3.1187 \(\int \frac{1}{(b d+2 c d x) (a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=112 \[ \frac{16 c^2 \log \left (a+b x+c x^2\right )}{d \left (b^2-4 a c\right )^3}-\frac{32 c^2 \log (b+2 c x)}{d \left (b^2-4 a c\right )^3}+\frac{4 c}{d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

[Out]

-1/(2*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^2) + (4*c)/((b^2 - 4*a*c)^2*d*(a + b*x + c*x^2)) - (32*c^2*Log[b + 2*c
*x])/((b^2 - 4*a*c)^3*d) + (16*c^2*Log[a + b*x + c*x^2])/((b^2 - 4*a*c)^3*d)

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Rubi [A]  time = 0.0596157, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {687, 681, 31, 628} \[ \frac{16 c^2 \log \left (a+b x+c x^2\right )}{d \left (b^2-4 a c\right )^3}-\frac{32 c^2 \log (b+2 c x)}{d \left (b^2-4 a c\right )^3}+\frac{4 c}{d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac{1}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^2) + (4*c)/((b^2 - 4*a*c)^2*d*(a + b*x + c*x^2)) - (32*c^2*Log[b + 2*c
*x])/((b^2 - 4*a*c)^3*d) + (16*c^2*Log[a + b*x + c*x^2])/((b^2 - 4*a*c)^3*d)

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 681

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 31

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

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{1}{(b d+2 c d x) \left (a+b x+c x^2\right )^3} \, dx &=-\frac{1}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac{(4 c) \int \frac{1}{(b d+2 c d x) \left (a+b x+c x^2\right )^2} \, dx}{b^2-4 a c}\\ &=-\frac{1}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{4 c}{\left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac{\left (16 c^2\right ) \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 )^2}\\ &=-\frac{1}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{4 c}{\left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac{\left (16 c^2\right ) \int \frac{b d+2 c d x}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3 d^2}-\frac{\left (64 c^3\right ) \int \frac{1}{b+2 c x} \, dx}{\left (b^2-4 a c\right )^3 d}\\ &=-\frac{1}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac{4 c}{\left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac{32 c^2 \log (b+2 c x)}{\left (b^2-4 a c\right )^3 d}+\frac{16 c^2 \log \left (a+b x+c x^2\right )}{\left (b^2-4 a c\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 0.0747044, size = 90, normalized size = 0.8 \[ \frac{\frac{8 c \left (b^2-4 a c\right )}{a+x (b+c x)}-\frac{\left (b^2-4 a c\right )^2}{(a+x (b+c x))^2}+32 c^2 \log (a+x (b+c x))-64 c^2 \log (b+2 c x)}{2 d \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b^2 - 4*a*c)^2/(a + x*(b + c*x))^2) + (8*c*(b^2 - 4*a*c))/(a + x*(b + c*x)) - 64*c^2*Log[b + 2*c*x] + 32*c
^2*Log[a + x*(b + c*x)])/(2*(b^2 - 4*a*c)^3*d)

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Maple [B]  time = 0.057, size = 304, normalized size = 2.7 \begin{align*} 16\,{\frac{a{x}^{2}{c}^{3}}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}-4\,{\frac{{b}^{2}{x}^{2}{c}^{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}+16\,{\frac{ba{c}^{2}x}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}-4\,{\frac{{b}^{3}cx}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}+24\,{\frac{{a}^{2}{c}^{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}-8\,{\frac{ac{b}^{2}}{d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}+{\frac{{b}^{4}}{2\,d \left ( 4\,ac-{b}^{2} \right ) ^{3} \left ( c{x}^{2}+bx+a \right ) ^{2}}}-16\,{\frac{{c}^{2}\ln \left ( c{x}^{2}+bx+a \right ) }{d \left ( 4\,ac-{b}^{2} \right ) ^{3}}}+32\,{\frac{{c}^{2}\ln \left ( 2\,cx+b \right ) }{d \left ( 4\,ac-{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/d/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*x^2*a*c^3-4/d/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*x^2*b^2*c^2+16/d/(4*a*c-b^2)^3/
(c*x^2+b*x+a)^2*b*a*c^2*x-4/d/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*b^3*c*x+24/d/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*a^2*c^2
-8/d/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*a*c*b^2+1/2/d/(4*a*c-b^2)^3/(c*x^2+b*x+a)^2*b^4-16/d/(4*a*c-b^2)^3*c^2*ln(c
*x^2+b*x+a)+32/d*c^2/(4*a*c-b^2)^3*ln(2*c*x+b)

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Maxima [B]  time = 1.26214, size = 359, normalized size = 3.21 \begin{align*} \frac{16 \, c^{2} \log \left (c x^{2} + b x + a\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d} - \frac{32 \, c^{2} \log \left (2 \, c x + b\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d} + \frac{8 \, c^{2} x^{2} + 8 \, b c x - b^{2} + 12 \, a c}{2 \,{\left ({\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} d x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} d x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} d x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} d x +{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*c^2*log(c*x^2 + b*x + a)/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d) - 32*c^2*log(2*c*x + b)/((b^6
- 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d) + 1/2*(8*c^2*x^2 + 8*b*c*x - b^2 + 12*a*c)/((b^4*c^2 - 8*a*b^2
*c^3 + 16*a^2*c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d*x^2
+ 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d)

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Fricas [B]  time = 2.16355, size = 825, normalized size = 7.37 \begin{align*} -\frac{b^{4} - 16 \, a b^{2} c + 48 \, a^{2} c^{2} - 8 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 8 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x - 32 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + 2 \, a b c^{2} x + a^{2} c^{2} +{\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{2}\right )} \log \left (c x^{2} + b x + a\right ) + 64 \,{\left (c^{4} x^{4} + 2 \, b c^{3} x^{3} + 2 \, a b c^{2} x + a^{2} c^{2} +{\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{2}\right )} \log \left (2 \, c x + b\right )}{2 \,{\left ({\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} d x^{4} + 2 \,{\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} d x^{3} +{\left (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}\right )} d x^{2} + 2 \,{\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} d x +{\left (a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^4 - 16*a*b^2*c + 48*a^2*c^2 - 8*(b^2*c^2 - 4*a*c^3)*x^2 - 8*(b^3*c - 4*a*b*c^2)*x - 32*(c^4*x^4 + 2*b*
c^3*x^3 + 2*a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*c^3)*x^2)*log(c*x^2 + b*x + a) + 64*(c^4*x^4 + 2*b*c^3*x^3 +
2*a*b*c^2*x + a^2*c^2 + (b^2*c^2 + 2*a*c^3)*x^2)*log(2*c*x + b))/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 6
4*a^3*c^5)*d*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)*d*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)*d*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)*d*x
+ (a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*d)

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Sympy [B]  time = 6.37589, size = 246, normalized size = 2.2 \begin{align*} \frac{32 c^{2} \log{\left (\frac{b}{2 c} + x \right )}}{d \left (4 a c - b^{2}\right )^{3}} - \frac{16 c^{2} \log{\left (\frac{a}{c} + \frac{b x}{c} + x^{2} \right )}}{d \left (4 a c - b^{2}\right )^{3}} + \frac{12 a c - b^{2} + 8 b c x + 8 c^{2} x^{2}}{32 a^{4} c^{2} d - 16 a^{3} b^{2} c d + 2 a^{2} b^{4} d + x^{4} \left (32 a^{2} c^{4} d - 16 a b^{2} c^{3} d + 2 b^{4} c^{2} d\right ) + x^{3} \left (64 a^{2} b c^{3} d - 32 a b^{3} c^{2} d + 4 b^{5} c d\right ) + x^{2} \left (64 a^{3} c^{3} d - 12 a b^{4} c d + 2 b^{6} d\right ) + x \left (64 a^{3} b c^{2} d - 32 a^{2} b^{3} c d + 4 a b^{5} d\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32*c**2*log(b/(2*c) + x)/(d*(4*a*c - b**2)**3) - 16*c**2*log(a/c + b*x/c + x**2)/(d*(4*a*c - b**2)**3) + (12*a
*c - b**2 + 8*b*c*x + 8*c**2*x**2)/(32*a**4*c**2*d - 16*a**3*b**2*c*d + 2*a**2*b**4*d + x**4*(32*a**2*c**4*d -
16*a*b**2*c**3*d + 2*b**4*c**2*d) + x**3*(64*a**2*b*c**3*d - 32*a*b**3*c**2*d + 4*b**5*c*d) + x**2*(64*a**3*c
**3*d - 12*a*b**4*c*d + 2*b**6*d) + x*(64*a**3*b*c**2*d - 32*a**2*b**3*c*d + 4*a*b**5*d))

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Giac [A]  time = 1.18831, size = 254, normalized size = 2.27 \begin{align*} -\frac{32 \, c^{3} \log \left ({\left | 2 \, c x + b \right |}\right )}{b^{6} c d - 12 \, a b^{4} c^{2} d + 48 \, a^{2} b^{2} c^{3} d - 64 \, a^{3} c^{4} d} + \frac{16 \, c^{2} \log \left (c x^{2} + b x + a\right )}{b^{6} d - 12 \, a b^{4} c d + 48 \, a^{2} b^{2} c^{2} d - 64 \, a^{3} c^{3} d} - \frac{b^{4} - 16 \, a b^{2} c + 48 \, a^{2} c^{2} - 8 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 8 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} x}{2 \,{\left (c x^{2} + b x + a\right )}^{2}{\left (b^{2} - 4 \, a c\right )}^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32*c^3*log(abs(2*c*x + b))/(b^6*c*d - 12*a*b^4*c^2*d + 48*a^2*b^2*c^3*d - 64*a^3*c^4*d) + 16*c^2*log(c*x^2 +
b*x + a)/(b^6*d - 12*a*b^4*c*d + 48*a^2*b^2*c^2*d - 64*a^3*c^3*d) - 1/2*(b^4 - 16*a*b^2*c + 48*a^2*c^2 - 8*(b^
2*c^2 - 4*a*c^3)*x^2 - 8*(b^3*c - 4*a*b*c^2)*x)/((c*x^2 + b*x + a)^2*(b^2 - 4*a*c)^3*d)