\(\int \frac {x^3 (c+d x^2)}{(a+b x^2)^2} \, dx\) [263]

Optimal result

Integrand size = 20, antiderivative size = 60 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {d x^2}{2 b^2}+\frac {a (b c-a d)}{2 b^3 \left (a+b x^2\right )}+\frac {(b c-2 a d) \log \left (a+b x^2\right )}{2 b^3} \]

[Out]

1/2*d*x^2/b^2+1/2*a*(-a*d+b*c)/b^3/(b*x^2+a)+1/2*(-2*a*d+b*c)*ln(b*x^2+a)/b^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, 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} \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {a (b c-a d)}{2 b^3 \left (a+b x^2\right )}+\frac {(b c-2 a d) \log \left (a+b x^2\right )}{2 b^3}+\frac {d x^2}{2 b^2} \]

[In]

Int[(x^3*(c + d*x^2))/(a + b*x^2)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {b d x^2+\frac {a (b c-a d)}{a+b x^2}+(b c-2 a d) \log \left (a+b x^2\right )}{2 b^3} \]

[In]

Integrate[(x^3*(c + d*x^2))/(a + b*x^2)^2,x]

[Out]

(b*d*x^2 + (a*(b*c - a*d))/(a + b*x^2) + (b*c - 2*a*d)*Log[a + b*x^2])/(2*b^3)

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98

method	result	size
norman	\(\frac {-\frac {a \left (2 a d -b c \right )}{2 b^{3}}+\frac {d \,x^{4}}{2 b}}{b \,x^{2}+a}-\frac {\left (2 a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(59\)
default	\(\frac {d \,x^{2}}{2 b^{2}}-\frac {\frac {\left (2 a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{b}+\frac {\left (a d -b c \right ) a}{b \left (b \,x^{2}+a \right )}}{2 b^{2}}\)	\(60\)
risch	\(\frac {d \,x^{2}}{2 b^{2}}-\frac {a^{2} d}{2 b^{3} \left (b \,x^{2}+a \right )}+\frac {a c}{2 b^{2} \left (b \,x^{2}+a \right )}-\frac {\ln \left (b \,x^{2}+a \right ) a d}{b^{3}}+\frac {c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(74\)
parallelrisch	\(-\frac {-b^{2} d \,x^{4}+2 \ln \left (b \,x^{2}+a \right ) x^{2} a b d -\ln \left (b \,x^{2}+a \right ) x^{2} b^{2} c +2 \ln \left (b \,x^{2}+a \right ) a^{2} d -\ln \left (b \,x^{2}+a \right ) a b c +2 a^{2} d -a b c}{2 b^{3} \left (b \,x^{2}+a \right )}\)	\(96\)

[In]

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

[Out]

(-1/2*a*(2*a*d-b*c)/b^3+1/2*d*x^4/b)/(b*x^2+a)-1/2*(2*a*d-b*c)/b^3*ln(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {b^{2} d x^{4} + a b d x^{2} + a b c - a^{2} d + {\left (a b c - 2 \, a^{2} d + {\left (b^{2} c - 2 \, a b d\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

[In]

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

[Out]

1/2*(b^2*d*x^4 + a*b*d*x^2 + a*b*c - a^2*d + (a*b*c - 2*a^2*d + (b^2*c - 2*a*b*d)*x^2)*log(b*x^2 + a))/(b^4*x^
2 + a*b^3)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {- a^{2} d + a b c}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {d x^{2}}{2 b^{2}} - \frac {\left (2 a d - b c\right ) \log {\left (a + b x^{2} \right )}}{2 b^{3}} \]

[In]

integrate(x**3*(d*x**2+c)/(b*x**2+a)**2,x)

[Out]

(-a**2*d + a*b*c)/(2*a*b**3 + 2*b**4*x**2) + d*x**2/(2*b**2) - (2*a*d - b*c)*log(a + b*x**2)/(2*b**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {d x^{2}}{2 \, b^{2}} + \frac {a b c - a^{2} d}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {{\left (b c - 2 \, a d\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} \]

[In]

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

[Out]

1/2*d*x^2/b^2 + 1/2*(a*b*c - a^2*d)/(b^4*x^2 + a*b^3) + 1/2*(b*c - 2*a*d)*log(b*x^2 + a)/b^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.50 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {\frac {{\left (b x^{2} + a\right )} d}{b^{2}} - \frac {{\left (b c - 2 \, a d\right )} \log \left (\frac {{\left | b x^{2} + a \right |}}{{\left (b x^{2} + a\right )}^{2} {\left | b \right |}}\right )}{b^{2}} + \frac {\frac {a b^{2} c}{b x^{2} + a} - \frac {a^{2} b d}{b x^{2} + a}}{b^{3}}}{2 \, b} \]

[In]

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

[Out]

1/2*((b*x^2 + a)*d/b^2 - (b*c - 2*a*d)*log(abs(b*x^2 + a)/((b*x^2 + a)^2*abs(b)))/b^2 + (a*b^2*c/(b*x^2 + a) -
a^2*b*d/(b*x^2 + a))/b^3)/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {d\,x^2}{2\,b^2}-\frac {\ln \left (b\,x^2+a\right )\,\left (2\,a\,d-b\,c\right )}{2\,b^3}-\frac {a^2\,d-a\,b\,c}{2\,b\,\left (b^3\,x^2+a\,b^2\right )} \]

[In]

int((x^3*(c + d*x^2))/(a + b*x^2)^2,x)

[Out]

(d*x^2)/(2*b^2) - (log(a + b*x^2)*(2*a*d - b*c))/(2*b^3) - (a^2*d - a*b*c)/(2*b*(a*b^2 + b^3*x^2))