3.1535 \(\int \frac {1}{(a-b x) (a+b x) (c+d x)} \, dx\)

Optimal. Leaf size=74 \[ -\frac {d \log (c+d x)}{b^2 c^2-a^2 d^2}-\frac {\log (a-b x)}{2 a (a d+b c)}+\frac {\log (a+b x)}{2 a (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {72} \[ -\frac {d \log (c+d x)}{b^2 c^2-a^2 d^2}-\frac {\log (a-b x)}{2 a (a d+b c)}+\frac {\log (a+b x)}{2 a (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a - b*x)*(a + b*x)*(c + d*x)),x]

[Out]

-Log[a - b*x]/(2*a*(b*c + a*d)) + Log[a + b*x]/(2*a*(b*c - a*d)) - (d*Log[c + d*x])/(b^2*c^2 - a^2*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}{(a-b x) (a+b x) (c+d x)} \, dx &=\int \left (\frac {b}{2 a (b c+a d) (a-b x)}-\frac {b}{2 a (-b c+a d) (a+b x)}-\frac {d^2}{(b c-a d) (b c+a d) (c+d x)}\right ) \, dx\\ &=-\frac {\log (a-b x)}{2 a (b c+a d)}+\frac {\log (a+b x)}{2 a (b c-a d)}-\frac {d \log (c+d x)}{b^2 c^2-a^2 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 68, normalized size = 0.92 \[ \frac {(b c-a d) \log (a-b x)-(a d+b c) \log (a+b x)+2 a d \log (c+d x)}{2 a (a d-b c) (a d+b c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.95, size = 64, normalized size = 0.86 \[ -\frac {2 \, a d \log \left (d x + c\right ) - {\left (b c + a d\right )} \log \left (b x + a\right ) + {\left (b c - a d\right )} \log \left (b x - a\right )}{2 \, {\left (a b^{2} c^{2} - a^{3} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.91, size = 93, normalized size = 1.26 \[ \frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )}} - \frac {b^{2} \log \left ({\left | b x - a \right |}\right )}{2 \, {\left (a b^{3} c + a^{2} b^{2} d\right )}} - \frac {d^{2} \log \left ({\left | d x + c \right |}\right )}{b^{2} c^{2} d - a^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 72, normalized size = 0.97 \[ \frac {d \ln \left (d x +c \right )}{\left (a d +b c \right ) \left (a d -b c \right )}-\frac {\ln \left (b x -a \right )}{2 \left (a d +b c \right ) a}-\frac {\ln \left (b x +a \right )}{2 \left (a d -b c \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 71, normalized size = 0.96 \[ -\frac {d \log \left (d x + c\right )}{b^{2} c^{2} - a^{2} d^{2}} + \frac {\log \left (b x + a\right )}{2 \, {\left (a b c - a^{2} d\right )}} - \frac {\log \left (b x - a\right )}{2 \, {\left (a b c + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.33, size = 72, normalized size = 0.97 \[ \frac {d\,\ln \left (c+d\,x\right )}{a^2\,d^2-b^2\,c^2}-\frac {\ln \left (a-b\,x\right )}{2\,d\,a^2+2\,b\,c\,a}-\frac {\ln \left (a+b\,x\right )}{2\,a^2\,d-2\,a\,b\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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