\(\int \frac {(A+B x) (d+e x)^5}{(a+c x^2)^2} \, dx\) [87]

Optimal result

Integrand size = 22, antiderivative size = 297 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx=-\frac {e^2 \left (3 A c d \left (2 c d^2-5 a e^2\right )-5 a B e \left (6 c d^2-a e^2\right )\right ) x}{2 a c^3}-\frac {e^3 \left (2 A c d^2-5 a B d e-a A e^2\right ) x^2}{a c^2}-\frac {e^4 (3 A c d-5 a B e) x^3}{6 a c^2}-\frac {(d+e x)^4 (a (B d+A e)-(A c d-a B e) x)}{2 a c \left (a+c x^2\right )}+\frac {\left (A c d \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right )+5 a B e \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{7/2}}+\frac {e^2 \left (5 B c d^3+5 A c d^2 e-5 a B d e^2-a A e^3\right ) \log \left (a+c x^2\right )}{c^3} \] Output:

-1/2*e^2*(3*A*c*d*(-5*a*e^2+2*c*d^2)-5*a*B*e*(-a*e^2+6*c*d^2))*x/a/c^3-e^3 
*(-A*a*e^2+2*A*c*d^2-5*B*a*d*e)*x^2/a/c^2-1/6*e^4*(3*A*c*d-5*B*a*e)*x^3/a/ 
c^2-1/2*(e*x+d)^4*(a*(A*e+B*d)-(A*c*d-B*a*e)*x)/a/c/(c*x^2+a)+1/2*(A*c*d*( 
-15*a^2*e^4+10*a*c*d^2*e^2+c^2*d^4)+5*a*B*e*(a^2*e^4-6*a*c*d^2*e^2+c^2*d^4 
))*arctan(c^(1/2)*x/a^(1/2))/a^(3/2)/c^(7/2)+e^2*(-A*a*e^3+5*A*c*d^2*e-5*B 
*a*d*e^2+5*B*c*d^3)*ln(c*x^2+a)/c^3
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\frac {e^3 \left (10 B c d^2+5 A c d e-2 a B e^2\right ) x}{c^3}+\frac {e^4 (5 B d+A e) x^2}{2 c^2}+\frac {B e^5 x^3}{3 c^2}+\frac {A c^3 d^5 x-a^3 e^4 (5 B d+A e+B e x)+5 a^2 c d e^2 (2 B d (d+e x)+A e (2 d+e x))-a c^2 d^3 (5 A e (d+2 e x)+B d (d+5 e x))}{2 a c^3 \left (a+c x^2\right )}+\frac {\left (A c d \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right )+5 a B e \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{7/2}}+\frac {e^2 \left (5 B c d^3+5 A c d^2 e-5 a B d e^2-a A e^3\right ) \log \left (a+c x^2\right )}{c^3} \] Input:

Integrate[((A + B*x)*(d + e*x)^5)/(a + c*x^2)^2,x]
 

Output:

(e^3*(10*B*c*d^2 + 5*A*c*d*e - 2*a*B*e^2)*x)/c^3 + (e^4*(5*B*d + A*e)*x^2) 
/(2*c^2) + (B*e^5*x^3)/(3*c^2) + (A*c^3*d^5*x - a^3*e^4*(5*B*d + A*e + B*e 
*x) + 5*a^2*c*d*e^2*(2*B*d*(d + e*x) + A*e*(2*d + e*x)) - a*c^2*d^3*(5*A*e 
*(d + 2*e*x) + B*d*(d + 5*e*x)))/(2*a*c^3*(a + c*x^2)) + ((A*c*d*(c^2*d^4 
+ 10*a*c*d^2*e^2 - 15*a^2*e^4) + 5*a*B*e*(c^2*d^4 - 6*a*c*d^2*e^2 + a^2*e^ 
4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*c^(7/2)) + (e^2*(5*B*c*d^3 + 5 
*A*c*d^2*e - 5*a*B*d*e^2 - a*A*e^3)*Log[a + c*x^2])/c^3
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {684, 657, 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.

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

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 657

\(\displaystyle \frac {\int \left (-\frac {(3 A c d-5 a B e) x^2 e^4}{c}-\frac {4 \left (2 A c d^2-5 a B e d-a A e^2\right ) x e^3}{c}-\frac {\left (3 A c d \left (2 c d^2-5 a e^2\right )-5 a B e \left (6 c d^2-a e^2\right )\right ) e^2}{c^2}+\frac {4 a c \left (5 B c d^3+5 A c e d^2-5 a B e^2 d-a A e^3\right ) x e^2+5 a B \left (c^2 d^4-6 a c e^2 d^2+a^2 e^4\right ) e+A c d \left (c^2 d^4+10 a c e^2 d^2-15 a^2 e^4\right )}{c^2 \left (c x^2+a\right )}\right )dx}{2 a c}-\frac {(d+e x)^4 (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (-15 a^2 e^4+10 a c d^2 e^2+c^2 d^4\right )+5 a B e \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )\right )}{\sqrt {a} c^{5/2}}-\frac {e^2 x \left (3 A c d \left (2 c d^2-5 a e^2\right )-5 a B e \left (6 c d^2-a e^2\right )\right )}{c^2}+\frac {2 a e^2 \log \left (a+c x^2\right ) \left (-a A e^3-5 a B d e^2+5 A c d^2 e+5 B c d^3\right )}{c^2}-\frac {2 e^3 x^2 \left (-a A e^2-5 a B d e+2 A c d^2\right )}{c}-\frac {e^4 x^3 (3 A c d-5 a B e)}{3 c}}{2 a c}-\frac {(d+e x)^4 (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )}\)

Input:

Int[((A + B*x)*(d + e*x)^5)/(a + c*x^2)^2,x]
 

Output:

-1/2*((d + e*x)^4*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(a*c*(a + c*x^2)) + 
 (-((e^2*(3*A*c*d*(2*c*d^2 - 5*a*e^2) - 5*a*B*e*(6*c*d^2 - a*e^2))*x)/c^2) 
 - (2*e^3*(2*A*c*d^2 - 5*a*B*d*e - a*A*e^2)*x^2)/c - (e^4*(3*A*c*d - 5*a*B 
*e)*x^3)/(3*c) + ((A*c*d*(c^2*d^4 + 10*a*c*d^2*e^2 - 15*a^2*e^4) + 5*a*B*e 
*(c^2*d^4 - 6*a*c*d^2*e^2 + a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a 
]*c^(5/2)) + (2*a*e^2*(5*B*c*d^3 + 5*A*c*d^2*e - 5*a*B*d*e^2 - a*A*e^3)*Lo 
g[a + c*x^2])/c^2)/(2*a*c)
 

Defintions of rubi rules used

rule 657

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

rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g 
) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Simp[1/(2*a*c*(p + 1))   Int[ 
(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^ 
2*f*(2*p + 3) + e*(a*e*g*m - c*d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a 
, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] 
 && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])
 

rule 2009

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

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {e^{3} \left (\frac {1}{3} B c \,x^{3} e^{2}+\frac {1}{2} A c \,e^{2} x^{2}+\frac {5}{2} B c d e \,x^{2}+5 A c d e x -2 B a \,e^{2} x +10 B c \,d^{2} x \right )}{c^{3}}-\frac {\frac {-\frac {\left (5 A \,a^{2} c d \,e^{4}-10 A a \,c^{2} d^{3} e^{2}+A \,d^{5} c^{3}-B \,e^{5} a^{3}+10 B \,a^{2} c \,d^{2} e^{3}-5 B a \,c^{2} d^{4} e \right ) x}{2 a}+\frac {A \,a^{2} e^{5}}{2}-5 A a c \,d^{2} e^{3}+\frac {5 A \,c^{2} d^{4} e}{2}+\frac {5 B \,a^{2} d \,e^{4}}{2}-5 B a c \,d^{3} e^{2}+\frac {B \,c^{2} d^{5}}{2}}{c \,x^{2}+a}+\frac {\frac {\left (4 A \,a^{2} c \,e^{5}-20 A a \,c^{2} d^{2} e^{3}+20 B \,a^{2} c d \,e^{4}-20 B a \,c^{2} d^{3} e^{2}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (15 A \,a^{2} c d \,e^{4}-10 A a \,c^{2} d^{3} e^{2}-A \,d^{5} c^{3}-5 B \,e^{5} a^{3}+30 B \,a^{2} c \,d^{2} e^{3}-5 B a \,c^{2} d^{4} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a}}{c^{3}}\)	\(360\)
risch	\(\text {Expression too large to display}\)	\(1971\)

Input:

int((B*x+A)*(e*x+d)^5/(c*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

e^3/c^3*(1/3*B*c*x^3*e^2+1/2*A*c*e^2*x^2+5/2*B*c*d*e*x^2+5*A*c*d*e*x-2*B*a 
*e^2*x+10*B*c*d^2*x)-1/c^3*((-1/2*(5*A*a^2*c*d*e^4-10*A*a*c^2*d^3*e^2+A*c^ 
3*d^5-B*a^3*e^5+10*B*a^2*c*d^2*e^3-5*B*a*c^2*d^4*e)/a*x+1/2*A*a^2*e^5-5*A* 
a*c*d^2*e^3+5/2*A*c^2*d^4*e+5/2*B*a^2*d*e^4-5*B*a*c*d^3*e^2+1/2*B*c^2*d^5) 
/(c*x^2+a)+1/2/a*(1/2*(4*A*a^2*c*e^5-20*A*a*c^2*d^2*e^3+20*B*a^2*c*d*e^4-2 
0*B*a*c^2*d^3*e^2)/c*ln(c*x^2+a)+(15*A*a^2*c*d*e^4-10*A*a*c^2*d^3*e^2-A*c^ 
3*d^5-5*B*a^3*e^5+30*B*a^2*c*d^2*e^3-5*B*a*c^2*d^4*e)/(a*c)^(1/2)*arctan(c 
*x/(a*c)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (281) = 562\).

Time = 0.10 (sec) , antiderivative size = 1190, normalized size of antiderivative = 4.01 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)^5/(c*x^2+a)^2,x, algorithm="fricas")
 

Output:

[1/12*(4*B*a^2*c^3*e^5*x^5 - 6*B*a^2*c^3*d^5 - 30*A*a^2*c^3*d^4*e + 60*B*a 
^3*c^2*d^3*e^2 + 60*A*a^3*c^2*d^2*e^3 - 30*B*a^4*c*d*e^4 - 6*A*a^4*c*e^5 + 
 6*(5*B*a^2*c^3*d*e^4 + A*a^2*c^3*e^5)*x^4 + 20*(6*B*a^2*c^3*d^2*e^3 + 3*A 
*a^2*c^3*d*e^4 - B*a^3*c^2*e^5)*x^3 + 6*(5*B*a^3*c^2*d*e^4 + A*a^3*c^2*e^5 
)*x^2 - 3*(A*a*c^3*d^5 + 5*B*a^2*c^2*d^4*e + 10*A*a^2*c^2*d^3*e^2 - 30*B*a 
^3*c*d^2*e^3 - 15*A*a^3*c*d*e^4 + 5*B*a^4*e^5 + (A*c^4*d^5 + 5*B*a*c^3*d^4 
*e + 10*A*a*c^3*d^3*e^2 - 30*B*a^2*c^2*d^2*e^3 - 15*A*a^2*c^2*d*e^4 + 5*B* 
a^3*c*e^5)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 
 6*(A*a*c^4*d^5 - 5*B*a^2*c^3*d^4*e - 10*A*a^2*c^3*d^3*e^2 + 30*B*a^3*c^2* 
d^2*e^3 + 15*A*a^3*c^2*d*e^4 - 5*B*a^4*c*e^5)*x + 12*(5*B*a^3*c^2*d^3*e^2 
+ 5*A*a^3*c^2*d^2*e^3 - 5*B*a^4*c*d*e^4 - A*a^4*c*e^5 + (5*B*a^2*c^3*d^3*e 
^2 + 5*A*a^2*c^3*d^2*e^3 - 5*B*a^3*c^2*d*e^4 - A*a^3*c^2*e^5)*x^2)*log(c*x 
^2 + a))/(a^2*c^5*x^2 + a^3*c^4), 1/6*(2*B*a^2*c^3*e^5*x^5 - 3*B*a^2*c^3*d 
^5 - 15*A*a^2*c^3*d^4*e + 30*B*a^3*c^2*d^3*e^2 + 30*A*a^3*c^2*d^2*e^3 - 15 
*B*a^4*c*d*e^4 - 3*A*a^4*c*e^5 + 3*(5*B*a^2*c^3*d*e^4 + A*a^2*c^3*e^5)*x^4 
 + 10*(6*B*a^2*c^3*d^2*e^3 + 3*A*a^2*c^3*d*e^4 - B*a^3*c^2*e^5)*x^3 + 3*(5 
*B*a^3*c^2*d*e^4 + A*a^3*c^2*e^5)*x^2 + 3*(A*a*c^3*d^5 + 5*B*a^2*c^2*d^4*e 
 + 10*A*a^2*c^2*d^3*e^2 - 30*B*a^3*c*d^2*e^3 - 15*A*a^3*c*d*e^4 + 5*B*a^4* 
e^5 + (A*c^4*d^5 + 5*B*a*c^3*d^4*e + 10*A*a*c^3*d^3*e^2 - 30*B*a^2*c^2*d^2 
*e^3 - 15*A*a^2*c^2*d*e^4 + 5*B*a^3*c*e^5)*x^2)*sqrt(a*c)*arctan(sqrt(a...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1091 vs. \(2 (294) = 588\).

Time = 6.67 (sec) , antiderivative size = 1091, normalized size of antiderivative = 3.67 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)**5/(c*x**2+a)**2,x)
 

Output:

B*e**5*x**3/(3*c**2) + x**2*(A*e**5/(2*c**2) + 5*B*d*e**4/(2*c**2)) + x*(5 
*A*d*e**4/c**2 - 2*B*a*e**5/c**3 + 10*B*d**2*e**3/c**2) + (-e**2*(A*a*e**3 
 - 5*A*c*d**2*e + 5*B*a*d*e**2 - 5*B*c*d**3)/c**3 - sqrt(-a**3*c**7)*(-15* 
A*a**2*c*d*e**4 + 10*A*a*c**2*d**3*e**2 + A*c**3*d**5 + 5*B*a**3*e**5 - 30 
*B*a**2*c*d**2*e**3 + 5*B*a*c**2*d**4*e)/(4*a**3*c**7))*log(x + (4*A*a**3* 
e**5 - 20*A*a**2*c*d**2*e**3 + 20*B*a**3*d*e**4 - 20*B*a**2*c*d**3*e**2 + 
4*a**2*c**3*(-e**2*(A*a*e**3 - 5*A*c*d**2*e + 5*B*a*d*e**2 - 5*B*c*d**3)/c 
**3 - sqrt(-a**3*c**7)*(-15*A*a**2*c*d*e**4 + 10*A*a*c**2*d**3*e**2 + A*c* 
*3*d**5 + 5*B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 + 5*B*a*c**2*d**4*e)/(4*a* 
*3*c**7)))/(-15*A*a**2*c*d*e**4 + 10*A*a*c**2*d**3*e**2 + A*c**3*d**5 + 5* 
B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 + 5*B*a*c**2*d**4*e)) + (-e**2*(A*a*e* 
*3 - 5*A*c*d**2*e + 5*B*a*d*e**2 - 5*B*c*d**3)/c**3 + sqrt(-a**3*c**7)*(-1 
5*A*a**2*c*d*e**4 + 10*A*a*c**2*d**3*e**2 + A*c**3*d**5 + 5*B*a**3*e**5 - 
30*B*a**2*c*d**2*e**3 + 5*B*a*c**2*d**4*e)/(4*a**3*c**7))*log(x + (4*A*a** 
3*e**5 - 20*A*a**2*c*d**2*e**3 + 20*B*a**3*d*e**4 - 20*B*a**2*c*d**3*e**2 
+ 4*a**2*c**3*(-e**2*(A*a*e**3 - 5*A*c*d**2*e + 5*B*a*d*e**2 - 5*B*c*d**3) 
/c**3 + sqrt(-a**3*c**7)*(-15*A*a**2*c*d*e**4 + 10*A*a*c**2*d**3*e**2 + A* 
c**3*d**5 + 5*B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 + 5*B*a*c**2*d**4*e)/(4* 
a**3*c**7)))/(-15*A*a**2*c*d*e**4 + 10*A*a*c**2*d**3*e**2 + A*c**3*d**5 + 
5*B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 + 5*B*a*c**2*d**4*e)) + (-A*a**3*...
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx=-\frac {B a c^{2} d^{5} + 5 \, A a c^{2} d^{4} e - 10 \, B a^{2} c d^{3} e^{2} - 10 \, A a^{2} c d^{2} e^{3} + 5 \, B a^{3} d e^{4} + A a^{3} e^{5} - {\left (A c^{3} d^{5} - 5 \, B a c^{2} d^{4} e - 10 \, A a c^{2} d^{3} e^{2} + 10 \, B a^{2} c d^{2} e^{3} + 5 \, A a^{2} c d e^{4} - B a^{3} e^{5}\right )} x}{2 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}} + \frac {{\left (5 \, B c d^{3} e^{2} + 5 \, A c d^{2} e^{3} - 5 \, B a d e^{4} - A a e^{5}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {2 \, B c e^{5} x^{3} + 3 \, {\left (5 \, B c d e^{4} + A c e^{5}\right )} x^{2} + 6 \, {\left (10 \, B c d^{2} e^{3} + 5 \, A c d e^{4} - 2 \, B a e^{5}\right )} x}{6 \, c^{3}} + \frac {{\left (A c^{3} d^{5} + 5 \, B a c^{2} d^{4} e + 10 \, A a c^{2} d^{3} e^{2} - 30 \, B a^{2} c d^{2} e^{3} - 15 \, A a^{2} c d e^{4} + 5 \, B a^{3} e^{5}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{3}} \] Input:

integrate((B*x+A)*(e*x+d)^5/(c*x^2+a)^2,x, algorithm="maxima")
 

Output:

-1/2*(B*a*c^2*d^5 + 5*A*a*c^2*d^4*e - 10*B*a^2*c*d^3*e^2 - 10*A*a^2*c*d^2* 
e^3 + 5*B*a^3*d*e^4 + A*a^3*e^5 - (A*c^3*d^5 - 5*B*a*c^2*d^4*e - 10*A*a*c^ 
2*d^3*e^2 + 10*B*a^2*c*d^2*e^3 + 5*A*a^2*c*d*e^4 - B*a^3*e^5)*x)/(a*c^4*x^ 
2 + a^2*c^3) + (5*B*c*d^3*e^2 + 5*A*c*d^2*e^3 - 5*B*a*d*e^4 - A*a*e^5)*log 
(c*x^2 + a)/c^3 + 1/6*(2*B*c*e^5*x^3 + 3*(5*B*c*d*e^4 + A*c*e^5)*x^2 + 6*( 
10*B*c*d^2*e^3 + 5*A*c*d*e^4 - 2*B*a*e^5)*x)/c^3 + 1/2*(A*c^3*d^5 + 5*B*a* 
c^2*d^4*e + 10*A*a*c^2*d^3*e^2 - 30*B*a^2*c*d^2*e^3 - 15*A*a^2*c*d*e^4 + 5 
*B*a^3*e^5)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\frac {{\left (5 \, B c d^{3} e^{2} + 5 \, A c d^{2} e^{3} - 5 \, B a d e^{4} - A a e^{5}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (A c^{3} d^{5} + 5 \, B a c^{2} d^{4} e + 10 \, A a c^{2} d^{3} e^{2} - 30 \, B a^{2} c d^{2} e^{3} - 15 \, A a^{2} c d e^{4} + 5 \, B a^{3} e^{5}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{3}} - \frac {B a c^{2} d^{5} + 5 \, A a c^{2} d^{4} e - 10 \, B a^{2} c d^{3} e^{2} - 10 \, A a^{2} c d^{2} e^{3} + 5 \, B a^{3} d e^{4} + A a^{3} e^{5} - {\left (A c^{3} d^{5} - 5 \, B a c^{2} d^{4} e - 10 \, A a c^{2} d^{3} e^{2} + 10 \, B a^{2} c d^{2} e^{3} + 5 \, A a^{2} c d e^{4} - B a^{3} e^{5}\right )} x}{2 \, {\left (c x^{2} + a\right )} a c^{3}} + \frac {2 \, B c^{4} e^{5} x^{3} + 15 \, B c^{4} d e^{4} x^{2} + 3 \, A c^{4} e^{5} x^{2} + 60 \, B c^{4} d^{2} e^{3} x + 30 \, A c^{4} d e^{4} x - 12 \, B a c^{3} e^{5} x}{6 \, c^{6}} \] Input:

integrate((B*x+A)*(e*x+d)^5/(c*x^2+a)^2,x, algorithm="giac")
 

Output:

(5*B*c*d^3*e^2 + 5*A*c*d^2*e^3 - 5*B*a*d*e^4 - A*a*e^5)*log(c*x^2 + a)/c^3 
 + 1/2*(A*c^3*d^5 + 5*B*a*c^2*d^4*e + 10*A*a*c^2*d^3*e^2 - 30*B*a^2*c*d^2* 
e^3 - 15*A*a^2*c*d*e^4 + 5*B*a^3*e^5)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c 
^3) - 1/2*(B*a*c^2*d^5 + 5*A*a*c^2*d^4*e - 10*B*a^2*c*d^3*e^2 - 10*A*a^2*c 
*d^2*e^3 + 5*B*a^3*d*e^4 + A*a^3*e^5 - (A*c^3*d^5 - 5*B*a*c^2*d^4*e - 10*A 
*a*c^2*d^3*e^2 + 10*B*a^2*c*d^2*e^3 + 5*A*a^2*c*d*e^4 - B*a^3*e^5)*x)/((c* 
x^2 + a)*a*c^3) + 1/6*(2*B*c^4*e^5*x^3 + 15*B*c^4*d*e^4*x^2 + 3*A*c^4*e^5* 
x^2 + 60*B*c^4*d^2*e^3*x + 30*A*c^4*d*e^4*x - 12*B*a*c^3*e^5*x)/c^6
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\frac {x^2\,\left (A\,e^5+5\,B\,d\,e^4\right )}{2\,c^2}-\frac {\frac {A\,a^2\,e^5}{2}+\frac {B\,c^2\,d^5}{2}-\frac {x\,\left (-B\,a^3\,e^5+10\,B\,a^2\,c\,d^2\,e^3+5\,A\,a^2\,c\,d\,e^4-5\,B\,a\,c^2\,d^4\,e-10\,A\,a\,c^2\,d^3\,e^2+A\,c^3\,d^5\right )}{2\,a}+\frac {5\,B\,a^2\,d\,e^4}{2}+\frac {5\,A\,c^2\,d^4\,e}{2}-5\,A\,a\,c\,d^2\,e^3-5\,B\,a\,c\,d^3\,e^2}{c^4\,x^2+a\,c^3}-x\,\left (\frac {2\,B\,a\,e^5}{c^3}-\frac {5\,d\,e^3\,\left (A\,e+2\,B\,d\right )}{c^2}\right )-\frac {\ln \left (c\,x^2+a\right )\,\left (160\,B\,a^4\,c^4\,d\,e^4+32\,A\,a^4\,c^4\,e^5-160\,B\,a^3\,c^5\,d^3\,e^2-160\,A\,a^3\,c^5\,d^2\,e^3\right )}{32\,a^3\,c^7}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (5\,B\,a^3\,e^5-30\,B\,a^2\,c\,d^2\,e^3-15\,A\,a^2\,c\,d\,e^4+5\,B\,a\,c^2\,d^4\,e+10\,A\,a\,c^2\,d^3\,e^2+A\,c^3\,d^5\right )}{2\,a^{3/2}\,c^{7/2}}+\frac {B\,e^5\,x^3}{3\,c^2} \] Input:

int(((A + B*x)*(d + e*x)^5)/(a + c*x^2)^2,x)
 

Output:

(x^2*(A*e^5 + 5*B*d*e^4))/(2*c^2) - ((A*a^2*e^5)/2 + (B*c^2*d^5)/2 - (x*(A 
*c^3*d^5 - B*a^3*e^5 - 10*A*a*c^2*d^3*e^2 + 10*B*a^2*c*d^2*e^3 + 5*A*a^2*c 
*d*e^4 - 5*B*a*c^2*d^4*e))/(2*a) + (5*B*a^2*d*e^4)/2 + (5*A*c^2*d^4*e)/2 - 
 5*A*a*c*d^2*e^3 - 5*B*a*c*d^3*e^2)/(a*c^3 + c^4*x^2) - x*((2*B*a*e^5)/c^3 
 - (5*d*e^3*(A*e + 2*B*d))/c^2) - (log(a + c*x^2)*(32*A*a^4*c^4*e^5 + 160* 
B*a^4*c^4*d*e^4 - 160*A*a^3*c^5*d^2*e^3 - 160*B*a^3*c^5*d^3*e^2))/(32*a^3* 
c^7) + (atan((c^(1/2)*x)/a^(1/2))*(A*c^3*d^5 + 5*B*a^3*e^5 + 10*A*a*c^2*d^ 
3*e^2 - 30*B*a^2*c*d^2*e^3 - 15*A*a^2*c*d*e^4 + 5*B*a*c^2*d^4*e))/(2*a^(3/ 
2)*c^(7/2)) + (B*e^5*x^3)/(3*c^2)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 787, normalized size of antiderivative = 2.65 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(e*x+d)^5/(c*x^2+a)^2,x)
 

Output:

(15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**3*b*e**5 - 45*sqrt(c) 
*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**3*c*d*e**4 - 90*sqrt(c)*sqrt(a)* 
atan((c*x)/(sqrt(c)*sqrt(a)))*a**2*b*c*d**2*e**3 + 15*sqrt(c)*sqrt(a)*atan 
((c*x)/(sqrt(c)*sqrt(a)))*a**2*b*c*e**5*x**2 + 30*sqrt(c)*sqrt(a)*atan((c* 
x)/(sqrt(c)*sqrt(a)))*a**2*c**2*d**3*e**2 - 45*sqrt(c)*sqrt(a)*atan((c*x)/ 
(sqrt(c)*sqrt(a)))*a**2*c**2*d*e**4*x**2 + 15*sqrt(c)*sqrt(a)*atan((c*x)/( 
sqrt(c)*sqrt(a)))*a*b*c**2*d**4*e - 90*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c) 
*sqrt(a)))*a*b*c**2*d**2*e**3*x**2 + 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c) 
*sqrt(a)))*a*c**3*d**5 + 30*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))* 
a*c**3*d**3*e**2*x**2 + 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*b 
*c**3*d**4*e*x**2 + 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c**4*d 
**5*x**2 - 6*log(a + c*x**2)*a**4*c*e**5 - 30*log(a + c*x**2)*a**3*b*c*d*e 
**4 + 30*log(a + c*x**2)*a**3*c**2*d**2*e**3 - 6*log(a + c*x**2)*a**3*c**2 
*e**5*x**2 + 30*log(a + c*x**2)*a**2*b*c**2*d**3*e**2 - 30*log(a + c*x**2) 
*a**2*b*c**2*d*e**4*x**2 + 30*log(a + c*x**2)*a**2*c**3*d**2*e**3*x**2 + 3 
0*log(a + c*x**2)*a*b*c**3*d**3*e**2*x**2 - 15*a**3*b*c*e**5*x + 45*a**3*c 
**2*d*e**4*x + 6*a**3*c**2*e**5*x**2 + 90*a**2*b*c**2*d**2*e**3*x + 30*a** 
2*b*c**2*d*e**4*x**2 - 10*a**2*b*c**2*e**5*x**3 - 30*a**2*c**3*d**3*e**2*x 
 - 30*a**2*c**3*d**2*e**3*x**2 + 30*a**2*c**3*d*e**4*x**3 + 3*a**2*c**3*e* 
*5*x**4 - 15*a*b*c**3*d**4*e*x - 30*a*b*c**3*d**3*e**2*x**2 + 60*a*b*c*...