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

Optimal. Leaf size=87 \[ -\frac {b^2 \log (a+b x)}{a (b c-a d)^2}+\frac {d (2 b c-a d) \log (c+d x)}{c^2 (b c-a d)^2}-\frac {d}{c (c+d x) (b c-a d)}+\frac {\log (x)}{a c^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 87, 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 = {72} \begin {gather*} -\frac {b^2 \log (a+b x)}{a (b c-a d)^2}+\frac {d (2 b c-a d) \log (c+d x)}{c^2 (b c-a d)^2}-\frac {d}{c (c+d x) (b c-a d)}+\frac {\log (x)}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x)*(c + d*x)^2),x]

[Out]

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 83, normalized size = 0.95 \begin {gather*} \frac {\frac {a d ((c+d x) (2 b c-a d) \log (c+d x)+c (a d-b c))-b^2 c^2 (c+d x) \log (a+b x)}{(c+d x) (b c-a d)^2}+\log (x)}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 3.46, size = 207, normalized size = 2.38 \begin {gather*} -\frac {a b c^{2} d - a^{2} c d^{2} + {\left (b^{2} c^{2} d x + b^{2} c^{3}\right )} \log \left (b x + a\right ) - {\left (2 \, a b c^{2} d - a^{2} c d^{2} + {\left (2 \, a b c d^{2} - a^{2} d^{3}\right )} x\right )} \log \left (d x + c\right ) - {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} + {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x\right )} \log \relax (x)}{a b^{2} c^{5} - 2 \, a^{2} b c^{4} d + a^{3} c^{3} d^{2} + {\left (a b^{2} c^{4} d - 2 \, a^{2} b c^{3} d^{2} + a^{3} c^{2} d^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.06, size = 282, normalized size = 3.24 \begin {gather*} -\frac {1}{2} \, d {\left (\frac {{\left (2 \, b c - a d\right )} \log \left ({\left | -b + \frac {2 \, b c}{d x + c} - \frac {b c^{2}}{{\left (d x + c\right )}^{2}} - \frac {a d}{d x + c} + \frac {a c d}{{\left (d x + c\right )}^{2}} \right |}\right )}{b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}} + \frac {2 \, d^{2}}{{\left (b c^{2} d^{2} - a c d^{3}\right )} {\left (d x + c\right )}} + \frac {{\left (2 \, b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \log \left (\frac {{\left | -2 \, b c d + \frac {2 \, b c^{2} d}{d x + c} + a d^{2} - \frac {2 \, a c d^{2}}{d x + c} - d^{2} {\left | a \right |} \right |}}{{\left | -2 \, b c d + \frac {2 \, b c^{2} d}{d x + c} + a d^{2} - \frac {2 \, a c d^{2}}{d x + c} + d^{2} {\left | a \right |} \right |}}\right )}{{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}\right )} d^{2} {\left | a \right |}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 105, normalized size = 1.21 \begin {gather*} -\frac {a \,d^{2} \ln \left (d x +c \right )}{\left (a d -b c \right )^{2} c^{2}}-\frac {b^{2} \ln \left (b x +a \right )}{\left (a d -b c \right )^{2} a}+\frac {2 b d \ln \left (d x +c \right )}{\left (a d -b c \right )^{2} c}+\frac {d}{\left (a d -b c \right ) \left (d x +c \right ) c}+\frac {\ln \relax (x )}{a \,c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.09, size = 128, normalized size = 1.47 \begin {gather*} -\frac {b^{2} \log \left (b x + a\right )}{a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \log \left (d x + c\right )}{b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2}} - \frac {d}{b c^{3} - a c^{2} d + {\left (b c^{2} d - a c d^{2}\right )} x} + \frac {\log \relax (x)}{a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.65, size = 116, normalized size = 1.33 \begin {gather*} \frac {\ln \relax (x)}{a\,c^2}-\frac {\ln \left (c+d\,x\right )\,\left (a\,d^2-2\,b\,c\,d\right )}{a^2\,c^2\,d^2-2\,a\,b\,c^3\,d+b^2\,c^4}-\frac {b^2\,\ln \left (a+b\,x\right )}{a^3\,d^2-2\,a^2\,b\,c\,d+a\,b^2\,c^2}+\frac {d}{c\,\left (a\,d-b\,c\right )\,\left (c+d\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a + b*x)*(c + d*x)^2),x)

[Out]

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

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

[Out]

Timed out

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