\(\int \frac {(A+B x) (b x+c x^2)}{(d+e x)^4} \, dx\) [1111]

Optimal result

Integrand size = 22, antiderivative size = 111 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {d (B d-A e) (c d-b e)}{3 e^4 (d+e x)^3}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{2 e^4 (d+e x)^2}+\frac {3 B c d-b B e-A c e}{e^4 (d+e x)}+\frac {B c \log (d+e x)}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {785} \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {d (B d-A e) (c d-b e)}{3 e^4 (d+e x)^3}+\frac {-A c e-b B e+3 B c d}{e^4 (d+e x)}-\frac {B d (3 c d-2 b e)-A e (2 c d-b e)}{2 e^4 (d+e x)^2}+\frac {B c \log (d+e x)}{e^4} \]

[In]

Int[((A + B*x)*(b*x + c*x^2))/(d + e*x)^4,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {-A e \left (b e (d+3 e x)+2 c \left (d^2+3 d e x+3 e^2 x^2\right )\right )+B \left (-2 b e \left (d^2+3 d e x+3 e^2 x^2\right )+c d \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )+6 B c (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \]

[In]

Integrate[((A + B*x)*(b*x + c*x^2))/(d + e*x)^4,x]

[Out]

(-(A*e*(b*e*(d + 3*e*x) + 2*c*(d^2 + 3*d*e*x + 3*e^2*x^2))) + B*(-2*b*e*(d^2 + 3*d*e*x + 3*e^2*x^2) + c*d*(11*
d^2 + 27*d*e*x + 18*e^2*x^2)) + 6*B*c*(d + e*x)^3*Log[d + e*x])/(6*e^4*(d + e*x)^3)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.98

method	result	size
norman	\(\frac {-\frac {d \left (A b \,e^{2}+2 A c d e +2 B b d e -11 B c \,d^{2}\right )}{6 e^{4}}-\frac {\left (A c e +B b e -3 B c d \right ) x^{2}}{e^{2}}-\frac {\left (A b \,e^{2}+2 A c d e +2 B b d e -9 B c \,d^{2}\right ) x}{2 e^{3}}}{\left (e x +d \right )^{3}}+\frac {B c \ln \left (e x +d \right )}{e^{4}}\)	\(109\)
risch	\(\frac {-\frac {d \left (A b \,e^{2}+2 A c d e +2 B b d e -11 B c \,d^{2}\right )}{6 e^{4}}-\frac {\left (A c e +B b e -3 B c d \right ) x^{2}}{e^{2}}-\frac {\left (A b \,e^{2}+2 A c d e +2 B b d e -9 B c \,d^{2}\right ) x}{2 e^{3}}}{\left (e x +d \right )^{3}}+\frac {B c \ln \left (e x +d \right )}{e^{4}}\)	\(109\)
default	\(\frac {d \left (A b \,e^{2}-A c d e -B b d e +B c \,d^{2}\right )}{3 e^{4} \left (e x +d \right )^{3}}-\frac {A c e +B b e -3 B c d}{e^{4} \left (e x +d \right )}+\frac {B c \ln \left (e x +d \right )}{e^{4}}-\frac {A b \,e^{2}-2 A c d e -2 B b d e +3 B c \,d^{2}}{2 e^{4} \left (e x +d \right )^{2}}\)	\(116\)
parallelrisch	\(-\frac {-6 B \ln \left (e x +d \right ) x^{3} c \,e^{3}-18 B \ln \left (e x +d \right ) x^{2} c d \,e^{2}+6 A c \,e^{3} x^{2}-18 B \ln \left (e x +d \right ) x c \,d^{2} e +6 B \,x^{2} b \,e^{3}-18 B \,x^{2} c d \,e^{2}+3 A b \,e^{3} x +6 A c d \,e^{2} x -6 B \ln \left (e x +d \right ) c \,d^{3}+6 B x b d \,e^{2}-27 B c \,d^{2} e x +A b d \,e^{2}+2 A c \,d^{2} e +2 B b \,d^{2} e -11 B c \,d^{3}}{6 e^{4} \left (e x +d \right )^{3}}\)	\(171\)

[In]

int((B*x+A)*(c*x^2+b*x)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

(-1/6*d*(A*b*e^2+2*A*c*d*e+2*B*b*d*e-11*B*c*d^2)/e^4-(A*c*e+B*b*e-3*B*c*d)/e^2*x^2-1/2*(A*b*e^2+2*A*c*d*e+2*B*
b*d*e-9*B*c*d^2)/e^3*x)/(e*x+d)^3+B*c*ln(e*x+d)/e^4

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.51 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {11 \, B c d^{3} - A b d e^{2} - 2 \, {\left (B b + A c\right )} d^{2} e + 6 \, {\left (3 \, B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} e - A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x + 6 \, {\left (B c e^{3} x^{3} + 3 \, B c d e^{2} x^{2} + 3 \, B c d^{2} e x + B c d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/6*(11*B*c*d^3 - A*b*d*e^2 - 2*(B*b + A*c)*d^2*e + 6*(3*B*c*d*e^2 - (B*b + A*c)*e^3)*x^2 + 3*(9*B*c*d^2*e - A
*b*e^3 - 2*(B*b + A*c)*d*e^2)*x + 6*(B*c*e^3*x^3 + 3*B*c*d*e^2*x^2 + 3*B*c*d^2*e*x + B*c*d^3)*log(e*x + d))/(e
^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

Sympy [A] (verification not implemented)

Time = 1.22 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {B c \log {\left (d + e x \right )}}{e^{4}} + \frac {- A b d e^{2} - 2 A c d^{2} e - 2 B b d^{2} e + 11 B c d^{3} + x^{2} \left (- 6 A c e^{3} - 6 B b e^{3} + 18 B c d e^{2}\right ) + x \left (- 3 A b e^{3} - 6 A c d e^{2} - 6 B b d e^{2} + 27 B c d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \]

[In]

integrate((B*x+A)*(c*x**2+b*x)/(e*x+d)**4,x)

[Out]

B*c*log(d + e*x)/e**4 + (-A*b*d*e**2 - 2*A*c*d**2*e - 2*B*b*d**2*e + 11*B*c*d**3 + x**2*(-6*A*c*e**3 - 6*B*b*e
**3 + 18*B*c*d*e**2) + x*(-3*A*b*e**3 - 6*A*c*d*e**2 - 6*B*b*d*e**2 + 27*B*c*d**2*e))/(6*d**3*e**4 + 18*d**2*e
**5*x + 18*d*e**6*x**2 + 6*e**7*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {11 \, B c d^{3} - A b d e^{2} - 2 \, {\left (B b + A c\right )} d^{2} e + 6 \, {\left (3 \, B c d e^{2} - {\left (B b + A c\right )} e^{3}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} e - A b e^{3} - 2 \, {\left (B b + A c\right )} d e^{2}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac {B c \log \left (e x + d\right )}{e^{4}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*(11*B*c*d^3 - A*b*d*e^2 - 2*(B*b + A*c)*d^2*e + 6*(3*B*c*d*e^2 - (B*b + A*c)*e^3)*x^2 + 3*(9*B*c*d^2*e - A
*b*e^3 - 2*(B*b + A*c)*d*e^2)*x)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) + B*c*log(e*x + d)/e^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {B c \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {6 \, {\left (3 \, B c d e - B b e^{2} - A c e^{2}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} - 2 \, B b d e - 2 \, A c d e - A b e^{2}\right )} x + \frac {11 \, B c d^{3} - 2 \, B b d^{2} e - 2 \, A c d^{2} e - A b d e^{2}}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \]

[In]

integrate((B*x+A)*(c*x^2+b*x)/(e*x+d)^4,x, algorithm="giac")

[Out]

B*c*log(abs(e*x + d))/e^4 + 1/6*(6*(3*B*c*d*e - B*b*e^2 - A*c*e^2)*x^2 + 3*(9*B*c*d^2 - 2*B*b*d*e - 2*A*c*d*e
- A*b*e^2)*x + (11*B*c*d^3 - 2*B*b*d^2*e - 2*A*c*d^2*e - A*b*d*e^2)/e)/((e*x + d)^3*e^3)

Mupad [B] (verification not implemented)

Time = 10.80 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) \left (b x+c x^2\right )}{(d+e x)^4} \, dx=\frac {B\,c\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {A\,b\,d\,e^2-11\,B\,c\,d^3+2\,A\,c\,d^2\,e+2\,B\,b\,d^2\,e}{6\,e^4}+\frac {x\,\left (A\,b\,e^2-9\,B\,c\,d^2+2\,A\,c\,d\,e+2\,B\,b\,d\,e\right )}{2\,e^3}+\frac {x^2\,\left (A\,c\,e+B\,b\,e-3\,B\,c\,d\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \]

[In]

int(((b*x + c*x^2)*(A + B*x))/(d + e*x)^4,x)

[Out]

(B*c*log(d + e*x))/e^4 - ((A*b*d*e^2 - 11*B*c*d^3 + 2*A*c*d^2*e + 2*B*b*d^2*e)/(6*e^4) + (x*(A*b*e^2 - 9*B*c*d
^2 + 2*A*c*d*e + 2*B*b*d*e))/(2*e^3) + (x^2*(A*c*e + B*b*e - 3*B*c*d))/e^2)/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d
^2*e*x)