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

Optimal. Leaf size=110 \[ \frac {b^3 \log (a+b x)}{a^2 (b c-a d)^2}-\frac {\log (x) (2 a d+b c)}{a^2 c^3}-\frac {d^2 (3 b c-2 a d) \log (c+d x)}{c^3 (b c-a d)^2}+\frac {d^2}{c^2 (c+d x) (b c-a d)}-\frac {1}{a c^2 x} \]

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Rubi [A]  time = 0.10, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {88} \begin {gather*} \frac {b^3 \log (a+b x)}{a^2 (b c-a d)^2}-\frac {\log (x) (2 a d+b c)}{a^2 c^3}+\frac {d^2}{c^2 (c+d x) (b c-a d)}-\frac {d^2 (3 b c-2 a d) \log (c+d x)}{c^3 (b c-a d)^2}-\frac {1}{a c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)*(c + d*x)^2),x]

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 111, normalized size = 1.01 \begin {gather*} \frac {b^3 \log (a+b x)}{a^2 (a d-b c)^2}+\frac {\log (x) (-2 a d-b c)}{a^2 c^3}+\frac {\left (2 a d^3-3 b c d^2\right ) \log (c+d x)}{c^3 (b c-a d)^2}+\frac {d^2}{c^2 (c+d x) (b c-a d)}-\frac {1}{a c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 12.60, size = 287, normalized size = 2.61 \begin {gather*} -\frac {a b^{2} c^{4} - 2 \, a^{2} b c^{3} d + a^{3} c^{2} d^{2} + {\left (a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x - {\left (b^{3} c^{3} d x^{2} + b^{3} c^{4} x\right )} \log \left (b x + a\right ) + {\left ({\left (3 \, a^{2} b c d^{3} - 2 \, a^{3} d^{4}\right )} x^{2} + {\left (3 \, a^{2} b c^{2} d^{2} - 2 \, a^{3} c d^{3}\right )} x\right )} \log \left (d x + c\right ) + {\left ({\left (b^{3} c^{3} d - 3 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{2} + {\left (b^{3} c^{4} - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x\right )} \log \relax (x)}{{\left (a^{2} b^{2} c^{5} d - 2 \, a^{3} b c^{4} d^{2} + a^{4} c^{3} d^{3}\right )} x^{2} + {\left (a^{2} b^{2} c^{6} - 2 \, a^{3} b c^{5} d + a^{4} c^{4} d^{2}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*b^2*c^4 - 2*a^2*b*c^3*d + a^3*c^2*d^2 + (a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*x - (b^3*c^3*d*x^2 +
b^3*c^4*x)*log(b*x + a) + ((3*a^2*b*c*d^3 - 2*a^3*d^4)*x^2 + (3*a^2*b*c^2*d^2 - 2*a^3*c*d^3)*x)*log(d*x + c)
+ ((b^3*c^3*d - 3*a^2*b*c*d^3 + 2*a^3*d^4)*x^2 + (b^3*c^4 - 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*x)*log(x))/((a^2*b^
2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^2 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4*c^4*d^2)*x)

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giac [A]  time = 1.04, size = 150, normalized size = 1.36 \begin {gather*} \frac {b^{3} d \log \left ({\left | b - \frac {b c}{d x + c} + \frac {a d}{d x + c} \right |}\right )}{a^{2} b^{2} c^{2} d - 2 \, a^{3} b c d^{2} + a^{4} d^{3}} + \frac {d^{5}}{{\left (b c^{3} d^{3} - a c^{2} d^{4}\right )} {\left (d x + c\right )}} + \frac {d}{a c^{3} {\left (\frac {c}{d x + c} - 1\right )}} - \frac {{\left (b c d + 2 \, a d^{2}\right )} \log \left ({\left | -\frac {c}{d x + c} + 1 \right |}\right )}{a^{2} c^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*d*log(abs(b - b*c/(d*x + c) + a*d/(d*x + c)))/(a^2*b^2*c^2*d - 2*a^3*b*c*d^2 + a^4*d^3) + d^5/((b*c^3*d^3
- a*c^2*d^4)*(d*x + c)) + d/(a*c^3*(c/(d*x + c) - 1)) - (b*c*d + 2*a*d^2)*log(abs(-c/(d*x + c) + 1))/(a^2*c^3*
d)

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maple [A]  time = 0.01, size = 133, normalized size = 1.21 \begin {gather*} \frac {2 a \,d^{3} \ln \left (d x +c \right )}{\left (a d -b c \right )^{2} c^{3}}+\frac {b^{3} \ln \left (b x +a \right )}{\left (a d -b c \right )^{2} a^{2}}-\frac {3 b \,d^{2} \ln \left (d x +c \right )}{\left (a d -b c \right )^{2} c^{2}}-\frac {d^{2}}{\left (a d -b c \right ) \left (d x +c \right ) c^{2}}-\frac {2 d \ln \relax (x )}{a \,c^{3}}-\frac {b \ln \relax (x )}{a^{2} c^{2}}-\frac {1}{a \,c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.14, size = 177, normalized size = 1.61 \begin {gather*} \frac {b^{3} \log \left (b x + a\right )}{a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}} - \frac {{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} \log \left (d x + c\right )}{b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}} - \frac {b c^{2} - a c d + {\left (b c d - 2 \, a d^{2}\right )} x}{{\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{2} + {\left (a b c^{4} - a^{2} c^{3} d\right )} x} - \frac {{\left (b c + 2 \, a d\right )} \log \relax (x)}{a^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*log(b*x + a)/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2) - (3*b*c*d^2 - 2*a*d^3)*log(d*x + c)/(b^2*c^5 - 2*a*b*c
^4*d + a^2*c^3*d^2) - (b*c^2 - a*c*d + (b*c*d - 2*a*d^2)*x)/((a*b*c^3*d - a^2*c^2*d^2)*x^2 + (a*b*c^4 - a^2*c^
3*d)*x) - (b*c + 2*a*d)*log(x)/(a^2*c^3)

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mupad [B]  time = 0.71, size = 157, normalized size = 1.43 \begin {gather*} \frac {b^3\,\ln \left (a+b\,x\right )}{a^4\,d^2-2\,a^3\,b\,c\,d+a^2\,b^2\,c^2}-\frac {\frac {1}{a\,c}+\frac {x\,\left (2\,a\,d^2-b\,c\,d\right )}{a\,c^2\,\left (a\,d-b\,c\right )}}{d\,x^2+c\,x}+\frac {\ln \left (c+d\,x\right )\,\left (2\,a\,d^3-3\,b\,c\,d^2\right )}{a^2\,c^3\,d^2-2\,a\,b\,c^4\,d+b^2\,c^5}-\frac {\ln \relax (x)\,\left (2\,a\,d+b\,c\right )}{a^2\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x)*(c + d*x)^2),x)

[Out]

(b^3*log(a + b*x))/(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d) - (1/(a*c) + (x*(2*a*d^2 - b*c*d))/(a*c^2*(a*d - b*c)
))/(c*x + d*x^2) + (log(c + d*x)*(2*a*d^3 - 3*b*c*d^2))/(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d) - (log(x)*(2*a*d
+ b*c))/(a^2*c^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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