\(\int \frac {x^5 (A+B x^2)}{a+b x^2} \, dx\) [57]

Optimal result

Integrand size = 20, antiderivative size = 75 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=-\frac {a (A b-a B) x^2}{2 b^3}+\frac {(A b-a B) x^4}{4 b^2}+\frac {B x^6}{6 b}+\frac {a^2 (A b-a B) \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 75, 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, 78} \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {a^2 (A b-a B) \log \left (a+b x^2\right )}{2 b^4}-\frac {a x^2 (A b-a B)}{2 b^3}+\frac {x^4 (A b-a B)}{4 b^2}+\frac {B x^6}{6 b} \]

[In]

Int[(x^5*(A + B*x^2))/(a + b*x^2),x]

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {b x^2 \left (6 a^2 B-3 a b \left (2 A+B x^2\right )+b^2 x^2 \left (3 A+2 B x^2\right )\right )+6 a^2 (A b-a B) \log \left (a+b x^2\right )}{12 b^4} \]

[In]

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

[Out]

(b*x^2*(6*a^2*B - 3*a*b*(2*A + B*x^2) + b^2*x^2*(3*A + 2*B*x^2)) + 6*a^2*(A*b - a*B)*Log[a + b*x^2])/(12*b^4)

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91

method	result	size
norman	\(-\frac {a \left (A b -B a \right ) x^{2}}{2 b^{3}}+\frac {\left (A b -B a \right ) x^{4}}{4 b^{2}}+\frac {B \,x^{6}}{6 b}+\frac {a^{2} \left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(68\)
default	\(-\frac {-\frac {1}{3} b^{2} B \,x^{6}-\frac {1}{2} A \,b^{2} x^{4}+\frac {1}{2} B a b \,x^{4}+a A b \,x^{2}-a^{2} B \,x^{2}}{2 b^{3}}+\frac {a^{2} \left (A b -B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(74\)
parallelrisch	\(\frac {2 b^{3} B \,x^{6}+3 A \,x^{4} b^{3}-3 B \,x^{4} a \,b^{2}-6 a A \,b^{2} x^{2}+6 B \,a^{2} b \,x^{2}+6 A \ln \left (b \,x^{2}+a \right ) a^{2} b -6 B \ln \left (b \,x^{2}+a \right ) a^{3}}{12 b^{4}}\)	\(84\)
risch	\(\frac {B \,x^{6}}{6 b}+\frac {A \,x^{4}}{4 b}-\frac {B a \,x^{4}}{4 b^{2}}-\frac {a A \,x^{2}}{2 b^{2}}+\frac {a^{2} B \,x^{2}}{2 b^{3}}+\frac {a^{2} \ln \left (b \,x^{2}+a \right ) A}{2 b^{3}}-\frac {a^{3} \ln \left (b \,x^{2}+a \right ) B}{2 b^{4}}\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 \, B b^{3} x^{6} - 3 \, {\left (B a b^{2} - A b^{3}\right )} x^{4} + 6 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} - 6 \, {\left (B a^{3} - A a^{2} b\right )} \log \left (b x^{2} + a\right )}{12 \, b^{4}} \]

[In]

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

[Out]

1/12*(2*B*b^3*x^6 - 3*(B*a*b^2 - A*b^3)*x^4 + 6*(B*a^2*b - A*a*b^2)*x^2 - 6*(B*a^3 - A*a^2*b)*log(b*x^2 + a))/
b^4

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {B x^{6}}{6 b} - \frac {a^{2} \left (- A b + B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{4}} + x^{4} \left (\frac {A}{4 b} - \frac {B a}{4 b^{2}}\right ) + x^{2} \left (- \frac {A a}{2 b^{2}} + \frac {B a^{2}}{2 b^{3}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 \, B b^{2} x^{6} - 3 \, {\left (B a b - A b^{2}\right )} x^{4} + 6 \, {\left (B a^{2} - A a b\right )} x^{2}}{12 \, b^{3}} - \frac {{\left (B a^{3} - A a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

[In]

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

[Out]

1/12*(2*B*b^2*x^6 - 3*(B*a*b - A*b^2)*x^4 + 6*(B*a^2 - A*a*b)*x^2)/b^3 - 1/2*(B*a^3 - A*a^2*b)*log(b*x^2 + a)/
b^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=\frac {2 \, B b^{2} x^{6} - 3 \, B a b x^{4} + 3 \, A b^{2} x^{4} + 6 \, B a^{2} x^{2} - 6 \, A a b x^{2}}{12 \, b^{3}} - \frac {{\left (B a^{3} - A a^{2} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} \]

[In]

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

[Out]

1/12*(2*B*b^2*x^6 - 3*B*a*b*x^4 + 3*A*b^2*x^4 + 6*B*a^2*x^2 - 6*A*a*b*x^2)/b^3 - 1/2*(B*a^3 - A*a^2*b)*log(abs
(b*x^2 + a))/b^4

Mupad [B] (verification not implemented)

Time = 4.89 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {x^5 \left (A+B x^2\right )}{a+b x^2} \, dx=x^4\,\left (\frac {A}{4\,b}-\frac {B\,a}{4\,b^2}\right )+\frac {B\,x^6}{6\,b}-\frac {\ln \left (b\,x^2+a\right )\,\left (B\,a^3-A\,a^2\,b\right )}{2\,b^4}-\frac {a\,x^2\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{2\,b} \]

[In]

int((x^5*(A + B*x^2))/(a + b*x^2),x)

[Out]

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