\(\int \frac {x^8 (A+B x^2)}{(a+b x^2)^3} \, dx\) [105]

Optimal result

Integrand size = 20, antiderivative size = 138 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {3 a (A b-2 a B) x}{b^5}+\frac {(A b-3 a B) x^3}{3 b^4}+\frac {B x^5}{5 b^3}+\frac {a^3 (A b-a B) x}{4 b^5 \left (a+b x^2\right )^2}-\frac {a^2 (13 A b-17 a B) x}{8 b^5 \left (a+b x^2\right )}+\frac {7 a^{3/2} (5 A b-9 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \] Output:

-3*a*(A*b-2*B*a)*x/b^5+1/3*(A*b-3*B*a)*x^3/b^4+1/5*B*x^5/b^3+1/4*a^3*(A*b- 
B*a)*x/b^5/(b*x^2+a)^2-1/8*a^2*(13*A*b-17*B*a)*x/b^5/(b*x^2+a)+7/8*a^(3/2) 
*(5*A*b-9*B*a)*arctan(b^(1/2)*x/a^(1/2))/b^(11/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (945 a^4 B-525 a^3 b \left (A-3 B x^2\right )+8 b^4 x^6 \left (5 A+3 B x^2\right )-8 a b^3 x^4 \left (35 A+9 B x^2\right )+7 a^2 b^2 x^2 \left (-125 A+72 B x^2\right )\right )}{120 b^5 \left (a+b x^2\right )^2}-\frac {7 a^{3/2} (-5 A b+9 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{11/2}} \] Input:

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

Output:

(x*(945*a^4*B - 525*a^3*b*(A - 3*B*x^2) + 8*b^4*x^6*(5*A + 3*B*x^2) - 8*a* 
b^3*x^4*(35*A + 9*B*x^2) + 7*a^2*b^2*x^2*(-125*A + 72*B*x^2)))/(120*b^5*(a 
 + b*x^2)^2) - (7*a^(3/2)*(-5*A*b + 9*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8 
*b^(11/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {360, 2345, 2341, 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.

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

\(\Big \downarrow \) 360

\(\displaystyle \frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {\int \frac {-4 b^4 B x^8-4 b^3 (A b-a B) x^6+4 a b^2 (A b-a B) x^4-4 a^2 b (A b-a B) x^2+a^3 (A b-a B)}{\left (b x^2+a\right )^2}dx}{4 b^5}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {\frac {a^2 x (13 A b-17 a B)}{2 \left (a+b x^2\right )}-\frac {\int \frac {8 a b^3 B x^6+8 a b^2 (A b-2 a B) x^4-8 a^2 b (2 A b-3 a B) x^2+a^3 (11 A b-15 a B)}{b x^2+a}dx}{2 a}}{4 b^5}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {\frac {a^2 x (13 A b-17 a B)}{2 \left (a+b x^2\right )}-\frac {\int \left (8 a b^2 B x^4+8 a b (A b-3 a B) x^2-24 a^2 (A b-2 a B)-\frac {7 \left (9 a^4 B-5 a^3 A b\right )}{b x^2+a}\right )dx}{2 a}}{4 b^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 x (A b-a B)}{4 b^5 \left (a+b x^2\right )^2}-\frac {\frac {a^2 x (13 A b-17 a B)}{2 \left (a+b x^2\right )}-\frac {\frac {7 a^{5/2} (5 A b-9 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-24 a^2 x (A b-2 a B)+\frac {8}{3} a b x^3 (A b-3 a B)+\frac {8}{5} a b^2 B x^5}{2 a}}{4 b^5}\)

Input:

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

Output:

(a^3*(A*b - a*B)*x)/(4*b^5*(a + b*x^2)^2) - ((a^2*(13*A*b - 17*a*B)*x)/(2* 
(a + b*x^2)) - (-24*a^2*(A*b - 2*a*B)*x + (8*a*b*(A*b - 3*a*B)*x^3)/3 + (8 
*a*b^2*B*x^5)/5 + (7*a^(5/2)*(5*A*b - 9*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/ 
Sqrt[b])/(2*a))/(4*b^5)
 

Defintions of rubi rules used

rule 360

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[(a + b*x^2)^(p + 1)*Expan 
dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 
- 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; 
FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & 
& (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 2009

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

rule 2341

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

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {-\frac {1}{5} b^{2} B \,x^{5}-\frac {1}{3} A \,b^{2} x^{3}+B a b \,x^{3}+3 a A b x -6 B \,a^{2} x}{b^{5}}+\frac {a^{2} \left (\frac {\left (-\frac {13}{8} b^{2} A +\frac {17}{8} a b B \right ) x^{3}-\frac {a \left (11 A b -15 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {7 \left (5 A b -9 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{5}}\)	\(119\)
risch	\(\frac {B \,x^{5}}{5 b^{3}}+\frac {A \,x^{3}}{3 b^{3}}-\frac {B a \,x^{3}}{b^{4}}-\frac {3 a A x}{b^{4}}+\frac {6 B \,a^{2} x}{b^{5}}+\frac {\left (-\frac {13}{8} A \,a^{2} b^{2}+\frac {17}{8} B \,a^{3} b \right ) x^{3}-\frac {a^{3} \left (11 A b -15 B a \right ) x}{8}}{b^{5} \left (b \,x^{2}+a \right )^{2}}+\frac {35 \sqrt {-a b}\, a \ln \left (-\sqrt {-a b}\, x +a \right ) A}{16 b^{5}}-\frac {63 \sqrt {-a b}\, a^{2} \ln \left (-\sqrt {-a b}\, x +a \right ) B}{16 b^{6}}-\frac {35 \sqrt {-a b}\, a \ln \left (\sqrt {-a b}\, x +a \right ) A}{16 b^{5}}+\frac {63 \sqrt {-a b}\, a^{2} \ln \left (\sqrt {-a b}\, x +a \right ) B}{16 b^{6}}\)	\(200\)

Input:

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

Output:

-1/b^5*(-1/5*b^2*B*x^5-1/3*A*b^2*x^3+B*a*b*x^3+3*a*A*b*x-6*B*a^2*x)+a^2/b^ 
5*(((-13/8*b^2*A+17/8*a*b*B)*x^3-1/8*a*(11*A*b-15*B*a)*x)/(b*x^2+a)^2+7/8* 
(5*A*b-9*B*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.01 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {48 \, B b^{4} x^{9} - 16 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{7} + 112 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + 350 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 210 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b\right )} x}{240 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac {24 \, B b^{4} x^{9} - 8 \, {\left (9 \, B a b^{3} - 5 \, A b^{4}\right )} x^{7} + 56 \, {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{5} + 175 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{3} - 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b + {\left (9 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 2 \, {\left (9 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 105 \, {\left (9 \, B a^{4} - 5 \, A a^{3} b\right )} x}{120 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \] Input:

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

Output:

[1/240*(48*B*b^4*x^9 - 16*(9*B*a*b^3 - 5*A*b^4)*x^7 + 112*(9*B*a^2*b^2 - 5 
*A*a*b^3)*x^5 + 350*(9*B*a^3*b - 5*A*a^2*b^2)*x^3 - 105*(9*B*a^4 - 5*A*a^3 
*b + (9*B*a^2*b^2 - 5*A*a*b^3)*x^4 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*x^2)*sqrt 
(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 210*(9*B*a^4 - 5* 
A*a^3*b)*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5), 1/120*(24*B*b^4*x^9 - 8*(9* 
B*a*b^3 - 5*A*b^4)*x^7 + 56*(9*B*a^2*b^2 - 5*A*a*b^3)*x^5 + 175*(9*B*a^3*b 
 - 5*A*a^2*b^2)*x^3 - 105*(9*B*a^4 - 5*A*a^3*b + (9*B*a^2*b^2 - 5*A*a*b^3) 
*x^4 + 2*(9*B*a^3*b - 5*A*a^2*b^2)*x^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) 
+ 105*(9*B*a^4 - 5*A*a^3*b)*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5)]
 

Sympy [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.83 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B x^{5}}{5 b^{3}} + x^{3} \left (\frac {A}{3 b^{3}} - \frac {B a}{b^{4}}\right ) + x \left (- \frac {3 A a}{b^{4}} + \frac {6 B a^{2}}{b^{5}}\right ) + \frac {7 \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right ) \log {\left (- \frac {7 b^{5} \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right )}{- 35 A a b + 63 B a^{2}} + x \right )}}{16} - \frac {7 \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right ) \log {\left (\frac {7 b^{5} \sqrt {- \frac {a^{3}}{b^{11}}} \left (- 5 A b + 9 B a\right )}{- 35 A a b + 63 B a^{2}} + x \right )}}{16} + \frac {x^{3} \left (- 13 A a^{2} b^{2} + 17 B a^{3} b\right ) + x \left (- 11 A a^{3} b + 15 B a^{4}\right )}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} \] Input:

integrate(x**8*(B*x**2+A)/(b*x**2+a)**3,x)
 

Output:

B*x**5/(5*b**3) + x**3*(A/(3*b**3) - B*a/b**4) + x*(-3*A*a/b**4 + 6*B*a**2 
/b**5) + 7*sqrt(-a**3/b**11)*(-5*A*b + 9*B*a)*log(-7*b**5*sqrt(-a**3/b**11 
)*(-5*A*b + 9*B*a)/(-35*A*a*b + 63*B*a**2) + x)/16 - 7*sqrt(-a**3/b**11)*( 
-5*A*b + 9*B*a)*log(7*b**5*sqrt(-a**3/b**11)*(-5*A*b + 9*B*a)/(-35*A*a*b + 
 63*B*a**2) + x)/16 + (x**3*(-13*A*a**2*b**2 + 17*B*a**3*b) + x*(-11*A*a** 
3*b + 15*B*a**4))/(8*a**2*b**5 + 16*a*b**6*x**2 + 8*b**7*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (17 \, B a^{3} b - 13 \, A a^{2} b^{2}\right )} x^{3} + {\left (15 \, B a^{4} - 11 \, A a^{3} b\right )} x}{8 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} - \frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {3 \, B b^{2} x^{5} - 5 \, {\left (3 \, B a b - A b^{2}\right )} x^{3} + 45 \, {\left (2 \, B a^{2} - A a b\right )} x}{15 \, b^{5}} \] Input:

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

Output:

1/8*((17*B*a^3*b - 13*A*a^2*b^2)*x^3 + (15*B*a^4 - 11*A*a^3*b)*x)/(b^7*x^4 
 + 2*a*b^6*x^2 + a^2*b^5) - 7/8*(9*B*a^3 - 5*A*a^2*b)*arctan(b*x/sqrt(a*b) 
)/(sqrt(a*b)*b^5) + 1/15*(3*B*b^2*x^5 - 5*(3*B*a*b - A*b^2)*x^3 + 45*(2*B* 
a^2 - A*a*b)*x)/b^5
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {7 \, {\left (9 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {17 \, B a^{3} b x^{3} - 13 \, A a^{2} b^{2} x^{3} + 15 \, B a^{4} x - 11 \, A a^{3} b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{5}} + \frac {3 \, B b^{12} x^{5} - 15 \, B a b^{11} x^{3} + 5 \, A b^{12} x^{3} + 90 \, B a^{2} b^{10} x - 45 \, A a b^{11} x}{15 \, b^{15}} \] Input:

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

Output:

-7/8*(9*B*a^3 - 5*A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/8*(17 
*B*a^3*b*x^3 - 13*A*a^2*b^2*x^3 + 15*B*a^4*x - 11*A*a^3*b*x)/((b*x^2 + a)^ 
2*b^5) + 1/15*(3*B*b^12*x^5 - 15*B*a*b^11*x^3 + 5*A*b^12*x^3 + 90*B*a^2*b^ 
10*x - 45*A*a*b^11*x)/b^15
 

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.28 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x\,\left (\frac {15\,B\,a^4}{8}-\frac {11\,A\,a^3\,b}{8}\right )-x^3\,\left (\frac {13\,A\,a^2\,b^2}{8}-\frac {17\,B\,a^3\,b}{8}\right )}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}-x\,\left (\frac {3\,a\,\left (\frac {A}{b^3}-\frac {3\,B\,a}{b^4}\right )}{b}+\frac {3\,B\,a^2}{b^5}\right )+x^3\,\left (\frac {A}{3\,b^3}-\frac {B\,a}{b^4}\right )+\frac {B\,x^5}{5\,b^3}-\frac {7\,a^{3/2}\,\mathrm {atan}\left (\frac {a^{3/2}\,\sqrt {b}\,x\,\left (5\,A\,b-9\,B\,a\right )}{9\,B\,a^3-5\,A\,a^2\,b}\right )\,\left (5\,A\,b-9\,B\,a\right )}{8\,b^{11/2}} \] Input:

int((x^8*(A + B*x^2))/(a + b*x^2)^3,x)
 

Output:

(x*((15*B*a^4)/8 - (11*A*a^3*b)/8) - x^3*((13*A*a^2*b^2)/8 - (17*B*a^3*b)/ 
8))/(a^2*b^5 + b^7*x^4 + 2*a*b^6*x^2) - x*((3*a*(A/b^3 - (3*B*a)/b^4))/b + 
 (3*B*a^2)/b^5) + x^3*(A/(3*b^3) - (B*a)/b^4) + (B*x^5)/(5*b^3) - (7*a^(3/ 
2)*atan((a^(3/2)*b^(1/2)*x*(5*A*b - 9*B*a))/(9*B*a^3 - 5*A*a^2*b))*(5*A*b 
- 9*B*a))/(8*b^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.70 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {-105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}-105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{2}+105 a^{3} b x +70 a^{2} b^{2} x^{3}-14 a \,b^{3} x^{5}+6 b^{4} x^{7}}{30 b^{5} \left (b \,x^{2}+a \right )} \] Input:

int(x^8*(B*x^2+A)/(b*x^2+a)^3,x)
 

Output:

( - 105*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3 - 105*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*x**2 + 105*a**3*b*x + 70*a**2* 
b**2*x**3 - 14*a*b**3*x**5 + 6*b**4*x**7)/(30*b**5*(a + b*x**2))