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

Optimal. Leaf size=119 \[ \frac {A e (2 c d-b e)-B \left (3 c d^2-e (2 b d-a e)\right )}{e^4 (d+e x)}+\frac {(B d-A e) \left (a e^2-b d e+c d^2\right )}{2 e^4 (d+e x)^2}-\frac {\log (d+e x) (-A c e-b B e+3 B c d)}{e^4}+\frac {B c x}{e^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {(B d-A e) \left (a e^2-b d e+c d^2\right )}{2 e^4 (d+e x)^2}-\frac {-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2}{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)*(a + b*x + c*x^2))/(d + e*x)^3,x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 135, normalized size = 1.13 \begin {gather*} \frac {-a B e^2-A b e^2+2 A c d e+2 b B d e-3 B c d^2}{e^4 (d+e x)}+\frac {-a A e^3+a B d e^2+A b d e^2-A c d^2 e-b B d^2 e+B c d^3}{2 e^4 (d+e x)^2}+\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]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a+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)*(a + b*x + c*x^2))/(d + e*x)^3,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.39, size = 203, normalized size = 1.71 \begin {gather*} \frac {2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} - A a e^{3} + 3 \, {\left (B b + A c\right )} d^{2} e - {\left (B a + A b\right )} d e^{2} - 2 \, {\left (2 \, B c d^{2} e - 2 \, {\left (B b + A c\right )} d e^{2} + {\left (B a + A b\right )} e^{3}\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+a)/(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*a*e^3 + 3*(B*b + A*c)*d^2*e - (B*a + A*b)*d*e^2 - 2*(2*B*
c*d^2*e - 2*(B*b + A*c)*d*e^2 + (B*a + A*b)*e^3)*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)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 129, normalized size = 1.08 \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 + B a d e^{2} + A b d e^{2} + A a e^{3} + 2 \, {\left (3 \, B c d^{2} e - 2 \, B b d e^{2} - 2 \, A c d e^{2} + B a e^{3} + 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+a)/(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
+ B*a*d*e^2 + A*b*d*e^2 + A*a*e^3 + 2*(3*B*c*d^2*e - 2*B*b*d*e^2 - 2*A*c*d*e^2 + B*a*e^3 + A*b*e^3)*x)*e^(-4)
/(x*e + d)^2

________________________________________________________________________________________

maple [A]  time = 0.05, size = 217, normalized size = 1.82 \begin {gather*} -\frac {A a}{2 \left (e x +d \right )^{2} e}+\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 a d}{2 \left (e x +d \right )^{2} e^{2}}-\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 {B a}{\left (e x +d \right ) e^{2}}+\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+a)/(e*x+d)^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 136, normalized size = 1.14 \begin {gather*} -\frac {5 \, B c d^{3} + A a e^{3} - 3 \, {\left (B b + A c\right )} d^{2} e + {\left (B a + A b\right )} d e^{2} + 2 \, {\left (3 \, B c d^{2} e - 2 \, {\left (B b + A c\right )} d e^{2} + {\left (B a + A b\right )} e^{3}\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+a)/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.12, size = 142, normalized size = 1.19 \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+B\,a\,e^2+3\,B\,c\,d^2-2\,A\,c\,d\,e-2\,B\,b\,d\,e\right )+\frac {A\,a\,e^3+5\,B\,c\,d^3+A\,b\,d\,e^2+B\,a\,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(((A + B*x)*(a + b*x + c*x^2))/(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 + B*a*e^2 + 3*B*c*d^2 - 2*A*c*d*e - 2*B*b*d*e) + (A
*a*e^3 + 5*B*c*d^3 + A*b*d*e^2 + B*a*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

________________________________________________________________________________________

sympy [A]  time = 4.70, size = 162, normalized size = 1.36 \begin {gather*} \frac {B c x}{e^{3}} + \frac {- A a e^{3} - A b d e^{2} + 3 A c d^{2} e - B a d e^{2} + 3 B b d^{2} e - 5 B c d^{3} + x \left (- 2 A b e^{3} + 4 A c d e^{2} - 2 B a e^{3} + 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+a)/(e*x+d)**3,x)

[Out]

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

________________________________________________________________________________________