3.1144 \(\int \frac{A+B x}{(d+e x)^3 (b x+c x^2)} \, dx\)

Optimal. Leaf size=171 \[ -\frac{\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac{c^2 (b B-A c) \log (b+c x)}{b (c d-b e)^3}+\frac{B c d^2-A e (2 c d-b e)}{d^2 (d+e x) (c d-b e)^2}+\frac{B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac{A \log (x)}{b d^3} \]

[Out]

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

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Rubi [A]  time = 0.225122, antiderivative size = 171, normalized size of antiderivative = 1., 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} \[ -\frac{\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac{c^2 (b B-A c) \log (b+c x)}{b (c d-b e)^3}+\frac{B c d^2-A e (2 c d-b e)}{d^2 (d+e x) (c d-b e)^2}+\frac{B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac{A \log (x)}{b d^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^3*(b*x + c*x^2)),x]

[Out]

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

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

Mathematica [A]  time = 0.222899, size = 169, normalized size = 0.99 \[ -\frac{\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac{c^2 (A c-b B) \log (b+c x)}{b (b e-c d)^3}+\frac{A e (b e-2 c d)+B c d^2}{d^2 (d+e x) (c d-b e)^2}+\frac{B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac{A \log (x)}{b d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((d + e*x)^3*(b*x + c*x^2)),x]

[Out]

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

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Maple [A]  time = 0.014, size = 275, normalized size = 1.6 \begin{align*}{\frac{A\ln \left ( x \right ) }{{d}^{3}b}}+{\frac{Ab{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{2} \left ( ex+d \right ) }}-2\,{\frac{Ace}{d \left ( be-cd \right ) ^{2} \left ( ex+d \right ) }}+{\frac{Bc}{ \left ( be-cd \right ) ^{2} \left ( ex+d \right ) }}-{\frac{\ln \left ( ex+d \right ) A{b}^{2}{e}^{3}}{{d}^{3} \left ( be-cd \right ) ^{3}}}+3\,{\frac{\ln \left ( ex+d \right ) Abc{e}^{2}}{{d}^{2} \left ( be-cd \right ) ^{3}}}-3\,{\frac{\ln \left ( ex+d \right ) A{c}^{2}e}{d \left ( be-cd \right ) ^{3}}}+{\frac{\ln \left ( ex+d \right ) B{c}^{2}}{ \left ( be-cd \right ) ^{3}}}+{\frac{Ae}{2\,d \left ( be-cd \right ) \left ( ex+d \right ) ^{2}}}-{\frac{B}{ \left ( 2\,be-2\,cd \right ) \left ( ex+d \right ) ^{2}}}+{\frac{{c}^{3}\ln \left ( cx+b \right ) A}{b \left ( be-cd \right ) ^{3}}}-{\frac{{c}^{2}\ln \left ( cx+b \right ) B}{ \left ( be-cd \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^3/(c*x^2+b*x),x)

[Out]

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

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Maxima [A]  time = 1.15893, size = 421, normalized size = 2.46 \begin{align*} \frac{{\left (B b c^{2} - A c^{3}\right )} \log \left (c x + b\right )}{b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}} - \frac{{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, A b c d e^{2} - A b^{2} e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}} + \frac{3 \, B c d^{3} + 3 \, A b d e^{2} -{\left (B b + 5 \, A c\right )} d^{2} e + 2 \,{\left (B c d^{2} e - 2 \, A c d e^{2} + A b e^{3}\right )} x}{2 \,{\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} +{\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}} + \frac{A \log \left (x\right )}{b d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x),x, algorithm="maxima")

[Out]

(B*b*c^2 - A*c^3)*log(c*x + b)/(b*c^3*d^3 - 3*b^2*c^2*d^2*e + 3*b^3*c*d*e^2 - b^4*e^3) - (B*c^2*d^3 - 3*A*c^2*
d^2*e + 3*A*b*c*d*e^2 - A*b^2*e^3)*log(e*x + d)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3) + 1/
2*(3*B*c*d^3 + 3*A*b*d*e^2 - (B*b + 5*A*c)*d^2*e + 2*(B*c*d^2*e - 2*A*c*d*e^2 + A*b*e^3)*x)/(c^2*d^6 - 2*b*c*d
^5*e + b^2*d^4*e^2 + (c^2*d^4*e^2 - 2*b*c*d^3*e^3 + b^2*d^2*e^4)*x^2 + 2*(c^2*d^5*e - 2*b*c*d^4*e^2 + b^2*d^3*
e^3)*x) + A*log(x)/(b*d^3)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**3/(c*x**2+b*x),x)

[Out]

Timed out

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Giac [A]  time = 1.24209, size = 416, normalized size = 2.43 \begin{align*} \frac{{\left (B b c^{3} - A c^{4}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{4} d^{3} - 3 \, b^{2} c^{3} d^{2} e + 3 \, b^{3} c^{2} d e^{2} - b^{4} c e^{3}} - \frac{{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} + 3 \, A b c d e^{3} - A b^{2} e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e - 3 \, b c^{2} d^{5} e^{2} + 3 \, b^{2} c d^{4} e^{3} - b^{3} d^{3} e^{4}} + \frac{A \log \left ({\left | x \right |}\right )}{b d^{3}} + \frac{3 \, B c^{2} d^{5} - 4 \, B b c d^{4} e - 5 \, A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 8 \, A b c d^{3} e^{2} - 3 \, A b^{2} d^{2} e^{3} + 2 \,{\left (B c^{2} d^{4} e - B b c d^{3} e^{2} - 2 \, A c^{2} d^{3} e^{2} + 3 \, A b c d^{2} e^{3} - A b^{2} d e^{4}\right )} x}{2 \,{\left (c d - b e\right )}^{3}{\left (x e + d\right )}^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/(c*x^2+b*x),x, algorithm="giac")

[Out]

(B*b*c^3 - A*c^4)*log(abs(c*x + b))/(b*c^4*d^3 - 3*b^2*c^3*d^2*e + 3*b^3*c^2*d*e^2 - b^4*c*e^3) - (B*c^2*d^3*e
- 3*A*c^2*d^2*e^2 + 3*A*b*c*d*e^3 - A*b^2*e^4)*log(abs(x*e + d))/(c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e
^3 - b^3*d^3*e^4) + A*log(abs(x))/(b*d^3) + 1/2*(3*B*c^2*d^5 - 4*B*b*c*d^4*e - 5*A*c^2*d^4*e + B*b^2*d^3*e^2 +
8*A*b*c*d^3*e^2 - 3*A*b^2*d^2*e^3 + 2*(B*c^2*d^4*e - B*b*c*d^3*e^2 - 2*A*c^2*d^3*e^2 + 3*A*b*c*d^2*e^3 - A*b^
2*d*e^4)*x)/((c*d - b*e)^3*(x*e + d)^2*d^3)