\(\int \frac {x^3 (c+d x^2)}{a+b x^2} \, dx\) [200]

Optimal result

Integrand size = 20, antiderivative size = 54 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=\frac {(b c-a d) x^2}{2 b^2}+\frac {d x^4}{4 b}-\frac {a (b c-a d) \log \left (a+b x^2\right )}{2 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, 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} \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=-\frac {a (b c-a d) \log \left (a+b x^2\right )}{2 b^3}+\frac {x^2 (b c-a d)}{2 b^2}+\frac {d x^4}{4 b} \]

[In]

Int[(x^3*(c + d*x^2))/(a + b*x^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=\frac {b x^2 \left (2 b c-2 a d+b d x^2\right )+2 a (-b c+a d) \log \left (a+b x^2\right )}{4 b^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.64 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91

method	result	size
norman	\(\frac {d \,x^{4}}{4 b}-\frac {\left (a d -b c \right ) x^{2}}{2 b^{2}}+\frac {a \left (a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(49\)
default	\(-\frac {-\frac {1}{2} b d \,x^{4}+a d \,x^{2}-c b \,x^{2}}{2 b^{2}}+\frac {a \left (a d -b c \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{3}}\)	\(50\)
parallelrisch	\(\frac {b^{2} d \,x^{4}-2 x^{2} a b d +2 b^{2} c \,x^{2}+2 \ln \left (b \,x^{2}+a \right ) a^{2} d -2 \ln \left (b \,x^{2}+a \right ) a b c}{4 b^{3}}\)	\(59\)
risch	\(\frac {d \,x^{4}}{4 b}-\frac {d \,x^{2} a}{2 b^{2}}+\frac {c \,x^{2}}{2 b}+\frac {d \,a^{2}}{4 b^{3}}-\frac {a c}{2 b^{2}}+\frac {c^{2}}{4 b d}+\frac {a^{2} \ln \left (b \,x^{2}+a \right ) d}{2 b^{3}}-\frac {a c \ln \left (b \,x^{2}+a \right )}{2 b^{2}}\)	\(89\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=\frac {b^{2} d x^{4} + 2 \, {\left (b^{2} c - a b d\right )} x^{2} - 2 \, {\left (a b c - a^{2} d\right )} \log \left (b x^{2} + a\right )}{4 \, b^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=\frac {a \left (a d - b c\right ) \log {\left (a + b x^{2} \right )}}{2 b^{3}} + x^{2} \left (- \frac {a d}{2 b^{2}} + \frac {c}{2 b}\right ) + \frac {d x^{4}}{4 b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=\frac {b d x^{4} + 2 \, {\left (b c - a d\right )} x^{2}}{4 \, b^{2}} - \frac {{\left (a b c - a^{2} d\right )} \log \left (b x^{2} + a\right )}{2 \, b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=\frac {b d x^{4} + 2 \, b c x^{2} - 2 \, a d x^{2}}{4 \, b^{2}} - \frac {{\left (a b c - a^{2} d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 \left (c+d x^2\right )}{a+b x^2} \, dx=x^2\,\left (\frac {c}{2\,b}-\frac {a\,d}{2\,b^2}\right )+\frac {\ln \left (b\,x^2+a\right )\,\left (a^2\,d-a\,b\,c\right )}{2\,b^3}+\frac {d\,x^4}{4\,b} \]

[In]

int((x^3*(c + d*x^2))/(a + b*x^2),x)

[Out]

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