3.2311 \(\int \frac {(A+B x) (a+b x+c x^2)}{d+e x} \, dx\)

Optimal. Leaf size=111 \[ -\frac {x \left (A e (c d-b e)-B \left (c d^2-e (b d-a e)\right )\right )}{e^3}-\frac {(B d-A e) \log (d+e x) \left (a e^2-b d e+c d^2\right )}{e^4}-\frac {x^2 (-A c e-b B e+B c d)}{2 e^2}+\frac {B c x^3}{3 e} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 109, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {771} \[ \frac {x \left (-B e (b d-a e)-A e (c d-b e)+B c d^2\right )}{e^3}-\frac {(B d-A e) \log (d+e x) \left (a e^2-b d e+c d^2\right )}{e^4}-\frac {x^2 (-A c e-b B e+B c d)}{2 e^2}+\frac {B c x^3}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 100, normalized size = 0.90 \[ \frac {e x \left (3 B e (2 a e-2 b d+b e x)+3 A e (2 b e-2 c d+c e x)+B c \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 (B d-A e) \log (d+e x) \left (e (a e-b d)+c d^2\right )}{6 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 123, normalized size = 1.11 \[ \frac {2 \, B c e^{3} x^{3} - 3 \, {\left (B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} + 6 \, {\left (B c d^{2} e - {\left (B b + A c\right )} d e^{2} + {\left (B a + A b\right )} e^{3}\right )} x - 6 \, {\left (B c d^{3} - A a e^{3} - {\left (B b + A c\right )} d^{2} e + {\left (B a + A b\right )} d e^{2}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 136, normalized size = 1.23 \[ -{\left (B c d^{3} - B b d^{2} e - A c d^{2} e + B a d e^{2} + A b d e^{2} - A a e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, B c x^{3} e^{2} - 3 \, B c d x^{2} e + 6 \, B c d^{2} x + 3 \, B b x^{2} e^{2} + 3 \, A c x^{2} e^{2} - 6 \, B b d x e - 6 \, A c d x e + 6 \, B a x e^{2} + 6 \, A b x e^{2}\right )} e^{\left (-3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 171, normalized size = 1.54 \[ \frac {B c \,x^{3}}{3 e}+\frac {A c \,x^{2}}{2 e}+\frac {B b \,x^{2}}{2 e}-\frac {B c d \,x^{2}}{2 e^{2}}+\frac {A a \ln \left (e x +d \right )}{e}-\frac {A b d \ln \left (e x +d \right )}{e^{2}}+\frac {A b x}{e}+\frac {A c \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {A c d x}{e^{2}}-\frac {B a d \ln \left (e x +d \right )}{e^{2}}+\frac {B a x}{e}+\frac {B b \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {B b d x}{e^{2}}-\frac {B c \,d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {B c \,d^{2} x}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 122, normalized size = 1.10 \[ \frac {2 \, B c e^{2} x^{3} - 3 \, {\left (B c d e - {\left (B b + A c\right )} e^{2}\right )} x^{2} + 6 \, {\left (B c d^{2} - {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} x}{6 \, e^{3}} - \frac {{\left (B c d^{3} - A a e^{3} - {\left (B b + A c\right )} d^{2} e + {\left (B a + A b\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.37, size = 130, normalized size = 1.17 \[ x^2\,\left (\frac {A\,c+B\,b}{2\,e}-\frac {B\,c\,d}{2\,e^2}\right )+x\,\left (\frac {A\,b+B\,a}{e}-\frac {d\,\left (\frac {A\,c+B\,b}{e}-\frac {B\,c\,d}{e^2}\right )}{e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (A\,a\,e^3-B\,c\,d^3-A\,b\,d\,e^2-B\,a\,d\,e^2+A\,c\,d^2\,e+B\,b\,d^2\,e\right )}{e^4}+\frac {B\,c\,x^3}{3\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.43, size = 107, normalized size = 0.96 \[ \frac {B c x^{3}}{3 e} + x^{2} \left (\frac {A c}{2 e} + \frac {B b}{2 e} - \frac {B c d}{2 e^{2}}\right ) + x \left (\frac {A b}{e} - \frac {A c d}{e^{2}} + \frac {B a}{e} - \frac {B b d}{e^{2}} + \frac {B c d^{2}}{e^{3}}\right ) - \frac {\left (- A e + B d\right ) \left (a e^{2} - b d e + c d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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