\(\int \frac {x^3 (a c+b c x^2)}{a+b x^2} \, dx\) [120]

Optimal result

Integrand size = 23, antiderivative size = 8 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {c x^4}{4} \]

[Out]

1/4*c*x^4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {21, 30} \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {c x^4}{4} \]

[In]

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

[Out]

(c*x^4)/4

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = c \int x^3 \, dx \\ & = \frac {c x^4}{4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {c x^4}{4} \]

[In]

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

[Out]

(c*x^4)/4

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {x^{4} c}{4}\)	\(7\)
default	\(\frac {x^{4} c}{4}\)	\(7\)
norman	\(\frac {x^{4} c}{4}\)	\(7\)
risch	\(\frac {x^{4} c}{4}\)	\(7\)
parallelrisch	\(\frac {x^{4} c}{4}\)	\(7\)

[In]

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

[Out]

1/4*x^4*c

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {1}{4} \, c x^{4} \]

[In]

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

[Out]

1/4*c*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {c x^{4}}{4} \]

[In]

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

[Out]

c*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {1}{4} \, c x^{4} \]

[In]

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

[Out]

1/4*c*x^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {1}{4} \, c x^{4} \]

[In]

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

[Out]

1/4*c*x^4

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {x^3 \left (a c+b c x^2\right )}{a+b x^2} \, dx=\frac {c\,x^4}{4} \]

[In]

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

[Out]

(c*x^4)/4