3.24.12 \(\int \frac {(A+B x) (a+b x+c x^2)}{(d+e x)^2} \, dx\) [2312]

Optimal. Leaf size=116 \[ -\frac {(2 B c d-b B e-A c e) x}{e^3}+\frac {B c x^2}{2 e^2}+\frac {(B d-A e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)}-\frac {\left (A e (2 c d-b e)-B \left (3 c d^2-e (2 b d-a e)\right )\right ) \log (d+e x)}{e^4} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 114, 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 = {785} \begin {gather*} \frac {\log (d+e x) \left (-B e (2 b d-a e)-A e (2 c d-b e)+3 B c d^2\right )}{e^4}+\frac {(B d-A e) \left (a e^2-b d e+c d^2\right )}{e^4 (d+e x)}-\frac {x (-A c e-b B e+2 B c d)}{e^3}+\frac {B c x^2}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

Rule 785

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

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Mathematica [A]
time = 0.06, size = 106, normalized size = 0.91 \begin {gather*} \frac {2 e (-2 B c d+b B e+A c e) x+B c e^2 x^2+\frac {2 (B d-A e) \left (c d^2+e (-b d+a e)\right )}{d+e x}+2 \left (3 B c d^2+B e (-2 b d+a e)+A e (-2 c d+b e)\right ) \log (d+e x)}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [A]
time = 0.02, size = 129, normalized size = 1.11

method	result	size
default	\(\frac {\frac {1}{2} B c e \,x^{2}+A c e x +b B e x -2 B c d x}{e^{3}}+\frac {\left (A b \,e^{2}-2 A c d e +a B \,e^{2}-2 B b d e +3 B c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{4}}-\frac {a A \,e^{3}-A b d \,e^{2}+A c \,d^{2} e -a B d \,e^{2}+B b \,d^{2} e -B c \,d^{3}}{e^{4} \left (e x +d \right )}\)	\(129\)
norman	\(\frac {\frac {\left (a A \,e^{3}-A b d \,e^{2}+2 A c \,d^{2} e -a B d \,e^{2}+2 B b \,d^{2} e -3 B c \,d^{3}\right ) x}{e^{3} d}+\frac {\left (2 A c e +2 b B e -3 B c d \right ) x^{2}}{2 e^{2}}+\frac {B c \,x^{3}}{2 e}}{e x +d}+\frac {\left (A b \,e^{2}-2 A c d e +a B \,e^{2}-2 B b d e +3 B c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{4}}\)	\(141\)
risch	\(\frac {B c \,x^{2}}{2 e^{2}}+\frac {A c x}{e^{2}}+\frac {b B x}{e^{2}}-\frac {2 B c d x}{e^{3}}+\frac {\ln \left (e x +d \right ) A b}{e^{2}}-\frac {2 \ln \left (e x +d \right ) A c d}{e^{3}}+\frac {\ln \left (e x +d \right ) a B}{e^{2}}-\frac {2 \ln \left (e x +d \right ) B b d}{e^{3}}+\frac {3 \ln \left (e x +d \right ) B c \,d^{2}}{e^{4}}-\frac {a A}{e \left (e x +d \right )}+\frac {A b d}{e^{2} \left (e x +d \right )}-\frac {A c \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {a B d}{e^{2} \left (e x +d \right )}-\frac {B b \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {B c \,d^{3}}{e^{4} \left (e x +d \right )}\)	\(195\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 131, normalized size = 1.13 \begin {gather*} {\left (3 \, B c d^{2} + B a e^{2} + A b e^{2} - 2 \, {\left (B b e + A c e\right )} d\right )} e^{\left (-4\right )} \log \left (x e + d\right ) + \frac {1}{2} \, {\left (B c x^{2} e - 2 \, {\left (2 \, B c d - B b e - A c e\right )} x\right )} e^{\left (-3\right )} + \frac {B c d^{3} - {\left (B b e + A c e\right )} d^{2} - A a e^{3} + {\left (B a e^{2} + A b e^{2}\right )} d}{x e^{5} + d 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.42, size = 183, normalized size = 1.58 \begin {gather*} \frac {2 \, B c d^{3} + {\left (B c x^{3} + 2 \, {\left (B b + A c\right )} x^{2} - 2 \, A a\right )} e^{3} - {\left (3 \, B c d x^{2} - 2 \, {\left (B b + A c\right )} d x - 2 \, {\left (B a + A b\right )} d\right )} e^{2} - 2 \, {\left (2 \, B c d^{2} x + {\left (B b + A c\right )} d^{2}\right )} e + 2 \, {\left (3 \, B c d^{3} + {\left (B a + A b\right )} x e^{3} - {\left (2 \, {\left (B b + A c\right )} d x - {\left (B a + A b\right )} d\right )} e^{2} + {\left (3 \, B c d^{2} x - 2 \, {\left (B b + A c\right )} d^{2}\right )} e\right )} \log \left (x e + d\right )}{2 \, {\left (x e^{5} + 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+a)/(e*x+d)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.52, size = 143, normalized size = 1.23 \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 a e^{3} + A b d e^{2} - A c d^{2} e + B a d e^{2} - 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 + B a e^{2} - 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+a)/(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*a*e**3 + A*b*d*e**2 - A*c*d**2*e + B*a*d*e**2
- B*b*d**2*e + B*c*d**3)/(d*e**4 + e**5*x) + (A*b*e**2 - 2*A*c*d*e + B*a*e**2 - 2*B*b*d*e + 3*B*c*d**2)*log(d
+ e*x)/e**4

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Giac [A]
time = 1.26, size = 200, normalized size = 1.72 \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 + B a e^{2} + 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 {B a d e^{4}}{x e + d} + \frac {A b d e^{4}}{x e + d} - \frac {A a e^{5}}{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+a)/(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 + B*a*e^2 + 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) + B*a*d*e^4/(x*e + d) + A*b*d*e^4/(x*e + d) - A*a*e^5/(x*e + d))*e^(-6)

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Mupad [B]
time = 0.09, size = 137, normalized size = 1.18 \begin {gather*} x\,\left (\frac {A\,c+B\,b}{e^2}-\frac {2\,B\,c\,d}{e^3}\right )-\frac {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\,\left (x\,e^4+d\,e^3\right )}+\frac {\ln \left (d+e\,x\right )\,\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 )}{e^4}+\frac {B\,c\,x^2}{2\,e^2} \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)^2,x)

[Out]

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

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