\(\int \frac {x^8}{(c+d x)^2 (a+b x^2)^3} \, dx\) [252]

Optimal result

Integrand size = 20, antiderivative size = 345 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\frac {x}{b^3 d^2}-\frac {c^8}{d^3 \left (b c^2+a d^2\right )^3 (c+d x)}+\frac {a^3 \left (2 a c d+\left (b c^2-a d^2\right ) x\right )}{4 b^3 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^2}-\frac {a^2 \left (16 a c d \left (2 b c^2+a d^2\right )+\left (13 b^2 c^4-12 a b c^2 d^2-9 a^2 d^4\right ) x\right )}{8 b^3 \left (b c^2+a d^2\right )^3 \left (a+b x^2\right )}+\frac {a^{3/2} \left (35 b^3 c^6-49 a b^2 c^4 d^2-51 a^2 b c^2 d^4-15 a^3 d^6\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{7/2} \left (b c^2+a d^2\right )^4}-\frac {2 c^7 \left (b c^2+4 a d^2\right ) \log (c+d x)}{d^3 \left (b c^2+a d^2\right )^4}-\frac {a^2 c d \left (6 b^2 c^4+4 a b c^2 d^2+a^2 d^4\right ) \log \left (a+b x^2\right )}{b^3 \left (b c^2+a d^2\right )^4} \] Output:

x/b^3/d^2-c^8/d^3/(a*d^2+b*c^2)^3/(d*x+c)+1/4*a^3*(2*a*c*d+(-a*d^2+b*c^2)* 
x)/b^3/(a*d^2+b*c^2)^2/(b*x^2+a)^2-1/8*a^2*(16*a*c*d*(a*d^2+2*b*c^2)+(-9*a 
^2*d^4-12*a*b*c^2*d^2+13*b^2*c^4)*x)/b^3/(a*d^2+b*c^2)^3/(b*x^2+a)+1/8*a^( 
3/2)*(-15*a^3*d^6-51*a^2*b*c^2*d^4-49*a*b^2*c^4*d^2+35*b^3*c^6)*arctan(b^( 
1/2)*x/a^(1/2))/b^(7/2)/(a*d^2+b*c^2)^4-2*c^7*(4*a*d^2+b*c^2)*ln(d*x+c)/d^ 
3/(a*d^2+b*c^2)^4-a^2*c*d*(a^2*d^4+4*a*b*c^2*d^2+6*b^2*c^4)*ln(b*x^2+a)/b^ 
3/(a*d^2+b*c^2)^4
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 333, normalized size of antiderivative = 0.97 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\frac {1}{8} \left (\frac {8 x}{b^3 d^2}-\frac {8 c^8}{\left (b c^2 d+a d^3\right )^3 (c+d x)}+\frac {2 a^3 \left (b c^2 x+a d (2 c-d x)\right )}{b^3 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^2}+\frac {a^2 \left (-13 b^2 c^4 x-4 a b c^2 d (8 c-3 d x)+a^2 d^3 (-16 c+9 d x)\right )}{b^3 \left (b c^2+a d^2\right )^3 \left (a+b x^2\right )}-\frac {a^{3/2} \left (-35 b^3 c^6+49 a b^2 c^4 d^2+51 a^2 b c^2 d^4+15 a^3 d^6\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2} \left (b c^2+a d^2\right )^4}-\frac {16 \left (b c^9+4 a c^7 d^2\right ) \log (c+d x)}{d^3 \left (b c^2+a d^2\right )^4}-\frac {8 a^2 c d \left (6 b^2 c^4+4 a b c^2 d^2+a^2 d^4\right ) \log \left (a+b x^2\right )}{b^3 \left (b c^2+a d^2\right )^4}\right ) \] Input:

Integrate[x^8/((c + d*x)^2*(a + b*x^2)^3),x]
 

Output:

((8*x)/(b^3*d^2) - (8*c^8)/((b*c^2*d + a*d^3)^3*(c + d*x)) + (2*a^3*(b*c^2 
*x + a*d*(2*c - d*x)))/(b^3*(b*c^2 + a*d^2)^2*(a + b*x^2)^2) + (a^2*(-13*b 
^2*c^4*x - 4*a*b*c^2*d*(8*c - 3*d*x) + a^2*d^3*(-16*c + 9*d*x)))/(b^3*(b*c 
^2 + a*d^2)^3*(a + b*x^2)) - (a^(3/2)*(-35*b^3*c^6 + 49*a*b^2*c^4*d^2 + 51 
*a^2*b*c^2*d^4 + 15*a^3*d^6)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b^(7/2)*(b*c^2 
+ a*d^2)^4) - (16*(b*c^9 + 4*a*c^7*d^2)*Log[c + d*x])/(d^3*(b*c^2 + a*d^2) 
^4) - (8*a^2*c*d*(6*b^2*c^4 + 4*a*b*c^2*d^2 + a^2*d^4)*Log[a + b*x^2])/(b^ 
3*(b*c^2 + a*d^2)^4))/8
 

Rubi [A] (verified)

Time = 3.47 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {601, 2178, 2160, 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.

\(\displaystyle \int \frac {x^8}{\left (a+b x^2\right )^3 (c+d x)^2} \, dx\)

\(\Big \downarrow \) 601

\(\displaystyle \frac {a^3 \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}-\frac {\int \frac {-\frac {4 a x^6}{b}+\frac {4 a^2 x^4}{b^2}-\frac {a^3 \left (4 b^2 c^4+11 a b d^2 c^2+a^2 d^4\right ) x^2}{b^3 \left (b c^2+a d^2\right )^2}-\frac {2 a^4 c d \left (3 b c^2+a d^2\right ) x}{b^3 \left (b c^2+a d^2\right )^2}+\frac {a^4 c^2 \left (b c^2-a d^2\right )}{b^3 \left (b c^2+a d^2\right )^2}}{(c+d x)^2 \left (b x^2+a\right )^2}dx}{4 a}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {a^3 \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}-\frac {\frac {a^3 \left (x \left (-9 a^2 d^4-12 a b c^2 d^2+13 b^2 c^4\right )+16 a c d \left (a d^2+2 b c^2\right )\right )}{2 b^3 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}-\frac {\int \frac {\frac {c^2 \left (11 b^2 c^4-12 a b d^2 c^2-7 a^2 d^4\right ) a^4}{b^2 \left (b c^2+a d^2\right )^3}-\frac {2 c d \left (13 b c^2+7 a d^2\right ) x a^4}{b^2 \left (b c^2+a d^2\right )^2}-\frac {\left (16 b^3 c^6+61 a b^2 d^2 c^4+36 a^2 b d^4 c^2+7 a^3 d^6\right ) x^2 a^3}{b^2 \left (b c^2+a d^2\right )^3}+\frac {8 x^4 a^2}{b}}{(c+d x)^2 \left (b x^2+a\right )}dx}{2 a b}}{4 a}\)

\(\Big \downarrow \) 2160

\(\displaystyle \frac {a^3 \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}-\frac {\frac {a^3 \left (x \left (-9 a^2 d^4-12 a b c^2 d^2+13 b^2 c^4\right )+16 a c d \left (a d^2+2 b c^2\right )\right )}{2 b^3 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}-\frac {\int \left (\frac {8 a^2 b c^8}{d^2 \left (b c^2+a d^2\right )^3 (c+d x)^2}-\frac {16 a^2 b \left (b c^2+4 a d^2\right ) c^7}{d^2 \left (b c^2+a d^2\right )^4 (c+d x)}+\frac {a^4 \left (35 b^3 c^6-49 a b^2 d^2 c^4-51 a^2 b d^4 c^2-16 b d \left (6 b^2 c^4+4 a b d^2 c^2+a^2 d^4\right ) x c-15 a^3 d^6\right )}{b^2 \left (b c^2+a d^2\right )^4 \left (b x^2+a\right )}+\frac {8 a^2}{b^2 d^2}\right )dx}{2 a b}}{4 a}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 \left (x \left (b c^2-a d^2\right )+2 a c d\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^2}-\frac {\frac {a^3 \left (x \left (-9 a^2 d^4-12 a b c^2 d^2+13 b^2 c^4\right )+16 a c d \left (a d^2+2 b c^2\right )\right )}{2 b^3 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^3}-\frac {\frac {8 a^2 x}{b^2 d^2}-\frac {8 a^2 b c^8}{d^3 (c+d x) \left (a d^2+b c^2\right )^3}-\frac {16 a^2 b c^7 \left (4 a d^2+b c^2\right ) \log (c+d x)}{d^3 \left (a d^2+b c^2\right )^4}-\frac {8 a^4 c d \left (a^2 d^4+4 a b c^2 d^2+6 b^2 c^4\right ) \log \left (a+b x^2\right )}{b^2 \left (a d^2+b c^2\right )^4}+\frac {a^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-15 a^3 d^6-51 a^2 b c^2 d^4-49 a b^2 c^4 d^2+35 b^3 c^6\right )}{b^{5/2} \left (a d^2+b c^2\right )^4}}{2 a b}}{4 a}\)

Input:

Int[x^8/((c + d*x)^2*(a + b*x^2)^3),x]
 

Output:

(a^3*(2*a*c*d + (b*c^2 - a*d^2)*x))/(4*b^3*(b*c^2 + a*d^2)^2*(a + b*x^2)^2 
) - ((a^3*(16*a*c*d*(2*b*c^2 + a*d^2) + (13*b^2*c^4 - 12*a*b*c^2*d^2 - 9*a 
^2*d^4)*x))/(2*b^3*(b*c^2 + a*d^2)^3*(a + b*x^2)) - ((8*a^2*x)/(b^2*d^2) - 
 (8*a^2*b*c^8)/(d^3*(b*c^2 + a*d^2)^3*(c + d*x)) + (a^(7/2)*(35*b^3*c^6 - 
49*a*b^2*c^4*d^2 - 51*a^2*b*c^2*d^4 - 15*a^3*d^6)*ArcTan[(Sqrt[b]*x)/Sqrt[ 
a]])/(b^(5/2)*(b*c^2 + a*d^2)^4) - (16*a^2*b*c^7*(b*c^2 + 4*a*d^2)*Log[c + 
 d*x])/(d^3*(b*c^2 + a*d^2)^4) - (8*a^4*c*d*(6*b^2*c^4 + 4*a*b*c^2*d^2 + a 
^2*d^4)*Log[a + b*x^2])/(b^2*(b*c^2 + a*d^2)^4))/(2*a*b))/(4*a)
 

Defintions of rubi rules used

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 2009

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

rule 2160

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, 
 d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {x}{b^{3} d^{2}}-\frac {a^{2} \left (\frac {\left (-\frac {9}{8} a^{3} b \,d^{6}-\frac {21}{8} a^{2} b^{2} c^{2} d^{4}+\frac {1}{8} a \,b^{3} c^{4} d^{2}+\frac {13}{8} b^{4} c^{6}\right ) x^{3}+\left (2 a^{3} b c \,d^{5}+6 a^{2} b^{2} c^{3} d^{3}+4 a \,b^{3} c^{5} d \right ) x^{2}-\frac {a \left (7 a^{3} d^{6}+19 a^{2} b \,c^{2} d^{4}+a \,b^{2} c^{4} d^{2}-11 b^{3} c^{6}\right ) x}{8}+\frac {a^{2} c d \left (3 a^{2} d^{4}+10 b \,c^{2} d^{2} a +7 b^{2} c^{4}\right )}{2}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (16 d^{5} a^{2} c b +64 d^{3} a \,c^{3} b^{2}+96 c^{5} d \,b^{3}\right ) \ln \left (b \,x^{2}+a \right )}{16 b}+\frac {\left (15 a^{3} d^{6}+51 a^{2} b \,c^{2} d^{4}+49 a \,b^{2} c^{4} d^{2}-35 b^{3} c^{6}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{\left (a \,d^{2}+b \,c^{2}\right )^{4} b^{3}}-\frac {c^{8}}{d^{3} \left (a \,d^{2}+b \,c^{2}\right )^{3} \left (d x +c \right )}-\frac {2 c^{7} \left (4 a \,d^{2}+b \,c^{2}\right ) \ln \left (d x +c \right )}{d^{3} \left (a \,d^{2}+b \,c^{2}\right )^{4}}\)	\(377\)
risch	\(\text {Expression too large to display}\)	\(1288\)

Input:

int(x^8/(d*x+c)^2/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

1/b^3/d^2*x-a^2/(a*d^2+b*c^2)^4/b^3*(((-9/8*a^3*b*d^6-21/8*a^2*b^2*c^2*d^4 
+1/8*a*b^3*c^4*d^2+13/8*b^4*c^6)*x^3+(2*a^3*b*c*d^5+6*a^2*b^2*c^3*d^3+4*a* 
b^3*c^5*d)*x^2-1/8*a*(7*a^3*d^6+19*a^2*b*c^2*d^4+a*b^2*c^4*d^2-11*b^3*c^6) 
*x+1/2*a^2*c*d*(3*a^2*d^4+10*a*b*c^2*d^2+7*b^2*c^4))/(b*x^2+a)^2+1/16*(16* 
a^2*b*c*d^5+64*a*b^2*c^3*d^3+96*b^3*c^5*d)/b*ln(b*x^2+a)+1/8*(15*a^3*d^6+5 
1*a^2*b*c^2*d^4+49*a*b^2*c^4*d^2-35*b^3*c^6)/(a*b)^(1/2)*arctan(b*x/(a*b)^ 
(1/2)))-c^8/d^3/(a*d^2+b*c^2)^3/(d*x+c)-2*c^7*(4*a*d^2+b*c^2)*ln(d*x+c)/d^ 
3/(a*d^2+b*c^2)^4
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1669 vs. \(2 (331) = 662\).

Time = 16.99 (sec) , antiderivative size = 3359, normalized size of antiderivative = 9.74 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(x^8/(d*x+c)^2/(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**8/(d*x+c)**2/(b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 854 vs. \(2 (331) = 662\).

Time = 0.14 (sec) , antiderivative size = 854, normalized size of antiderivative = 2.48 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^8/(d*x+c)^2/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

-(6*a^2*b^2*c^5*d + 4*a^3*b*c^3*d^3 + a^4*c*d^5)*log(b*x^2 + a)/(b^7*c^8 + 
 4*a*b^6*c^6*d^2 + 6*a^2*b^5*c^4*d^4 + 4*a^3*b^4*c^2*d^6 + a^4*b^3*d^8) - 
2*(b*c^9 + 4*a*c^7*d^2)*log(d*x + c)/(b^4*c^8*d^3 + 4*a*b^3*c^6*d^5 + 6*a^ 
2*b^2*c^4*d^7 + 4*a^3*b*c^2*d^9 + a^4*d^11) + 1/8*(35*a^2*b^3*c^6 - 49*a^3 
*b^2*c^4*d^2 - 51*a^4*b*c^2*d^4 - 15*a^5*d^6)*arctan(b*x/sqrt(a*b))/((b^7* 
c^8 + 4*a*b^6*c^6*d^2 + 6*a^2*b^5*c^4*d^4 + 4*a^3*b^4*c^2*d^6 + a^4*b^3*d^ 
8)*sqrt(a*b)) - 1/8*(8*a^2*b^3*c^8 + 28*a^4*b*c^4*d^4 + 12*a^5*c^2*d^6 + ( 
8*b^5*c^8 + 13*a^2*b^3*c^4*d^4 - 12*a^3*b^2*c^2*d^6 - 9*a^4*b*d^8)*x^4 + ( 
13*a^2*b^3*c^5*d^3 + 20*a^3*b^2*c^3*d^5 + 7*a^4*b*c*d^7)*x^3 + (16*a*b^4*c 
^8 + 43*a^3*b^2*c^4*d^4 + 4*a^4*b*c^2*d^6 - 7*a^5*d^8)*x^2 + (11*a^3*b^2*c 
^5*d^3 + 16*a^4*b*c^3*d^5 + 5*a^5*c*d^7)*x)/(a^2*b^6*c^7*d^3 + 3*a^3*b^5*c 
^5*d^5 + 3*a^4*b^4*c^3*d^7 + a^5*b^3*c*d^9 + (b^8*c^6*d^4 + 3*a*b^7*c^4*d^ 
6 + 3*a^2*b^6*c^2*d^8 + a^3*b^5*d^10)*x^5 + (b^8*c^7*d^3 + 3*a*b^7*c^5*d^5 
 + 3*a^2*b^6*c^3*d^7 + a^3*b^5*c*d^9)*x^4 + 2*(a*b^7*c^6*d^4 + 3*a^2*b^6*c 
^4*d^6 + 3*a^3*b^5*c^2*d^8 + a^4*b^4*d^10)*x^3 + 2*(a*b^7*c^7*d^3 + 3*a^2* 
b^6*c^5*d^5 + 3*a^3*b^5*c^3*d^7 + a^4*b^4*c*d^9)*x^2 + (a^2*b^6*c^6*d^4 + 
3*a^3*b^5*c^4*d^6 + 3*a^4*b^4*c^2*d^8 + a^5*b^3*d^10)*x) + x/(b^3*d^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 850 vs. \(2 (331) = 662\).

Time = 0.14 (sec) , antiderivative size = 850, normalized size of antiderivative = 2.46 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^8/(d*x+c)^2/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

-c^8*d^7/((b^3*c^6*d^10 + 3*a*b^2*c^4*d^12 + 3*a^2*b*c^2*d^14 + a^3*d^16)* 
(d*x + c)) - (6*a^2*b^2*c^5*d + 4*a^3*b*c^3*d^3 + a^4*c*d^5)*log(b - 2*b*c 
/(d*x + c) + b*c^2/(d*x + c)^2 + a*d^2/(d*x + c)^2)/(b^7*c^8 + 4*a*b^6*c^6 
*d^2 + 6*a^2*b^5*c^4*d^4 + 4*a^3*b^4*c^2*d^6 + a^4*b^3*d^8) + 1/8*(35*a^2* 
b^3*c^6*d^2 - 49*a^3*b^2*c^4*d^4 - 51*a^4*b*c^2*d^6 - 15*a^5*d^8)*arctan(( 
b*c - b*c^2/(d*x + c) - a*d^2/(d*x + c))/(sqrt(a*b)*d))/((b^7*c^8 + 4*a*b^ 
6*c^6*d^2 + 6*a^2*b^5*c^4*d^4 + 4*a^3*b^4*c^2*d^6 + a^4*b^3*d^8)*sqrt(a*b) 
*d^2) + 2*c*log(abs(d*x + c)/((d*x + c)^2*abs(d)))/(b^3*d^3) + 1/8*(8*b^6* 
c^8 + 32*a*b^5*c^6*d^2 + 48*a^2*b^4*c^4*d^4 + 32*a^3*b^3*c^2*d^6 + 8*a^4*b 
^2*d^8 - (32*b^6*c^9*d + 128*a*b^5*c^7*d^3 + 205*a^2*b^4*c^5*d^5 + 82*a^3* 
b^3*c^3*d^7 + 13*a^4*b^2*c*d^9)/((d*x + c)*d) + (48*b^6*c^10*d^2 + 208*a*b 
^5*c^8*d^4 + 391*a^2*b^4*c^6*d^6 + 103*a^3*b^3*c^4*d^8 + 57*a^4*b^2*c^2*d^ 
10 + 25*a^5*b*d^12)/((d*x + c)^2*d^2) - (32*b^6*c^11*d^3 + 160*a*b^5*c^9*d 
^5 + 359*a^2*b^4*c^7*d^7 + 99*a^3*b^3*c^5*d^9 + 65*a^4*b^2*c^3*d^11 + 37*a 
^5*b*c*d^13)/((d*x + c)^3*d^3) + (8*b^6*c^12*d^4 + 48*a*b^5*c^10*d^6 + 133 
*a^2*b^4*c^8*d^8 + 78*a^3*b^3*c^6*d^10 + 20*a^4*b^2*c^4*d^12 + 50*a^5*b*c^ 
2*d^14 + 15*a^6*d^16)/((d*x + c)^4*d^4))*(d*x + c)/((b*c^2 + a*d^2)^4*(b - 
 2*b*c/(d*x + c) + b*c^2/(d*x + c)^2 + a*d^2/(d*x + c)^2)^2*b^3*d^3)
 

Mupad [B] (verification not implemented)

Time = 8.65 (sec) , antiderivative size = 939, normalized size of antiderivative = 2.72 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx=\frac {x}{b^3\,d^2}-\frac {\ln \left (\sqrt {-a^3\,b^7}+a\,b^4\,x\right )\,\left (15\,a^3\,d^6\,\sqrt {-a^3\,b^7}-35\,b^3\,c^6\,\sqrt {-a^3\,b^7}+96\,a^2\,b^6\,c^5\,d+16\,a^4\,b^4\,c\,d^5+64\,a^3\,b^5\,c^3\,d^3+49\,a\,b^2\,c^4\,d^2\,\sqrt {-a^3\,b^7}+51\,a^2\,b\,c^2\,d^4\,\sqrt {-a^3\,b^7}\right )}{16\,b^{11}\,c^8+a^3\,b^7\,\left (64\,b\,c^2\,d^6+16\,a\,d^8\right )+64\,a\,b^{10}\,c^6\,d^2+96\,a^2\,b^9\,c^4\,d^4}-\frac {\ln \left (\sqrt {-a^3\,b^7}-a\,b^4\,x\right )\,\left (35\,b^3\,c^6\,\sqrt {-a^3\,b^7}-15\,a^3\,d^6\,\sqrt {-a^3\,b^7}+96\,a^2\,b^6\,c^5\,d+16\,a^4\,b^4\,c\,d^5+64\,a^3\,b^5\,c^3\,d^3-49\,a\,b^2\,c^4\,d^2\,\sqrt {-a^3\,b^7}-51\,a^2\,b\,c^2\,d^4\,\sqrt {-a^3\,b^7}\right )}{16\,b^{11}\,c^8+a^3\,b^7\,\left (64\,b\,c^2\,d^6+16\,a\,d^8\right )+64\,a\,b^{10}\,c^6\,d^2+96\,a^2\,b^9\,c^4\,d^4}-\frac {\frac {d\,x\,\left (5\,a^4\,c\,d^3+11\,b\,a^3\,c^3\,d\right )}{8\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}+\frac {d\,x^3\,\left (7\,a^3\,b\,c\,d^3+13\,a^2\,b^2\,c^3\,d\right )}{8\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}+\frac {x^4\,\left (-9\,a^4\,b\,d^8-12\,a^3\,b^2\,c^2\,d^6+13\,a^2\,b^3\,c^4\,d^4+8\,b^5\,c^8\right )}{8\,d\,\left (a^3\,d^6+3\,a^2\,b\,c^2\,d^4+3\,a\,b^2\,c^4\,d^2+b^3\,c^6\right )}+\frac {c\,\left (3\,a^5\,c\,d^6+7\,a^4\,b\,c^3\,d^4+2\,a^2\,b^3\,c^7\right )}{2\,d\,\left (b\,c^2+a\,d^2\right )\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}+\frac {x^2\,\left (-7\,a^5\,d^8+4\,a^4\,b\,c^2\,d^6+43\,a^3\,b^2\,c^4\,d^4+16\,a\,b^4\,c^8\right )}{8\,d\,\left (b\,c^2+a\,d^2\right )\,\left (a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4\right )}}{a^2\,b^3\,d^3\,x+c\,a^2\,b^3\,d^2+2\,a\,b^4\,d^3\,x^3+2\,c\,a\,b^4\,d^2\,x^2+b^5\,d^3\,x^5+c\,b^5\,d^2\,x^4}-\frac {\ln \left (c+d\,x\right )\,\left (2\,b\,c^9+8\,a\,c^7\,d^2\right )}{a^4\,d^{11}+4\,a^3\,b\,c^2\,d^9+6\,a^2\,b^2\,c^4\,d^7+4\,a\,b^3\,c^6\,d^5+b^4\,c^8\,d^3} \] Input:

int(x^8/((a + b*x^2)^3*(c + d*x)^2),x)
 

Output:

x/(b^3*d^2) - (log((-a^3*b^7)^(1/2) + a*b^4*x)*(15*a^3*d^6*(-a^3*b^7)^(1/2 
) - 35*b^3*c^6*(-a^3*b^7)^(1/2) + 96*a^2*b^6*c^5*d + 16*a^4*b^4*c*d^5 + 64 
*a^3*b^5*c^3*d^3 + 49*a*b^2*c^4*d^2*(-a^3*b^7)^(1/2) + 51*a^2*b*c^2*d^4*(- 
a^3*b^7)^(1/2)))/(16*b^11*c^8 + a^3*b^7*(16*a*d^8 + 64*b*c^2*d^6) + 64*a*b 
^10*c^6*d^2 + 96*a^2*b^9*c^4*d^4) - (log((-a^3*b^7)^(1/2) - a*b^4*x)*(35*b 
^3*c^6*(-a^3*b^7)^(1/2) - 15*a^3*d^6*(-a^3*b^7)^(1/2) + 96*a^2*b^6*c^5*d + 
 16*a^4*b^4*c*d^5 + 64*a^3*b^5*c^3*d^3 - 49*a*b^2*c^4*d^2*(-a^3*b^7)^(1/2) 
 - 51*a^2*b*c^2*d^4*(-a^3*b^7)^(1/2)))/(16*b^11*c^8 + a^3*b^7*(16*a*d^8 + 
64*b*c^2*d^6) + 64*a*b^10*c^6*d^2 + 96*a^2*b^9*c^4*d^4) - ((d*x*(5*a^4*c*d 
^3 + 11*a^3*b*c^3*d))/(8*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) + (d*x^3*(13 
*a^2*b^2*c^3*d + 7*a^3*b*c*d^3))/(8*(a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)) + 
 (x^4*(8*b^5*c^8 - 9*a^4*b*d^8 + 13*a^2*b^3*c^4*d^4 - 12*a^3*b^2*c^2*d^6)) 
/(8*d*(a^3*d^6 + b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4)) + (c*(3*a^5 
*c*d^6 + 2*a^2*b^3*c^7 + 7*a^4*b*c^3*d^4))/(2*d*(a*d^2 + b*c^2)*(a^2*d^4 + 
 b^2*c^4 + 2*a*b*c^2*d^2)) + (x^2*(16*a*b^4*c^8 - 7*a^5*d^8 + 4*a^4*b*c^2* 
d^6 + 43*a^3*b^2*c^4*d^4))/(8*d*(a*d^2 + b*c^2)*(a^2*d^4 + b^2*c^4 + 2*a*b 
*c^2*d^2)))/(b^5*d^3*x^5 + a^2*b^3*c*d^2 + a^2*b^3*d^3*x + 2*a*b^4*d^3*x^3 
 + b^5*c*d^2*x^4 + 2*a*b^4*c*d^2*x^2) - (log(c + d*x)*(2*b*c^9 + 8*a*c^7*d 
^2))/(a^4*d^11 + b^4*c^8*d^3 + 4*a*b^3*c^6*d^5 + 4*a^3*b*c^2*d^9 + 6*a^2*b 
^2*c^4*d^7)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 2337, normalized size of antiderivative = 6.77 \[ \int \frac {x^8}{(c+d x)^2 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^8/(d*x+c)^2/(b*x^2+a)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**6*c**2*d**9 - 15*s 
qrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**6*c*d**10*x - 51*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**4*d**7 - 51*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**3*d**8*x - 30*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**2*d**9*x**2 - 30*sqrt(b)*sqrt(a)*atan 
((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c*d**10*x**3 - 49*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**6*d**5 - 49*sqrt(b)*sqrt(a)*atan((b*x) 
/(sqrt(b)*sqrt(a)))*a**4*b**2*c**5*d**6*x - 102*sqrt(b)*sqrt(a)*atan((b*x) 
/(sqrt(b)*sqrt(a)))*a**4*b**2*c**4*d**7*x**2 - 102*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**3*d**8*x**3 - 15*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**2*d**9*x**4 - 15*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c*d**10*x**5 + 35*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**8*d**3 + 35*sqrt(b)*sqrt(a)*atan(( 
b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**7*d**4*x - 98*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**6*d**5*x**2 - 98*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**5*d**6*x**3 - 51*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**4*d**7*x**4 - 51*sqrt(b)*sqrt(a)*a 
tan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**3*d**8*x**5 + 70*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**4*c**8*d**3*x**2 + 70*sqrt(b)*sqrt( 
a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**4*c**7*d**4*x**3 - 49*sqrt(b)*...