\(\int \frac {1}{(b d+2 c d x)^3 (a+b x+c x^2)} \, dx\) [1164]

Optimal result

Integrand size = 24, antiderivative size = 70 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2}-\frac {2 \log (b+2 c x)}{\left (b^2-4 a c\right )^2 d^3}+\frac {\log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {707, 695, 31, 642} \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {\log \left (a+b x+c x^2\right )}{d^3 \left (b^2-4 a c\right )^2}+\frac {1}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2}-\frac {2 \log (b+2 c x)}{d^3 \left (b^2-4 a c\right )^2} \]

[In]

Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x]

[Out]

1/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2) - (2*Log[b + 2*c*x])/((b^2 - 4*a*c)^2*d^3) + Log[a + b*x + c*x^2]/((b^2 -
4*a*c)^2*d^3)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 695

Int[1/(((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[-4*b*(c/(d*(b^2 - 4*a*c))),
Int[1/(b + 2*c*x), x], x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e, 0]

Rule 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> 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] + Dist[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] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2}+\frac {\int \frac {1}{(b d+2 c d x) \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) d^2} \\ & = \frac {1}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2}+\frac {\int \frac {b d+2 c d x}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d^4}-\frac {(4 c) \int \frac {1}{b+2 c x} \, dx}{\left (b^2-4 a c\right )^2 d^3} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {\frac {1}{\left (b^2-4 a c\right ) (b+2 c x)^2}-\frac {2 \log (b+2 c x)}{\left (b^2-4 a c\right )^2}+\frac {\log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^2}}{d^3} \]

[In]

Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x]

[Out]

(1/((b^2 - 4*a*c)*(b + 2*c*x)^2) - (2*Log[b + 2*c*x])/(b^2 - 4*a*c)^2 + Log[a + b*x + c*x^2]/(b^2 - 4*a*c)^2)/
d^3

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {\ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right )^{2}}-\frac {2 \ln \left (2 c x +b \right )}{\left (4 a c -b^{2}\right )^{2}}-\frac {1}{\left (4 a c -b^{2}\right ) \left (2 c x +b \right )^{2}}}{d^{3}}\)	\(73\)
risch	\(-\frac {1}{\left (4 a c -b^{2}\right ) d^{3} \left (2 c x +b \right )^{2}}-\frac {2 \ln \left (2 c x +b \right )}{d^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}+\frac {\ln \left (-c \,x^{2}-b x -a \right )}{d^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(100\)
norman	\(\frac {\frac {4 c x}{d b \left (4 a c -b^{2}\right )}+\frac {4 c^{2} x^{2}}{d \,b^{2} \left (4 a c -b^{2}\right )}}{d^{2} \left (2 c x +b \right )^{2}}+\frac {\ln \left (c \,x^{2}+b x +a \right )}{d^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {2 \ln \left (2 c x +b \right )}{d^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(132\)
parallelrisch	\(\frac {4 b^{3} c -16 b \,c^{2} a -32 \ln \left (\frac {b}{2}+c x \right ) x^{2} b \,c^{3}+4 \ln \left (c \,x^{2}+b x +a \right ) b^{3} c -8 \ln \left (\frac {b}{2}+c x \right ) b^{3} c +16 \ln \left (c \,x^{2}+b x +a \right ) x^{2} b \,c^{3}-32 \ln \left (\frac {b}{2}+c x \right ) x \,b^{2} c^{2}+16 \ln \left (c \,x^{2}+b x +a \right ) x \,b^{2} c^{2}}{4 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \left (2 c x +b \right )^{2} b c \,d^{3}}\)	\(160\)

[In]

int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/d^3*(1/(4*a*c-b^2)^2*ln(c*x^2+b*x+a)-2/(4*a*c-b^2)^2*ln(2*c*x+b)-1/(4*a*c-b^2)/(2*c*x+b)^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (70) = 140\).

Time = 0.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.23 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {b^{2} - 4 \, a c + {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left (4 \, c^{2} x^{2} + 4 \, b c x + b^{2}\right )} \log \left (2 \, c x + b\right )}{4 \, {\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d^{3} x^{2} + 4 \, {\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d^{3} x + {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} d^{3}} \]

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

(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)

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=- \frac {1}{4 a b^{2} c d^{3} - b^{4} d^{3} + x^{2} \cdot \left (16 a c^{3} d^{3} - 4 b^{2} c^{2} d^{3}\right ) + x \left (16 a b c^{2} d^{3} - 4 b^{3} c d^{3}\right )} - \frac {2 \log {\left (\frac {b}{2 c} + x \right )}}{d^{3} \left (4 a c - b^{2}\right )^{2}} + \frac {\log {\left (\frac {a}{c} + \frac {b x}{c} + x^{2} \right )}}{d^{3} \left (4 a c - b^{2}\right )^{2}} \]

[In]

integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a),x)

[Out]

-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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.84 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {1}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{3} x^{2} + 4 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{3} x + {\left (b^{4} - 4 \, a b^{2} c\right )} d^{3}} + \frac {\log \left (c x^{2} + b x + a\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{3}} - \frac {2 \, \log \left (2 \, c x + b\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{3}} \]

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.59 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=-\frac {2 \, c \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{4} c d^{3} - 8 \, a b^{2} c^{2} d^{3} + 16 \, a^{2} c^{3} d^{3}} + \frac {\log \left (c x^{2} + b x + a\right )}{b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}} + \frac {1}{{\left (b^{2} - 4 \, a c\right )} {\left (2 \, c x + b\right )}^{2} d^{3}} \]

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )} \, dx=\frac {\ln \left (c\,x^2+b\,x+a\right )}{16\,a^2\,c^2\,d^3-8\,a\,b^2\,c\,d^3+b^4\,d^3}-\frac {2\,\ln \left (b+2\,c\,x\right )}{16\,a^2\,c^2\,d^3-8\,a\,b^2\,c\,d^3+b^4\,d^3}+\frac {1}{b^4\,d^3+4\,b^3\,c\,d^3\,x+4\,b^2\,c^2\,d^3\,x^2-4\,a\,b^2\,c\,d^3-16\,a\,b\,c^2\,d^3\,x-16\,a\,c^3\,d^3\,x^2} \]

[In]

int(1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)),x)

[Out]

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)