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

Optimal. Leaf size=99 \[ \frac {d (B d-A e) (c d-b e)}{e^4 (d+e x)}+\frac {\log (d+e x) (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4}-\frac {x (-A c e-b B e+2 B c d)}{e^3}+\frac {B c x^2}{2 e^2} \]

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Rubi [A]  time = 0.11, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {771} \begin {gather*} -\frac {x (-A c e-b B e+2 B c d)}{e^3}+\frac {d (B d-A e) (c d-b e)}{e^4 (d+e x)}+\frac {\log (d+e x) (B d (3 c d-2 b e)-A e (2 c d-b e))}{e^4}+\frac {B c x^2}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 93, normalized size = 0.94 \begin {gather*} \frac {2 e x (A c e+b B e-2 B c d)+\frac {2 d (B d-A e) (c d-b e)}{d+e x}+2 \log (d+e x) (A e (b e-2 c d)+B d (3 c d-2 b e))+B c e^2 x^2}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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 )}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 170, normalized size = 1.72 \begin {gather*} \frac {B c e^{3} x^{3} + 2 \, B c d^{3} + 2 \, A b d e^{2} - 2 \, {\left (B b + A c\right )} d^{2} e - {\left (3 \, B c d e^{2} - 2 \, {\left (B b + A c\right )} e^{3}\right )} x^{2} - 2 \, {\left (2 \, B c d^{2} e - {\left (B b + A c\right )} d e^{2}\right )} x + 2 \, {\left (3 \, B c d^{3} + A b d e^{2} - 2 \, {\left (B b + A c\right )} d^{2} e + {\left (3 \, B c d^{2} e + A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x + d e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 167, normalized size = 1.69 \begin {gather*} \frac {1}{2} \, {\left (B c - \frac {2 \, {\left (3 \, B c d e - B b e^{2} - A c e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )} {\left (x e + d\right )}^{2} e^{\left (-4\right )} - {\left (3 \, B c d^{2} - 2 \, B b d e - 2 \, A c d e + A b e^{2}\right )} e^{\left (-4\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (\frac {B c d^{3} e^{2}}{x e + d} - \frac {B b d^{2} e^{3}}{x e + d} - \frac {A c d^{2} e^{3}}{x e + d} + \frac {A b d e^{4}}{x e + d}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 155, normalized size = 1.57 \begin {gather*} \frac {B c \,x^{2}}{2 e^{2}}+\frac {A b d}{\left (e x +d \right ) e^{2}}+\frac {A b \ln \left (e x +d \right )}{e^{2}}-\frac {A c \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 A c d \ln \left (e x +d \right )}{e^{3}}+\frac {A c x}{e^{2}}-\frac {B b \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 B b d \ln \left (e x +d \right )}{e^{3}}+\frac {B b x}{e^{2}}+\frac {B c \,d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 B c \,d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {2 B c d x}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 109, normalized size = 1.10 \begin {gather*} \frac {B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e}{e^{5} x + d e^{4}} + \frac {B c e x^{2} - 2 \, {\left (2 \, B c d - {\left (B b + A c\right )} e\right )} x}{2 \, e^{3}} + \frac {{\left (3 \, B c d^{2} + A b e^{2} - 2 \, {\left (B b + A c\right )} d e\right )} \log \left (e x + d\right )}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.39, size = 116, normalized size = 1.17 \begin {gather*} x\,\left (\frac {A\,c+B\,b}{e^2}-\frac {2\,B\,c\,d}{e^3}\right )+\frac {\ln \left (d+e\,x\right )\,\left (A\,b\,e^2+3\,B\,c\,d^2-2\,A\,c\,d\,e-2\,B\,b\,d\,e\right )}{e^4}+\frac {B\,c\,d^3+A\,b\,d\,e^2-A\,c\,d^2\,e-B\,b\,d^2\,e}{e\,\left (x\,e^4+d\,e^3\right )}+\frac {B\,c\,x^2}{2\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.66, size = 121, normalized size = 1.22 \begin {gather*} \frac {B c x^{2}}{2 e^{2}} + x \left (\frac {A c}{e^{2}} + \frac {B b}{e^{2}} - \frac {2 B c d}{e^{3}}\right ) + \frac {A b d e^{2} - A c d^{2} e - B b d^{2} e + B c d^{3}}{d e^{4} + e^{5} x} - \frac {\left (- A b e^{2} + 2 A c d e + 2 B b d e - 3 B c d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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