Integrand size = 16, antiderivative size = 203 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {b x}{2 c^2 \left (b^2-4 a c\right )}+\frac {x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {a \left (b^4-11 a b^2 c+16 a^2 c^2\right )+b \left (b^4-12 a b^2 c+26 a^2 c^2\right ) x}{2 c^3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {b \left (b^4-10 a b^2 c+30 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \left (b^2-4 a c\right )^{5/2}}+\frac {\log \left (a+b x+c x^2\right )}{2 c^3} \] Output:
-1/2*b*x/c^2/(-4*a*c+b^2)+1/2*x^4*(b*x+2*a)/(-4*a*c+b^2)/(c*x^2+b*x+a)^2+1 /2*(a*(16*a^2*c^2-11*a*b^2*c+b^4)+b*(26*a^2*c^2-12*a*b^2*c+b^4)*x)/c^3/(-4 *a*c+b^2)^2/(c*x^2+b*x+a)+b*(30*a^2*c^2-10*a*b^2*c+b^4)*arctanh((2*c*x+b)/ (-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(5/2)+1/2*ln(c*x^2+b*x+a)/c^3
Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.09 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=\frac {\frac {-b^6+11 a b^4 c-39 a^2 b^2 c^2+32 a^3 c^3+4 b^5 c x-30 a b^3 c^2 x+50 a^2 b c^3 x}{\left (b^2-4 a c\right )^2 (a+x (b+c x))}+\frac {2 a^3 c^2+b^5 x+a b^3 (b-5 c x)+a^2 b c (-4 b+5 c x)}{\left (b^2-4 a c\right ) (a+x (b+c x))^2}-\frac {2 b c \left (b^4-10 a b^2 c+30 a^2 c^2\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{5/2}}+c \log (a+x (b+c x))}{2 c^4} \] Input:
Integrate[x^5/(a + b*x + c*x^2)^3,x]
Output:
((-b^6 + 11*a*b^4*c - 39*a^2*b^2*c^2 + 32*a^3*c^3 + 4*b^5*c*x - 30*a*b^3*c ^2*x + 50*a^2*b*c^3*x)/((b^2 - 4*a*c)^2*(a + x*(b + c*x))) + (2*a^3*c^2 + b^5*x + a*b^3*(b - 5*c*x) + a^2*b*c*(-4*b + 5*c*x))/((b^2 - 4*a*c)*(a + x* (b + c*x))^2) - (2*b*c*(b^4 - 10*a*b^2*c + 30*a^2*c^2)*ArcTan[(b + 2*c*x)/ Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2) + c*Log[a + x*(b + c*x)])/(2*c^4 )
Time = 0.75 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1164, 1233, 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.
| \(\displaystyle \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1164 |
| \(\displaystyle \frac {x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {\int \frac {x^3 (8 a+b x)}{\left (c x^2+b x+a\right )^2}dx}{2 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle \frac {x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {\frac {\int \frac {2 x \left (a \left (b^2-16 a c\right )+b \left (b^2-7 a c\right ) x\right )}{c x^2+b x+a}dx}{c \left (b^2-4 a c\right )}-\frac {x^2 \left (b x \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}}{2 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {\frac {2 \int \frac {x \left (a \left (b^2-16 a c\right )+b \left (b^2-7 a c\right ) x\right )}{c x^2+b x+a}dx}{c \left (b^2-4 a c\right )}-\frac {x^2 \left (b x \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}}{2 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {\frac {2 \int \left (-b \left (7 a-\frac {b^2}{c}\right )-\frac {x \left (b^2-4 a c\right )^2+a b \left (b^2-7 a c\right )}{c \left (c x^2+b x+a\right )}\right )dx}{c \left (b^2-4 a c\right )}-\frac {x^2 \left (b x \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}}{2 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^4 (2 a+b x)}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {\frac {2 \left (-\frac {b \left (30 a^2 c^2-10 a b^2 c+b^4\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {\left (b^2-4 a c\right )^2 \log \left (a+b x+c x^2\right )}{2 c^2}-b x \left (7 a-\frac {b^2}{c}\right )\right )}{c \left (b^2-4 a c\right )}-\frac {x^2 \left (b x \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}}{2 \left (b^2-4 a c\right )}\) |
Input:
Int[x^5/(a + b*x + c*x^2)^3,x]
Output:
(x^4*(2*a + b*x))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) - (-((x^2*(a*(b^2 - 16*a*c) + b*(b^2 - 10*a*c)*x))/(c*(b^2 - 4*a*c)*(a + b*x + c*x^2))) + (2 *(-(b*(7*a - b^2/c)*x) - (b*(b^4 - 10*a*b^2*c + 30*a^2*c^2)*ArcTanh[(b + 2 *c*x)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]) - ((b^2 - 4*a*c)^2*Log[a + b*x + c*x^2])/(2*c^2)))/(c*(b^2 - 4*a*c)))/(2*(b^2 - 4*a*c))
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)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* c)) Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int QuadraticQ[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))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 0])
Time = 0.79 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.76
| method | result | size |
| default | \(\frac {\frac {b \left (25 a^{2} c^{2}-15 c a \,b^{2}+2 b^{4}\right ) x^{3}}{c^{2} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}+\frac {\left (32 a^{3} c^{3}+11 a^{2} b^{2} c^{2}-19 a \,b^{4} c +3 b^{6}\right ) x^{2}}{2 c^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}+\frac {a b \left (31 a^{2} c^{2}-22 c a \,b^{2}+3 b^{4}\right ) x}{\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) c^{3}}+\frac {3 a^{2} \left (8 a^{2} c^{2}-7 c a \,b^{2}+b^{4}\right )}{2 c^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}}{\left (c \,x^{2}+b x +a \right )^{2}}+\frac {\frac {\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-7 c \,a^{2} b +a \,b^{3}-\frac {\left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{2} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(357\) |
| risch | \(\text {Expression too large to display}\) | \(1641\) |
Input:
int(x^5/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
(1/c^2*b*(25*a^2*c^2-15*a*b^2*c+2*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+1/2* (32*a^3*c^3+11*a^2*b^2*c^2-19*a*b^4*c+3*b^6)/c^3/(16*a^2*c^2-8*a*b^2*c+b^4 )*x^2+a*b*(31*a^2*c^2-22*a*b^2*c+3*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/c^3*x+3 /2*a^2*(8*a^2*c^2-7*a*b^2*c+b^4)/c^3/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^2+b* x+a)^2+1/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)*(1/2*(16*a^2*c^2-8*a*b^2*c+b^4)/c* ln(c*x^2+b*x+a)+2*(-7*c*a^2*b+a*b^3-1/2*(16*a^2*c^2-8*a*b^2*c+b^4)*b/c)/(4 *a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 792 vs. \(2 (191) = 382\).
Time = 0.11 (sec) , antiderivative size = 1603, normalized size of antiderivative = 7.90 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(x^5/(c*x^2+b*x+a)^3,x, algorithm="fricas")
Output:
[1/2*(3*a^2*b^6 - 33*a^3*b^4*c + 108*a^4*b^2*c^2 - 96*a^5*c^3 + 2*(2*b^7*c - 23*a*b^5*c^2 + 85*a^2*b^3*c^3 - 100*a^3*b*c^4)*x^3 + (3*b^8 - 31*a*b^6* c + 87*a^2*b^4*c^2 - 12*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + (a^2*b^5 - 10*a^3 *b^3*c + 30*a^4*b*c^2 + (b^5*c^2 - 10*a*b^3*c^3 + 30*a^2*b*c^4)*x^4 + 2*(b ^6*c - 10*a*b^4*c^2 + 30*a^2*b^2*c^3)*x^3 + (b^7 - 8*a*b^5*c + 10*a^2*b^3* c^2 + 60*a^3*b*c^3)*x^2 + 2*(a*b^6 - 10*a^2*b^4*c + 30*a^3*b^2*c^2)*x)*sqr t(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)* (2*c*x + b))/(c*x^2 + b*x + a)) + 2*(3*a*b^7 - 34*a^2*b^5*c + 119*a^3*b^3* c^2 - 124*a^4*b*c^3)*x + (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5 *c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7 *c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2* b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)*log(c*x^2 + b*x + a))/(a^2*b^6*c ^3 - 12*a^3*b^4*c^4 + 48*a^4*b^2*c^5 - 64*a^5*c^6 + (b^6*c^5 - 12*a*b^4*c^ 6 + 48*a^2*b^2*c^7 - 64*a^3*c^8)*x^4 + 2*(b^7*c^4 - 12*a*b^5*c^5 + 48*a^2* b^3*c^6 - 64*a^3*b*c^7)*x^3 + (b^8*c^3 - 10*a*b^6*c^4 + 24*a^2*b^4*c^5 + 3 2*a^3*b^2*c^6 - 128*a^4*c^7)*x^2 + 2*(a*b^7*c^3 - 12*a^2*b^5*c^4 + 48*a^3* b^3*c^5 - 64*a^4*b*c^6)*x), 1/2*(3*a^2*b^6 - 33*a^3*b^4*c + 108*a^4*b^2*c^ 2 - 96*a^5*c^3 + 2*(2*b^7*c - 23*a*b^5*c^2 + 85*a^2*b^3*c^3 - 100*a^3*b*c^ 4)*x^3 + (3*b^8 - 31*a*b^6*c + 87*a^2*b^4*c^2 - 12*a^3*b^2*c^3 - 128*a^...
Leaf count of result is larger than twice the leaf count of optimal. 1510 vs. \(2 (196) = 392\).
Time = 1.67 (sec) , antiderivative size = 1510, normalized size of antiderivative = 7.44 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(x**5/(c*x**2+b*x+a)**3,x)
Output:
(-b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*( 1024*a**5*c**5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6* c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3))*log(x + (-64*a**3*c**5*(-b*sqrt (-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a** 5*c**5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 2 0*a*b**8*c - b**10)) + 1/(2*c**3)) + 32*a**3*c**2 + 48*a**2*b**2*c**4*(-b* sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024 *a**5*c**5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) - 9*a**2*b**2*c - 12*a*b**4*c**3*(- b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(10 24*a**5*c**5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c* *2 + 20*a*b**8*c - b**10)) + 1/(2*c**3)) + a*b**4 + b**6*c**2*(-b*sqrt(-(4 *a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c* *5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 20*a* b**8*c - b**10)) + 1/(2*c**3)))/(30*a**2*b*c**2 - 10*a*b**3*c + b**5)) + ( b*sqrt(-(4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(10 24*a**5*c**5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c* *2 + 20*a*b**8*c - b**10)) + 1/(2*c**3))*log(x + (-64*a**3*c**5*(b*sqrt(-( 4*a*c - b**2)**5)*(30*a**2*c**2 - 10*a*b**2*c + b**4)/(2*c**3*(1024*a**5*c **5 - 1280*a**4*b**2*c**4 + 640*a**3*b**4*c**3 - 160*a**2*b**6*c**2 + 2...
Exception generated. \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^5/(c*x^2+b*x+a)^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.15 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.21 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {{\left (b^{5} - 10 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {\log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} + \frac {3 \, a^{2} b^{4} - 21 \, a^{3} b^{2} c + 24 \, a^{4} c^{2} + 2 \, {\left (2 \, b^{5} c - 15 \, a b^{3} c^{2} + 25 \, a^{2} b c^{3}\right )} x^{3} + {\left (3 \, b^{6} - 19 \, a b^{4} c + 11 \, a^{2} b^{2} c^{2} + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (3 \, a b^{5} - 22 \, a^{2} b^{3} c + 31 \, a^{3} b c^{2}\right )} x}{2 \, {\left (c x^{2} + b x + a\right )}^{2} {\left (b^{2} - 4 \, a c\right )}^{2} c^{3}} \] Input:
integrate(x^5/(c*x^2+b*x+a)^3,x, algorithm="giac")
Output:
-(b^5 - 10*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/ ((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(-b^2 + 4*a*c)) + 1/2*log(c*x^2 + b*x + a)/c^3 + 1/2*(3*a^2*b^4 - 21*a^3*b^2*c + 24*a^4*c^2 + 2*(2*b^5*c - 15*a*b^3*c^2 + 25*a^2*b*c^3)*x^3 + (3*b^6 - 19*a*b^4*c + 11*a^2*b^2*c^2 + 32*a^3*c^3)*x^2 + 2*(3*a*b^5 - 22*a^2*b^3*c + 31*a^3*b*c^2)*x)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c)^2*c^3)
Time = 9.77 (sec) , antiderivative size = 620, normalized size of antiderivative = 3.05 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx=\frac {\frac {3\,a^2\,\left (8\,a^2\,c^2-7\,a\,b^2\,c+b^4\right )}{2\,c^3\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (32\,a^3\,c^3+11\,a^2\,b^2\,c^2-19\,a\,b^4\,c+3\,b^6\right )}{2\,c^3\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {b\,x^3\,\left (25\,a^2\,c^2-15\,a\,b^2\,c+2\,b^4\right )}{c^2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {a\,b\,x\,\left (31\,a^2\,c^2-22\,a\,b^2\,c+3\,b^4\right )}{c^3\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3}-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (-1024\,a^5\,c^5+1280\,a^4\,b^2\,c^4-640\,a^3\,b^4\,c^3+160\,a^2\,b^6\,c^2-20\,a\,b^8\,c+b^{10}\right )}{2\,\left (1024\,a^5\,c^8-1280\,a^4\,b^2\,c^7+640\,a^3\,b^4\,c^6-160\,a^2\,b^6\,c^5+20\,a\,b^8\,c^4-b^{10}\,c^3\right )}-\frac {b\,\mathrm {atan}\left (\frac {\left (\frac {b\,x\,\left (30\,a^2\,c^2-10\,a\,b^2\,c+b^4\right )}{c^2\,{\left (4\,a\,c-b^2\right )}^5}+\frac {b^2\,\left (16\,a^2\,c^4-8\,a\,b^2\,c^3+b^4\,c^2\right )\,\left (30\,a^2\,c^2-10\,a\,b^2\,c+b^4\right )}{2\,c^5\,{\left (4\,a\,c-b^2\right )}^5\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,\left (32\,a^2\,c^5\,{\left (4\,a\,c-b^2\right )}^{5/2}+2\,b^4\,c^3\,{\left (4\,a\,c-b^2\right )}^{5/2}-16\,a\,b^2\,c^4\,{\left (4\,a\,c-b^2\right )}^{5/2}\right )}{30\,a^2\,b\,c^2-10\,a\,b^3\,c+b^5}\right )\,\left (30\,a^2\,c^2-10\,a\,b^2\,c+b^4\right )}{c^3\,{\left (4\,a\,c-b^2\right )}^{5/2}} \] Input:
int(x^5/(a + b*x + c*x^2)^3,x)
Output:
((3*a^2*(b^4 + 8*a^2*c^2 - 7*a*b^2*c))/(2*c^3*(b^4 + 16*a^2*c^2 - 8*a*b^2* c)) + (x^2*(3*b^6 + 32*a^3*c^3 + 11*a^2*b^2*c^2 - 19*a*b^4*c))/(2*c^3*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (b*x^3*(2*b^4 + 25*a^2*c^2 - 15*a*b^2*c))/(c ^2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (a*b*x*(3*b^4 + 31*a^2*c^2 - 22*a*b^2 *c))/(c^3*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^2*(2*a*c + b^2) + a^2 + c^2* x^4 + 2*a*b*x + 2*b*c*x^3) - (log(a + b*x + c*x^2)*(b^10 - 1024*a^5*c^5 + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 20*a*b^8*c))/(2*(10 24*a^5*c^8 - b^10*c^3 + 20*a*b^8*c^4 - 160*a^2*b^6*c^5 + 640*a^3*b^4*c^6 - 1280*a^4*b^2*c^7)) - (b*atan((((b*x*(b^4 + 30*a^2*c^2 - 10*a*b^2*c))/(c^2 *(4*a*c - b^2)^5) + (b^2*(16*a^2*c^4 + b^4*c^2 - 8*a*b^2*c^3)*(b^4 + 30*a^ 2*c^2 - 10*a*b^2*c))/(2*c^5*(4*a*c - b^2)^5*(b^4 + 16*a^2*c^2 - 8*a*b^2*c) ))*(32*a^2*c^5*(4*a*c - b^2)^(5/2) + 2*b^4*c^3*(4*a*c - b^2)^(5/2) - 16*a* b^2*c^4*(4*a*c - b^2)^(5/2)))/(b^5 + 30*a^2*b*c^2 - 10*a*b^3*c))*(b^4 + 30 *a^2*c^2 - 10*a*b^2*c))/(c^3*(4*a*c - b^2)^(5/2))
Time = 0.20 (sec) , antiderivative size = 1555, normalized size of antiderivative = 7.66 \[ \int \frac {x^5}{\left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:
int(x^5/(c*x^2+b*x+a)^3,x)
Output:
( - 60*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**4*b*c**2 + 20*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b**3*c - 120*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b**2*c* *2*x - 120*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b* c**3*x**2 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2 *b**5 + 40*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b* *4*c*x - 20*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b **3*c**2*x**2 - 120*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2) )*a**2*b**2*c**3*x**3 - 60*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b*c**4*x**4 - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a *c - b**2))*a*b**6*x + 16*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**5*c*x**2 + 40*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**4*c**2*x**3 + 20*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4* a*c - b**2))*a*b**3*c**3*x**4 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt (4*a*c - b**2))*b**7*x**2 - 4*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a *c - b**2))*b**6*c*x**3 - 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**5*c**2*x**4 + 64*log(a + b*x + c*x**2)*a**5*c**3 - 48*log(a + b*x + c*x**2)*a**4*b**2*c**2 + 128*log(a + b*x + c*x**2)*a**4*b*c**3*x + 128*log(a + b*x + c*x**2)*a**4*c**4*x**2 + 12*log(a + b*x + c*x**2)*a**3*b **4*c - 96*log(a + b*x + c*x**2)*a**3*b**3*c**2*x - 32*log(a + b*x + c*...