3.1147 \(\int \frac{(A+B x) (d+e x)^3}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=128 \[ \frac{(b B-A c) (c d-b e)^3}{b^2 c^3 (b+c x)}+\frac{(c d-b e)^2 \log (b+c x) \left (-b c (B d-A e)+2 A c^2 d-2 b^2 B e\right )}{b^3 c^3}+\frac{d^2 \log (x) (3 A b e-2 A c d+b B d)}{b^3}-\frac{A d^3}{b^2 x}+\frac{B e^3 x}{c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.158671, antiderivative size = 128, 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} \[ \frac{(b B-A c) (c d-b e)^3}{b^2 c^3 (b+c x)}+\frac{(c d-b e)^2 \log (b+c x) \left (-b c (B d-A e)+2 A c^2 d-2 b^2 B e\right )}{b^3 c^3}+\frac{d^2 \log (x) (3 A b e-2 A c d+b B d)}{b^3}-\frac{A d^3}{b^2 x}+\frac{B e^3 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0778754, size = 128, normalized size = 1. \[ -\frac{(b B-A c) (b e-c d)^3}{b^2 c^3 (b+c x)}+\frac{(c d-b e)^2 \log (b+c x) \left (A b c e+2 A c^2 d-2 b^2 B e-b B c d\right )}{b^3 c^3}+\frac{d^2 \log (x) (3 A b e-2 A c d+b B d)}{b^3}-\frac{A d^3}{b^2 x}+\frac{B e^3 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.013, size = 286, normalized size = 2.2 \begin{align*}{\frac{B{e}^{3}x}{{c}^{2}}}-{\frac{A{d}^{3}}{{b}^{2}x}}+3\,{\frac{{d}^{2}\ln \left ( x \right ) Ae}{{b}^{2}}}-2\,{\frac{{d}^{3}\ln \left ( x \right ) Ac}{{b}^{3}}}+{\frac{{d}^{3}\ln \left ( x \right ) B}{{b}^{2}}}+{\frac{\ln \left ( cx+b \right ) A{e}^{3}}{{c}^{2}}}-3\,{\frac{\ln \left ( cx+b \right ) A{d}^{2}e}{{b}^{2}}}+2\,{\frac{c\ln \left ( cx+b \right ) A{d}^{3}}{{b}^{3}}}-2\,{\frac{b\ln \left ( cx+b \right ) B{e}^{3}}{{c}^{3}}}+3\,{\frac{\ln \left ( cx+b \right ) Bd{e}^{2}}{{c}^{2}}}-{\frac{\ln \left ( cx+b \right ) B{d}^{3}}{{b}^{2}}}+{\frac{Ab{e}^{3}}{{c}^{2} \left ( cx+b \right ) }}-3\,{\frac{Ad{e}^{2}}{c \left ( cx+b \right ) }}+3\,{\frac{A{d}^{2}e}{b \left ( cx+b \right ) }}-{\frac{A{d}^{3}c}{{b}^{2} \left ( cx+b \right ) }}-{\frac{B{e}^{3}{b}^{2}}{{c}^{3} \left ( cx+b \right ) }}+3\,{\frac{Bbd{e}^{2}}{{c}^{2} \left ( cx+b \right ) }}-3\,{\frac{B{d}^{2}e}{c \left ( cx+b \right ) }}+{\frac{B{d}^{3}}{b \left ( cx+b \right ) }} \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)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.09645, size = 304, normalized size = 2.38 \begin{align*} \frac{B e^{3} x}{c^{2}} - \frac{A b c^{3} d^{3} -{\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} - 3 \,{\left (B b^{2} c^{2} - A b c^{3}\right )} d^{2} e + 3 \,{\left (B b^{3} c - A b^{2} c^{2}\right )} d e^{2} -{\left (B b^{4} - A b^{3} c\right )} e^{3}\right )} x}{b^{2} c^{4} x^{2} + b^{3} c^{3} x} + \frac{{\left (3 \, A b d^{2} e +{\left (B b - 2 \, A c\right )} d^{3}\right )} \log \left (x\right )}{b^{3}} - \frac{{\left (3 \, A b c^{3} d^{2} e - 3 \, B b^{3} c d e^{2} +{\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} +{\left (2 \, B b^{4} - A b^{3} c\right )} e^{3}\right )} \log \left (c x + b\right )}{b^{3} c^{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)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.50293, size = 717, normalized size = 5.6 \begin{align*} \frac{B b^{3} c^{2} e^{3} x^{3} + B b^{4} c e^{3} x^{2} - A b^{2} c^{3} d^{3} +{\left ({\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3} - 3 \,{\left (B b^{3} c^{2} - A b^{2} c^{3}\right )} d^{2} e + 3 \,{\left (B b^{4} c - A b^{3} c^{2}\right )} d e^{2} -{\left (B b^{5} - A b^{4} c\right )} e^{3}\right )} x -{\left ({\left (3 \, A b c^{4} d^{2} e - 3 \, B b^{3} c^{2} d e^{2} +{\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} +{\left (2 \, B b^{4} c - A b^{3} c^{2}\right )} e^{3}\right )} x^{2} +{\left (3 \, A b^{2} c^{3} d^{2} e - 3 \, B b^{4} c d e^{2} +{\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3} +{\left (2 \, B b^{5} - A b^{4} c\right )} e^{3}\right )} x\right )} \log \left (c x + b\right ) +{\left ({\left (3 \, A b c^{4} d^{2} e +{\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3}\right )} x^{2} +{\left (3 \, A b^{2} c^{3} d^{2} e +{\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3}\right )} x\right )} \log \left (x\right )}{b^{3} c^{4} x^{2} + b^{4} c^{3} x} \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)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 11.9035, size = 502, normalized size = 3.92 \begin{align*} \frac{B e^{3} x}{c^{2}} - \frac{A b c^{3} d^{3} + x \left (- A b^{3} c e^{3} + 3 A b^{2} c^{2} d e^{2} - 3 A b c^{3} d^{2} e + 2 A c^{4} d^{3} + B b^{4} e^{3} - 3 B b^{3} c d e^{2} + 3 B b^{2} c^{2} d^{2} e - B b c^{3} d^{3}\right )}{b^{3} c^{3} x + b^{2} c^{4} x^{2}} + \frac{d^{2} \left (3 A b e - 2 A c d + B b d\right ) \log{\left (x + \frac{3 A b^{2} c^{2} d^{2} e - 2 A b c^{3} d^{3} + B b^{2} c^{2} d^{3} - b c^{2} d^{2} \left (3 A b e - 2 A c d + B b d\right )}{- A b^{3} c e^{3} + 6 A b c^{3} d^{2} e - 4 A c^{4} d^{3} + 2 B b^{4} e^{3} - 3 B b^{3} c d e^{2} + 2 B b c^{3} d^{3}} \right )}}{b^{3}} - \frac{\left (b e - c d\right )^{2} \left (- A b c e - 2 A c^{2} d + 2 B b^{2} e + B b c d\right ) \log{\left (x + \frac{3 A b^{2} c^{2} d^{2} e - 2 A b c^{3} d^{3} + B b^{2} c^{2} d^{3} + \frac{b \left (b e - c d\right )^{2} \left (- A b c e - 2 A c^{2} d + 2 B b^{2} e + B b c d\right )}{c}}{- A b^{3} c e^{3} + 6 A b c^{3} d^{2} e - 4 A c^{4} d^{3} + 2 B b^{4} e^{3} - 3 B b^{3} c d e^{2} + 2 B b c^{3} d^{3}} \right )}}{b^{3} c^{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)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.31509, size = 309, normalized size = 2.41 \begin{align*} \frac{B x e^{3}}{c^{2}} + \frac{{\left (B b d^{3} - 2 \, A c d^{3} + 3 \, A b d^{2} e\right )} \log \left ({\left | x \right |}\right )}{b^{3}} - \frac{{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} + 3 \, A b c^{3} d^{2} e - 3 \, B b^{3} c d e^{2} + 2 \, B b^{4} e^{3} - A b^{3} c e^{3}\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c^{3}} - \frac{A b c^{2} d^{3} - \frac{{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 3 \, A b^{2} c^{2} d e^{2} - B b^{4} e^{3} + A b^{3} c e^{3}\right )} x}{c}}{{\left (c x + b\right )} b^{2} c^{2} x} \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)^2,x, algorithm="giac")

[Out]

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