Integrand size = 23, antiderivative size = 338 \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {\left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right ) n x}{4 c^3}-\frac {e \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) n x^2}{8 c^2}-\frac {e^2 (8 c d-b e) n x^3}{12 c}-\frac {1}{8} e^3 n x^4+\frac {\sqrt {b^2-4 a c} (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{4 c^4}-\frac {\left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) n \log \left (a+b x+c x^2\right )}{8 c^4 e}+\frac {(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e} \] Output:
-1/4*(8*c^3*d^3-b^3*e^3+b*c*e^2*(3*a*e+4*b*d)-2*c^2*d*e*(4*a*e+3*b*d))*n*x /c^3-1/8*e*(12*c^2*d^2+b^2*e^2-2*c*e*(a*e+2*b*d))*n*x^2/c^2-1/12*e^2*(-b*e +8*c*d)*n*x^3/c-1/8*e^3*n*x^4+1/4*(-4*a*c+b^2)^(1/2)*(-b*e+2*c*d)*(2*c^2*d ^2+b^2*e^2-2*c*e*(a*e+b*d))*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/c^4-1/ 8*(2*c^4*d^4+b^4*e^4-4*b^2*c*e^3*(a*e+b*d)-4*c^3*d^2*e*(3*a*e+b*d)+2*c^2*e ^2*(a^2*e^2+6*a*b*d*e+3*b^2*d^2))*n*ln(c*x^2+b*x+a)/c^4/e+1/4*(e*x+d)^4*ln (d*(c*x^2+b*x+a)^n)/e
Time = 0.56 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.96 \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\frac {-\frac {n \left (6 c e \left (8 c^3 d^3-b^3 e^3+b c e^2 (4 b d+3 a e)-2 c^2 d e (3 b d+4 a e)\right ) x+3 c^2 e^2 \left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) x^2+2 c^3 e^3 (8 c d-b e) x^3+3 c^4 e^4 x^4-6 \sqrt {b^2-4 a c} e (2 c d-b e) \left (2 c^2 d^2+b^2 e^2-2 c e (b d+a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+3 \left (2 c^4 d^4+b^4 e^4-4 b^2 c e^3 (b d+a e)-4 c^3 d^2 e (b d+3 a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b d e+a^2 e^2\right )\right ) \log (a+x (b+c x))\right )}{6 c^4}+(d+e x)^4 \log \left (d (a+x (b+c x))^n\right )}{4 e} \] Input:
Integrate[(d + e*x)^3*Log[d*(a + b*x + c*x^2)^n],x]
Output:
(-1/6*(n*(6*c*e*(8*c^3*d^3 - b^3*e^3 + b*c*e^2*(4*b*d + 3*a*e) - 2*c^2*d*e *(3*b*d + 4*a*e))*x + 3*c^2*e^2*(12*c^2*d^2 + b^2*e^2 - 2*c*e*(2*b*d + a*e ))*x^2 + 2*c^3*e^3*(8*c*d - b*e)*x^3 + 3*c^4*e^4*x^4 - 6*Sqrt[b^2 - 4*a*c] *e*(2*c*d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*ArcTanh[(b + 2* c*x)/Sqrt[b^2 - 4*a*c]] + 3*(2*c^4*d^4 + b^4*e^4 - 4*b^2*c*e^3*(b*d + a*e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a*b*d*e + a^2*e^2) )*Log[a + x*(b + c*x)]))/c^4 + (d + e*x)^4*Log[d*(a + x*(b + c*x))^n])/(4* e)
Time = 0.79 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3005, 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.
| \(\displaystyle \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx\) |
| \(\Big \downarrow \) 3005 |
| \(\displaystyle \frac {(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac {n \int \frac {(b+2 c x) (d+e x)^4}{c x^2+b x+a}dx}{4 e}\) |
| \(\Big \downarrow \) 1200 |
| \(\displaystyle \frac {(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac {n \int \left (2 x^3 e^4+\frac {(8 c d-b e) x^2 e^3}{c}+\frac {\left (12 c^2 d^2+b^2 e^2-2 c e (2 b d+a e)\right ) x e^2}{c^2}+\frac {\left (8 c^3 d^3-2 c^2 e (3 b d+4 a e) d-b^3 e^3+b c e^2 (4 b d+3 a e)\right ) e}{c^3}+\frac {a b^3 e^4-4 a b^2 c d e^3-8 a c^2 d \left (c d^2-a e^2\right ) e+b c \left (c^2 d^4+6 a c e^2 d^2-3 a^2 e^4\right )+\left (2 c^4 d^4-4 c^3 e (b d+3 a e) d^2+b^4 e^4-4 b^2 c e^3 (b d+a e)+2 c^2 e^2 \left (3 b^2 d^2+6 a b e d+a^2 e^2\right )\right ) x}{c^3 \left (c x^2+b x+a\right )}\right )dx}{4 e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(d+e x)^4 \log \left (d \left (a+b x+c x^2\right )^n\right )}{4 e}-\frac {n \left (\frac {\left (2 c^2 e^2 \left (a^2 e^2+6 a b d e+3 b^2 d^2\right )-4 b^2 c e^3 (a e+b d)-4 c^3 d^2 e (3 a e+b d)+b^4 e^4+2 c^4 d^4\right ) \log \left (a+b x+c x^2\right )}{2 c^4}-\frac {e \sqrt {b^2-4 a c} (2 c d-b e) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2\right )}{c^4}+\frac {e x \left (-2 c^2 d e (4 a e+3 b d)+b c e^2 (3 a e+4 b d)-b^3 e^3+8 c^3 d^3\right )}{c^3}+\frac {e^2 x^2 \left (-2 c e (a e+2 b d)+b^2 e^2+12 c^2 d^2\right )}{2 c^2}+\frac {e^3 x^3 (8 c d-b e)}{3 c}+\frac {e^4 x^4}{2}\right )}{4 e}\) |
Input:
Int[(d + e*x)^3*Log[d*(a + b*x + c*x^2)^n],x]
Output:
-1/4*(n*((e*(8*c^3*d^3 - b^3*e^3 + b*c*e^2*(4*b*d + 3*a*e) - 2*c^2*d*e*(3* b*d + 4*a*e))*x)/c^3 + (e^2*(12*c^2*d^2 + b^2*e^2 - 2*c*e*(2*b*d + a*e))*x ^2)/(2*c^2) + (e^3*(8*c*d - b*e)*x^3)/(3*c) + (e^4*x^4)/2 - (Sqrt[b^2 - 4* a*c]*e*(2*c*d - b*e)*(2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/c^4 + ((2*c^4*d^4 + b^4*e^4 - 4*b^2*c*e^3*(b* d + a*e) - 4*c^3*d^2*e*(b*d + 3*a*e) + 2*c^2*e^2*(3*b^2*d^2 + 6*a*b*d*e + a^2*e^2))*Log[a + b*x + c*x^2])/(2*c^4)))/e + ((d + e*x)^4*Log[d*(a + b*x + c*x^2)^n])/(4*e)
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[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_. ), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))) , x] - Simp[b*n*(p/(e*(m + 1))) Int[SimplifyIntegrand[(d + e*x)^(m + 1)*( a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]
Time = 1.76 (sec) , antiderivative size = 594, normalized size of antiderivative = 1.76
| method | result | size |
| parts | \(\frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) e^{3} x^{4}}{4}+\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) e^{2} d \,x^{3}+\frac {3 \ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) e \,d^{2} x^{2}}{2}+\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) d^{3} x +\frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right ) d^{4}}{4 e}-\frac {n \left (\frac {e \left (\frac {1}{2} c^{3} e^{3} x^{4}-\frac {1}{3} b \,c^{2} e^{3} x^{3}+\frac {8}{3} c^{3} d \,e^{2} x^{3}-a \,c^{2} e^{3} x^{2}+\frac {1}{2} b^{2} c \,e^{3} x^{2}-2 b \,c^{2} d \,e^{2} x^{2}+6 c^{3} d^{2} e \,x^{2}+3 a b x c \,e^{3}-8 a \,c^{2} x d \,e^{2}-b^{3} x \,e^{3}+4 b^{2} x c d \,e^{2}-6 x b \,c^{2} d^{2} e +8 c^{3} x \,d^{3}\right )}{c^{3}}+\frac {\frac {\left (2 a^{2} c^{2} e^{4}-4 a \,b^{2} c \,e^{4}+12 a b \,c^{2} d \,e^{3}-12 a \,c^{3} d^{2} e^{2}+b^{4} e^{4}-4 b^{3} c d \,e^{3}+6 b^{2} c^{2} d^{2} e^{2}-4 b \,c^{3} d^{3} e +2 c^{4} d^{4}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-3 a^{2} b c \,e^{4}+8 a^{2} c^{2} d \,e^{3}+a \,b^{3} e^{4}-4 a \,b^{2} c d \,e^{3}+6 a b \,c^{2} d^{2} e^{2}-8 a \,c^{3} d^{3} e +b \,c^{3} d^{4}-\frac {\left (2 a^{2} c^{2} e^{4}-4 a \,b^{2} c \,e^{4}+12 a b \,c^{2} d \,e^{3}-12 a \,c^{3} d^{2} e^{2}+b^{4} e^{4}-4 b^{3} c d \,e^{3}+6 b^{2} c^{2} d^{2} e^{2}-4 b \,c^{3} d^{3} e +2 c^{4} d^{4}\right ) b}{2 c}\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{c^{3}}\right )}{4 e}\) | \(594\) |
| risch | \(\text {Expression too large to display}\) | \(16059\) |
Input:
int((e*x+d)^3*ln(d*(c*x^2+b*x+a)^n),x,method=_RETURNVERBOSE)
Output:
1/4*ln(d*(c*x^2+b*x+a)^n)*e^3*x^4+ln(d*(c*x^2+b*x+a)^n)*e^2*d*x^3+3/2*ln(d *(c*x^2+b*x+a)^n)*e*d^2*x^2+ln(d*(c*x^2+b*x+a)^n)*d^3*x+1/4*ln(d*(c*x^2+b* x+a)^n)/e*d^4-1/4/e*n*(e/c^3*(1/2*c^3*e^3*x^4-1/3*b*c^2*e^3*x^3+8/3*c^3*d* e^2*x^3-a*c^2*e^3*x^2+1/2*b^2*c*e^3*x^2-2*b*c^2*d*e^2*x^2+6*c^3*d^2*e*x^2+ 3*a*b*x*c*e^3-8*a*c^2*x*d*e^2-b^3*x*e^3+4*b^2*x*c*d*e^2-6*x*b*c^2*d^2*e+8* c^3*x*d^3)+1/c^3*(1/2*(2*a^2*c^2*e^4-4*a*b^2*c*e^4+12*a*b*c^2*d*e^3-12*a*c ^3*d^2*e^2+b^4*e^4-4*b^3*c*d*e^3+6*b^2*c^2*d^2*e^2-4*b*c^3*d^3*e+2*c^4*d^4 )/c*ln(c*x^2+b*x+a)+2*(-3*a^2*b*c*e^4+8*a^2*c^2*d*e^3+a*b^3*e^4-4*a*b^2*c* d*e^3+6*a*b*c^2*d^2*e^2-8*a*c^3*d^3*e+b*c^3*d^4-1/2*(2*a^2*c^2*e^4-4*a*b^2 *c*e^4+12*a*b*c^2*d*e^3-12*a*c^3*d^2*e^2+b^4*e^4-4*b^3*c*d*e^3+6*b^2*c^2*d ^2*e^2-4*b*c^3*d^3*e+2*c^4*d^4)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4 *a*c-b^2)^(1/2))))
Time = 0.10 (sec) , antiderivative size = 880, normalized size of antiderivative = 2.60 \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx =\text {Too large to display} \] Input:
integrate((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),x, algorithm="fricas")
Output:
[-1/24*(3*c^4*e^3*n*x^4 + 2*(8*c^4*d*e^2 - b*c^3*e^3)*n*x^3 + 3*(12*c^4*d^ 2*e - 4*b*c^3*d*e^2 + (b^2*c^2 - 2*a*c^3)*e^3)*n*x^2 - 3*(4*c^3*d^3 - 6*b* c^2*d^2*e + 4*(b^2*c - a*c^2)*d*e^2 - (b^3 - 2*a*b*c)*e^3)*sqrt(b^2 - 4*a* c)*n*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)) + 6*(8*c^4*d^3 - 6*b*c^3*d^2*e + 4*(b^2*c^2 - 2*a*c^ 3)*d*e^2 - (b^3*c - 3*a*b*c^2)*e^3)*n*x - 3*(2*c^4*e^3*n*x^4 + 8*c^4*d*e^2 *n*x^3 + 12*c^4*d^2*e*n*x^2 + 8*c^4*d^3*n*x + (4*b*c^3*d^3 - 6*(b^2*c^2 - 2*a*c^3)*d^2*e + 4*(b^3*c - 3*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c + 2*a^2*c^ 2)*e^3)*n)*log(c*x^2 + b*x + a) - 6*(c^4*e^3*x^4 + 4*c^4*d*e^2*x^3 + 6*c^4 *d^2*e*x^2 + 4*c^4*d^3*x)*log(d))/c^4, -1/24*(3*c^4*e^3*n*x^4 + 2*(8*c^4*d *e^2 - b*c^3*e^3)*n*x^3 + 3*(12*c^4*d^2*e - 4*b*c^3*d*e^2 + (b^2*c^2 - 2*a *c^3)*e^3)*n*x^2 - 6*(4*c^3*d^3 - 6*b*c^2*d^2*e + 4*(b^2*c - a*c^2)*d*e^2 - (b^3 - 2*a*b*c)*e^3)*sqrt(-b^2 + 4*a*c)*n*arctan(-sqrt(-b^2 + 4*a*c)*(2* c*x + b)/(b^2 - 4*a*c)) + 6*(8*c^4*d^3 - 6*b*c^3*d^2*e + 4*(b^2*c^2 - 2*a* c^3)*d*e^2 - (b^3*c - 3*a*b*c^2)*e^3)*n*x - 3*(2*c^4*e^3*n*x^4 + 8*c^4*d*e ^2*n*x^3 + 12*c^4*d^2*e*n*x^2 + 8*c^4*d^3*n*x + (4*b*c^3*d^3 - 6*(b^2*c^2 - 2*a*c^3)*d^2*e + 4*(b^3*c - 3*a*b*c^2)*d*e^2 - (b^4 - 4*a*b^2*c + 2*a^2* c^2)*e^3)*n)*log(c*x^2 + b*x + a) - 6*(c^4*e^3*x^4 + 4*c^4*d*e^2*x^3 + 6*c ^4*d^2*e*x^2 + 4*c^4*d^3*x)*log(d))/c^4]
Timed out. \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\text {Timed out} \] Input:
integrate((e*x+d)**3*ln(d*(c*x**2+b*x+a)**n),x)
Output:
Timed out
Exception generated. \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),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 = 0.16 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.46 \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=-\frac {1}{8} \, {\left (e^{3} n - 2 \, e^{3} \log \left (d\right )\right )} x^{4} - \frac {{\left (8 \, c d e^{2} n - b e^{3} n - 12 \, c d e^{2} \log \left (d\right )\right )} x^{3}}{12 \, c} + \frac {1}{4} \, {\left (e^{3} n x^{4} + 4 \, d e^{2} n x^{3} + 6 \, d^{2} e n x^{2} + 4 \, d^{3} n x\right )} \log \left (c x^{2} + b x + a\right ) - \frac {{\left (12 \, c^{2} d^{2} e n - 4 \, b c d e^{2} n + b^{2} e^{3} n - 2 \, a c e^{3} n - 12 \, c^{2} d^{2} e \log \left (d\right )\right )} x^{2}}{8 \, c^{2}} - \frac {{\left (8 \, c^{3} d^{3} n - 6 \, b c^{2} d^{2} e n + 4 \, b^{2} c d e^{2} n - 8 \, a c^{2} d e^{2} n - b^{3} e^{3} n + 3 \, a b c e^{3} n - 4 \, c^{3} d^{3} \log \left (d\right )\right )} x}{4 \, c^{3}} + \frac {{\left (4 \, b c^{3} d^{3} n - 6 \, b^{2} c^{2} d^{2} e n + 12 \, a c^{3} d^{2} e n + 4 \, b^{3} c d e^{2} n - 12 \, a b c^{2} d e^{2} n - b^{4} e^{3} n + 4 \, a b^{2} c e^{3} n - 2 \, a^{2} c^{2} e^{3} n\right )} \log \left (c x^{2} + b x + a\right )}{8 \, c^{4}} - \frac {{\left (4 \, b^{2} c^{3} d^{3} n - 16 \, a c^{4} d^{3} n - 6 \, b^{3} c^{2} d^{2} e n + 24 \, a b c^{3} d^{2} e n + 4 \, b^{4} c d e^{2} n - 20 \, a b^{2} c^{2} d e^{2} n + 16 \, a^{2} c^{3} d e^{2} n - b^{5} e^{3} n + 6 \, a b^{3} c e^{3} n - 8 \, a^{2} b c^{2} e^{3} n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{4 \, \sqrt {-b^{2} + 4 \, a c} c^{4}} \] Input:
integrate((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),x, algorithm="giac")
Output:
-1/8*(e^3*n - 2*e^3*log(d))*x^4 - 1/12*(8*c*d*e^2*n - b*e^3*n - 12*c*d*e^2 *log(d))*x^3/c + 1/4*(e^3*n*x^4 + 4*d*e^2*n*x^3 + 6*d^2*e*n*x^2 + 4*d^3*n* x)*log(c*x^2 + b*x + a) - 1/8*(12*c^2*d^2*e*n - 4*b*c*d*e^2*n + b^2*e^3*n - 2*a*c*e^3*n - 12*c^2*d^2*e*log(d))*x^2/c^2 - 1/4*(8*c^3*d^3*n - 6*b*c^2* d^2*e*n + 4*b^2*c*d*e^2*n - 8*a*c^2*d*e^2*n - b^3*e^3*n + 3*a*b*c*e^3*n - 4*c^3*d^3*log(d))*x/c^3 + 1/8*(4*b*c^3*d^3*n - 6*b^2*c^2*d^2*e*n + 12*a*c^ 3*d^2*e*n + 4*b^3*c*d*e^2*n - 12*a*b*c^2*d*e^2*n - b^4*e^3*n + 4*a*b^2*c*e ^3*n - 2*a^2*c^2*e^3*n)*log(c*x^2 + b*x + a)/c^4 - 1/4*(4*b^2*c^3*d^3*n - 16*a*c^4*d^3*n - 6*b^3*c^2*d^2*e*n + 24*a*b*c^3*d^2*e*n + 4*b^4*c*d*e^2*n - 20*a*b^2*c^2*d*e^2*n + 16*a^2*c^3*d*e^2*n - b^5*e^3*n + 6*a*b^3*c*e^3*n - 8*a^2*b*c^2*e^3*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4 *a*c)*c^4)
Time = 25.96 (sec) , antiderivative size = 775, normalized size of antiderivative = 2.29 \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx=\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )\,\left (d^3\,x+\frac {3\,d^2\,e\,x^2}{2}+d\,e^2\,x^3+\frac {e^3\,x^4}{4}\right )-x^3\,\left (\frac {e^2\,n\,\left (b\,e+8\,c\,d\right )}{12\,c}-\frac {b\,e^3\,n}{6\,c}\right )-x\,\left (\frac {b\,\left (\frac {b\,\left (\frac {e^2\,n\,\left (b\,e+8\,c\,d\right )}{4\,c}-\frac {b\,e^3\,n}{2\,c}\right )}{c}+\frac {a\,e^3\,n}{2\,c}-\frac {d\,e\,n\,\left (b\,e+3\,c\,d\right )}{c}\right )}{c}-\frac {a\,\left (\frac {e^2\,n\,\left (b\,e+8\,c\,d\right )}{4\,c}-\frac {b\,e^3\,n}{2\,c}\right )}{c}+\frac {d^2\,n\,\left (3\,b\,e+4\,c\,d\right )}{2\,c}\right )+x^2\,\left (\frac {b\,\left (\frac {e^2\,n\,\left (b\,e+8\,c\,d\right )}{4\,c}-\frac {b\,e^3\,n}{2\,c}\right )}{2\,c}+\frac {a\,e^3\,n}{4\,c}-\frac {d\,e\,n\,\left (b\,e+3\,c\,d\right )}{2\,c}\right )-\frac {\ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^4\,e^3\,n+2\,a^2\,c^2\,e^3\,n-4\,b\,c^3\,d^3\,n+b^3\,e^3\,n\,\sqrt {b^2-4\,a\,c}-4\,c^3\,d^3\,n\,\sqrt {b^2-4\,a\,c}-4\,a\,b^2\,c\,e^3\,n-12\,a\,c^3\,d^2\,e\,n-4\,b^3\,c\,d\,e^2\,n+6\,b^2\,c^2\,d^2\,e\,n-2\,a\,b\,c\,e^3\,n\,\sqrt {b^2-4\,a\,c}+12\,a\,b\,c^2\,d\,e^2\,n+4\,a\,c^2\,d\,e^2\,n\,\sqrt {b^2-4\,a\,c}+6\,b\,c^2\,d^2\,e\,n\,\sqrt {b^2-4\,a\,c}-4\,b^2\,c\,d\,e^2\,n\,\sqrt {b^2-4\,a\,c}\right )}{8\,c^4}-\frac {e^3\,n\,x^4}{8}-\frac {\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^4\,e^3\,n+2\,a^2\,c^2\,e^3\,n-4\,b\,c^3\,d^3\,n-b^3\,e^3\,n\,\sqrt {b^2-4\,a\,c}+4\,c^3\,d^3\,n\,\sqrt {b^2-4\,a\,c}-4\,a\,b^2\,c\,e^3\,n-12\,a\,c^3\,d^2\,e\,n-4\,b^3\,c\,d\,e^2\,n+6\,b^2\,c^2\,d^2\,e\,n+2\,a\,b\,c\,e^3\,n\,\sqrt {b^2-4\,a\,c}+12\,a\,b\,c^2\,d\,e^2\,n-4\,a\,c^2\,d\,e^2\,n\,\sqrt {b^2-4\,a\,c}-6\,b\,c^2\,d^2\,e\,n\,\sqrt {b^2-4\,a\,c}+4\,b^2\,c\,d\,e^2\,n\,\sqrt {b^2-4\,a\,c}\right )}{8\,c^4} \] Input:
int(log(d*(a + b*x + c*x^2)^n)*(d + e*x)^3,x)
Output:
log(d*(a + b*x + c*x^2)^n)*(d^3*x + (e^3*x^4)/4 + (3*d^2*e*x^2)/2 + d*e^2* x^3) - x^3*((e^2*n*(b*e + 8*c*d))/(12*c) - (b*e^3*n)/(6*c)) - x*((b*((b*(( e^2*n*(b*e + 8*c*d))/(4*c) - (b*e^3*n)/(2*c)))/c + (a*e^3*n)/(2*c) - (d*e* n*(b*e + 3*c*d))/c))/c - (a*((e^2*n*(b*e + 8*c*d))/(4*c) - (b*e^3*n)/(2*c) ))/c + (d^2*n*(3*b*e + 4*c*d))/(2*c)) + x^2*((b*((e^2*n*(b*e + 8*c*d))/(4* c) - (b*e^3*n)/(2*c)))/(2*c) + (a*e^3*n)/(4*c) - (d*e*n*(b*e + 3*c*d))/(2* c)) - (log(b*(b^2 - 4*a*c)^(1/2) - 4*a*c + b^2 + 2*c*x*(b^2 - 4*a*c)^(1/2) )*(b^4*e^3*n + 2*a^2*c^2*e^3*n - 4*b*c^3*d^3*n + b^3*e^3*n*(b^2 - 4*a*c)^( 1/2) - 4*c^3*d^3*n*(b^2 - 4*a*c)^(1/2) - 4*a*b^2*c*e^3*n - 12*a*c^3*d^2*e* n - 4*b^3*c*d*e^2*n + 6*b^2*c^2*d^2*e*n - 2*a*b*c*e^3*n*(b^2 - 4*a*c)^(1/2 ) + 12*a*b*c^2*d*e^2*n + 4*a*c^2*d*e^2*n*(b^2 - 4*a*c)^(1/2) + 6*b*c^2*d^2 *e*n*(b^2 - 4*a*c)^(1/2) - 4*b^2*c*d*e^2*n*(b^2 - 4*a*c)^(1/2)))/(8*c^4) - (e^3*n*x^4)/8 - (log(4*a*c + b*(b^2 - 4*a*c)^(1/2) - b^2 + 2*c*x*(b^2 - 4 *a*c)^(1/2))*(b^4*e^3*n + 2*a^2*c^2*e^3*n - 4*b*c^3*d^3*n - b^3*e^3*n*(b^2 - 4*a*c)^(1/2) + 4*c^3*d^3*n*(b^2 - 4*a*c)^(1/2) - 4*a*b^2*c*e^3*n - 12*a *c^3*d^2*e*n - 4*b^3*c*d*e^2*n + 6*b^2*c^2*d^2*e*n + 2*a*b*c*e^3*n*(b^2 - 4*a*c)^(1/2) + 12*a*b*c^2*d*e^2*n - 4*a*c^2*d*e^2*n*(b^2 - 4*a*c)^(1/2) - 6*b*c^2*d^2*e*n*(b^2 - 4*a*c)^(1/2) + 4*b^2*c*d*e^2*n*(b^2 - 4*a*c)^(1/2)) )/(8*c^4)
Time = 0.15 (sec) , antiderivative size = 727, normalized size of antiderivative = 2.15 \[ \int (d+e x)^3 \log \left (d \left (a+b x+c x^2\right )^n\right ) \, dx =\text {Too large to display} \] Input:
int((e*x+d)^3*log(d*(c*x^2+b*x+a)^n),x)
Output:
(12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c*e**3*n - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c**2*d*e**2* n - 6*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**3*e**3*n + 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2*c*d*e**2 *n - 36*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c**2*d** 2*e*n + 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*d* *3*n - 6*log((a + b*x + c*x**2)**n*d)*a**2*c**2*e**3 + 12*log((a + b*x + c *x**2)**n*d)*a*b**2*c*e**3 - 36*log((a + b*x + c*x**2)**n*d)*a*b*c**2*d*e* *2 + 36*log((a + b*x + c*x**2)**n*d)*a*c**3*d**2*e - 3*log((a + b*x + c*x* *2)**n*d)*b**4*e**3 + 12*log((a + b*x + c*x**2)**n*d)*b**3*c*d*e**2 - 18*l og((a + b*x + c*x**2)**n*d)*b**2*c**2*d**2*e + 12*log((a + b*x + c*x**2)** n*d)*b*c**3*d**3 + 24*log((a + b*x + c*x**2)**n*d)*c**4*d**3*x + 36*log((a + b*x + c*x**2)**n*d)*c**4*d**2*e*x**2 + 24*log((a + b*x + c*x**2)**n*d)* c**4*d*e**2*x**3 + 6*log((a + b*x + c*x**2)**n*d)*c**4*e**3*x**4 - 18*a*b* c**2*e**3*n*x + 48*a*c**3*d*e**2*n*x + 6*a*c**3*e**3*n*x**2 + 6*b**3*c*e** 3*n*x - 24*b**2*c**2*d*e**2*n*x - 3*b**2*c**2*e**3*n*x**2 + 36*b*c**3*d**2 *e*n*x + 12*b*c**3*d*e**2*n*x**2 + 2*b*c**3*e**3*n*x**3 - 48*c**4*d**3*n*x - 36*c**4*d**2*e*n*x**2 - 16*c**4*d*e**2*n*x**3 - 3*c**4*e**3*n*x**4)/(24 *c**4)