\(\int \frac {(A+B x) (d+e x)^5}{(a+c x^2)^3} \, dx\) [95]

Optimal result

Integrand size = 22, antiderivative size = 306 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx=-\frac {e^2 \left (\frac {5 a B e \left (c d^2-3 a e^2\right )}{c}+A \left (3 c d^3+7 a d e^2\right )\right ) x}{8 a^2 c^2}-\frac {(d+e x)^4 (a (B d+A e)-(A c d-a B e) x)}{4 a c \left (a+c x^2\right )^2}-\frac {(d+e x)^2 \left (2 a e \left (A c d^2+5 a B d e+2 a A e^2\right )-\left (5 a B e \left (c d^2-a e^2\right )+A c d \left (3 c d^2+5 a e^2\right )\right ) x\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (5 a B e \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+A c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{7/2}}+\frac {e^4 (5 B d+A e) \log \left (a+c x^2\right )}{2 c^3} \] Output:

-1/8*e^2*(5*a*B*e*(-3*a*e^2+c*d^2)/c+A*(7*a*d*e^2+3*c*d^3))*x/a^2/c^2-1/4* 
(e*x+d)^4*(a*(A*e+B*d)-(A*c*d-B*a*e)*x)/a/c/(c*x^2+a)^2-1/8*(e*x+d)^2*(2*a 
*e*(2*A*a*e^2+A*c*d^2+5*B*a*d*e)-(5*a*B*e*(-a*e^2+c*d^2)+A*c*d*(5*a*e^2+3* 
c*d^2))*x)/a^2/c^2/(c*x^2+a)+1/8*(5*a*B*e*(-3*a^2*e^4+6*a*c*d^2*e^2+c^2*d^ 
4)+A*c*d*(15*a^2*e^4+10*a*c*d^2*e^2+3*c^2*d^4))*arctan(c^(1/2)*x/a^(1/2))/ 
a^(5/2)/c^(7/2)+1/2*e^4*(A*e+5*B*d)*ln(c*x^2+a)/c^3
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx=\frac {8 B \sqrt {c} e^5 x+\frac {2 \sqrt {c} \left (A c^3 d^5 x-a^3 e^4 (5 B d+A e+B e x)+5 a^2 c d e^2 (2 B d (d+e x)+A e (2 d+e x))-a c^2 d^3 (5 A e (d+2 e x)+B d (d+5 e x))\right )}{a \left (a+c x^2\right )^2}+\frac {\sqrt {c} \left (3 A c^3 d^5 x+5 a c^2 d^3 e (B d+2 A e) x+a^3 e^4 (40 B d+8 A e+9 B e x)-5 a^2 c d e^2 (2 B d (4 d+5 e x)+A e (8 d+5 e x))\right )}{a^2 \left (a+c x^2\right )}+\frac {\left (5 a B e \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+A c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{5/2}}+4 \sqrt {c} e^4 (5 B d+A e) \log \left (a+c x^2\right )}{8 c^{7/2}} \] Input:

Integrate[((A + B*x)*(d + e*x)^5)/(a + c*x^2)^3,x]
 

Output:

(8*B*Sqrt[c]*e^5*x + (2*Sqrt[c]*(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))))/(a*(a + c*x^2)^2) + (Sqrt[c]*(3*A*c^3*d^5 
*x + 5*a*c^2*d^3*e*(B*d + 2*A*e)*x + a^3*e^4*(40*B*d + 8*A*e + 9*B*e*x) - 
5*a^2*c*d*e^2*(2*B*d*(4*d + 5*e*x) + A*e*(8*d + 5*e*x))))/(a^2*(a + c*x^2) 
) + ((5*a*B*e*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4) + A*c*d*(3*c^2*d^4 + 1 
0*a*c*d^2*e^2 + 15*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(5/2) + 4*Sqrt 
[c]*e^4*(5*B*d + A*e)*Log[a + c*x^2])/(8*c^(7/2))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {684, 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.

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

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 684

\(\displaystyle \frac {\frac {\int \frac {(d+e x) \left (5 a B d e \left (c d^2+5 a e^2\right )+A \left (3 c^2 d^4+7 a c e^2 d^2+8 a^2 e^4\right )-e \left (5 a B e \left (c d^2-3 a e^2\right )+A c d \left (3 c d^2+7 a e^2\right )\right ) x\right )}{c x^2+a}dx}{2 a c}-\frac {(d+e x)^2 \left (2 a e \left (2 a A e^2+5 a B d e+A c d^2\right )-x \left (A c d \left (5 a e^2+3 c d^2\right )+5 a B e \left (c d^2-a e^2\right )\right )\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^4 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {\frac {\int \left (\frac {8 a^2 c (5 B d+A e) x e^4+5 a B \left (c^2 d^4+6 a c e^2 d^2-3 a^2 e^4\right ) e+A c d \left (3 c^2 d^4+10 a c e^2 d^2+15 a^2 e^4\right )}{c \left (c x^2+a\right )}-e^2 \left (3 A c d^3+5 a B e d^2+7 a A e^2 d-\frac {15 a^2 B e^3}{c}\right )\right )dx}{2 a c}-\frac {(d+e x)^2 \left (2 a e \left (2 a A e^2+5 a B d e+A c d^2\right )-x \left (A c d \left (5 a e^2+3 c d^2\right )+5 a B e \left (c d^2-a e^2\right )\right )\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^4 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right )+5 a B e \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )\right )}{\sqrt {a} c^{3/2}}+\frac {4 a^2 e^4 \log \left (a+c x^2\right ) (A e+5 B d)}{c}-e^2 x \left (A \left (7 a d e^2+3 c d^3\right )+\frac {5 a B e \left (c d^2-3 a e^2\right )}{c}\right )}{2 a c}-\frac {(d+e x)^2 \left (2 a e \left (2 a A e^2+5 a B d e+A c d^2\right )-x \left (A c d \left (5 a e^2+3 c d^2\right )+5 a B e \left (c d^2-a e^2\right )\right )\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^4 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}\)

Input:

Int[((A + B*x)*(d + e*x)^5)/(a + c*x^2)^3,x]
 

Output:

-1/4*((d + e*x)^4*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(a*c*(a + c*x^2)^2) 
 + (-1/2*((d + e*x)^2*(2*a*e*(A*c*d^2 + 5*a*B*d*e + 2*a*A*e^2) - (5*a*B*e* 
(c*d^2 - a*e^2) + A*c*d*(3*c*d^2 + 5*a*e^2))*x))/(a*c*(a + c*x^2)) + (-(e^ 
2*((5*a*B*e*(c*d^2 - 3*a*e^2))/c + A*(3*c*d^3 + 7*a*d*e^2))*x) + ((5*a*B*e 
*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4) + A*c*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 
 + 15*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2)) + (4*a^2*e^ 
4*(5*B*d + A*e)*Log[a + c*x^2])/c)/(2*a*c))/(4*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 = 1.04 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.32

method	result	size
default	\(\frac {B \,e^{5} x}{c^{3}}+\frac {\frac {-\frac {c \left (25 A \,a^{2} c d \,e^{4}-10 A a \,c^{2} d^{3} e^{2}-3 A \,d^{5} c^{3}-9 B \,e^{5} a^{3}+50 B \,a^{2} c \,d^{2} e^{3}-5 B a \,c^{2} d^{4} e \right ) x^{3}}{8 a^{2}}+\left (A a c \,e^{5}-5 A \,c^{2} d^{2} e^{3}+5 B a c d \,e^{4}-5 B \,c^{2} d^{3} e^{2}\right ) x^{2}-\frac {\left (15 A \,a^{2} c d \,e^{4}+10 A a \,c^{2} d^{3} e^{2}-5 A \,d^{5} c^{3}-7 B \,e^{5} a^{3}+30 B \,a^{2} c \,d^{2} e^{3}+5 B a \,c^{2} d^{4} e \right ) x}{8 a}+\frac {3 A \,a^{2} e^{5}}{4}-\frac {5 A a c \,d^{2} e^{3}}{2}-\frac {5 A \,c^{2} d^{4} e}{4}+\frac {15 B \,a^{2} d \,e^{4}}{4}-\frac {5 B a c \,d^{3} e^{2}}{2}-\frac {B \,c^{2} d^{5}}{4}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\frac {\left (8 A \,a^{2} c \,e^{5}+40 B \,a^{2} c d \,e^{4}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (15 A \,a^{2} c d \,e^{4}+10 A a \,c^{2} d^{3} e^{2}+3 A \,d^{5} c^{3}-15 B \,e^{5} a^{3}+30 B \,a^{2} c \,d^{2} e^{3}+5 B a \,c^{2} d^{4} e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{8 a^{2}}}{c^{3}}\)	\(404\)
risch	\(\text {Expression too large to display}\)	\(1386\)

Input:

int((B*x+A)*(e*x+d)^5/(c*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

B*e^5/c^3*x+1/c^3*((-1/8*c*(25*A*a^2*c*d*e^4-10*A*a*c^2*d^3*e^2-3*A*c^3*d^ 
5-9*B*a^3*e^5+50*B*a^2*c*d^2*e^3-5*B*a*c^2*d^4*e)/a^2*x^3+(A*a*c*e^5-5*A*c 
^2*d^2*e^3+5*B*a*c*d*e^4-5*B*c^2*d^3*e^2)*x^2-1/8*(15*A*a^2*c*d*e^4+10*A*a 
*c^2*d^3*e^2-5*A*c^3*d^5-7*B*a^3*e^5+30*B*a^2*c*d^2*e^3+5*B*a*c^2*d^4*e)/a 
*x+3/4*A*a^2*e^5-5/2*A*a*c*d^2*e^3-5/4*A*c^2*d^4*e+15/4*B*a^2*d*e^4-5/2*B* 
a*c*d^3*e^2-1/4*B*c^2*d^5)/(c*x^2+a)^2+1/8/a^2*(1/2*(8*A*a^2*c*e^5+40*B*a^ 
2*c*d*e^4)/c*ln(c*x^2+a)+(15*A*a^2*c*d*e^4+10*A*a*c^2*d^3*e^2+3*A*c^3*d^5- 
15*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. 691 vs. \(2 (288) = 576\).

Time = 0.11 (sec) , antiderivative size = 1403, normalized size of antiderivative = 4.58 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)^5/(c*x^2+a)^3,x, algorithm="fricas")
 

Output:

[1/16*(16*B*a^3*c^3*e^5*x^5 - 4*B*a^3*c^3*d^5 - 20*A*a^3*c^3*d^4*e - 40*B* 
a^4*c^2*d^3*e^2 - 40*A*a^4*c^2*d^2*e^3 + 60*B*a^5*c*d*e^4 + 12*A*a^5*c*e^5 
 + 2*(3*A*a*c^5*d^5 + 5*B*a^2*c^4*d^4*e + 10*A*a^2*c^4*d^3*e^2 - 50*B*a^3* 
c^3*d^2*e^3 - 25*A*a^3*c^3*d*e^4 + 25*B*a^4*c^2*e^5)*x^3 - 16*(5*B*a^3*c^3 
*d^3*e^2 + 5*A*a^3*c^3*d^2*e^3 - 5*B*a^4*c^2*d*e^4 - A*a^4*c^2*e^5)*x^2 + 
(3*A*a^2*c^3*d^5 + 5*B*a^3*c^2*d^4*e + 10*A*a^3*c^2*d^3*e^2 + 30*B*a^4*c*d 
^2*e^3 + 15*A*a^4*c*d*e^4 - 15*B*a^5*e^5 + (3*A*c^5*d^5 + 5*B*a*c^4*d^4*e 
+ 10*A*a*c^4*d^3*e^2 + 30*B*a^2*c^3*d^2*e^3 + 15*A*a^2*c^3*d*e^4 - 15*B*a^ 
3*c^2*e^5)*x^4 + 2*(3*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 - 15*B*a^4*c*e^5)*x^2)*sqrt 
(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 10*(A*a^2*c^4*d^5 - 
 B*a^3*c^3*d^4*e - 2*A*a^3*c^3*d^3*e^2 - 6*B*a^4*c^2*d^2*e^3 - 3*A*a^4*c^2 
*d*e^4 + 3*B*a^5*c*e^5)*x + 8*(5*B*a^5*c*d*e^4 + A*a^5*c*e^5 + (5*B*a^3*c^ 
3*d*e^4 + A*a^3*c^3*e^5)*x^4 + 2*(5*B*a^4*c^2*d*e^4 + A*a^4*c^2*e^5)*x^2)* 
log(c*x^2 + a))/(a^3*c^6*x^4 + 2*a^4*c^5*x^2 + a^5*c^4), 1/8*(8*B*a^3*c^3* 
e^5*x^5 - 2*B*a^3*c^3*d^5 - 10*A*a^3*c^3*d^4*e - 20*B*a^4*c^2*d^3*e^2 - 20 
*A*a^4*c^2*d^2*e^3 + 30*B*a^5*c*d*e^4 + 6*A*a^5*c*e^5 + (3*A*a*c^5*d^5 + 5 
*B*a^2*c^4*d^4*e + 10*A*a^2*c^4*d^3*e^2 - 50*B*a^3*c^3*d^2*e^3 - 25*A*a^3* 
c^3*d*e^4 + 25*B*a^4*c^2*e^5)*x^3 - 8*(5*B*a^3*c^3*d^3*e^2 + 5*A*a^3*c^3*d 
^2*e^3 - 5*B*a^4*c^2*d*e^4 - A*a^4*c^2*e^5)*x^2 + (3*A*a^2*c^3*d^5 + 5*...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1044 vs. \(2 (299) = 598\).

Time = 56.41 (sec) , antiderivative size = 1044, normalized size of antiderivative = 3.41 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)**5/(c*x**2+a)**3,x)
 

Output:

B*e**5*x/c**3 + (e**4*(A*e + 5*B*d)/(2*c**3) - sqrt(-a**5*c**7)*(-15*A*a** 
2*c*d*e**4 - 10*A*a*c**2*d**3*e**2 - 3*A*c**3*d**5 + 15*B*a**3*e**5 - 30*B 
*a**2*c*d**2*e**3 - 5*B*a*c**2*d**4*e)/(16*a**5*c**7))*log(x + (8*A*a**3*e 
**5 + 40*B*a**3*d*e**4 - 16*a**3*c**3*(e**4*(A*e + 5*B*d)/(2*c**3) - sqrt( 
-a**5*c**7)*(-15*A*a**2*c*d*e**4 - 10*A*a*c**2*d**3*e**2 - 3*A*c**3*d**5 + 
 15*B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 - 5*B*a*c**2*d**4*e)/(16*a**5*c**7 
)))/(-15*A*a**2*c*d*e**4 - 10*A*a*c**2*d**3*e**2 - 3*A*c**3*d**5 + 15*B*a* 
*3*e**5 - 30*B*a**2*c*d**2*e**3 - 5*B*a*c**2*d**4*e)) + (e**4*(A*e + 5*B*d 
)/(2*c**3) + sqrt(-a**5*c**7)*(-15*A*a**2*c*d*e**4 - 10*A*a*c**2*d**3*e**2 
 - 3*A*c**3*d**5 + 15*B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 - 5*B*a*c**2*d** 
4*e)/(16*a**5*c**7))*log(x + (8*A*a**3*e**5 + 40*B*a**3*d*e**4 - 16*a**3*c 
**3*(e**4*(A*e + 5*B*d)/(2*c**3) + sqrt(-a**5*c**7)*(-15*A*a**2*c*d*e**4 - 
 10*A*a*c**2*d**3*e**2 - 3*A*c**3*d**5 + 15*B*a**3*e**5 - 30*B*a**2*c*d**2 
*e**3 - 5*B*a*c**2*d**4*e)/(16*a**5*c**7)))/(-15*A*a**2*c*d*e**4 - 10*A*a* 
c**2*d**3*e**2 - 3*A*c**3*d**5 + 15*B*a**3*e**5 - 30*B*a**2*c*d**2*e**3 - 
5*B*a*c**2*d**4*e)) + (6*A*a**4*e**5 - 20*A*a**3*c*d**2*e**3 - 10*A*a**2*c 
**2*d**4*e + 30*B*a**4*d*e**4 - 20*B*a**3*c*d**3*e**2 - 2*B*a**2*c**2*d**5 
 + x**3*(-25*A*a**2*c**2*d*e**4 + 10*A*a*c**3*d**3*e**2 + 3*A*c**4*d**5 + 
9*B*a**3*c*e**5 - 50*B*a**2*c**2*d**2*e**3 + 5*B*a*c**3*d**4*e) + x**2*(8* 
A*a**3*c*e**5 - 40*A*a**2*c**2*d**2*e**3 + 40*B*a**3*c*d*e**4 - 40*B*a*...
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx=\frac {B e^{5} x}{c^{3}} - \frac {2 \, B a^{2} c^{2} d^{5} + 10 \, A a^{2} c^{2} d^{4} e + 20 \, B a^{3} c d^{3} e^{2} + 20 \, A a^{3} c d^{2} e^{3} - 30 \, B a^{4} d e^{4} - 6 \, A a^{4} e^{5} - {\left (3 \, A c^{4} d^{5} + 5 \, B a c^{3} d^{4} e + 10 \, A a c^{3} d^{3} e^{2} - 50 \, B a^{2} c^{2} d^{2} e^{3} - 25 \, A a^{2} c^{2} d e^{4} + 9 \, B a^{3} c e^{5}\right )} x^{3} + 8 \, {\left (5 \, B a^{2} c^{2} d^{3} e^{2} + 5 \, A a^{2} c^{2} d^{2} e^{3} - 5 \, B a^{3} c d e^{4} - A a^{3} c e^{5}\right )} x^{2} - {\left (5 \, A a c^{3} d^{5} - 5 \, B a^{2} c^{2} d^{4} e - 10 \, A a^{2} c^{2} d^{3} e^{2} - 30 \, B a^{3} c d^{2} e^{3} - 15 \, A a^{3} c d e^{4} + 7 \, B a^{4} e^{5}\right )} x}{8 \, {\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}} + \frac {{\left (5 \, B d e^{4} + A e^{5}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {{\left (3 \, A c^{3} d^{5} + 5 \, B a c^{2} d^{4} e + 10 \, A a c^{2} d^{3} e^{2} + 30 \, B a^{2} c d^{2} e^{3} + 15 \, A a^{2} c d e^{4} - 15 \, B a^{3} e^{5}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{3}} \] Input:

integrate((B*x+A)*(e*x+d)^5/(c*x^2+a)^3,x, algorithm="maxima")
 

Output:

B*e^5*x/c^3 - 1/8*(2*B*a^2*c^2*d^5 + 10*A*a^2*c^2*d^4*e + 20*B*a^3*c*d^3*e 
^2 + 20*A*a^3*c*d^2*e^3 - 30*B*a^4*d*e^4 - 6*A*a^4*e^5 - (3*A*c^4*d^5 + 5* 
B*a*c^3*d^4*e + 10*A*a*c^3*d^3*e^2 - 50*B*a^2*c^2*d^2*e^3 - 25*A*a^2*c^2*d 
*e^4 + 9*B*a^3*c*e^5)*x^3 + 8*(5*B*a^2*c^2*d^3*e^2 + 5*A*a^2*c^2*d^2*e^3 - 
 5*B*a^3*c*d*e^4 - A*a^3*c*e^5)*x^2 - (5*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 + 7*B*a^4*e^ 
5)*x)/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3) + 1/2*(5*B*d*e^4 + A*e^5)*lo 
g(c*x^2 + a)/c^3 + 1/8*(3*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 - 15*B*a^3*e^5)*arctan(c*x/sqrt(a 
*c))/(sqrt(a*c)*a^2*c^3)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx=\frac {B e^{5} x}{c^{3}} + \frac {{\left (5 \, B d e^{4} + A e^{5}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {{\left (3 \, A c^{3} d^{5} + 5 \, B a c^{2} d^{4} e + 10 \, A a c^{2} d^{3} e^{2} + 30 \, B a^{2} c d^{2} e^{3} + 15 \, A a^{2} c d e^{4} - 15 \, B a^{3} e^{5}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{3}} - \frac {2 \, B a^{2} c^{2} d^{5} + 10 \, A a^{2} c^{2} d^{4} e + 20 \, B a^{3} c d^{3} e^{2} + 20 \, A a^{3} c d^{2} e^{3} - 30 \, B a^{4} d e^{4} - 6 \, A a^{4} e^{5} - {\left (3 \, A c^{4} d^{5} + 5 \, B a c^{3} d^{4} e + 10 \, A a c^{3} d^{3} e^{2} - 50 \, B a^{2} c^{2} d^{2} e^{3} - 25 \, A a^{2} c^{2} d e^{4} + 9 \, B a^{3} c e^{5}\right )} x^{3} + 8 \, {\left (5 \, B a^{2} c^{2} d^{3} e^{2} + 5 \, A a^{2} c^{2} d^{2} e^{3} - 5 \, B a^{3} c d e^{4} - A a^{3} c e^{5}\right )} x^{2} - {\left (5 \, A a c^{3} d^{5} - 5 \, B a^{2} c^{2} d^{4} e - 10 \, A a^{2} c^{2} d^{3} e^{2} - 30 \, B a^{3} c d^{2} e^{3} - 15 \, A a^{3} c d e^{4} + 7 \, B a^{4} e^{5}\right )} x}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{3}} \] Input:

integrate((B*x+A)*(e*x+d)^5/(c*x^2+a)^3,x, algorithm="giac")
 

Output:

B*e^5*x/c^3 + 1/2*(5*B*d*e^4 + A*e^5)*log(c*x^2 + a)/c^3 + 1/8*(3*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 - 15*B*a^3*e^5)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c^3) - 1/8*(2* 
B*a^2*c^2*d^5 + 10*A*a^2*c^2*d^4*e + 20*B*a^3*c*d^3*e^2 + 20*A*a^3*c*d^2*e 
^3 - 30*B*a^4*d*e^4 - 6*A*a^4*e^5 - (3*A*c^4*d^5 + 5*B*a*c^3*d^4*e + 10*A* 
a*c^3*d^3*e^2 - 50*B*a^2*c^2*d^2*e^3 - 25*A*a^2*c^2*d*e^4 + 9*B*a^3*c*e^5) 
*x^3 + 8*(5*B*a^2*c^2*d^3*e^2 + 5*A*a^2*c^2*d^2*e^3 - 5*B*a^3*c*d*e^4 - A* 
a^3*c*e^5)*x^2 - (5*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 + 7*B*a^4*e^5)*x)/((c*x^2 + a)^2* 
a^2*c^3)
 

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx=\frac {\ln \left (c\,x^2+a\right )\,\left (256\,A\,a^5\,c^4\,e^5+1280\,B\,d\,a^5\,c^4\,e^4\right )}{512\,a^5\,c^7}-\frac {\frac {B\,c^2\,d^5}{4}-\frac {3\,A\,a^2\,e^5}{4}-x^2\,\left (-5\,B\,c^2\,d^3\,e^2-5\,A\,c^2\,d^2\,e^3+5\,B\,a\,c\,d\,e^4+A\,a\,c\,e^5\right )-\frac {x^3\,\left (9\,B\,a^3\,c\,e^5-50\,B\,a^2\,c^2\,d^2\,e^3-25\,A\,a^2\,c^2\,d\,e^4+5\,B\,a\,c^3\,d^4\,e+10\,A\,a\,c^3\,d^3\,e^2+3\,A\,c^4\,d^5\right )}{8\,a^2}+\frac {x\,\left (-7\,B\,a^3\,e^5+30\,B\,a^2\,c\,d^2\,e^3+15\,A\,a^2\,c\,d\,e^4+5\,B\,a\,c^2\,d^4\,e+10\,A\,a\,c^2\,d^3\,e^2-5\,A\,c^3\,d^5\right )}{8\,a}-\frac {15\,B\,a^2\,d\,e^4}{4}+\frac {5\,A\,c^2\,d^4\,e}{4}+\frac {5\,A\,a\,c\,d^2\,e^3}{2}+\frac {5\,B\,a\,c\,d^3\,e^2}{2}}{a^2\,c^3+2\,a\,c^4\,x^2+c^5\,x^4}+\frac {B\,e^5\,x}{c^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-15\,B\,a^3\,e^5+30\,B\,a^2\,c\,d^2\,e^3+15\,A\,a^2\,c\,d\,e^4+5\,B\,a\,c^2\,d^4\,e+10\,A\,a\,c^2\,d^3\,e^2+3\,A\,c^3\,d^5\right )}{8\,a^{5/2}\,c^{7/2}} \] Input:

int(((A + B*x)*(d + e*x)^5)/(a + c*x^2)^3,x)
 

Output:

(log(a + c*x^2)*(256*A*a^5*c^4*e^5 + 1280*B*a^5*c^4*d*e^4))/(512*a^5*c^7) 
- ((B*c^2*d^5)/4 - (3*A*a^2*e^5)/4 - x^2*(A*a*c*e^5 - 5*A*c^2*d^2*e^3 - 5* 
B*c^2*d^3*e^2 + 5*B*a*c*d*e^4) - (x^3*(3*A*c^4*d^5 + 9*B*a^3*c*e^5 + 10*A* 
a*c^3*d^3*e^2 - 25*A*a^2*c^2*d*e^4 - 50*B*a^2*c^2*d^2*e^3 + 5*B*a*c^3*d^4* 
e))/(8*a^2) + (x*(10*A*a*c^2*d^3*e^2 - 7*B*a^3*e^5 - 5*A*c^3*d^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))/(8*a) - (15*B*a^2*d*e^4 
)/4 + (5*A*c^2*d^4*e)/4 + (5*A*a*c*d^2*e^3)/2 + (5*B*a*c*d^3*e^2)/2)/(a^2* 
c^3 + c^5*x^4 + 2*a*c^4*x^2) + (B*e^5*x)/c^3 + (atan((c^(1/2)*x)/a^(1/2))* 
(3*A*c^3*d^5 - 15*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))/(8*a^(5/2)*c^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 983, normalized size of antiderivative = 3.21 \[ \int \frac {(A+B x) (d+e x)^5}{\left (a+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(e*x+d)^5/(c*x^2+a)^3,x)
 

Output:

( - 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**4*b*e**5 + 15*sqrt 
(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**4*c*d*e**4 + 30*sqrt(c)*sqrt( 
a)*atan((c*x)/(sqrt(c)*sqrt(a)))*a**3*b*c*d**2*e**3 - 30*sqrt(c)*sqrt(a)*a 
tan((c*x)/(sqrt(c)*sqrt(a)))*a**3*b*c*e**5*x**2 + 10*sqrt(c)*sqrt(a)*atan( 
(c*x)/(sqrt(c)*sqrt(a)))*a**3*c**2*d**3*e**2 + 30*sqrt(c)*sqrt(a)*atan((c* 
x)/(sqrt(c)*sqrt(a)))*a**3*c**2*d*e**4*x**2 + 5*sqrt(c)*sqrt(a)*atan((c*x) 
/(sqrt(c)*sqrt(a)))*a**2*b*c**2*d**4*e + 60*sqrt(c)*sqrt(a)*atan((c*x)/(sq 
rt(c)*sqrt(a)))*a**2*b*c**2*d**2*e**3*x**2 - 15*sqrt(c)*sqrt(a)*atan((c*x) 
/(sqrt(c)*sqrt(a)))*a**2*b*c**2*e**5*x**4 + 3*sqrt(c)*sqrt(a)*atan((c*x)/( 
sqrt(c)*sqrt(a)))*a**2*c**3*d**5 + 20*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)* 
sqrt(a)))*a**2*c**3*d**3*e**2*x**2 + 15*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c 
)*sqrt(a)))*a**2*c**3*d*e**4*x**4 + 10*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c) 
*sqrt(a)))*a*b*c**3*d**4*e*x**2 + 30*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*s 
qrt(a)))*a*b*c**3*d**2*e**3*x**4 + 6*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*s 
qrt(a)))*a*c**4*d**5*x**2 + 10*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a) 
))*a*c**4*d**3*e**2*x**4 + 5*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a))) 
*b*c**4*d**4*e*x**4 + 3*sqrt(c)*sqrt(a)*atan((c*x)/(sqrt(c)*sqrt(a)))*c**5 
*d**5*x**4 + 4*log(a + c*x**2)*a**5*c*e**5 + 20*log(a + c*x**2)*a**4*b*c*d 
*e**4 + 8*log(a + c*x**2)*a**4*c**2*e**5*x**2 + 40*log(a + c*x**2)*a**3*b* 
c**2*d*e**4*x**2 + 4*log(a + c*x**2)*a**3*c**3*e**5*x**4 + 20*log(a + c...