\(\int \frac {(a+b x^2)^5 (A+B x^2)}{x^3} \, dx\) [35]

Optimal result

Integrand size = 20, antiderivative size = 113 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=-\frac {a^5 A}{2 x^2}+\frac {5}{2} a^3 b (2 A b+a B) x^2+\frac {5}{2} a^2 b^2 (A b+a B) x^4+\frac {5}{6} a b^3 (A b+2 a B) x^6+\frac {1}{8} b^4 (A b+5 a B) x^8+\frac {1}{10} b^5 B x^{10}+a^4 (5 A b+a B) \log (x) \]

[Out]

-1/2*a^5*A/x^2+5/2*a^3*b*(2*A*b+B*a)*x^2+5/2*a^2*b^2*(A*b+B*a)*x^4+5/6*a*b^3*(A*b+2*B*a)*x^6+1/8*b^4*(A*b+5*B*
a)*x^8+1/10*b^5*B*x^10+a^4*(5*A*b+B*a)*ln(x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {457, 77} \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=-\frac {a^5 A}{2 x^2}+a^4 \log (x) (a B+5 A b)+\frac {5}{2} a^3 b x^2 (a B+2 A b)+\frac {5}{2} a^2 b^2 x^4 (a B+A b)+\frac {1}{8} b^4 x^8 (5 a B+A b)+\frac {5}{6} a b^3 x^6 (2 a B+A b)+\frac {1}{10} b^5 B x^{10} \]

[In]

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

[Out]

-1/2*(a^5*A)/x^2 + (5*a^3*b*(2*A*b + a*B)*x^2)/2 + (5*a^2*b^2*(A*b + a*B)*x^4)/2 + (5*a*b^3*(A*b + 2*a*B)*x^6)
/6 + (b^4*(A*b + 5*a*B)*x^8)/8 + (b^5*B*x^10)/10 + a^4*(5*A*b + a*B)*Log[x]

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=-\frac {a^5 A}{2 x^2}+\frac {5}{2} a^3 b (2 A b+a B) x^2+\frac {5}{2} a^2 b^2 (A b+a B) x^4+\frac {5}{6} a b^3 (A b+2 a B) x^6+\frac {1}{8} b^4 (A b+5 a B) x^8+\frac {1}{10} b^5 B x^{10}+\left (5 a^4 A b+a^5 B\right ) \log (x) \]

[In]

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

[Out]

-1/2*(a^5*A)/x^2 + (5*a^3*b*(2*A*b + a*B)*x^2)/2 + (5*a^2*b^2*(A*b + a*B)*x^4)/2 + (5*a*b^3*(A*b + 2*a*B)*x^6)
/6 + (b^4*(A*b + 5*a*B)*x^8)/8 + (b^5*B*x^10)/10 + (5*a^4*A*b + a^5*B)*Log[x]

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {b^{5} B \,x^{10}}{10}+\frac {A \,b^{5} x^{8}}{8}+\frac {5 B a \,b^{4} x^{8}}{8}+\frac {5 A a \,b^{4} x^{6}}{6}+\frac {5 B \,a^{2} b^{3} x^{6}}{3}+\frac {5 A \,a^{2} b^{3} x^{4}}{2}+\frac {5 B \,a^{3} b^{2} x^{4}}{2}+5 a^{3} A \,b^{2} x^{2}+\frac {5 B \,a^{4} b \,x^{2}}{2}+a^{4} \left (5 A b +B a \right ) \ln \left (x \right )-\frac {a^{5} A}{2 x^{2}}\)	\(121\)
norman	\(\frac {\left (\frac {1}{8} b^{5} A +\frac {5}{8} a \,b^{4} B \right ) x^{10}+\left (\frac {5}{6} a \,b^{4} A +\frac {5}{3} a^{2} b^{3} B \right ) x^{8}+\left (\frac {5}{2} a^{2} b^{3} A +\frac {5}{2} a^{3} b^{2} B \right ) x^{6}+\left (5 a^{3} b^{2} A +\frac {5}{2} a^{4} b B \right ) x^{4}-\frac {a^{5} A}{2}+\frac {b^{5} B \,x^{12}}{10}}{x^{2}}+\left (5 a^{4} b A +a^{5} B \right ) \ln \left (x \right )\)	\(121\)
risch	\(\frac {b^{5} B \,x^{10}}{10}+\frac {A \,b^{5} x^{8}}{8}+\frac {5 B a \,b^{4} x^{8}}{8}+\frac {5 A a \,b^{4} x^{6}}{6}+\frac {5 B \,a^{2} b^{3} x^{6}}{3}+\frac {5 A \,a^{2} b^{3} x^{4}}{2}+\frac {5 B \,a^{3} b^{2} x^{4}}{2}+5 a^{3} A \,b^{2} x^{2}+\frac {5 B \,a^{4} b \,x^{2}}{2}-\frac {a^{5} A}{2 x^{2}}+5 A \ln \left (x \right ) a^{4} b +B \ln \left (x \right ) a^{5}\)	\(123\)
parallelrisch	\(\frac {12 b^{5} B \,x^{12}+15 A \,b^{5} x^{10}+75 B a \,b^{4} x^{10}+100 a A \,b^{4} x^{8}+200 B \,a^{2} b^{3} x^{8}+300 a^{2} A \,b^{3} x^{6}+300 B \,a^{3} b^{2} x^{6}+600 a^{3} A \,b^{2} x^{4}+300 B \,a^{4} b \,x^{4}+600 A \ln \left (x \right ) x^{2} a^{4} b +120 B \ln \left (x \right ) x^{2} a^{5}-60 a^{5} A}{120 x^{2}}\)	\(132\)

[In]

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

[Out]

1/10*b^5*B*x^10+1/8*A*b^5*x^8+5/8*B*a*b^4*x^8+5/6*A*a*b^4*x^6+5/3*B*a^2*b^3*x^6+5/2*A*a^2*b^3*x^4+5/2*B*a^3*b^
2*x^4+5*a^3*A*b^2*x^2+5/2*B*a^4*b*x^2+a^4*(5*A*b+B*a)*ln(x)-1/2*a^5*A/x^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=\frac {12 \, B b^{5} x^{12} + 15 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + 100 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} - 60 \, A a^{5} + 300 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + 120 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2} \log \left (x\right )}{120 \, x^{2}} \]

[In]

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

[Out]

1/120*(12*B*b^5*x^12 + 15*(5*B*a*b^4 + A*b^5)*x^10 + 100*(2*B*a^2*b^3 + A*a*b^4)*x^8 + 300*(B*a^3*b^2 + A*a^2*
b^3)*x^6 - 60*A*a^5 + 300*(B*a^4*b + 2*A*a^3*b^2)*x^4 + 120*(B*a^5 + 5*A*a^4*b)*x^2*log(x))/x^2

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=- \frac {A a^{5}}{2 x^{2}} + \frac {B b^{5} x^{10}}{10} + a^{4} \cdot \left (5 A b + B a\right ) \log {\left (x \right )} + x^{8} \left (\frac {A b^{5}}{8} + \frac {5 B a b^{4}}{8}\right ) + x^{6} \cdot \left (\frac {5 A a b^{4}}{6} + \frac {5 B a^{2} b^{3}}{3}\right ) + x^{4} \cdot \left (\frac {5 A a^{2} b^{3}}{2} + \frac {5 B a^{3} b^{2}}{2}\right ) + x^{2} \cdot \left (5 A a^{3} b^{2} + \frac {5 B a^{4} b}{2}\right ) \]

[In]

integrate((b*x**2+a)**5*(B*x**2+A)/x**3,x)

[Out]

-A*a**5/(2*x**2) + B*b**5*x**10/10 + a**4*(5*A*b + B*a)*log(x) + x**8*(A*b**5/8 + 5*B*a*b**4/8) + x**6*(5*A*a*
b**4/6 + 5*B*a**2*b**3/3) + x**4*(5*A*a**2*b**3/2 + 5*B*a**3*b**2/2) + x**2*(5*A*a**3*b**2 + 5*B*a**4*b/2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=\frac {1}{10} \, B b^{5} x^{10} + \frac {1}{8} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{8} + \frac {5}{6} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{6} + \frac {5}{2} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{4} - \frac {A a^{5}}{2 \, x^{2}} + \frac {5}{2} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + \frac {1}{2} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} \log \left (x^{2}\right ) \]

[In]

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

[Out]

1/10*B*b^5*x^10 + 1/8*(5*B*a*b^4 + A*b^5)*x^8 + 5/6*(2*B*a^2*b^3 + A*a*b^4)*x^6 + 5/2*(B*a^3*b^2 + A*a^2*b^3)*
x^4 - 1/2*A*a^5/x^2 + 5/2*(B*a^4*b + 2*A*a^3*b^2)*x^2 + 1/2*(B*a^5 + 5*A*a^4*b)*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=\frac {1}{10} \, B b^{5} x^{10} + \frac {5}{8} \, B a b^{4} x^{8} + \frac {1}{8} \, A b^{5} x^{8} + \frac {5}{3} \, B a^{2} b^{3} x^{6} + \frac {5}{6} \, A a b^{4} x^{6} + \frac {5}{2} \, B a^{3} b^{2} x^{4} + \frac {5}{2} \, A a^{2} b^{3} x^{4} + \frac {5}{2} \, B a^{4} b x^{2} + 5 \, A a^{3} b^{2} x^{2} + \frac {1}{2} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} \log \left (x^{2}\right ) - \frac {B a^{5} x^{2} + 5 \, A a^{4} b x^{2} + A a^{5}}{2 \, x^{2}} \]

[In]

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

[Out]

1/10*B*b^5*x^10 + 5/8*B*a*b^4*x^8 + 1/8*A*b^5*x^8 + 5/3*B*a^2*b^3*x^6 + 5/6*A*a*b^4*x^6 + 5/2*B*a^3*b^2*x^4 +
5/2*A*a^2*b^3*x^4 + 5/2*B*a^4*b*x^2 + 5*A*a^3*b^2*x^2 + 1/2*(B*a^5 + 5*A*a^4*b)*log(x^2) - 1/2*(B*a^5*x^2 + 5*
A*a^4*b*x^2 + A*a^5)/x^2

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^3} \, dx=x^8\,\left (\frac {A\,b^5}{8}+\frac {5\,B\,a\,b^4}{8}\right )+\ln \left (x\right )\,\left (B\,a^5+5\,A\,b\,a^4\right )-\frac {A\,a^5}{2\,x^2}+\frac {B\,b^5\,x^{10}}{10}+\frac {5\,a^2\,b^2\,x^4\,\left (A\,b+B\,a\right )}{2}+\frac {5\,a^3\,b\,x^2\,\left (2\,A\,b+B\,a\right )}{2}+\frac {5\,a\,b^3\,x^6\,\left (A\,b+2\,B\,a\right )}{6} \]

[In]

int(((A + B*x^2)*(a + b*x^2)^5)/x^3,x)

[Out]

x^8*((A*b^5)/8 + (5*B*a*b^4)/8) + log(x)*(B*a^5 + 5*A*a^4*b) - (A*a^5)/(2*x^2) + (B*b^5*x^10)/10 + (5*a^2*b^2*
x^4*(A*b + B*a))/2 + (5*a^3*b*x^2*(2*A*b + B*a))/2 + (5*a*b^3*x^6*(A*b + 2*B*a))/6