\(\int \frac {x^3 (A+B x^2)}{(a+b x^2)^3} \, dx\) [92]

Optimal result

Integrand size = 20, antiderivative size = 66 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {a (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}-\frac {A b-2 a B}{2 b^3 \left (a+b x^2\right )}+\frac {B \log \left (a+b x^2\right )}{2 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, 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} \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {A b-2 a B}{2 b^3 \left (a+b x^2\right )}+\frac {a (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}+\frac {B \log \left (a+b x^2\right )}{2 b^3} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {3 a^2 B-2 A b^2 x^2-a b \left (A-4 B x^2\right )+2 B \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 b^3 \left (a+b x^2\right )^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86

method	result	size
norman	\(\frac {-\frac {a \left (A b -3 B a \right )}{4 b^{3}}-\frac {\left (A b -2 B a \right ) x^{2}}{2 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(57\)
risch	\(\frac {-\frac {a \left (A b -3 B a \right )}{4 b^{3}}-\frac {\left (A b -2 B a \right ) x^{2}}{2 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(57\)
default	\(\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{3}}+\frac {a \left (A b -B a \right )}{4 b^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {A b -2 B a}{2 b^{3} \left (b \,x^{2}+a \right )}\)	\(61\)
parallelrisch	\(-\frac {-2 B \ln \left (b \,x^{2}+a \right ) x^{4} b^{2}-4 B \ln \left (b \,x^{2}+a \right ) x^{2} a b +2 A \,b^{2} x^{2}-4 B a b \,x^{2}-2 B \ln \left (b \,x^{2}+a \right ) a^{2}+a b A -3 a^{2} B}{4 b^{3} \left (b \,x^{2}+a \right )^{2}}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.35 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {3 \, B a^{2} - A a b + 2 \, {\left (2 \, B a b - A b^{2}\right )} x^{2} + 2 \, {\left (B b^{2} x^{4} + 2 \, B a b x^{2} + B a^{2}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B \log {\left (a + b x^{2} \right )}}{2 b^{3}} + \frac {- A a b + 3 B a^{2} + x^{2} \left (- 2 A b^{2} + 4 B a b\right )}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.09 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {3 \, B a^{2} - A a b + 2 \, {\left (2 \, B a b - A b^{2}\right )} x^{2}}{4 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {B \log \left (b x^{2} + a\right )}{2 \, b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {B \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{3}} + \frac {2 \, {\left (2 \, B a - A b\right )} x^{2} + \frac {3 \, B a^{2} - A a b}{b}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {3\,B\,a^2-A\,a\,b}{4\,b^3}-\frac {x^2\,\left (A\,b-2\,B\,a\right )}{2\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {B\,\ln \left (b\,x^2+a\right )}{2\,b^3} \]

[In]

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

[Out]

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