Integrand size = 24, antiderivative size = 70 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2}-\frac {2 \log (b+2 c x)}{\left (b^2-4 a c\right )^2 d^3}+\frac {\log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2 d^3} \] Output:
1/(-4*a*c+b^2)/d^3/(2*c*x+b)^2-2*ln(2*c*x+b)/(-4*a*c+b^2)^2/d^3+ln(c*x^2+b *x+a)/(-4*a*c+b^2)^2/d^3
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^2}-\frac {2 \log (b+2 c x)}{\left (b^2-4 a c\right )^2}+\frac {\log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2}}{d^3} \] Input:
Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x]
Output:
(1/((b^2 - 4*a*c)*(b + 2*c*x)^2) - (2*Log[b + 2*c*x])/(b^2 - 4*a*c)^2 + Lo g[a + b*x + c*x^2]/(b^2 - 4*a*c)^2)/d^3
Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1117, 27, 1105, 16, 1103}
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.
\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right ) (b d+2 c d x)^3} \, dx\) |
\(\Big \downarrow \) 1117 |
\(\displaystyle \frac {\int \frac {1}{d (b+2 c x) \left (c x^2+b x+a\right )}dx}{d^2 \left (b^2-4 a c\right )}+\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{(b+2 c x) \left (c x^2+b x+a\right )}dx}{d^3 \left (b^2-4 a c\right )}+\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}\) |
\(\Big \downarrow \) 1105 |
\(\displaystyle \frac {\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {4 c \int \frac {1}{b+2 c x}dx}{b^2-4 a c}}{d^3 \left (b^2-4 a c\right )}+\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {\int \frac {b+2 c x}{c x^2+b x+a}dx}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}}{d^3 \left (b^2-4 a c\right )}+\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\frac {\log \left (a+b x+c x^2\right )}{b^2-4 a c}-\frac {2 \log (b+2 c x)}{b^2-4 a c}}{d^3 \left (b^2-4 a c\right )}+\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}\) |
Input:
Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x]
Output:
1/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2) + ((-2*Log[b + 2*c*x])/(b^2 - 4*a*c) + Log[a + b*x + c*x^2]/(b^2 - 4*a*c))/((b^2 - 4*a*c)*d^3)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[1/(((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[-4*b*(c/(d*(b^2 - 4*a*c))) Int[1/(b + 2*c*x), x], x] + Simp[b^2/( d^2*(b^2 - 4*a*c)) Int[(d + e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b , c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* c))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & & (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) /2])
Time = 0.96 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {-\frac {2 \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{2}}-\frac {1}{\left (4 a c -b^{2}\right ) \left (2 c x +b \right )^{2}}+\frac {\ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{2}}}{d^{3}}\) | \(73\) |
risch | \(-\frac {1}{\left (4 a c -b^{2}\right ) d^{3} \left (2 c x +b \right )^{2}}-\frac {2 \ln \left (2 c x +b \right )}{d^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}+\frac {\ln \left (-c \,x^{2}-b x -a \right )}{d^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(100\) |
norman | \(\frac {\frac {4 c x}{d b \left (4 a c -b^{2}\right )}+\frac {4 c^{2} x^{2}}{d \,b^{2} \left (4 a c -b^{2}\right )}}{d^{2} \left (2 c x +b \right )^{2}}+\frac {\ln \left (c \,x^{2}+b x +a \right )}{d^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}-\frac {2 \ln \left (2 c x +b \right )}{d^{3} \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right )}\) | \(132\) |
parallelrisch | \(\frac {-16 a b \,c^{2}+4 b^{3} c -32 \ln \left (\frac {b}{2}+c x \right ) x^{2} b \,c^{3}+16 \ln \left (c \,x^{2}+b x +a \right ) x^{2} b \,c^{3}-32 \ln \left (\frac {b}{2}+c x \right ) x \,b^{2} c^{2}+16 \ln \left (c \,x^{2}+b x +a \right ) x \,b^{2} c^{2}+4 \ln \left (c \,x^{2}+b x +a \right ) b^{3} c -8 \ln \left (\frac {b}{2}+c x \right ) b^{3} c}{4 \left (16 a^{2} c^{2}-8 c a \,b^{2}+b^{4}\right ) \left (2 c x +b \right )^{2} b c \,d^{3}}\) | \(160\) |
Input:
int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
Output:
1/d^3*(-2/(4*a*c-b^2)^2*ln(2*c*x+b)-1/(4*a*c-b^2)/(2*c*x+b)^2+1/(4*a*c-b^2 )^2*ln(c*x^2+b*x+a))
Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (70) = 140\).
Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.23 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {b^{2} - 4 \, a c + {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (2 \, c x + b\right )}{4 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{3} x^{2} + 4 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{3} x + {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} d^{3}} \] Input:
integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="fricas")
Output:
(b^2 - 4*a*c + (4*c^2*x^2 + 4*b*c*x + b^2)*log(c*x^2 + b*x + a) - 2*(4*c^2 *x^2 + 4*b*c*x + b^2)*log(2*c*x + b))/(4*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c ^4)*d^3*x^2 + 4*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^3*x + (b^6 - 8*a*b^ 4*c + 16*a^2*b^2*c^2)*d^3)
Time = 0.83 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=- \frac {1}{4 a b^{2} c d^{3} - b^{4} d^{3} + x^{2} \cdot \left (16 a c^{3} d^{3} - 4 b^{2} c^{2} d^{3}\right ) + x \left (16 a b c^{2} d^{3} - 4 b^{3} c d^{3}\right )} - \frac {2 \log {\left (\frac {b}{2 c} + x \right )}}{d^{3} \left (4 a c - b^{2}\right )^{2}} + \frac {\log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d^{3} \left (4 a c - b^{2}\right )^{2}} \] Input:
integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a),x)
Output:
-1/(4*a*b**2*c*d**3 - b**4*d**3 + x**2*(16*a*c**3*d**3 - 4*b**2*c**2*d**3) + x*(16*a*b*c**2*d**3 - 4*b**3*c*d**3)) - 2*log(b/(2*c) + x)/(d**3*(4*a*c - b**2)**2) + log(a/c + b*x/c + x**2)/(d**3*(4*a*c - b**2)**2)
Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.84 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {1}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{3} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{3} x + {\left (b^{4} - 4 \, a b^{2} c\right )} d^{3}} + \frac {\log \left (c x^{2} + b x + a\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{3}} - \frac {2 \, \log \left (2 \, c x + b\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{3}} \] Input:
integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="maxima")
Output:
1/(4*(b^2*c^2 - 4*a*c^3)*d^3*x^2 + 4*(b^3*c - 4*a*b*c^2)*d^3*x + (b^4 - 4* a*b^2*c)*d^3) + log(c*x^2 + b*x + a)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3) - 2*log(2*c*x + b)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*d^3)
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=-\frac {2 \, c \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{4} c d^{3} - 8 \, a b^{2} c^{2} d^{3} + 16 \, a^{2} c^{3} d^{3}} + \frac {\log \left (c x^{2} + b x + a\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} + \frac {1}{{\left (b^{2} - 4 \, a c\right )} {\left (2 \, c x + b\right )}^{2} d^{3}} \] Input:
integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="giac")
Output:
-2*c*log(abs(2*c*x + b))/(b^4*c*d^3 - 8*a*b^2*c^2*d^3 + 16*a^2*c^3*d^3) + log(c*x^2 + b*x + a)/(b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3) + 1/((b^2 - 4*a*c)*(2*c*x + b)^2*d^3)
Time = 5.70 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {\ln \left (c\,x^2+b\,x+a\right )}{16\,a^2\,c^2\,d^3-8\,a\,b^2\,c\,d^3+b^4\,d^3}-\frac {2\,\ln \left (b+2\,c\,x\right )}{16\,a^2\,c^2\,d^3-8\,a\,b^2\,c\,d^3+b^4\,d^3}+\frac {1}{b^4\,d^3+4\,b^3\,c\,d^3\,x+4\,b^2\,c^2\,d^3\,x^2-4\,a\,b^2\,c\,d^3-16\,a\,b\,c^2\,d^3\,x-16\,a\,c^3\,d^3\,x^2} \] Input:
int(1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x)
Output:
log(a + b*x + c*x^2)/(b^4*d^3 + 16*a^2*c^2*d^3 - 8*a*b^2*c*d^3) - (2*log(b + 2*c*x))/(b^4*d^3 + 16*a^2*c^2*d^3 - 8*a*b^2*c*d^3) + 1/(b^4*d^3 - 16*a* c^3*d^3*x^2 + 4*b^2*c^2*d^3*x^2 - 4*a*b^2*c*d^3 + 4*b^3*c*d^3*x - 16*a*b*c ^2*d^3*x)
Time = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.66 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{2}+4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b c x +4 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) c^{2} x^{2}-2 \,\mathrm {log}\left (2 c x +b \right ) b^{2}-8 \,\mathrm {log}\left (2 c x +b \right ) b c x -8 \,\mathrm {log}\left (2 c x +b \right ) c^{2} x^{2}-4 a c +b^{2}}{d^{3} \left (64 a^{2} c^{4} x^{2}-32 a \,b^{2} c^{3} x^{2}+4 b^{4} c^{2} x^{2}+64 a^{2} b \,c^{3} x -32 a \,b^{3} c^{2} x +4 b^{5} c x +16 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}\right )} \] Input:
int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x)
Output:
(log(a + b*x + c*x**2)*b**2 + 4*log(a + b*x + c*x**2)*b*c*x + 4*log(a + b* x + c*x**2)*c**2*x**2 - 2*log(b + 2*c*x)*b**2 - 8*log(b + 2*c*x)*b*c*x - 8 *log(b + 2*c*x)*c**2*x**2 - 4*a*c + b**2)/(d**3*(16*a**2*b**2*c**2 + 64*a* *2*b*c**3*x + 64*a**2*c**4*x**2 - 8*a*b**4*c - 32*a*b**3*c**2*x - 32*a*b** 2*c**3*x**2 + b**6 + 4*b**5*c*x + 4*b**4*c**2*x**2))