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

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 161, 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 {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} \end {gather*}

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 = 0.14, size = 156, normalized size = 0.97 \begin {gather*} \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} \end {gather*}

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)

________________________________________________________________________________________

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

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 283, normalized size = 1.76 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

giac [B]  time = 0.16, size = 322, normalized size = 2.00 \begin {gather*} -{\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 {gather*}

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)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 369, normalized size = 2.29 \begin {gather*} \frac {B \,c^{2} x^{5}}{5 e}+\frac {A \,c^{2} x^{4}}{4 e}+\frac {B b c \,x^{4}}{2 e}-\frac {B \,c^{2} d \,x^{4}}{4 e^{2}}+\frac {2 A b c \,x^{3}}{3 e}-\frac {A \,c^{2} d \,x^{3}}{3 e^{2}}+\frac {B \,b^{2} x^{3}}{3 e}-\frac {2 B b c d \,x^{3}}{3 e^{2}}+\frac {B \,c^{2} d^{2} x^{3}}{3 e^{3}}+\frac {A \,b^{2} x^{2}}{2 e}-\frac {A b c d \,x^{2}}{e^{2}}+\frac {A \,c^{2} d^{2} x^{2}}{2 e^{3}}-\frac {B \,b^{2} d \,x^{2}}{2 e^{2}}+\frac {B b c \,d^{2} x^{2}}{e^{3}}-\frac {B \,c^{2} d^{3} x^{2}}{2 e^{4}}+\frac {A \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {A \,b^{2} d x}{e^{2}}-\frac {2 A b c \,d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {2 A b c \,d^{2} x}{e^{3}}+\frac {A \,c^{2} d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {A \,c^{2} d^{3} x}{e^{4}}-\frac {B \,b^{2} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {B \,b^{2} d^{2} x}{e^{3}}+\frac {2 B b c \,d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {2 B b c \,d^{3} x}{e^{4}}-\frac {B \,c^{2} d^{5} \ln \left (e x +d \right )}{e^{6}}+\frac {B \,c^{2} d^{4} x}{e^{5}} \end {gather*}

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

________________________________________________________________________________________

maxima [A]  time = 0.54, size = 282, normalized size = 1.75 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

mupad [B]  time = 1.37, size = 308, normalized size = 1.91 \begin {gather*} x^4\,\left (\frac {A\,c^2+2\,B\,b\,c}{4\,e}-\frac {B\,c^2\,d}{4\,e^2}\right )+x^3\,\left (\frac {B\,b^2+2\,A\,c\,b}{3\,e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{3\,e}\right )+x^2\,\left (\frac {A\,b^2}{2\,e}-\frac {d\,\left (\frac {B\,b^2+2\,A\,c\,b}{e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}\right )}{2\,e}\right )-\frac {\ln \left (d+e\,x\right )\,\left (B\,b^2\,d^3\,e^2-A\,b^2\,d^2\,e^3-2\,B\,b\,c\,d^4\,e+2\,A\,b\,c\,d^3\,e^2+B\,c^2\,d^5-A\,c^2\,d^4\,e\right )}{e^6}-\frac {d\,x\,\left (\frac {A\,b^2}{e}-\frac {d\,\left (\frac {B\,b^2+2\,A\,c\,b}{e}-\frac {d\,\left (\frac {A\,c^2+2\,B\,b\,c}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}\right )}{e}\right )}{e}+\frac {B\,c^2\,x^5}{5\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.62, size = 280, normalized size = 1.74 \begin {gather*} \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}} + x^{4} \left (\frac {A c^{2}}{4 e} + \frac {B b c}{2 e} - \frac {B c^{2} d}{4 e^{2}}\right ) + x^{3} \left (\frac {2 A b c}{3 e} - \frac {A c^{2} d}{3 e^{2}} + \frac {B b^{2}}{3 e} - \frac {2 B b c d}{3 e^{2}} + \frac {B c^{2} d^{2}}{3 e^{3}}\right ) + x^{2} \left (\frac {A b^{2}}{2 e} - \frac {A b c d}{e^{2}} + \frac {A c^{2} d^{2}}{2 e^{3}} - \frac {B b^{2} d}{2 e^{2}} + \frac {B b c d^{2}}{e^{3}} - \frac {B c^{2} d^{3}}{2 e^{4}}\right ) + x \left (- \frac {A b^{2} d}{e^{2}} + \frac {2 A b c d^{2}}{e^{3}} - \frac {A c^{2} d^{3}}{e^{4}} + \frac {B b^{2} d^{2}}{e^{3}} - \frac {2 B b c d^{3}}{e^{4}} + \frac {B c^{2} d^{4}}{e^{5}}\right ) \end {gather*}

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

________________________________________________________________________________________