3.11.5 \(\int \frac {A+B x}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=62 \[ \frac {\log (x) (b B-2 A c)}{b^3}-\frac {(b B-2 A c) \log (b+c x)}{b^3}+\frac {b B-A c}{b^2 (b+c x)}-\frac {A}{b^2 x} \]

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Rubi [A]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {631} \begin {gather*} \frac {b B-A c}{b^2 (b+c x)}+\frac {\log (x) (b B-2 A c)}{b^3}-\frac {(b B-2 A c) \log (b+c x)}{b^3}-\frac {A}{b^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(b*x + c*x^2)^2,x]

[Out]

-(A/(b^2*x)) + (b*B - A*c)/(b^2*(b + c*x)) + ((b*B - 2*A*c)*Log[x])/b^3 - ((b*B - 2*A*c)*Log[b + c*x])/b^3

Rule 631

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 56, normalized size = 0.90 \begin {gather*} \frac {\frac {b (b B-A c)}{b+c x}+\log (x) (b B-2 A c)+(2 A c-b B) \log (b+c x)-\frac {A b}{x}}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(b*x + c*x^2)^2,x]

[Out]

(-((A*b)/x) + (b*(b*B - A*c))/(b + c*x) + (b*B - 2*A*c)*Log[x] + (-(b*B) + 2*A*c)*Log[b + c*x])/b^3

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)/(b*x + c*x^2)^2,x]

[Out]

IntegrateAlgebraic[(A + B*x)/(b*x + c*x^2)^2, x]

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fricas [A]  time = 0.40, size = 107, normalized size = 1.73 \begin {gather*} -\frac {A b^{2} - {\left (B b^{2} - 2 \, A b c\right )} x + {\left ({\left (B b c - 2 \, A c^{2}\right )} x^{2} + {\left (B b^{2} - 2 \, A b c\right )} x\right )} \log \left (c x + b\right ) - {\left ({\left (B b c - 2 \, A c^{2}\right )} x^{2} + {\left (B b^{2} - 2 \, A b c\right )} x\right )} \log \relax (x)}{b^{3} c x^{2} + b^{4} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

-(A*b^2 - (B*b^2 - 2*A*b*c)*x + ((B*b*c - 2*A*c^2)*x^2 + (B*b^2 - 2*A*b*c)*x)*log(c*x + b) - ((B*b*c - 2*A*c^2
)*x^2 + (B*b^2 - 2*A*b*c)*x)*log(x))/(b^3*c*x^2 + b^4*x)

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giac [A]  time = 0.16, size = 71, normalized size = 1.15 \begin {gather*} \frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | x \right |}\right )}{b^{3}} + \frac {B b x - 2 \, A c x - A b}{{\left (c x^{2} + b x\right )} b^{2}} - \frac {{\left (B b c - 2 \, A c^{2}\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

(B*b - 2*A*c)*log(abs(x))/b^3 + (B*b*x - 2*A*c*x - A*b)/((c*x^2 + b*x)*b^2) - (B*b*c - 2*A*c^2)*log(abs(c*x +
b))/(b^3*c)

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maple [A]  time = 0.06, size = 78, normalized size = 1.26 \begin {gather*} -\frac {A c}{\left (c x +b \right ) b^{2}}-\frac {2 A c \ln \relax (x )}{b^{3}}+\frac {2 A c \ln \left (c x +b \right )}{b^{3}}+\frac {B}{\left (c x +b \right ) b}+\frac {B \ln \relax (x )}{b^{2}}-\frac {B \ln \left (c x +b \right )}{b^{2}}-\frac {A}{b^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+b*x)^2,x)

[Out]

2/b^3*ln(c*x+b)*A*c-1/b^2*ln(c*x+b)*B-1/b^2/(c*x+b)*A*c+1/b/(c*x+b)*B-A/b^2/x-2/b^3*ln(x)*A*c+1/b^2*ln(x)*B

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maxima [A]  time = 0.64, size = 67, normalized size = 1.08 \begin {gather*} -\frac {A b - {\left (B b - 2 \, A c\right )} x}{b^{2} c x^{2} + b^{3} x} - \frac {{\left (B b - 2 \, A c\right )} \log \left (c x + b\right )}{b^{3}} + \frac {{\left (B b - 2 \, A c\right )} \log \relax (x)}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

-(A*b - (B*b - 2*A*c)*x)/(b^2*c*x^2 + b^3*x) - (B*b - 2*A*c)*log(c*x + b)/b^3 + (B*b - 2*A*c)*log(x)/b^3

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mupad [B]  time = 1.42, size = 58, normalized size = 0.94 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {2\,c\,x}{b}+1\right )\,\left (2\,A\,c-B\,b\right )}{b^3}-\frac {\frac {A}{b}+\frac {x\,\left (2\,A\,c-B\,b\right )}{b^2}}{c\,x^2+b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(b*x + c*x^2)^2,x)

[Out]

(2*atanh((2*c*x)/b + 1)*(2*A*c - B*b))/b^3 - (A/b + (x*(2*A*c - B*b))/b^2)/(b*x + c*x^2)

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sympy [B]  time = 0.47, size = 128, normalized size = 2.06 \begin {gather*} \frac {- A b + x \left (- 2 A c + B b\right )}{b^{3} x + b^{2} c x^{2}} + \frac {\left (- 2 A c + B b\right ) \log {\left (x + \frac {- 2 A b c + B b^{2} - b \left (- 2 A c + B b\right )}{- 4 A c^{2} + 2 B b c} \right )}}{b^{3}} - \frac {\left (- 2 A c + B b\right ) \log {\left (x + \frac {- 2 A b c + B b^{2} + b \left (- 2 A c + B b\right )}{- 4 A c^{2} + 2 B b c} \right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(c*x**2+b*x)**2,x)

[Out]

(-A*b + x*(-2*A*c + B*b))/(b**3*x + b**2*c*x**2) + (-2*A*c + B*b)*log(x + (-2*A*b*c + B*b**2 - b*(-2*A*c + B*b
))/(-4*A*c**2 + 2*B*b*c))/b**3 - (-2*A*c + B*b)*log(x + (-2*A*b*c + B*b**2 + b*(-2*A*c + B*b))/(-4*A*c**2 + 2*
B*b*c))/b**3

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