\(\int \frac {(A+B x) (a+c x^2)^2}{(d+e x)^4} \, dx\) [55]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=-\frac {c^2 (4 B d-A e) x}{e^5}+\frac {B c^2 x^2}{2 e^4}+\frac {(B d-A e) \left (c d^2+a e^2\right )^2}{3 e^6 (d+e x)^3}-\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{2 e^6 (d+e x)^2}+\frac {2 c \left (5 B c d^3-3 A c d^2 e+3 a B d e^2-a A e^3\right )}{e^6 (d+e x)}+\frac {2 c \left (5 B c d^2-2 A c d e+a B e^2\right ) \log (d+e x)}{e^6} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {-2 A e \left (a^2 e^4+2 a c e^2 \left (d^2+3 d e x+3 e^2 x^2\right )+c^2 \left (13 d^4+27 d^3 e x+9 d^2 e^2 x^2-9 d e^3 x^3-3 e^4 x^4\right )\right )+B \left (-a^2 e^4 (d+3 e x)+2 a c d e^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )+c^2 \left (47 d^5+81 d^4 e x-9 d^3 e^2 x^2-63 d^2 e^3 x^3-15 d e^4 x^4+3 e^5 x^5\right )\right )+12 c \left (5 B c d^2-2 A c d e+a B e^2\right ) (d+e x)^3 \log (d+e x)}{6 e^6 (d+e x)^3} \] Input:

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

Output:

(-2*A*e*(a^2*e^4 + 2*a*c*e^2*(d^2 + 3*d*e*x + 3*e^2*x^2) + c^2*(13*d^4 + 2 
7*d^3*e*x + 9*d^2*e^2*x^2 - 9*d*e^3*x^3 - 3*e^4*x^4)) + B*(-(a^2*e^4*(d + 
3*e*x)) + 2*a*c*d*e^2*(11*d^2 + 27*d*e*x + 18*e^2*x^2) + c^2*(47*d^5 + 81* 
d^4*e*x - 9*d^3*e^2*x^2 - 63*d^2*e^3*x^3 - 15*d*e^4*x^4 + 3*e^5*x^5)) + 12 
*c*(5*B*c*d^2 - 2*A*c*d*e + a*B*e^2)*(d + e*x)^3*Log[d + e*x])/(6*e^6*(d + 
 e*x)^3)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {652, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+c x^2\right )^2 (A+B x)}{(d+e x)^4} \, dx\)

\(\Big \downarrow \) 652

\(\displaystyle \int \left (-\frac {2 c \left (-a B e^2+2 A c d e-5 B c d^2\right )}{e^5 (d+e x)}+\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^5 (d+e x)^3}+\frac {\left (a e^2+c d^2\right )^2 (A e-B d)}{e^5 (d+e x)^4}+\frac {2 c \left (a A e^3-3 a B d e^2+3 A c d^2 e-5 B c d^3\right )}{e^5 (d+e x)^2}+\frac {c^2 (A e-4 B d)}{e^5}+\frac {B c^2 x}{e^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{2 e^6 (d+e x)^2}+\frac {\left (a e^2+c d^2\right )^2 (B d-A e)}{3 e^6 (d+e x)^3}+\frac {2 c \log (d+e x) \left (a B e^2-2 A c d e+5 B c d^2\right )}{e^6}+\frac {2 c \left (-a A e^3+3 a B d e^2-3 A c d^2 e+5 B c d^3\right )}{e^6 (d+e x)}-\frac {c^2 x (4 B d-A e)}{e^5}+\frac {B c^2 x^2}{2 e^4}\)

Input:

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

Output:

-((c^2*(4*B*d - A*e)*x)/e^5) + (B*c^2*x^2)/(2*e^4) + ((B*d - A*e)*(c*d^2 + 
 a*e^2)^2)/(3*e^6*(d + e*x)^3) - ((c*d^2 + a*e^2)*(5*B*c*d^2 - 4*A*c*d*e + 
 a*B*e^2))/(2*e^6*(d + e*x)^2) + (2*c*(5*B*c*d^3 - 3*A*c*d^2*e + 3*a*B*d*e 
^2 - a*A*e^3))/(e^6*(d + e*x)) + (2*c*(5*B*c*d^2 - 2*A*c*d*e + a*B*e^2)*Lo 
g[d + e*x])/e^6
 

Defintions of rubi rules used

rule 652

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ 
)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c 
*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.23

method	result	size
default	\(\frac {c^{2} \left (\frac {1}{2} B e \,x^{2}+A e x -4 B d x \right )}{e^{5}}-\frac {A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}}{3 e^{6} \left (e x +d \right )^{3}}-\frac {2 c \left (2 A c d e -B a \,e^{2}-5 B c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{6}}-\frac {2 c \left (A a \,e^{3}+3 A c \,d^{2} e -3 B a d \,e^{2}-5 B c \,d^{3}\right )}{e^{6} \left (e x +d \right )}-\frac {-4 A a c d \,e^{3}-4 A \,c^{2} d^{3} e +B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}+5 B \,c^{2} d^{4}}{2 e^{6} \left (e x +d \right )^{2}}\)	\(233\)
norman	\(\frac {-\frac {2 A \,a^{2} e^{5}+4 A a c \,d^{2} e^{3}+44 A \,c^{2} d^{4} e +B \,a^{2} d \,e^{4}-22 B a c \,d^{3} e^{2}-110 B \,c^{2} d^{5}}{6 e^{6}}-\frac {\left (2 A a c \,e^{3}+12 A \,c^{2} d^{2} e -6 B a c d \,e^{2}-30 B \,c^{2} d^{3}\right ) x^{2}}{e^{4}}-\frac {\left (4 A a c d \,e^{3}+36 A \,c^{2} d^{3} e +B \,e^{4} a^{2}-18 B a c \,d^{2} e^{2}-90 B \,c^{2} d^{4}\right ) x}{2 e^{5}}+\frac {B \,c^{2} x^{5}}{2 e}+\frac {c^{2} \left (2 A e -5 B d \right ) x^{4}}{2 e^{2}}}{\left (e x +d \right )^{3}}-\frac {2 c \left (2 A c d e -B a \,e^{2}-5 B c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{6}}\)	\(240\)
risch	\(\frac {B \,c^{2} x^{2}}{2 e^{4}}+\frac {c^{2} A x}{e^{4}}-\frac {4 c^{2} B d x}{e^{5}}+\frac {\left (-2 A a c \,e^{4}-6 A \,c^{2} d^{2} e^{2}+6 B a c d \,e^{3}+10 B \,c^{2} d^{3} e \right ) x^{2}+\left (-2 A a c d \,e^{3}-10 A \,c^{2} d^{3} e -\frac {1}{2} B \,e^{4} a^{2}+9 B a c \,d^{2} e^{2}+\frac {35}{2} B \,c^{2} d^{4}\right ) x -\frac {2 A \,a^{2} e^{5}+4 A a c \,d^{2} e^{3}+26 A \,c^{2} d^{4} e +B \,a^{2} d \,e^{4}-22 B a c \,d^{3} e^{2}-47 B \,c^{2} d^{5}}{6 e}}{e^{5} \left (e x +d \right )^{3}}-\frac {4 c^{2} \ln \left (e x +d \right ) A d}{e^{5}}+\frac {2 c \ln \left (e x +d \right ) B a}{e^{4}}+\frac {10 c^{2} \ln \left (e x +d \right ) B \,d^{2}}{e^{6}}\)	\(254\)
parallelrisch	\(-\frac {-6 A \,x^{4} c^{2} e^{5}-60 B \ln \left (e x +d \right ) c^{2} d^{5}+3 B x \,a^{2} e^{5}+72 A \ln \left (e x +d \right ) x \,c^{2} d^{3} e^{2}-180 B \ln \left (e x +d \right ) x \,c^{2} d^{4} e +108 A x \,c^{2} d^{3} e^{2}+4 A a c \,d^{2} e^{3}-22 B a c \,d^{3} e^{2}+15 B \,x^{4} c^{2} d \,e^{4}+24 A \ln \left (e x +d \right ) x^{3} c^{2} d \,e^{4}-36 B \ln \left (e x +d \right ) x^{2} a c d \,e^{4}-36 B \ln \left (e x +d \right ) x a c \,d^{2} e^{3}-270 B x \,c^{2} d^{4} e -36 B \,x^{2} a c d \,e^{4}+12 A x a c d \,e^{4}+B \,a^{2} d \,e^{4}-180 B \,x^{2} c^{2} d^{3} e^{2}+24 A \ln \left (e x +d \right ) c^{2} d^{4} e +44 A \,c^{2} d^{4} e +12 A \,x^{2} a c \,e^{5}+72 A \,x^{2} c^{2} d^{2} e^{3}-12 B \ln \left (e x +d \right ) a c \,d^{3} e^{2}-54 B x a c \,d^{2} e^{3}+72 A \ln \left (e x +d \right ) x^{2} c^{2} d^{2} e^{3}-180 B \ln \left (e x +d \right ) x^{2} c^{2} d^{3} e^{2}-12 B \ln \left (e x +d \right ) x^{3} a c \,e^{5}-60 B \ln \left (e x +d \right ) x^{3} c^{2} d^{2} e^{3}-3 B \,x^{5} c^{2} e^{5}-110 B \,c^{2} d^{5}+2 A \,a^{2} e^{5}}{6 e^{6} \left (e x +d \right )^{3}}\)	\(438\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (183) = 366\).

Time = 0.08 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.17 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {3 \, B c^{2} e^{5} x^{5} + 47 \, B c^{2} d^{5} - 26 \, A c^{2} d^{4} e + 22 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - 2 \, A a^{2} e^{5} - 3 \, {\left (5 \, B c^{2} d e^{4} - 2 \, A c^{2} e^{5}\right )} x^{4} - 9 \, {\left (7 \, B c^{2} d^{2} e^{3} - 2 \, A c^{2} d e^{4}\right )} x^{3} - 3 \, {\left (3 \, B c^{2} d^{3} e^{2} + 6 \, A c^{2} d^{2} e^{3} - 12 \, B a c d e^{4} + 4 \, A a c e^{5}\right )} x^{2} + 3 \, {\left (27 \, B c^{2} d^{4} e - 18 \, A c^{2} d^{3} e^{2} + 18 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} - B a^{2} e^{5}\right )} x + 12 \, {\left (5 \, B c^{2} d^{5} - 2 \, A c^{2} d^{4} e + B a c d^{3} e^{2} + {\left (5 \, B c^{2} d^{2} e^{3} - 2 \, A c^{2} d e^{4} + B a c e^{5}\right )} x^{3} + 3 \, {\left (5 \, B c^{2} d^{3} e^{2} - 2 \, A c^{2} d^{2} e^{3} + B a c d e^{4}\right )} x^{2} + 3 \, {\left (5 \, B c^{2} d^{4} e - 2 \, A c^{2} d^{3} e^{2} + B a c d^{2} e^{3}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \] Input:

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

Output:

1/6*(3*B*c^2*e^5*x^5 + 47*B*c^2*d^5 - 26*A*c^2*d^4*e + 22*B*a*c*d^3*e^2 - 
4*A*a*c*d^2*e^3 - B*a^2*d*e^4 - 2*A*a^2*e^5 - 3*(5*B*c^2*d*e^4 - 2*A*c^2*e 
^5)*x^4 - 9*(7*B*c^2*d^2*e^3 - 2*A*c^2*d*e^4)*x^3 - 3*(3*B*c^2*d^3*e^2 + 6 
*A*c^2*d^2*e^3 - 12*B*a*c*d*e^4 + 4*A*a*c*e^5)*x^2 + 3*(27*B*c^2*d^4*e - 1 
8*A*c^2*d^3*e^2 + 18*B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 - B*a^2*e^5)*x + 12*(5* 
B*c^2*d^5 - 2*A*c^2*d^4*e + B*a*c*d^3*e^2 + (5*B*c^2*d^2*e^3 - 2*A*c^2*d*e 
^4 + B*a*c*e^5)*x^3 + 3*(5*B*c^2*d^3*e^2 - 2*A*c^2*d^2*e^3 + B*a*c*d*e^4)* 
x^2 + 3*(5*B*c^2*d^4*e - 2*A*c^2*d^3*e^2 + B*a*c*d^2*e^3)*x)*log(e*x + d)) 
/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^7*x + d^3*e^6)
 

Sympy [A] (verification not implemented)

Time = 3.83 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.56 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {B c^{2} x^{2}}{2 e^{4}} + \frac {2 c \left (- 2 A c d e + B a e^{2} + 5 B c d^{2}\right ) \log {\left (d + e x \right )}}{e^{6}} + x \left (\frac {A c^{2}}{e^{4}} - \frac {4 B c^{2} d}{e^{5}}\right ) + \frac {- 2 A a^{2} e^{5} - 4 A a c d^{2} e^{3} - 26 A c^{2} d^{4} e - B a^{2} d e^{4} + 22 B a c d^{3} e^{2} + 47 B c^{2} d^{5} + x^{2} \left (- 12 A a c e^{5} - 36 A c^{2} d^{2} e^{3} + 36 B a c d e^{4} + 60 B c^{2} d^{3} e^{2}\right ) + x \left (- 12 A a c d e^{4} - 60 A c^{2} d^{3} e^{2} - 3 B a^{2} e^{5} + 54 B a c d^{2} e^{3} + 105 B c^{2} d^{4} e\right )}{6 d^{3} e^{6} + 18 d^{2} e^{7} x + 18 d e^{8} x^{2} + 6 e^{9} x^{3}} \] Input:

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

Output:

B*c**2*x**2/(2*e**4) + 2*c*(-2*A*c*d*e + B*a*e**2 + 5*B*c*d**2)*log(d + e* 
x)/e**6 + x*(A*c**2/e**4 - 4*B*c**2*d/e**5) + (-2*A*a**2*e**5 - 4*A*a*c*d* 
*2*e**3 - 26*A*c**2*d**4*e - B*a**2*d*e**4 + 22*B*a*c*d**3*e**2 + 47*B*c** 
2*d**5 + x**2*(-12*A*a*c*e**5 - 36*A*c**2*d**2*e**3 + 36*B*a*c*d*e**4 + 60 
*B*c**2*d**3*e**2) + x*(-12*A*a*c*d*e**4 - 60*A*c**2*d**3*e**2 - 3*B*a**2* 
e**5 + 54*B*a*c*d**2*e**3 + 105*B*c**2*d**4*e))/(6*d**3*e**6 + 18*d**2*e** 
7*x + 18*d*e**8*x**2 + 6*e**9*x**3)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {47 \, B c^{2} d^{5} - 26 \, A c^{2} d^{4} e + 22 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - 2 \, A a^{2} e^{5} + 12 \, {\left (5 \, B c^{2} d^{3} e^{2} - 3 \, A c^{2} d^{2} e^{3} + 3 \, B a c d e^{4} - A a c e^{5}\right )} x^{2} + 3 \, {\left (35 \, B c^{2} d^{4} e - 20 \, A c^{2} d^{3} e^{2} + 18 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} - B a^{2} e^{5}\right )} x}{6 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} + \frac {B c^{2} e x^{2} - 2 \, {\left (4 \, B c^{2} d - A c^{2} e\right )} x}{2 \, e^{5}} + \frac {2 \, {\left (5 \, B c^{2} d^{2} - 2 \, A c^{2} d e + B a c e^{2}\right )} \log \left (e x + d\right )}{e^{6}} \] Input:

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

Output:

1/6*(47*B*c^2*d^5 - 26*A*c^2*d^4*e + 22*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 - 
B*a^2*d*e^4 - 2*A*a^2*e^5 + 12*(5*B*c^2*d^3*e^2 - 3*A*c^2*d^2*e^3 + 3*B*a* 
c*d*e^4 - A*a*c*e^5)*x^2 + 3*(35*B*c^2*d^4*e - 20*A*c^2*d^3*e^2 + 18*B*a*c 
*d^2*e^3 - 4*A*a*c*d*e^4 - B*a^2*e^5)*x)/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^2*e^ 
7*x + d^3*e^6) + 1/2*(B*c^2*e*x^2 - 2*(4*B*c^2*d - A*c^2*e)*x)/e^5 + 2*(5* 
B*c^2*d^2 - 2*A*c^2*d*e + B*a*c*e^2)*log(e*x + d)/e^6
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {2 \, {\left (5 \, B c^{2} d^{2} - 2 \, A c^{2} d e + B a c e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{e^{6}} + \frac {B c^{2} e^{4} x^{2} - 8 \, B c^{2} d e^{3} x + 2 \, A c^{2} e^{4} x}{2 \, e^{8}} + \frac {47 \, B c^{2} d^{5} - 26 \, A c^{2} d^{4} e + 22 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - 2 \, A a^{2} e^{5} + 12 \, {\left (5 \, B c^{2} d^{3} e^{2} - 3 \, A c^{2} d^{2} e^{3} + 3 \, B a c d e^{4} - A a c e^{5}\right )} x^{2} + 3 \, {\left (35 \, B c^{2} d^{4} e - 20 \, A c^{2} d^{3} e^{2} + 18 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} - B a^{2} e^{5}\right )} x}{6 \, {\left (e x + d\right )}^{3} e^{6}} \] Input:

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

Output:

2*(5*B*c^2*d^2 - 2*A*c^2*d*e + B*a*c*e^2)*log(abs(e*x + d))/e^6 + 1/2*(B*c 
^2*e^4*x^2 - 8*B*c^2*d*e^3*x + 2*A*c^2*e^4*x)/e^8 + 1/6*(47*B*c^2*d^5 - 26 
*A*c^2*d^4*e + 22*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 - B*a^2*d*e^4 - 2*A*a^2* 
e^5 + 12*(5*B*c^2*d^3*e^2 - 3*A*c^2*d^2*e^3 + 3*B*a*c*d*e^4 - A*a*c*e^5)*x 
^2 + 3*(35*B*c^2*d^4*e - 20*A*c^2*d^3*e^2 + 18*B*a*c*d^2*e^3 - 4*A*a*c*d*e 
^4 - B*a^2*e^5)*x)/((e*x + d)^3*e^6)
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=x\,\left (\frac {A\,c^2}{e^4}-\frac {4\,B\,c^2\,d}{e^5}\right )-\frac {x\,\left (\frac {B\,a^2\,e^4}{2}-9\,B\,a\,c\,d^2\,e^2+2\,A\,a\,c\,d\,e^3-\frac {35\,B\,c^2\,d^4}{2}+10\,A\,c^2\,d^3\,e\right )+\frac {B\,a^2\,d\,e^4+2\,A\,a^2\,e^5-22\,B\,a\,c\,d^3\,e^2+4\,A\,a\,c\,d^2\,e^3-47\,B\,c^2\,d^5+26\,A\,c^2\,d^4\,e}{6\,e}+x^2\,\left (-10\,B\,c^2\,d^3\,e+6\,A\,c^2\,d^2\,e^2-6\,B\,a\,c\,d\,e^3+2\,A\,a\,c\,e^4\right )}{d^3\,e^5+3\,d^2\,e^6\,x+3\,d\,e^7\,x^2+e^8\,x^3}+\frac {\ln \left (d+e\,x\right )\,\left (10\,B\,c^2\,d^2-4\,A\,c^2\,d\,e+2\,B\,a\,c\,e^2\right )}{e^6}+\frac {B\,c^2\,x^2}{2\,e^4} \] Input:

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

Output:

x*((A*c^2)/e^4 - (4*B*c^2*d)/e^5) - (x*((B*a^2*e^4)/2 - (35*B*c^2*d^4)/2 + 
 10*A*c^2*d^3*e + 2*A*a*c*d*e^3 - 9*B*a*c*d^2*e^2) + (2*A*a^2*e^5 - 47*B*c 
^2*d^5 + B*a^2*d*e^4 + 26*A*c^2*d^4*e + 4*A*a*c*d^2*e^3 - 22*B*a*c*d^3*e^2 
)/(6*e) + x^2*(2*A*a*c*e^4 - 10*B*c^2*d^3*e + 6*A*c^2*d^2*e^2 - 6*B*a*c*d* 
e^3))/(d^3*e^5 + e^8*x^3 + 3*d^2*e^6*x + 3*d*e^7*x^2) + (log(d + e*x)*(10* 
B*c^2*d^2 + 2*B*a*c*e^2 - 4*A*c^2*d*e))/e^6 + (B*c^2*x^2)/(2*e^4)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.41 \[ \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^4} \, dx=\frac {-12 a b c d \,e^{5} x^{3}-2 a^{3} d \,e^{5}+50 b \,c^{2} d^{6}+3 b \,c^{2} d \,e^{5} x^{5}-15 b \,c^{2} d^{2} e^{4} x^{4}-60 b \,c^{2} d^{3} e^{3} x^{3}+90 b \,c^{2} d^{5} e x +6 a \,c^{2} d \,e^{5} x^{4}+24 a \,c^{2} d^{2} e^{4} x^{3}-3 a^{2} b d \,e^{5} x -36 a \,c^{2} d^{4} e^{2} x -24 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{5} e -72 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{3} e^{3} x^{2}+180 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{4} e^{2} x^{2}+60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{6}-20 a \,c^{2} d^{5} e -24 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{2} e^{4} x^{3}+60 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{3} e^{3} x^{3}+10 a b c \,d^{4} e^{2}+36 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{3} e^{3} x -a^{2} b \,d^{2} e^{4}+4 a^{2} c \,e^{6} x^{3}+12 \,\mathrm {log}\left (e x +d \right ) a b c d \,e^{5} x^{3}+12 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{4} e^{2}-72 \,\mathrm {log}\left (e x +d \right ) a \,c^{2} d^{4} e^{2} x +180 \,\mathrm {log}\left (e x +d \right ) b \,c^{2} d^{5} e x +18 a b c \,d^{3} e^{3} x +36 \,\mathrm {log}\left (e x +d \right ) a b c \,d^{2} e^{4} x^{2}}{6 d \,e^{6} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )} \] Input:

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

Output:

(12*log(d + e*x)*a*b*c*d**4*e**2 + 36*log(d + e*x)*a*b*c*d**3*e**3*x + 36* 
log(d + e*x)*a*b*c*d**2*e**4*x**2 + 12*log(d + e*x)*a*b*c*d*e**5*x**3 - 24 
*log(d + e*x)*a*c**2*d**5*e - 72*log(d + e*x)*a*c**2*d**4*e**2*x - 72*log( 
d + e*x)*a*c**2*d**3*e**3*x**2 - 24*log(d + e*x)*a*c**2*d**2*e**4*x**3 + 6 
0*log(d + e*x)*b*c**2*d**6 + 180*log(d + e*x)*b*c**2*d**5*e*x + 180*log(d 
+ e*x)*b*c**2*d**4*e**2*x**2 + 60*log(d + e*x)*b*c**2*d**3*e**3*x**3 - 2*a 
**3*d*e**5 - a**2*b*d**2*e**4 - 3*a**2*b*d*e**5*x + 4*a**2*c*e**6*x**3 + 1 
0*a*b*c*d**4*e**2 + 18*a*b*c*d**3*e**3*x - 12*a*b*c*d*e**5*x**3 - 20*a*c** 
2*d**5*e - 36*a*c**2*d**4*e**2*x + 24*a*c**2*d**2*e**4*x**3 + 6*a*c**2*d*e 
**5*x**4 + 50*b*c**2*d**6 + 90*b*c**2*d**5*e*x - 60*b*c**2*d**3*e**3*x**3 
- 15*b*c**2*d**2*e**4*x**4 + 3*b*c**2*d*e**5*x**5)/(6*d*e**6*(d**3 + 3*d** 
2*e*x + 3*d*e**2*x**2 + e**3*x**3))