\(\int \frac {x^8}{(c+d x)^3 (a+b x^2)^3} \, dx\) [265]

Optimal result

Integrand size = 20, antiderivative size = 436 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx=-\frac {c^8}{2 d^3 \left (b c^2+a d^2\right )^3 (c+d x)^2}+\frac {2 c^7 \left (b c^2+4 a d^2\right )}{d^3 \left (b c^2+a d^2\right )^4 (c+d x)}+\frac {a^3 \left (a d \left (3 b c^2-a d^2\right )+b c \left (b c^2-3 a d^2\right ) x\right )}{4 b^3 \left (b c^2+a d^2\right )^3 \left (a+b x^2\right )^2}-\frac {a^2 \left (8 a d \left (2 b c^2-a d^2\right ) \left (3 b c^2+a d^2\right )+b c \left (13 b^2 c^4-50 a b c^2 d^2-15 a^2 d^4\right ) x\right )}{8 b^3 \left (b c^2+a d^2\right )^4 \left (a+b x^2\right )}+\frac {a^{3/2} c \left (35 b^3 c^6-203 a b^2 c^4 d^2-55 a^2 b c^2 d^4-9 a^3 d^6\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{5/2} \left (b c^2+a d^2\right )^5}+\frac {c^6 \left (b^2 c^4+5 a b c^2 d^2+28 a^2 d^4\right ) \log (c+d x)}{d^3 \left (b c^2+a d^2\right )^5}-\frac {a^2 d \left (18 b^3 c^6-10 a b^2 c^4 d^2-5 a^2 b c^2 d^4-a^3 d^6\right ) \log \left (a+b x^2\right )}{2 b^3 \left (b c^2+a d^2\right )^5} \] Output:

-1/2*c^8/d^3/(a*d^2+b*c^2)^3/(d*x+c)^2+2*c^7*(4*a*d^2+b*c^2)/d^3/(a*d^2+b* 
c^2)^4/(d*x+c)+1/4*a^3*(a*d*(-a*d^2+3*b*c^2)+b*c*(-3*a*d^2+b*c^2)*x)/b^3/( 
a*d^2+b*c^2)^3/(b*x^2+a)^2-1/8*a^2*(8*a*d*(-a*d^2+2*b*c^2)*(a*d^2+3*b*c^2) 
+b*c*(-15*a^2*d^4-50*a*b*c^2*d^2+13*b^2*c^4)*x)/b^3/(a*d^2+b*c^2)^4/(b*x^2 
+a)+1/8*a^(3/2)*c*(-9*a^3*d^6-55*a^2*b*c^2*d^4-203*a*b^2*c^4*d^2+35*b^3*c^ 
6)*arctan(b^(1/2)*x/a^(1/2))/b^(5/2)/(a*d^2+b*c^2)^5+c^6*(28*a^2*d^4+5*a*b 
*c^2*d^2+b^2*c^4)*ln(d*x+c)/d^3/(a*d^2+b*c^2)^5-1/2*a^2*d*(-a^3*d^6-5*a^2* 
b*c^2*d^4-10*a*b^2*c^4*d^2+18*b^3*c^6)*ln(b*x^2+a)/b^3/(a*d^2+b*c^2)^5
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 387, normalized size of antiderivative = 0.89 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx=\frac {-\frac {4 c^8 \left (b c^2+a d^2\right )^2}{d^3 (c+d x)^2}+\frac {16 c^7 \left (b c^2+a d^2\right ) \left (b c^2+4 a d^2\right )}{d^3 (c+d x)}+\frac {2 a^3 \left (b c^2+a d^2\right )^2 \left (-a^2 d^3+b^2 c^3 x+3 a b c d (c-d x)\right )}{b^3 \left (a+b x^2\right )^2}+\frac {a^2 \left (b c^2+a d^2\right ) \left (8 a^3 d^5-13 b^3 c^5 x+a^2 b c d^3 (8 c+15 d x)+2 a b^2 c^3 d (-24 c+25 d x)\right )}{b^3 \left (a+b x^2\right )}-\frac {a^{3/2} c \left (-35 b^3 c^6+203 a b^2 c^4 d^2+55 a^2 b c^2 d^4+9 a^3 d^6\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}+\frac {8 c^6 \left (b^2 c^4+5 a b c^2 d^2+28 a^2 d^4\right ) \log (c+d x)}{d^3}+\frac {4 a^2 d \left (-18 b^3 c^6+10 a b^2 c^4 d^2+5 a^2 b c^2 d^4+a^3 d^6\right ) \log \left (a+b x^2\right )}{b^3}}{8 \left (b c^2+a d^2\right )^5} \] Input:

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

Output:

((-4*c^8*(b*c^2 + a*d^2)^2)/(d^3*(c + d*x)^2) + (16*c^7*(b*c^2 + a*d^2)*(b 
*c^2 + 4*a*d^2))/(d^3*(c + d*x)) + (2*a^3*(b*c^2 + a*d^2)^2*(-(a^2*d^3) + 
b^2*c^3*x + 3*a*b*c*d*(c - d*x)))/(b^3*(a + b*x^2)^2) + (a^2*(b*c^2 + a*d^ 
2)*(8*a^3*d^5 - 13*b^3*c^5*x + a^2*b*c*d^3*(8*c + 15*d*x) + 2*a*b^2*c^3*d* 
(-24*c + 25*d*x)))/(b^3*(a + b*x^2)) - (a^(3/2)*c*(-35*b^3*c^6 + 203*a*b^2 
*c^4*d^2 + 55*a^2*b*c^2*d^4 + 9*a^3*d^6)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(5 
/2) + (8*c^6*(b^2*c^4 + 5*a*b*c^2*d^2 + 28*a^2*d^4)*Log[c + d*x])/d^3 + (4 
*a^2*d*(-18*b^3*c^6 + 10*a*b^2*c^4*d^2 + 5*a^2*b*c^2*d^4 + a^3*d^6)*Log[a 
+ b*x^2])/b^3)/(8*(b*c^2 + a*d^2)^5)
 

Rubi [A] (verified)

Time = 6.58 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.06, 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.

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

\(\Big \downarrow \) 601

\(\displaystyle \frac {a^3 \left (b c x \left (b c^2-3 a d^2\right )+a d \left (3 b c^2-a d^2\right )\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^3}-\frac {\int \frac {-\frac {4 a x^6}{b}+\frac {4 a^2 x^4}{b^2}-\frac {3 a^4 c d^3 \left (b c^2-3 a d^2\right ) x^3}{b^2 \left (b c^2+a d^2\right )^3}-\frac {a^3 c^2 \left (4 b^2 c^4+21 a b d^2 c^2-3 a^2 d^4\right ) x^2}{b^2 \left (b c^2+a d^2\right )^3}-\frac {a^4 c^3 d \left (9 b c^2+5 a d^2\right ) x}{b^2 \left (b c^2+a d^2\right )^3}+\frac {a^4 c^4 \left (b c^2-3 a d^2\right )}{b^2 \left (b c^2+a d^2\right )^3}}{(c+d x)^3 \left (b x^2+a\right )^2}dx}{4 a}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {a^3 \left (b c x \left (b c^2-3 a d^2\right )+a d \left (3 b c^2-a d^2\right )\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^3}-\frac {\frac {a^3 \left (b c x \left (-15 a^2 d^4-50 a b c^2 d^2+13 b^2 c^4\right )+8 a d \left (2 b c^2-a d^2\right ) \left (a d^2+3 b c^2\right )\right )}{2 b^3 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^4}-\frac {\int \frac {-\frac {c d^3 \left (13 b^2 c^4-50 a b d^2 c^2-15 a^2 d^4\right ) x^3 a^4}{b \left (b c^2+a d^2\right )^4}+\frac {c^4 \left (11 b^2 c^4-46 a b d^2 c^2-9 a^2 d^4\right ) a^4}{b \left (b c^2+a d^2\right )^4}-\frac {c^3 d \left (39 b^2 c^4+106 a b d^2 c^2+19 a^2 d^4\right ) x a^4}{b \left (b c^2+a d^2\right )^4}-\frac {c^2 \left (16 b^3 c^6+103 a b^2 d^2 c^4+42 a^2 b d^4 c^2+3 a^3 d^6\right ) x^2 a^3}{b \left (b c^2+a d^2\right )^4}+\frac {8 x^4 a^2}{b}}{(c+d x)^3 \left (b x^2+a\right )}dx}{2 a b}}{4 a}\)

\(\Big \downarrow \) 2160

\(\displaystyle \frac {a^3 \left (b c x \left (b c^2-3 a d^2\right )+a d \left (3 b c^2-a d^2\right )\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^3}-\frac {\frac {a^3 \left (b c x \left (-15 a^2 d^4-50 a b c^2 d^2+13 b^2 c^4\right )+8 a d \left (2 b c^2-a d^2\right ) \left (a d^2+3 b c^2\right )\right )}{2 b^3 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^4}-\frac {\int \left (\frac {8 a^2 b c^8}{d^2 \left (b c^2+a d^2\right )^3 (c+d x)^3}-\frac {16 a^2 b \left (b c^2+4 a d^2\right ) c^7}{d^2 \left (b c^2+a d^2\right )^4 (c+d x)^2}+\frac {8 a^2 b \left (b^2 c^4+5 a b d^2 c^2+28 a^2 d^4\right ) c^6}{d^2 \left (b c^2+a d^2\right )^5 (c+d x)}+\frac {a^4 \left (c \left (35 b^3 c^6-203 a b^2 d^2 c^4-55 a^2 b d^4 c^2-9 a^3 d^6\right )-8 d \left (18 b^3 c^6-10 a b^2 d^2 c^4-5 a^2 b d^4 c^2-a^3 d^6\right ) x\right )}{b \left (b c^2+a d^2\right )^5 \left (b x^2+a\right )}\right )dx}{2 a b}}{4 a}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 \left (b c x \left (b c^2-3 a d^2\right )+a d \left (3 b c^2-a d^2\right )\right )}{4 b^3 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )^3}-\frac {\frac {a^3 \left (b c x \left (-15 a^2 d^4-50 a b c^2 d^2+13 b^2 c^4\right )+8 a d \left (2 b c^2-a d^2\right ) \left (a d^2+3 b c^2\right )\right )}{2 b^3 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^4}-\frac {\frac {8 a^2 b c^6 \left (28 a^2 d^4+5 a b c^2 d^2+b^2 c^4\right ) \log (c+d x)}{d^3 \left (a d^2+b c^2\right )^5}-\frac {4 a^2 b c^8}{d^3 (c+d x)^2 \left (a d^2+b c^2\right )^3}+\frac {16 a^2 b c^7 \left (4 a d^2+b c^2\right )}{d^3 (c+d x) \left (a d^2+b c^2\right )^4}+\frac {a^{7/2} c \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-9 a^3 d^6-55 a^2 b c^2 d^4-203 a b^2 c^4 d^2+35 b^3 c^6\right )}{b^{3/2} \left (a d^2+b c^2\right )^5}-\frac {4 a^4 d \left (-a^3 d^6-5 a^2 b c^2 d^4-10 a b^2 c^4 d^2+18 b^3 c^6\right ) \log \left (a+b x^2\right )}{b^2 \left (a d^2+b c^2\right )^5}}{2 a b}}{4 a}\)

Input:

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

Output:

(a^3*(a*d*(3*b*c^2 - a*d^2) + b*c*(b*c^2 - 3*a*d^2)*x))/(4*b^3*(b*c^2 + a* 
d^2)^3*(a + b*x^2)^2) - ((a^3*(8*a*d*(2*b*c^2 - a*d^2)*(3*b*c^2 + a*d^2) + 
 b*c*(13*b^2*c^4 - 50*a*b*c^2*d^2 - 15*a^2*d^4)*x))/(2*b^3*(b*c^2 + a*d^2) 
^4*(a + b*x^2)) - ((-4*a^2*b*c^8)/(d^3*(b*c^2 + a*d^2)^3*(c + d*x)^2) + (1 
6*a^2*b*c^7*(b*c^2 + 4*a*d^2))/(d^3*(b*c^2 + a*d^2)^4*(c + d*x)) + (a^(7/2 
)*c*(35*b^3*c^6 - 203*a*b^2*c^4*d^2 - 55*a^2*b*c^2*d^4 - 9*a^3*d^6)*ArcTan 
[(Sqrt[b]*x)/Sqrt[a]])/(b^(3/2)*(b*c^2 + a*d^2)^5) + (8*a^2*b*c^6*(b^2*c^4 
 + 5*a*b*c^2*d^2 + 28*a^2*d^4)*Log[c + d*x])/(d^3*(b*c^2 + a*d^2)^5) - (4* 
a^4*d*(18*b^3*c^6 - 10*a*b^2*c^4*d^2 - 5*a^2*b*c^2*d^4 - a^3*d^6)*Log[a + 
b*x^2])/(b^2*(b*c^2 + a*d^2)^5))/(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 = 0.40 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {a^{2} \left (\frac {-\frac {c \left (15 a^{3} d^{6}+65 a^{2} b \,c^{2} d^{4}+37 a \,b^{2} c^{4} d^{2}-13 b^{3} c^{6}\right ) x^{3}}{8 b}-\frac {a d \left (a^{3} d^{6}+2 a^{2} b \,c^{2} d^{4}-5 a \,b^{2} c^{4} d^{2}-6 b^{3} c^{6}\right ) x^{2}}{b^{2}}-\frac {a c \left (9 a^{3} d^{6}+55 a^{2} b \,c^{2} d^{4}+35 a \,b^{2} c^{4} d^{2}-11 b^{3} c^{6}\right ) x}{8 b^{2}}-\frac {3 a^{2} d \left (a^{3} d^{6}+3 a^{2} b \,c^{2} d^{4}-5 a \,b^{2} c^{4} d^{2}-7 b^{3} c^{6}\right )}{4 b^{3}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\frac {\left (-8 a^{3} d^{7}-40 a^{2} b \,c^{2} d^{5}-80 a \,b^{2} c^{4} d^{3}+144 b^{3} c^{6} d \right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (9 d^{6} c \,a^{3}+55 a^{2} b \,c^{3} d^{4}+203 a \,b^{2} c^{5} d^{2}-35 b^{3} c^{7}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{8 b^{2}}\right )}{\left (a \,d^{2}+b \,c^{2}\right )^{5}}-\frac {c^{8}}{2 d^{3} \left (a \,d^{2}+b \,c^{2}\right )^{3} \left (d x +c \right )^{2}}+\frac {c^{6} \left (28 a^{2} d^{4}+5 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) \ln \left (d x +c \right )}{d^{3} \left (a \,d^{2}+b \,c^{2}\right )^{5}}+\frac {2 c^{7} \left (4 a \,d^{2}+b \,c^{2}\right )}{d^{3} \left (a \,d^{2}+b \,c^{2}\right )^{4} \left (d x +c \right )}\)	\(468\)
risch	\(\text {Expression too large to display}\)	\(1934\)

Input:

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

Output:

-a^2/(a*d^2+b*c^2)^5*((-1/8*c*(15*a^3*d^6+65*a^2*b*c^2*d^4+37*a*b^2*c^4*d^ 
2-13*b^3*c^6)/b*x^3-a*d*(a^3*d^6+2*a^2*b*c^2*d^4-5*a*b^2*c^4*d^2-6*b^3*c^6 
)/b^2*x^2-1/8*a*c*(9*a^3*d^6+55*a^2*b*c^2*d^4+35*a*b^2*c^4*d^2-11*b^3*c^6) 
/b^2*x-3/4*a^2*d*(a^3*d^6+3*a^2*b*c^2*d^4-5*a*b^2*c^4*d^2-7*b^3*c^6)/b^3)/ 
(b*x^2+a)^2+1/8/b^2*(1/2*(-8*a^3*d^7-40*a^2*b*c^2*d^5-80*a*b^2*c^4*d^3+144 
*b^3*c^6*d)/b*ln(b*x^2+a)+(9*a^3*c*d^6+55*a^2*b*c^3*d^4+203*a*b^2*c^5*d^2- 
35*b^3*c^7)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))-1/2*c^8/d^3/(a*d^2+b*c^2 
)^3/(d*x+c)^2+c^6*(28*a^2*d^4+5*a*b*c^2*d^2+b^2*c^4)*ln(d*x+c)/d^3/(a*d^2+ 
b*c^2)^5+2*c^7*(4*a*d^2+b*c^2)/d^3/(a*d^2+b*c^2)^4/(d*x+c)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2310 vs. \(2 (418) = 836\).

Time = 39.60 (sec) , antiderivative size = 4644, normalized size of antiderivative = 10.65 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

integrate(x**8/(d*x+c)**3/(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. 1299 vs. \(2 (418) = 836\).

Time = 0.16 (sec) , antiderivative size = 1299, normalized size of antiderivative = 2.98 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(18*a^2*b^3*c^6*d - 10*a^3*b^2*c^4*d^3 - 5*a^4*b*c^2*d^5 - a^5*d^7)*l 
og(b*x^2 + a)/(b^8*c^10 + 5*a*b^7*c^8*d^2 + 10*a^2*b^6*c^6*d^4 + 10*a^3*b^ 
5*c^4*d^6 + 5*a^4*b^4*c^2*d^8 + a^5*b^3*d^10) + (b^2*c^10 + 5*a*b*c^8*d^2 
+ 28*a^2*c^6*d^4)*log(d*x + c)/(b^5*c^10*d^3 + 5*a*b^4*c^8*d^5 + 10*a^2*b^ 
3*c^6*d^7 + 10*a^3*b^2*c^4*d^9 + 5*a^4*b*c^2*d^11 + a^5*d^13) + 1/8*(35*a^ 
2*b^3*c^7 - 203*a^3*b^2*c^5*d^2 - 55*a^4*b*c^3*d^4 - 9*a^5*c*d^6)*arctan(b 
*x/sqrt(a*b))/((b^7*c^10 + 5*a*b^6*c^8*d^2 + 10*a^2*b^5*c^6*d^4 + 10*a^3*b 
^4*c^4*d^6 + 5*a^4*b^3*c^2*d^8 + a^5*b^2*d^10)*sqrt(a*b)) + 1/8*(12*a^2*b^ 
4*c^10 + 60*a^3*b^3*c^8*d^2 - 42*a^4*b^2*c^6*d^4 + 12*a^5*b*c^4*d^6 + 6*a^ 
6*c^2*d^8 + (16*b^6*c^9*d + 64*a*b^5*c^7*d^3 - 13*a^2*b^4*c^5*d^5 + 50*a^3 
*b^3*c^3*d^7 + 15*a^4*b^2*c*d^9)*x^5 + 2*(6*b^6*c^10 + 30*a*b^5*c^8*d^2 - 
13*a^2*b^4*c^6*d^4 + 26*a^3*b^3*c^4*d^6 + 19*a^4*b^2*c^2*d^8 + 4*a^5*b*d^1 
0)*x^4 + (32*a*b^5*c^9*d + 115*a^2*b^4*c^7*d^3 - 57*a^3*b^3*c^5*d^5 + 77*a 
^4*b^2*c^3*d^7 + 25*a^5*b*c*d^9)*x^3 + 2*(12*a*b^5*c^10 + 60*a^2*b^4*c^8*d 
^2 - 35*a^3*b^3*c^6*d^4 + 29*a^4*b^2*c^4*d^6 + 19*a^5*b*c^2*d^8 + 3*a^6*d^ 
10)*x^2 + (16*a^2*b^4*c^9*d + 53*a^3*b^3*c^7*d^3 - 38*a^4*b^2*c^5*d^5 + 33 
*a^5*b*c^3*d^7 + 12*a^6*c*d^9)*x)/(a^2*b^7*c^10*d^3 + 4*a^3*b^6*c^8*d^5 + 
6*a^4*b^5*c^6*d^7 + 4*a^5*b^4*c^4*d^9 + a^6*b^3*c^2*d^11 + (b^9*c^8*d^5 + 
4*a*b^8*c^6*d^7 + 6*a^2*b^7*c^4*d^9 + 4*a^3*b^6*c^2*d^11 + a^4*b^5*d^13)*x 
^6 + 2*(b^9*c^9*d^4 + 4*a*b^8*c^7*d^6 + 6*a^2*b^7*c^5*d^8 + 4*a^3*b^6*c...
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 829, normalized size of antiderivative = 1.90 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(18*a^2*b^3*c^6*d - 10*a^3*b^2*c^4*d^3 - 5*a^4*b*c^2*d^5 - a^5*d^7)*l 
og(b*x^2 + a)/(b^8*c^10 + 5*a*b^7*c^8*d^2 + 10*a^2*b^6*c^6*d^4 + 10*a^3*b^ 
5*c^4*d^6 + 5*a^4*b^4*c^2*d^8 + a^5*b^3*d^10) + (b^2*c^10 + 5*a*b*c^8*d^2 
+ 28*a^2*c^6*d^4)*log(abs(d*x + c))/(b^5*c^10*d^3 + 5*a*b^4*c^8*d^5 + 10*a 
^2*b^3*c^6*d^7 + 10*a^3*b^2*c^4*d^9 + 5*a^4*b*c^2*d^11 + a^5*d^13) + 1/8*( 
35*a^2*b^3*c^7 - 203*a^3*b^2*c^5*d^2 - 55*a^4*b*c^3*d^4 - 9*a^5*c*d^6)*arc 
tan(b*x/sqrt(a*b))/((b^7*c^10 + 5*a*b^6*c^8*d^2 + 10*a^2*b^5*c^6*d^4 + 10* 
a^3*b^4*c^4*d^6 + 5*a^4*b^3*c^2*d^8 + a^5*b^2*d^10)*sqrt(a*b)) + 1/8*(12*a 
^2*b^4*c^10 + 60*a^3*b^3*c^8*d^2 - 42*a^4*b^2*c^6*d^4 + 12*a^5*b*c^4*d^6 + 
 6*a^6*c^2*d^8 + (16*b^6*c^9*d + 64*a*b^5*c^7*d^3 - 13*a^2*b^4*c^5*d^5 + 5 
0*a^3*b^3*c^3*d^7 + 15*a^4*b^2*c*d^9)*x^5 + 2*(6*b^6*c^10 + 30*a*b^5*c^8*d 
^2 - 13*a^2*b^4*c^6*d^4 + 26*a^3*b^3*c^4*d^6 + 19*a^4*b^2*c^2*d^8 + 4*a^5* 
b*d^10)*x^4 + (32*a*b^5*c^9*d + 115*a^2*b^4*c^7*d^3 - 57*a^3*b^3*c^5*d^5 + 
 77*a^4*b^2*c^3*d^7 + 25*a^5*b*c*d^9)*x^3 + 2*(12*a*b^5*c^10 + 60*a^2*b^4* 
c^8*d^2 - 35*a^3*b^3*c^6*d^4 + 29*a^4*b^2*c^4*d^6 + 19*a^5*b*c^2*d^8 + 3*a 
^6*d^10)*x^2 + (16*a^2*b^4*c^9*d + 53*a^3*b^3*c^7*d^3 - 38*a^4*b^2*c^5*d^5 
 + 33*a^5*b*c^3*d^7 + 12*a^6*c*d^9)*x)/((b*c^2 + a*d^2)^4*(b*x^2 + a)^2*(d 
*x + c)^2*b^3*d^3)
 

Mupad [B] (verification not implemented)

Time = 8.01 (sec) , antiderivative size = 1359, normalized size of antiderivative = 3.12 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((x^3*(32*a*b^4*c^9 + 25*a^5*c*d^8 + 77*a^4*b*c^3*d^6 + 115*a^2*b^3*c^7*d^ 
2 - 57*a^3*b^2*c^5*d^4))/(8*b^2*d^2*(a^4*d^8 + b^4*c^8 + 4*a*b^3*c^6*d^2 + 
 4*a^3*b*c^2*d^6 + 6*a^2*b^2*c^4*d^4)) + (x^2*(3*a^6*d^10 + 12*a*b^5*c^10 
+ 19*a^5*b*c^2*d^8 + 60*a^2*b^4*c^8*d^2 - 35*a^3*b^3*c^6*d^4 + 29*a^4*b^2* 
c^4*d^6))/(4*b^3*d^3*(a^4*d^8 + b^4*c^8 + 4*a*b^3*c^6*d^2 + 4*a^3*b*c^2*d^ 
6 + 6*a^2*b^2*c^4*d^4)) + (x^5*(16*b^4*c^9 + 15*a^4*c*d^8 + 64*a*b^3*c^7*d 
^2 + 50*a^3*b*c^3*d^6 - 13*a^2*b^2*c^5*d^4))/(8*b*d^2*(a^4*d^8 + b^4*c^8 + 
 4*a*b^3*c^6*d^2 + 4*a^3*b*c^2*d^6 + 6*a^2*b^2*c^4*d^4)) + (x^4*(4*a^5*d^1 
0 + 6*b^5*c^10 + 30*a*b^4*c^8*d^2 + 19*a^4*b*c^2*d^8 - 13*a^2*b^3*c^6*d^4 
+ 26*a^3*b^2*c^4*d^6))/(4*b^2*d^3*(a^4*d^8 + b^4*c^8 + 4*a*b^3*c^6*d^2 + 4 
*a^3*b*c^2*d^6 + 6*a^2*b^2*c^4*d^4)) + (3*a*c^2*(a^5*d^8 + 2*a*b^4*c^8 + 2 
*a^4*b*c^2*d^6 + 10*a^2*b^3*c^6*d^2 - 7*a^3*b^2*c^4*d^4))/(4*b^3*d^3*(a*d^ 
2 + b*c^2)*(a^3*d^6 + b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4)) + (a*c 
*x*(12*a^5*d^8 + 16*a*b^4*c^8 + 33*a^4*b*c^2*d^6 + 53*a^2*b^3*c^6*d^2 - 38 
*a^3*b^2*c^4*d^4))/(8*b^3*d^2*(a*d^2 + b*c^2)*(a^3*d^6 + b^3*c^6 + 3*a*b^2 
*c^4*d^2 + 3*a^2*b*c^2*d^4)))/(x^2*(a^2*d^2 + 2*a*b*c^2) + x^4*(b^2*c^2 + 
2*a*b*d^2) + a^2*c^2 + b^2*d^2*x^6 + 2*a^2*c*d*x + 2*b^2*c*d*x^5 + 4*a*b*c 
*d*x^3) + (log((-a^3*b^7)^(1/2) + a*b^4*x)*(35*b^3*c^7*(-a^3*b^7)^(1/2) + 
8*a^5*b^3*d^7 - 144*a^2*b^6*c^6*d + 80*a^3*b^5*c^4*d^3 + 40*a^4*b^4*c^2*d^ 
5 - 9*a^3*c*d^6*(-a^3*b^7)^(1/2) - 203*a*b^2*c^5*d^2*(-a^3*b^7)^(1/2) -...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 3302, normalized size of antiderivative = 7.57 \[ \int \frac {x^8}{(c+d x)^3 \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 18*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**6*c**3*d**9 - 36*s 
qrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**6*c**2*d**10*x - 18*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**6*c*d**11*x**2 - 110*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**5*d**7 - 220*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**4*d**8*x - 146*sqrt(b)*sqrt(a)*at 
an((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**3*d**9*x**2 - 72*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c**2*d**10*x**3 - 36*sqrt(b)*sqrt(a)*ata 
n((b*x)/(sqrt(b)*sqrt(a)))*a**5*b*c*d**11*x**4 - 406*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**7*d**5 - 812*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**6*d**6*x - 626*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**5*d**7*x**2 - 440*sqrt(b)*sqrt(a)*atan 
((b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**4*d**8*x**3 - 238*sqrt(b)*sqrt(a)*a 
tan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**3*d**9*x**4 - 36*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c**2*d**10*x**5 - 18*sqrt(b)*sqrt 
(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b**2*c*d**11*x**6 + 70*sqrt(b)*sqrt 
(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**9*d**3 + 140*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**8*d**4*x - 742*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**7*d**5*x**2 - 1624*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**6*d**6*x**3 - 922*sqrt(b) 
*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b**3*c**5*d**7*x**4 - 220*s...