\(\int \frac {(a+b x^2)^5 (A+B x^2)}{x^8} \, dx\) [40]

Optimal result

Integrand size = 20, antiderivative size = 111 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=-\frac {a^5 A}{7 x^7}-\frac {a^4 (5 A b+a B)}{5 x^5}-\frac {5 a^3 b (2 A b+a B)}{3 x^3}-\frac {10 a^2 b^2 (A b+a B)}{x}+5 a b^3 (A b+2 a B) x+\frac {1}{3} b^4 (A b+5 a B) x^3+\frac {1}{5} b^5 B x^5 \]

[Out]

-1/7*a^5*A/x^7-1/5*a^4*(5*A*b+B*a)/x^5-5/3*a^3*b*(2*A*b+B*a)/x^3-10*a^2*b^2*(A*b+B*a)/x+5*a*b^3*(A*b+2*B*a)*x+
1/3*b^4*(A*b+5*B*a)*x^3+1/5*b^5*B*x^5

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=-\frac {a^5 A}{7 x^7}-\frac {a^4 (a B+5 A b)}{5 x^5}-\frac {5 a^3 b (a B+2 A b)}{3 x^3}-\frac {10 a^2 b^2 (a B+A b)}{x}+\frac {1}{3} b^4 x^3 (5 a B+A b)+5 a b^3 x (2 a B+A b)+\frac {1}{5} b^5 B x^5 \]

[In]

Int[((a + b*x^2)^5*(A + B*x^2))/x^8,x]

[Out]

-1/7*(a^5*A)/x^7 - (a^4*(5*A*b + a*B))/(5*x^5) - (5*a^3*b*(2*A*b + a*B))/(3*x^3) - (10*a^2*b^2*(A*b + a*B))/x
+ 5*a*b^3*(A*b + 2*a*B)*x + (b^4*(A*b + 5*a*B)*x^3)/3 + (b^5*B*x^5)/5

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (5 a b^3 (A b+2 a B)+\frac {a^5 A}{x^8}+\frac {a^4 (5 A b+a B)}{x^6}+\frac {5 a^3 b (2 A b+a B)}{x^4}+\frac {10 a^2 b^2 (A b+a B)}{x^2}+b^4 (A b+5 a B) x^2+b^5 B x^4\right ) \, dx \\ & = -\frac {a^5 A}{7 x^7}-\frac {a^4 (5 A b+a B)}{5 x^5}-\frac {5 a^3 b (2 A b+a B)}{3 x^3}-\frac {10 a^2 b^2 (A b+a B)}{x}+5 a b^3 (A b+2 a B) x+\frac {1}{3} b^4 (A b+5 a B) x^3+\frac {1}{5} b^5 B x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=-\frac {a^5 A}{7 x^7}-\frac {a^4 (5 A b+a B)}{5 x^5}-\frac {5 a^3 b (2 A b+a B)}{3 x^3}-\frac {10 a^2 b^2 (A b+a B)}{x}+5 a b^3 (A b+2 a B) x+\frac {1}{3} b^4 (A b+5 a B) x^3+\frac {1}{5} b^5 B x^5 \]

[In]

Integrate[((a + b*x^2)^5*(A + B*x^2))/x^8,x]

[Out]

-1/7*(a^5*A)/x^7 - (a^4*(5*A*b + a*B))/(5*x^5) - (5*a^3*b*(2*A*b + a*B))/(3*x^3) - (10*a^2*b^2*(A*b + a*B))/x
+ 5*a*b^3*(A*b + 2*a*B)*x + (b^4*(A*b + 5*a*B)*x^3)/3 + (b^5*B*x^5)/5

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {b^{5} B \,x^{5}}{5}+\frac {A \,b^{5} x^{3}}{3}+\frac {5 B a \,b^{4} x^{3}}{3}+5 A a \,b^{4} x +10 B \,a^{2} b^{3} x -\frac {a^{5} A}{7 x^{7}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{3 x^{3}}-\frac {10 a^{2} b^{2} \left (A b +B a \right )}{x}-\frac {a^{4} \left (5 A b +B a \right )}{5 x^{5}}\)	\(108\)
risch	\(\frac {b^{5} B \,x^{5}}{5}+\frac {A \,b^{5} x^{3}}{3}+\frac {5 B a \,b^{4} x^{3}}{3}+5 A a \,b^{4} x +10 B \,a^{2} b^{3} x +\frac {\left (-10 a^{2} b^{3} A -10 a^{3} b^{2} B \right ) x^{6}+\left (-\frac {10}{3} a^{3} b^{2} A -\frac {5}{3} a^{4} b B \right ) x^{4}+\left (-a^{4} b A -\frac {1}{5} a^{5} B \right ) x^{2}-\frac {a^{5} A}{7}}{x^{7}}\)	\(121\)
norman	\(\frac {\frac {b^{5} B \,x^{12}}{5}+\left (\frac {1}{3} b^{5} A +\frac {5}{3} a \,b^{4} B \right ) x^{10}+\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{8}+\left (-10 a^{2} b^{3} A -10 a^{3} b^{2} B \right ) x^{6}+\left (-\frac {10}{3} a^{3} b^{2} A -\frac {5}{3} a^{4} b B \right ) x^{4}+\left (-a^{4} b A -\frac {1}{5} a^{5} B \right ) x^{2}-\frac {a^{5} A}{7}}{x^{7}}\)	\(122\)
gosper	\(-\frac {-21 b^{5} B \,x^{12}-35 A \,b^{5} x^{10}-175 B a \,b^{4} x^{10}-525 a A \,b^{4} x^{8}-1050 B \,a^{2} b^{3} x^{8}+1050 a^{2} A \,b^{3} x^{6}+1050 B \,a^{3} b^{2} x^{6}+350 a^{3} A \,b^{2} x^{4}+175 B \,a^{4} b \,x^{4}+105 a^{4} A b \,x^{2}+21 a^{5} B \,x^{2}+15 a^{5} A}{105 x^{7}}\)	\(128\)
parallelrisch	\(\frac {21 b^{5} B \,x^{12}+35 A \,b^{5} x^{10}+175 B a \,b^{4} x^{10}+525 a A \,b^{4} x^{8}+1050 B \,a^{2} b^{3} x^{8}-1050 a^{2} A \,b^{3} x^{6}-1050 B \,a^{3} b^{2} x^{6}-350 a^{3} A \,b^{2} x^{4}-175 B \,a^{4} b \,x^{4}-105 a^{4} A b \,x^{2}-21 a^{5} B \,x^{2}-15 a^{5} A}{105 x^{7}}\)	\(128\)

[In]

int((b*x^2+a)^5*(B*x^2+A)/x^8,x,method=_RETURNVERBOSE)

[Out]

1/5*b^5*B*x^5+1/3*A*b^5*x^3+5/3*B*a*b^4*x^3+5*A*a*b^4*x+10*B*a^2*b^3*x-1/7*a^5*A/x^7-5/3*a^3*b*(2*A*b+B*a)/x^3
-10*a^2*b^2*(A*b+B*a)/x-1/5*a^4*(5*A*b+B*a)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=\frac {21 \, B b^{5} x^{12} + 35 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + 525 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} - 1050 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} - 15 \, A a^{5} - 175 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} - 21 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{105 \, x^{7}} \]

[In]

integrate((b*x^2+a)^5*(B*x^2+A)/x^8,x, algorithm="fricas")

[Out]

1/105*(21*B*b^5*x^12 + 35*(5*B*a*b^4 + A*b^5)*x^10 + 525*(2*B*a^2*b^3 + A*a*b^4)*x^8 - 1050*(B*a^3*b^2 + A*a^2
*b^3)*x^6 - 15*A*a^5 - 175*(B*a^4*b + 2*A*a^3*b^2)*x^4 - 21*(B*a^5 + 5*A*a^4*b)*x^2)/x^7

Sympy [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=\frac {B b^{5} x^{5}}{5} + x^{3} \left (\frac {A b^{5}}{3} + \frac {5 B a b^{4}}{3}\right ) + x \left (5 A a b^{4} + 10 B a^{2} b^{3}\right ) + \frac {- 15 A a^{5} + x^{6} \left (- 1050 A a^{2} b^{3} - 1050 B a^{3} b^{2}\right ) + x^{4} \left (- 350 A a^{3} b^{2} - 175 B a^{4} b\right ) + x^{2} \left (- 105 A a^{4} b - 21 B a^{5}\right )}{105 x^{7}} \]

[In]

integrate((b*x**2+a)**5*(B*x**2+A)/x**8,x)

[Out]

B*b**5*x**5/5 + x**3*(A*b**5/3 + 5*B*a*b**4/3) + x*(5*A*a*b**4 + 10*B*a**2*b**3) + (-15*A*a**5 + x**6*(-1050*A
*a**2*b**3 - 1050*B*a**3*b**2) + x**4*(-350*A*a**3*b**2 - 175*B*a**4*b) + x**2*(-105*A*a**4*b - 21*B*a**5))/(1
05*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=\frac {1}{5} \, B b^{5} x^{5} + \frac {1}{3} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{3} + 5 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x - \frac {1050 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} + 15 \, A a^{5} + 175 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + 21 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{105 \, x^{7}} \]

[In]

integrate((b*x^2+a)^5*(B*x^2+A)/x^8,x, algorithm="maxima")

[Out]

1/5*B*b^5*x^5 + 1/3*(5*B*a*b^4 + A*b^5)*x^3 + 5*(2*B*a^2*b^3 + A*a*b^4)*x - 1/105*(1050*(B*a^3*b^2 + A*a^2*b^3
)*x^6 + 15*A*a^5 + 175*(B*a^4*b + 2*A*a^3*b^2)*x^4 + 21*(B*a^5 + 5*A*a^4*b)*x^2)/x^7

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=\frac {1}{5} \, B b^{5} x^{5} + \frac {5}{3} \, B a b^{4} x^{3} + \frac {1}{3} \, A b^{5} x^{3} + 10 \, B a^{2} b^{3} x + 5 \, A a b^{4} x - \frac {1050 \, B a^{3} b^{2} x^{6} + 1050 \, A a^{2} b^{3} x^{6} + 175 \, B a^{4} b x^{4} + 350 \, A a^{3} b^{2} x^{4} + 21 \, B a^{5} x^{2} + 105 \, A a^{4} b x^{2} + 15 \, A a^{5}}{105 \, x^{7}} \]

[In]

integrate((b*x^2+a)^5*(B*x^2+A)/x^8,x, algorithm="giac")

[Out]

1/5*B*b^5*x^5 + 5/3*B*a*b^4*x^3 + 1/3*A*b^5*x^3 + 10*B*a^2*b^3*x + 5*A*a*b^4*x - 1/105*(1050*B*a^3*b^2*x^6 + 1
050*A*a^2*b^3*x^6 + 175*B*a^4*b*x^4 + 350*A*a^3*b^2*x^4 + 21*B*a^5*x^2 + 105*A*a^4*b*x^2 + 15*A*a^5)/x^7

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^8} \, dx=x^3\,\left (\frac {A\,b^5}{3}+\frac {5\,B\,a\,b^4}{3}\right )-\frac {\frac {A\,a^5}{7}+x^4\,\left (\frac {5\,B\,a^4\,b}{3}+\frac {10\,A\,a^3\,b^2}{3}\right )+x^2\,\left (\frac {B\,a^5}{5}+A\,b\,a^4\right )+x^6\,\left (10\,B\,a^3\,b^2+10\,A\,a^2\,b^3\right )}{x^7}+\frac {B\,b^5\,x^5}{5}+5\,a\,b^3\,x\,\left (A\,b+2\,B\,a\right ) \]

[In]

int(((A + B*x^2)*(a + b*x^2)^5)/x^8,x)

[Out]

x^3*((A*b^5)/3 + (5*B*a*b^4)/3) - ((A*a^5)/7 + x^4*((10*A*a^3*b^2)/3 + (5*B*a^4*b)/3) + x^2*((B*a^5)/5 + A*a^4
*b) + x^6*(10*A*a^2*b^3 + 10*B*a^3*b^2))/x^7 + (B*b^5*x^5)/5 + 5*a*b^3*x*(A*b + 2*B*a)