3.1119 \(\int \frac{(A+B x) (b x+c x^2)^2}{d+e x} \, dx\)

Optimal. Leaf size=161 \[ -\frac{d^2 (B d-A e) (c d-b e)^2 \log (d+e x)}{e^6}-\frac{c x^4 (-A c e-2 b B e+B c d)}{4 e^2}-\frac{x^3 \left (A c e (c d-2 b e)-B (c d-b e)^2\right )}{3 e^3}-\frac{x^2 (B d-A e) (c d-b e)^2}{2 e^4}+\frac{d x (B d-A e) (c d-b e)^2}{e^5}+\frac{B c^2 x^5}{5 e} \]

[Out]

(d*(B*d - A*e)*(c*d - b*e)^2*x)/e^5 - ((B*d - A*e)*(c*d - b*e)^2*x^2)/(2*e^4) - ((A*c*e*(c*d - 2*b*e) - B*(c*d
- b*e)^2)*x^3)/(3*e^3) - (c*(B*c*d - 2*b*B*e - A*c*e)*x^4)/(4*e^2) + (B*c^2*x^5)/(5*e) - (d^2*(B*d - A*e)*(c*
d - b*e)^2*Log[d + e*x])/e^6

________________________________________________________________________________________

Rubi [A]  time = 0.225691, antiderivative size = 161, 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} \[ -\frac{d^2 (B d-A e) (c d-b e)^2 \log (d+e x)}{e^6}-\frac{c x^4 (-A c e-2 b B e+B c d)}{4 e^2}-\frac{x^3 \left (A c e (c d-2 b e)-B (c d-b e)^2\right )}{3 e^3}-\frac{x^2 (B d-A e) (c d-b e)^2}{2 e^4}+\frac{d x (B d-A e) (c d-b e)^2}{e^5}+\frac{B c^2 x^5}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(B*d - A*e)*(c*d - b*e)^2*x)/e^5 - ((B*d - A*e)*(c*d - b*e)^2*x^2)/(2*e^4) - ((A*c*e*(c*d - 2*b*e) - B*(c*d
- b*e)^2)*x^3)/(3*e^3) - (c*(B*c*d - 2*b*B*e - A*c*e)*x^4)/(4*e^2) + (B*c^2*x^5)/(5*e) - (d^2*(B*d - A*e)*(c*
d - b*e)^2*Log[d + e*x])/e^6

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) \left (b x+c x^2\right )^2}{d+e x} \, dx &=\int \left (\frac{d (B d-A e) (c d-b e)^2}{e^5}+\frac{(-B d+A e) (-c d+b e)^2 x}{e^4}+\frac{\left (-A c e (c d-2 b e)+B (c d-b e)^2\right ) x^2}{e^3}+\frac{c (-B c d+2 b B e+A c e) x^3}{e^2}+\frac{B c^2 x^4}{e}-\frac{d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)}\right ) \, dx\\ &=\frac{d (B d-A e) (c d-b e)^2 x}{e^5}-\frac{(B d-A e) (c d-b e)^2 x^2}{2 e^4}-\frac{\left (A c e (c d-2 b e)-B (c d-b e)^2\right ) x^3}{3 e^3}-\frac{c (B c d-2 b B e-A c e) x^4}{4 e^2}+\frac{B c^2 x^5}{5 e}-\frac{d^2 (B d-A e) (c d-b e)^2 \log (d+e x)}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.156124, size = 156, normalized size = 0.97 \[ \frac{-60 d^2 (B d-A e) (c d-b e)^2 \log (d+e x)+15 c e^4 x^4 (A c e+2 b B e-B c d)+20 e^3 x^3 \left (A c e (2 b e-c d)+B (c d-b e)^2\right )+30 e^2 x^2 (A e-B d) (c d-b e)^2+60 d e x (B d-A e) (c d-b e)^2+12 B c^2 e^5 x^5}{60 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*d*e*(B*d - A*e)*(c*d - b*e)^2*x + 30*e^2*(-(B*d) + A*e)*(c*d - b*e)^2*x^2 + 20*e^3*(B*(c*d - b*e)^2 + A*c*
e*(-(c*d) + 2*b*e))*x^3 + 15*c*e^4*(-(B*c*d) + 2*b*B*e + A*c*e)*x^4 + 12*B*c^2*e^5*x^5 - 60*d^2*(B*d - A*e)*(c
*d - b*e)^2*Log[d + e*x])/(60*e^6)

________________________________________________________________________________________

Maple [B]  time = 0.006, size = 369, normalized size = 2.3 \begin{align*}{\frac{B{x}^{4}bc}{2\,e}}-{\frac{B{c}^{2}{x}^{4}d}{4\,{e}^{2}}}+{\frac{2\,A{x}^{3}bc}{3\,e}}-{\frac{A{c}^{2}{x}^{3}d}{3\,{e}^{2}}}+{\frac{B{c}^{2}{x}^{3}{d}^{2}}{3\,{e}^{3}}}+{\frac{A{c}^{2}{x}^{2}{d}^{2}}{2\,{e}^{3}}}-{\frac{{b}^{2}B{x}^{2}d}{2\,{e}^{2}}}-{\frac{B{c}^{2}{x}^{2}{d}^{3}}{2\,{e}^{4}}}-{\frac{A{b}^{2}dx}{{e}^{2}}}+{\frac{{b}^{2}B{d}^{2}x}{{e}^{3}}}+{\frac{B{c}^{2}{d}^{4}x}{{e}^{5}}}+{\frac{{d}^{2}\ln \left ( ex+d \right ) A{b}^{2}}{{e}^{3}}}+{\frac{{d}^{4}\ln \left ( ex+d \right ) A{c}^{2}}{{e}^{5}}}-{\frac{{d}^{3}\ln \left ( ex+d \right ) B{b}^{2}}{{e}^{4}}}-{\frac{{d}^{5}\ln \left ( ex+d \right ) B{c}^{2}}{{e}^{6}}}-{\frac{A{d}^{3}{c}^{2}x}{{e}^{4}}}+{\frac{B{c}^{2}{x}^{5}}{5\,e}}+2\,{\frac{{d}^{4}\ln \left ( ex+d \right ) Bbc}{{e}^{5}}}-2\,{\frac{{d}^{3}\ln \left ( ex+d \right ) Abc}{{e}^{4}}}+2\,{\frac{A{d}^{2}bcx}{{e}^{3}}}-2\,{\frac{Bcb{d}^{3}x}{{e}^{4}}}-{\frac{2\,B{x}^{3}bcd}{3\,{e}^{2}}}-{\frac{A{x}^{2}bcd}{{e}^{2}}}+{\frac{B{x}^{2}bc{d}^{2}}{{e}^{3}}}+{\frac{A{c}^{2}{x}^{4}}{4\,e}}+{\frac{{b}^{2}B{x}^{3}}{3\,e}}+{\frac{A{b}^{2}{x}^{2}}{2\,e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.05081, size = 381, normalized size = 2.37 \begin{align*} \frac{12 \, B c^{2} e^{4} x^{5} - 15 \,{\left (B c^{2} d e^{3} -{\left (2 \, B b c + A c^{2}\right )} e^{4}\right )} x^{4} + 20 \,{\left (B c^{2} d^{2} e^{2} -{\left (2 \, B b c + A c^{2}\right )} d e^{3} +{\left (B b^{2} + 2 \, A b c\right )} e^{4}\right )} x^{3} - 30 \,{\left (B c^{2} d^{3} e - A b^{2} e^{4} -{\left (2 \, B b c + A c^{2}\right )} d^{2} e^{2} +{\left (B b^{2} + 2 \, A b c\right )} d e^{3}\right )} x^{2} + 60 \,{\left (B c^{2} d^{4} - A b^{2} d e^{3} -{\left (2 \, B b c + A c^{2}\right )} d^{3} e +{\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{2}\right )} x}{60 \, e^{5}} - \frac{{\left (B c^{2} d^{5} - A b^{2} d^{2} e^{3} -{\left (2 \, B b c + A c^{2}\right )} d^{4} e +{\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2}\right )} \log \left (e x + d\right )}{e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*c^2*e^4*x^5 - 15*(B*c^2*d*e^3 - (2*B*b*c + A*c^2)*e^4)*x^4 + 20*(B*c^2*d^2*e^2 - (2*B*b*c + A*c^2)*
d*e^3 + (B*b^2 + 2*A*b*c)*e^4)*x^3 - 30*(B*c^2*d^3*e - A*b^2*e^4 - (2*B*b*c + A*c^2)*d^2*e^2 + (B*b^2 + 2*A*b*
c)*d*e^3)*x^2 + 60*(B*c^2*d^4 - A*b^2*d*e^3 - (2*B*b*c + A*c^2)*d^3*e + (B*b^2 + 2*A*b*c)*d^2*e^2)*x)/e^5 - (B
*c^2*d^5 - A*b^2*d^2*e^3 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2)*log(e*x + d)/e^6

________________________________________________________________________________________

Fricas [A]  time = 1.39688, size = 585, normalized size = 3.63 \begin{align*} \frac{12 \, B c^{2} e^{5} x^{5} - 15 \,{\left (B c^{2} d e^{4} -{\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 20 \,{\left (B c^{2} d^{2} e^{3} -{\left (2 \, B b c + A c^{2}\right )} d e^{4} +{\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} - 30 \,{\left (B c^{2} d^{3} e^{2} - A b^{2} e^{5} -{\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} +{\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} + 60 \,{\left (B c^{2} d^{4} e - A b^{2} d e^{4} -{\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} +{\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x - 60 \,{\left (B c^{2} d^{5} - A b^{2} d^{2} e^{3} -{\left (2 \, B b c + A c^{2}\right )} d^{4} e +{\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*c^2*e^5*x^5 - 15*(B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4 + 20*(B*c^2*d^2*e^3 - (2*B*b*c + A*c^2)*
d*e^4 + (B*b^2 + 2*A*b*c)*e^5)*x^3 - 30*(B*c^2*d^3*e^2 - A*b^2*e^5 - (2*B*b*c + A*c^2)*d^2*e^3 + (B*b^2 + 2*A*
b*c)*d*e^4)*x^2 + 60*(B*c^2*d^4*e - A*b^2*d*e^4 - (2*B*b*c + A*c^2)*d^3*e^2 + (B*b^2 + 2*A*b*c)*d^2*e^3)*x - 6
0*(B*c^2*d^5 - A*b^2*d^2*e^3 - (2*B*b*c + A*c^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2)*log(e*x + d))/e^6

________________________________________________________________________________________

Sympy [A]  time = 1.38764, size = 265, normalized size = 1.65 \begin{align*} \frac{B c^{2} x^{5}}{5 e} - \frac{d^{2} \left (- A e + B d\right ) \left (b e - c d\right )^{2} \log{\left (d + e x \right )}}{e^{6}} + \frac{x^{4} \left (A c^{2} e + 2 B b c e - B c^{2} d\right )}{4 e^{2}} + \frac{x^{3} \left (2 A b c e^{2} - A c^{2} d e + B b^{2} e^{2} - 2 B b c d e + B c^{2} d^{2}\right )}{3 e^{3}} - \frac{x^{2} \left (- A b^{2} e^{3} + 2 A b c d e^{2} - A c^{2} d^{2} e + B b^{2} d e^{2} - 2 B b c d^{2} e + B c^{2} d^{3}\right )}{2 e^{4}} + \frac{x \left (- A b^{2} d e^{3} + 2 A b c d^{2} e^{2} - A c^{2} d^{3} e + B b^{2} d^{2} e^{2} - 2 B b c d^{3} e + B c^{2} d^{4}\right )}{e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.23143, size = 435, normalized size = 2.7 \begin{align*} -{\left (B c^{2} d^{5} - 2 \, B b c d^{4} e - A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 2 \, A b c d^{3} e^{2} - A b^{2} d^{2} e^{3}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{60} \,{\left (12 \, B c^{2} x^{5} e^{4} - 15 \, B c^{2} d x^{4} e^{3} + 20 \, B c^{2} d^{2} x^{3} e^{2} - 30 \, B c^{2} d^{3} x^{2} e + 60 \, B c^{2} d^{4} x + 30 \, B b c x^{4} e^{4} + 15 \, A c^{2} x^{4} e^{4} - 40 \, B b c d x^{3} e^{3} - 20 \, A c^{2} d x^{3} e^{3} + 60 \, B b c d^{2} x^{2} e^{2} + 30 \, A c^{2} d^{2} x^{2} e^{2} - 120 \, B b c d^{3} x e - 60 \, A c^{2} d^{3} x e + 20 \, B b^{2} x^{3} e^{4} + 40 \, A b c x^{3} e^{4} - 30 \, B b^{2} d x^{2} e^{3} - 60 \, A b c d x^{2} e^{3} + 60 \, B b^{2} d^{2} x e^{2} + 120 \, A b c d^{2} x e^{2} + 30 \, A b^{2} x^{2} e^{4} - 60 \, A b^{2} d x e^{3}\right )} e^{\left (-5\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*c^2*d^5 - 2*B*b*c*d^4*e - A*c^2*d^4*e + B*b^2*d^3*e^2 + 2*A*b*c*d^3*e^2 - A*b^2*d^2*e^3)*e^(-6)*log(abs(x*
e + d)) + 1/60*(12*B*c^2*x^5*e^4 - 15*B*c^2*d*x^4*e^3 + 20*B*c^2*d^2*x^3*e^2 - 30*B*c^2*d^3*x^2*e + 60*B*c^2*d
^4*x + 30*B*b*c*x^4*e^4 + 15*A*c^2*x^4*e^4 - 40*B*b*c*d*x^3*e^3 - 20*A*c^2*d*x^3*e^3 + 60*B*b*c*d^2*x^2*e^2 +
30*A*c^2*d^2*x^2*e^2 - 120*B*b*c*d^3*x*e - 60*A*c^2*d^3*x*e + 20*B*b^2*x^3*e^4 + 40*A*b*c*x^3*e^4 - 30*B*b^2*d
*x^2*e^3 - 60*A*b*c*d*x^2*e^3 + 60*B*b^2*d^2*x*e^2 + 120*A*b*c*d^2*x*e^2 + 30*A*b^2*x^2*e^4 - 60*A*b^2*d*x*e^3
)*e^(-5)