=_RETURNVERBOSE)
Output:
b^3/(a*f-b*e)^3/(a*d-b*c)*ln(b*x+a)-1/2*f/(a*f-b*e)/(c*f-d*e)/(f*x+e)^2+f* (a^2*d^2*f^2+a*b*c*d*f^2-3*a*b*d^2*e*f+b^2*c^2*f^2-3*b^2*c*d*e*f+3*b^2*d^2 *e^2)/(a*f-b*e)^3/(c*f-d*e)^3*ln(f*x+e)+f*(a*d*f+b*c*f-2*b*d*e)/(a*f-b*e)^ 2/(c*f-d*e)^2/(f*x+e)-d^3/(c*f-d*e)^3/(a*d-b*c)*ln(d*x+c)
Timed out. \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(f*x+e)^3/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(f*x+e)**3/(a*c+(a*d+b*c)*x+b*d*x**2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 771 vs. \(2 (218) = 436\).
Time = 0.08 (sec) , antiderivative size = 771, normalized size of antiderivative = 3.50 \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b^{3} \log \left (b x + a\right )}{{\left (b^{4} c - a b^{3} d\right )} e^{3} - 3 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} e^{2} f + 3 \, {\left (a^{2} b^{2} c - a^{3} b d\right )} e f^{2} - {\left (a^{3} b c - a^{4} d\right )} f^{3}} - \frac {d^{3} \log \left (d x + c\right )}{{\left (b c d^{3} - a d^{4}\right )} e^{3} - 3 \, {\left (b c^{2} d^{2} - a c d^{3}\right )} e^{2} f + 3 \, {\left (b c^{3} d - a c^{2} d^{2}\right )} e f^{2} - {\left (b c^{4} - a c^{3} d\right )} f^{3}} + \frac {{\left (3 \, b^{2} d^{2} e^{2} f - 3 \, {\left (b^{2} c d + a b d^{2}\right )} e f^{2} + {\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} f^{3}\right )} \log \left (f x + e\right )}{b^{3} d^{3} e^{6} + a^{3} c^{3} f^{6} - 3 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} e^{5} f + 3 \, {\left (b^{3} c^{2} d + 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} e^{4} f^{2} - {\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} e^{3} f^{3} + 3 \, {\left (a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} e^{2} f^{4} - 3 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} e f^{5}} - \frac {5 \, b d e^{2} f + a c f^{3} - 3 \, {\left (b c + a d\right )} e f^{2} + 2 \, {\left (2 \, b d e f^{2} - {\left (b c + a d\right )} f^{3}\right )} x}{2 \, {\left (b^{2} d^{2} e^{6} + a^{2} c^{2} e^{2} f^{4} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} e^{5} f + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} e^{4} f^{2} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} e^{3} f^{3} + {\left (b^{2} d^{2} e^{4} f^{2} + a^{2} c^{2} f^{6} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} e^{3} f^{3} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} e^{2} f^{4} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} e f^{5}\right )} x^{2} + 2 \, {\left (b^{2} d^{2} e^{5} f + a^{2} c^{2} e f^{5} - 2 \, {\left (b^{2} c d + a b d^{2}\right )} e^{4} f^{2} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} e^{3} f^{3} - 2 \, {\left (a b c^{2} + a^{2} c d\right )} e^{2} f^{4}\right )} x\right )}} \] Input:
integrate(1/(f*x+e)^3/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="maxima")
Output:
b^3*log(b*x + a)/((b^4*c - a*b^3*d)*e^3 - 3*(a*b^3*c - a^2*b^2*d)*e^2*f + 3*(a^2*b^2*c - a^3*b*d)*e*f^2 - (a^3*b*c - a^4*d)*f^3) - d^3*log(d*x + c)/ ((b*c*d^3 - a*d^4)*e^3 - 3*(b*c^2*d^2 - a*c*d^3)*e^2*f + 3*(b*c^3*d - a*c^ 2*d^2)*e*f^2 - (b*c^4 - a*c^3*d)*f^3) + (3*b^2*d^2*e^2*f - 3*(b^2*c*d + a* b*d^2)*e*f^2 + (b^2*c^2 + a*b*c*d + a^2*d^2)*f^3)*log(f*x + e)/(b^3*d^3*e^ 6 + a^3*c^3*f^6 - 3*(b^3*c*d^2 + a*b^2*d^3)*e^5*f + 3*(b^3*c^2*d + 3*a*b^2 *c*d^2 + a^2*b*d^3)*e^4*f^2 - (b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a ^3*d^3)*e^3*f^3 + 3*(a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d^2)*e^2*f^4 - 3*(a ^2*b*c^3 + a^3*c^2*d)*e*f^5) - 1/2*(5*b*d*e^2*f + a*c*f^3 - 3*(b*c + a*d)* e*f^2 + 2*(2*b*d*e*f^2 - (b*c + a*d)*f^3)*x)/(b^2*d^2*e^6 + a^2*c^2*e^2*f^ 4 - 2*(b^2*c*d + a*b*d^2)*e^5*f + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*e^4*f^2 - 2*(a*b*c^2 + a^2*c*d)*e^3*f^3 + (b^2*d^2*e^4*f^2 + a^2*c^2*f^6 - 2*(b^2* c*d + a*b*d^2)*e^3*f^3 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*e^2*f^4 - 2*(a*b* c^2 + a^2*c*d)*e*f^5)*x^2 + 2*(b^2*d^2*e^5*f + a^2*c^2*e*f^5 - 2*(b^2*c*d + a*b*d^2)*e^4*f^2 + (b^2*c^2 + 4*a*b*c*d + a^2*d^2)*e^3*f^3 - 2*(a*b*c^2 + a^2*c*d)*e^2*f^4)*x)
Leaf count of result is larger than twice the leaf count of optimal. 749 vs. \(2 (218) = 436\).
Time = 0.32 (sec) , antiderivative size = 749, normalized size of antiderivative = 3.40 \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b^{4} \log \left ({\left | b x + a \right |}\right )}{b^{5} c e^{3} - a b^{4} d e^{3} - 3 \, a b^{4} c e^{2} f + 3 \, a^{2} b^{3} d e^{2} f + 3 \, a^{2} b^{3} c e f^{2} - 3 \, a^{3} b^{2} d e f^{2} - a^{3} b^{2} c f^{3} + a^{4} b d f^{3}} - \frac {d^{4} \log \left ({\left | d x + c \right |}\right )}{b c d^{4} e^{3} - a d^{5} e^{3} - 3 \, b c^{2} d^{3} e^{2} f + 3 \, a c d^{4} e^{2} f + 3 \, b c^{3} d^{2} e f^{2} - 3 \, a c^{2} d^{3} e f^{2} - b c^{4} d f^{3} + a c^{3} d^{2} f^{3}} + \frac {{\left (3 \, b^{2} d^{2} e^{2} f^{2} - 3 \, b^{2} c d e f^{3} - 3 \, a b d^{2} e f^{3} + b^{2} c^{2} f^{4} + a b c d f^{4} + a^{2} d^{2} f^{4}\right )} \log \left ({\left | f x + e \right |}\right )}{b^{3} d^{3} e^{6} f - 3 \, b^{3} c d^{2} e^{5} f^{2} - 3 \, a b^{2} d^{3} e^{5} f^{2} + 3 \, b^{3} c^{2} d e^{4} f^{3} + 9 \, a b^{2} c d^{2} e^{4} f^{3} + 3 \, a^{2} b d^{3} e^{4} f^{3} - b^{3} c^{3} e^{3} f^{4} - 9 \, a b^{2} c^{2} d e^{3} f^{4} - 9 \, a^{2} b c d^{2} e^{3} f^{4} - a^{3} d^{3} e^{3} f^{4} + 3 \, a b^{2} c^{3} e^{2} f^{5} + 9 \, a^{2} b c^{2} d e^{2} f^{5} + 3 \, a^{3} c d^{2} e^{2} f^{5} - 3 \, a^{2} b c^{3} e f^{6} - 3 \, a^{3} c^{2} d e f^{6} + a^{3} c^{3} f^{7}} - \frac {5 \, b^{2} d^{2} e^{4} f - 8 \, b^{2} c d e^{3} f^{2} - 8 \, a b d^{2} e^{3} f^{2} + 3 \, b^{2} c^{2} e^{2} f^{3} + 12 \, a b c d e^{2} f^{3} + 3 \, a^{2} d^{2} e^{2} f^{3} - 4 \, a b c^{2} e f^{4} - 4 \, a^{2} c d e f^{4} + a^{2} c^{2} f^{5} + 2 \, {\left (2 \, b^{2} d^{2} e^{3} f^{2} - 3 \, b^{2} c d e^{2} f^{3} - 3 \, a b d^{2} e^{2} f^{3} + b^{2} c^{2} e f^{4} + 4 \, a b c d e f^{4} + a^{2} d^{2} e f^{4} - a b c^{2} f^{5} - a^{2} c d f^{5}\right )} x}{2 \, {\left (b e - a f\right )}^{3} {\left (d e - c f\right )}^{3} {\left (f x + e\right )}^{2}} \] Input:
integrate(1/(f*x+e)^3/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="giac")
Output:
b^4*log(abs(b*x + a))/(b^5*c*e^3 - a*b^4*d*e^3 - 3*a*b^4*c*e^2*f + 3*a^2*b ^3*d*e^2*f + 3*a^2*b^3*c*e*f^2 - 3*a^3*b^2*d*e*f^2 - a^3*b^2*c*f^3 + a^4*b *d*f^3) - d^4*log(abs(d*x + c))/(b*c*d^4*e^3 - a*d^5*e^3 - 3*b*c^2*d^3*e^2 *f + 3*a*c*d^4*e^2*f + 3*b*c^3*d^2*e*f^2 - 3*a*c^2*d^3*e*f^2 - b*c^4*d*f^3 + a*c^3*d^2*f^3) + (3*b^2*d^2*e^2*f^2 - 3*b^2*c*d*e*f^3 - 3*a*b*d^2*e*f^3 + b^2*c^2*f^4 + a*b*c*d*f^4 + a^2*d^2*f^4)*log(abs(f*x + e))/(b^3*d^3*e^6 *f - 3*b^3*c*d^2*e^5*f^2 - 3*a*b^2*d^3*e^5*f^2 + 3*b^3*c^2*d*e^4*f^3 + 9*a *b^2*c*d^2*e^4*f^3 + 3*a^2*b*d^3*e^4*f^3 - b^3*c^3*e^3*f^4 - 9*a*b^2*c^2*d *e^3*f^4 - 9*a^2*b*c*d^2*e^3*f^4 - a^3*d^3*e^3*f^4 + 3*a*b^2*c^3*e^2*f^5 + 9*a^2*b*c^2*d*e^2*f^5 + 3*a^3*c*d^2*e^2*f^5 - 3*a^2*b*c^3*e*f^6 - 3*a^3*c ^2*d*e*f^6 + a^3*c^3*f^7) - 1/2*(5*b^2*d^2*e^4*f - 8*b^2*c*d*e^3*f^2 - 8*a *b*d^2*e^3*f^2 + 3*b^2*c^2*e^2*f^3 + 12*a*b*c*d*e^2*f^3 + 3*a^2*d^2*e^2*f^ 3 - 4*a*b*c^2*e*f^4 - 4*a^2*c*d*e*f^4 + a^2*c^2*f^5 + 2*(2*b^2*d^2*e^3*f^2 - 3*b^2*c*d*e^2*f^3 - 3*a*b*d^2*e^2*f^3 + b^2*c^2*e*f^4 + 4*a*b*c*d*e*f^4 + a^2*d^2*e*f^4 - a*b*c^2*f^5 - a^2*c*d*f^5)*x)/((b*e - a*f)^3*(d*e - c*f )^3*(f*x + e)^2)
Time = 7.80 (sec) , antiderivative size = 779, normalized size of antiderivative = 3.54 \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b^3\,\ln \left (a+b\,x\right )}{d\,a^4\,f^3-3\,d\,a^3\,b\,e\,f^2-c\,a^3\,b\,f^3+3\,d\,a^2\,b^2\,e^2\,f+3\,c\,a^2\,b^2\,e\,f^2-d\,a\,b^3\,e^3-3\,c\,a\,b^3\,e^2\,f+c\,b^4\,e^3}-\frac {\frac {a\,c\,f^3-3\,a\,d\,e\,f^2-3\,b\,c\,e\,f^2+5\,b\,d\,e^2\,f}{2\,\left (a^2\,c^2\,f^4-2\,a^2\,c\,d\,e\,f^3+a^2\,d^2\,e^2\,f^2-2\,a\,b\,c^2\,e\,f^3+4\,a\,b\,c\,d\,e^2\,f^2-2\,a\,b\,d^2\,e^3\,f+b^2\,c^2\,e^2\,f^2-2\,b^2\,c\,d\,e^3\,f+b^2\,d^2\,e^4\right )}-\frac {x\,\left (a\,d\,f^3+b\,c\,f^3-2\,b\,d\,e\,f^2\right )}{a^2\,c^2\,f^4-2\,a^2\,c\,d\,e\,f^3+a^2\,d^2\,e^2\,f^2-2\,a\,b\,c^2\,e\,f^3+4\,a\,b\,c\,d\,e^2\,f^2-2\,a\,b\,d^2\,e^3\,f+b^2\,c^2\,e^2\,f^2-2\,b^2\,c\,d\,e^3\,f+b^2\,d^2\,e^4}}{e^2+2\,e\,f\,x+f^2\,x^2}+\frac {d^3\,\ln \left (c+d\,x\right )}{b\,c^4\,f^3-3\,b\,c^3\,d\,e\,f^2-a\,c^3\,d\,f^3+3\,b\,c^2\,d^2\,e^2\,f+3\,a\,c^2\,d^2\,e\,f^2-b\,c\,d^3\,e^3-3\,a\,c\,d^3\,e^2\,f+a\,d^4\,e^3}+\frac {\ln \left (e+f\,x\right )\,\left (f^3\,\left (a^2\,d^2+a\,b\,c\,d+b^2\,c^2\right )-f^2\,\left (3\,c\,e\,b^2\,d+3\,a\,e\,b\,d^2\right )+3\,b^2\,d^2\,e^2\,f\right )}{a^3\,c^3\,f^6-3\,a^3\,c^2\,d\,e\,f^5+3\,a^3\,c\,d^2\,e^2\,f^4-a^3\,d^3\,e^3\,f^3-3\,a^2\,b\,c^3\,e\,f^5+9\,a^2\,b\,c^2\,d\,e^2\,f^4-9\,a^2\,b\,c\,d^2\,e^3\,f^3+3\,a^2\,b\,d^3\,e^4\,f^2+3\,a\,b^2\,c^3\,e^2\,f^4-9\,a\,b^2\,c^2\,d\,e^3\,f^3+9\,a\,b^2\,c\,d^2\,e^4\,f^2-3\,a\,b^2\,d^3\,e^5\,f-b^3\,c^3\,e^3\,f^3+3\,b^3\,c^2\,d\,e^4\,f^2-3\,b^3\,c\,d^2\,e^5\,f+b^3\,d^3\,e^6} \] Input:
int(1/((e + f*x)^3*(a*c + x*(a*d + b*c) + b*d*x^2)),x)
Output:
(b^3*log(a + b*x))/(b^4*c*e^3 + a^4*d*f^3 - a*b^3*d*e^3 - a^3*b*c*f^3 - 3* a*b^3*c*e^2*f - 3*a^3*b*d*e*f^2 + 3*a^2*b^2*c*e*f^2 + 3*a^2*b^2*d*e^2*f) - ((a*c*f^3 - 3*a*d*e*f^2 - 3*b*c*e*f^2 + 5*b*d*e^2*f)/(2*(a^2*c^2*f^4 + b^ 2*d^2*e^4 + a^2*d^2*e^2*f^2 + b^2*c^2*e^2*f^2 - 2*a*b*c^2*e*f^3 - 2*a*b*d^ 2*e^3*f - 2*a^2*c*d*e*f^3 - 2*b^2*c*d*e^3*f + 4*a*b*c*d*e^2*f^2)) - (x*(a* d*f^3 + b*c*f^3 - 2*b*d*e*f^2))/(a^2*c^2*f^4 + b^2*d^2*e^4 + a^2*d^2*e^2*f ^2 + b^2*c^2*e^2*f^2 - 2*a*b*c^2*e*f^3 - 2*a*b*d^2*e^3*f - 2*a^2*c*d*e*f^3 - 2*b^2*c*d*e^3*f + 4*a*b*c*d*e^2*f^2))/(e^2 + f^2*x^2 + 2*e*f*x) + (d^3* log(c + d*x))/(a*d^4*e^3 + b*c^4*f^3 - a*c^3*d*f^3 - b*c*d^3*e^3 - 3*a*c*d ^3*e^2*f - 3*b*c^3*d*e*f^2 + 3*a*c^2*d^2*e*f^2 + 3*b*c^2*d^2*e^2*f) + (log (e + f*x)*(f^3*(a^2*d^2 + b^2*c^2 + a*b*c*d) - f^2*(3*a*b*d^2*e + 3*b^2*c* d*e) + 3*b^2*d^2*e^2*f))/(a^3*c^3*f^6 + b^3*d^3*e^6 - a^3*d^3*e^3*f^3 - b^ 3*c^3*e^3*f^3 - 3*a^2*b*c^3*e*f^5 - 3*a*b^2*d^3*e^5*f - 3*a^3*c^2*d*e*f^5 - 3*b^3*c*d^2*e^5*f + 3*a*b^2*c^3*e^2*f^4 + 3*a^2*b*d^3*e^4*f^2 + 3*a^3*c* d^2*e^2*f^4 + 3*b^3*c^2*d*e^4*f^2 + 9*a*b^2*c*d^2*e^4*f^2 - 9*a*b^2*c^2*d* e^3*f^3 - 9*a^2*b*c*d^2*e^3*f^3 + 9*a^2*b*c^2*d*e^2*f^4)
Time = 0.24 (sec) , antiderivative size = 2244, normalized size of antiderivative = 10.20 \[ \int \frac {1}{(e+f x)^3 \left (a c+(b c+a d) x+b d x^2\right )} \, dx =\text {Too large to display} \] Input:
int(1/(f*x+e)^3/(a*c+(a*d+b*c)*x+b*d*x^2),x)
Output:
(2*log(a + b*x)*b**3*c**3*e**3*f**3 + 4*log(a + b*x)*b**3*c**3*e**2*f**4*x + 2*log(a + b*x)*b**3*c**3*e*f**5*x**2 - 6*log(a + b*x)*b**3*c**2*d*e**4* f**2 - 12*log(a + b*x)*b**3*c**2*d*e**3*f**3*x - 6*log(a + b*x)*b**3*c**2* d*e**2*f**4*x**2 + 6*log(a + b*x)*b**3*c*d**2*e**5*f + 12*log(a + b*x)*b** 3*c*d**2*e**4*f**2*x + 6*log(a + b*x)*b**3*c*d**2*e**3*f**3*x**2 - 2*log(a + b*x)*b**3*d**3*e**6 - 4*log(a + b*x)*b**3*d**3*e**5*f*x - 2*log(a + b*x )*b**3*d**3*e**4*f**2*x**2 - 2*log(c + d*x)*a**3*d**3*e**3*f**3 - 4*log(c + d*x)*a**3*d**3*e**2*f**4*x - 2*log(c + d*x)*a**3*d**3*e*f**5*x**2 + 6*lo g(c + d*x)*a**2*b*d**3*e**4*f**2 + 12*log(c + d*x)*a**2*b*d**3*e**3*f**3*x + 6*log(c + d*x)*a**2*b*d**3*e**2*f**4*x**2 - 6*log(c + d*x)*a*b**2*d**3* e**5*f - 12*log(c + d*x)*a*b**2*d**3*e**4*f**2*x - 6*log(c + d*x)*a*b**2*d **3*e**3*f**3*x**2 + 2*log(c + d*x)*b**3*d**3*e**6 + 4*log(c + d*x)*b**3*d **3*e**5*f*x + 2*log(c + d*x)*b**3*d**3*e**4*f**2*x**2 + 2*log(e + f*x)*a* *3*d**3*e**3*f**3 + 4*log(e + f*x)*a**3*d**3*e**2*f**4*x + 2*log(e + f*x)* a**3*d**3*e*f**5*x**2 - 6*log(e + f*x)*a**2*b*d**3*e**4*f**2 - 12*log(e + f*x)*a**2*b*d**3*e**3*f**3*x - 6*log(e + f*x)*a**2*b*d**3*e**2*f**4*x**2 + 6*log(e + f*x)*a*b**2*d**3*e**5*f + 12*log(e + f*x)*a*b**2*d**3*e**4*f**2 *x + 6*log(e + f*x)*a*b**2*d**3*e**3*f**3*x**2 - 2*log(e + f*x)*b**3*c**3* e**3*f**3 - 4*log(e + f*x)*b**3*c**3*e**2*f**4*x - 2*log(e + f*x)*b**3*c** 3*e*f**5*x**2 + 6*log(e + f*x)*b**3*c**2*d*e**4*f**2 + 12*log(e + f*x)*...