3.1143 \(\int \frac {A+B x}{(d+e x)^2 (b x+c x^2)} \, dx\)

Optimal. Leaf size=110 \[ -\frac {\log (d+e x) \left (B c d^2-A e (2 c d-b e)\right )}{d^2 (c d-b e)^2}+\frac {B d-A e}{d (d+e x) (c d-b e)}+\frac {c (b B-A c) \log (b+c x)}{b (c d-b e)^2}+\frac {A \log (x)}{b d^2} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 110, 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} \[ -\frac {\log (d+e x) \left (B c d^2-A e (2 c d-b e)\right )}{d^2 (c d-b e)^2}+\frac {B d-A e}{d (d+e x) (c d-b e)}+\frac {c (b B-A c) \log (b+c x)}{b (c d-b e)^2}+\frac {A \log (x)}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.14, size = 106, normalized size = 0.96 \[ \frac {\frac {c d^2 (d+e x) (b B-A c) \log (b+c x)-b (d+e x) \log (d+e x) \left (A e (b e-2 c d)+B c d^2\right )+b d (B d-A e) (c d-b e)}{(d+e x) (c d-b e)^2}+A \log (x)}{b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 20.77, size = 260, normalized size = 2.36 \[ \frac {B b c d^{3} + A b^{2} d e^{2} - {\left (B b^{2} + A b c\right )} d^{2} e + {\left ({\left (B b c - A c^{2}\right )} d^{2} e x + {\left (B b c - A c^{2}\right )} d^{3}\right )} \log \left (c x + b\right ) - {\left (B b c d^{3} - 2 \, A b c d^{2} e + A b^{2} d e^{2} + {\left (B b c d^{2} e - 2 \, A b c d e^{2} + A b^{2} e^{3}\right )} x\right )} \log \left (e x + d\right ) + {\left (A c^{2} d^{3} - 2 \, A b c d^{2} e + A b^{2} d e^{2} + {\left (A c^{2} d^{2} e - 2 \, A b c d e^{2} + A b^{2} e^{3}\right )} x\right )} \log \relax (x)}{b c^{2} d^{5} - 2 \, b^{2} c d^{4} e + b^{3} d^{3} e^{2} + {\left (b c^{2} d^{4} e - 2 \, b^{2} c d^{3} e^{2} + b^{3} d^{2} e^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.24, size = 323, normalized size = 2.94 \[ -\frac {{\left (B b c d^{2} e^{2} - 2 \, A c^{2} d^{2} e^{2} + 2 \, A b c d e^{3} - A b^{2} e^{4}\right )} e^{\left (-2\right )} \log \left (\frac {{\left | 2 \, c d e - \frac {2 \, c d^{2} e}{x e + d} - b e^{2} + \frac {2 \, b d e^{2}}{x e + d} - {\left | b \right |} e^{2} \right |}}{{\left | 2 \, c d e - \frac {2 \, c d^{2} e}{x e + d} - b e^{2} + \frac {2 \, b d e^{2}}{x e + d} + {\left | b \right |} e^{2} \right |}}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} {\left | b \right |}} + \frac {{\left (B c d^{2} - 2 \, A c d e + A b e^{2}\right )} \log \left ({\left | c - \frac {2 \, c d}{x e + d} + \frac {c d^{2}}{{\left (x e + d\right )}^{2}} + \frac {b e}{x e + d} - \frac {b d e}{{\left (x e + d\right )}^{2}} \right |}\right )}{2 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )}} + \frac {\frac {B d e^{2}}{x e + d} - \frac {A e^{3}}{x e + d}}{c d^{2} e^{2} - b d e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 169, normalized size = 1.54 \[ -\frac {A b \,e^{2} \ln \left (e x +d \right )}{\left (b e -c d \right )^{2} d^{2}}-\frac {A \,c^{2} \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b}+\frac {2 A c e \ln \left (e x +d \right )}{\left (b e -c d \right )^{2} d}+\frac {B c \ln \left (c x +b \right )}{\left (b e -c d \right )^{2}}-\frac {B c \ln \left (e x +d \right )}{\left (b e -c d \right )^{2}}+\frac {A e}{\left (b e -c d \right ) \left (e x +d \right ) d}-\frac {B}{\left (b e -c d \right ) \left (e x +d \right )}+\frac {A \ln \relax (x )}{b \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2/(b*e-c*d)^2/b*ln(c*x+b)*A+c/(b*e-c*d)^2*ln(c*x+b)*B-1/(b*e-c*d)^2/d^2*ln(e*x+d)*A*b*e^2+2/(b*e-c*d)^2/d*l
n(e*x+d)*A*c*e-1/(b*e-c*d)^2*ln(e*x+d)*B*c+1/(b*e-c*d)/d/(e*x+d)*A*e-1/(b*e-c*d)/(e*x+d)*B+A*ln(x)/b/d^2

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maxima [A]  time = 0.63, size = 150, normalized size = 1.36 \[ \frac {{\left (B b c - A c^{2}\right )} \log \left (c x + b\right )}{b c^{2} d^{2} - 2 \, b^{2} c d e + b^{3} e^{2}} - \frac {{\left (B c d^{2} - 2 \, A c d e + A b e^{2}\right )} \log \left (e x + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} + \frac {B d - A e}{c d^{3} - b d^{2} e + {\left (c d^{2} e - b d e^{2}\right )} x} + \frac {A \log \relax (x)}{b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.81, size = 141, normalized size = 1.28 \[ \frac {A\,\ln \relax (x)}{b\,d^2}-\frac {\ln \left (d+e\,x\right )\,\left (c\,\left (B\,d^2-2\,A\,d\,e\right )+A\,b\,e^2\right )}{b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4}-\frac {\ln \left (b+c\,x\right )\,\left (A\,c^2-B\,b\,c\right )}{b^3\,e^2-2\,b^2\,c\,d\,e+b\,c^2\,d^2}+\frac {A\,e-B\,d}{d\,\left (b\,e-c\,d\right )\,\left (d+e\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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