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

Optimal. Leaf size=104 \[ \frac {d (B d-A e) (c d-b e)}{2 e^4 (d+e x)^2}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^4 (d+e x)}-\frac {\log (d+e x) (-A c e-b B e+3 B c d)}{e^4}+\frac {B c x}{e^3} \]

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Rubi [A]  time = 0.11, antiderivative size = 104, 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 {d (B d-A e) (c d-b e)}{2 e^4 (d+e x)^2}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{e^4 (d+e x)}-\frac {\log (d+e x) (-A c e-b B e+3 B c d)}{e^4}+\frac {B c x}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 96, normalized size = 0.92 \begin {gather*} \frac {\frac {-2 A b e^2+4 A c d e+4 b B d e-6 B c d^2}{d+e x}+\frac {d (B d-A e) (c d-b e)}{(d+e x)^2}+2 \log (d+e x) (A c e+b B e-3 B c d)+2 B c e x}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*c*e*x + (d*(B*d - A*e)*(c*d - b*e))/(d + e*x)^2 + (-6*B*c*d^2 + 4*b*B*d*e + 4*A*c*d*e - 2*A*b*e^2)/(d + e
*x) + 2*(-3*B*c*d + b*B*e + A*c*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)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 186, normalized size = 1.79 \begin {gather*} \frac {2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} - A b d e^{2} + 3 \, {\left (B b + A c\right )} d^{2} e - 2 \, {\left (2 \, B c d^{2} e + A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x - 2 \, {\left (3 \, B c d^{3} - {\left (B b + A c\right )} d^{2} e + {\left (3 \, B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} + 2 \, {\left (3 \, B c d^{2} e - {\left (B b + A c\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 113, normalized size = 1.09 \begin {gather*} B c x e^{\left (-3\right )} - {\left (3 \, B c d - B b e - A c e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (5 \, B c d^{3} - 3 \, B b d^{2} e - 3 \, A c d^{2} e + A b d e^{2} + 2 \, {\left (3 \, B c d^{2} e - 2 \, B b d e^{2} - 2 \, A c d e^{2} + A b e^{3}\right )} x\right )} e^{\left (-4\right )}}{2 \, {\left (x e + d\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 174, normalized size = 1.67 \begin {gather*} \frac {A b d}{2 \left (e x +d \right )^{2} e^{2}}-\frac {A c \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}-\frac {B b \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {B c \,d^{3}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {A b}{\left (e x +d \right ) e^{2}}+\frac {2 A c d}{\left (e x +d \right ) e^{3}}+\frac {A c \ln \left (e x +d \right )}{e^{3}}+\frac {2 B b d}{\left (e x +d \right ) e^{3}}+\frac {B b \ln \left (e x +d \right )}{e^{3}}-\frac {3 B c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {3 B c d \ln \left (e x +d \right )}{e^{4}}+\frac {B c 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)^3,x)

[Out]

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

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maxima [A]  time = 0.49, size = 120, normalized size = 1.15 \begin {gather*} -\frac {5 \, B c d^{3} + A b d e^{2} - 3 \, {\left (B b + A c\right )} d^{2} e + 2 \, {\left (3 \, B c d^{2} e + A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - {\left (B b + A c\right )} 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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.11, size = 123, normalized size = 1.18 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,c\,e+B\,b\,e-3\,B\,c\,d\right )}{e^4}-\frac {x\,\left (A\,b\,e^2+3\,B\,c\,d^2-2\,A\,c\,d\,e-2\,B\,b\,d\,e\right )+\frac {5\,B\,c\,d^3+A\,b\,d\,e^2-3\,A\,c\,d^2\,e-3\,B\,b\,d^2\,e}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,c\,x}{e^3} \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)^3,x)

[Out]

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

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sympy [A]  time = 1.49, size = 138, normalized size = 1.33 \begin {gather*} \frac {B c x}{e^{3}} + \frac {- A b d e^{2} + 3 A c d^{2} e + 3 B b d^{2} e - 5 B c d^{3} + x \left (- 2 A b e^{3} + 4 A c d e^{2} + 4 B b d e^{2} - 6 B c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} + \frac {\left (A c e + B b e - 3 B c d\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)**3,x)

[Out]

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

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