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

Optimal. Leaf size=68 \[ \frac {(b B-A c) \log (b+c x)}{b (c d-b e)}-\frac {(B d-A e) \log (d+e x)}{d (c d-b e)}+\frac {A \log (x)}{b d} \]

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Rubi [A]  time = 0.07, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {771} \begin {gather*} \frac {(b B-A c) \log (b+c x)}{b (c d-b e)}-\frac {(B d-A e) \log (d+e x)}{d (c d-b e)}+\frac {A \log (x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)*(b*x + c*x^2)),x]

[Out]

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

Rule 771

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 63, normalized size = 0.93 \begin {gather*} \frac {\log (b+c x) (A c d-b B d)+b (B d-A e) \log (d+e x)+A \log (x) (b e-c d)}{b d (b e-c d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.54, size = 65, normalized size = 0.96 \begin {gather*} \frac {{\left (B b - A c\right )} d \log \left (c x + b\right ) - {\left (B b d - A b e\right )} \log \left (e x + d\right ) + {\left (A c d - A 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((B*x+A)/(e*x+d)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 85, normalized size = 1.25 \begin {gather*} \frac {{\left (B b c - A c^{2}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{2} d - b^{2} c e} - \frac {{\left (B d e - A e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{c d^{2} e - b d e^{2}} + \frac {A \log \left ({\left | x \right |}\right )}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 94, normalized size = 1.38 \begin {gather*} \frac {A c \ln \left (c x +b \right )}{\left (b e -c d \right ) b}-\frac {A e \ln \left (e x +d \right )}{\left (b e -c d \right ) d}-\frac {B \ln \left (c x +b \right )}{b e -c d}+\frac {B \ln \left (e x +d \right )}{b e -c d}+\frac {A \ln \relax (x )}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(b*e-c*d)/b*ln(c*x+b)*A*c-1/(b*e-c*d)*ln(c*x+b)*B-1/(b*e-c*d)/d*ln(e*x+d)*A*e+1/(b*e-c*d)*ln(e*x+d)*B+A*ln(x
)/b/d

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maxima [A]  time = 0.53, size = 68, normalized size = 1.00 \begin {gather*} \frac {{\left (B b - A c\right )} \log \left (c x + b\right )}{b c d - b^{2} e} - \frac {{\left (B d - A e\right )} \log \left (e x + d\right )}{c d^{2} - b d e} + \frac {A \log \relax (x)}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.67, size = 67, normalized size = 0.99 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,e-B\,d\right )}{c\,d^2-b\,d\,e}+\frac {\ln \left (b+c\,x\right )\,\left (A\,c-B\,b\right )}{b^2\,e-b\,c\,d}+\frac {A\,\ln \relax (x)}{b\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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