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

Optimal. Leaf size=53 \[ -\frac {c \log (b+c x)}{b (c d-b e)}+\frac {e \log (d+e x)}{d (c d-b e)}+\frac {\log (x)}{b d} \]

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Rubi [A]  time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {698} \begin {gather*} -\frac {c \log (b+c x)}{b (c d-b e)}+\frac {e \log (d+e x)}{d (c d-b e)}+\frac {\log (x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(b*x + c*x^2)),x]

[Out]

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

Rule 698

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.91 \begin {gather*} \frac {-c d \log (b+c x)+b e \log (d+e x)-b e \log (x)+c d \log (x)}{b c d^2-b^2 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 50, normalized size = 0.94 \begin {gather*} -\frac {c d \log \left (c x + b\right ) - b e \log \left (e x + d\right ) - {\left (c d - b e\right )} \log \relax (x)}{b c d^{2} - b^{2} d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 67, normalized size = 1.26 \begin {gather*} -\frac {c^{2} \log \left ({\left | c x + b \right |}\right )}{b c^{2} d - b^{2} c e} + \frac {e^{2} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e - b d e^{2}} + \frac {\log \left ({\left | x \right |}\right )}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 54, normalized size = 1.02 \begin {gather*} \frac {c \ln \left (c x +b \right )}{\left (b e -c d \right ) b}-\frac {e \ln \left (e x +d \right )}{\left (b e -c d \right ) d}+\frac {\ln \relax (x )}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c/(b*e-c*d)/b*ln(c*x+b)-e/(b*e-c*d)/d*ln(e*x+d)+ln(x)/b/d

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maxima [A]  time = 1.39, size = 53, normalized size = 1.00 \begin {gather*} -\frac {c \log \left (c x + b\right )}{b c d - b^{2} e} + \frac {e \log \left (e x + d\right )}{c d^{2} - b d e} + \frac {\log \relax (x)}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.33, size = 93, normalized size = 1.75 \begin {gather*} \frac {e\,\ln \left (\frac {{\left (d+e\,x\right )}^2}{x\,\left (b+c\,x\right )}\right )}{2\,c\,d^2-2\,b\,d\,e}-\frac {\ln \left (\frac {b-\sqrt {b^2}+2\,c\,x}{b+\sqrt {b^2}+2\,c\,x}\right )\,\left (b\,e-2\,c\,d\right )}{\left (2\,c\,d^2-2\,b\,d\,e\right )\,\sqrt {b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x + c*x^2)*(d + e*x)),x)

[Out]

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

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

[Out]

Timed out

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