3.3.12 \(\int \frac {(a x^2+b x^3)^2}{x^2} \, dx\) [212]

Optimal. Leaf size=30 \[ \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \]

[Out]

1/3*a^2*x^3+1/2*a*b*x^4+1/5*b^2*x^5

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 45} \begin {gather*} \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^2/x^2,x]

[Out]

(a^2*x^3)/3 + (a*b*x^4)/2 + (b^2*x^5)/5

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\left (a x^2+b x^3\right )^2}{x^2} \, dx &=\int x^2 (a+b x)^2 \, dx\\ &=\int \left (a^2 x^2+2 a b x^3+b^2 x^4\right ) \, dx\\ &=\frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} \frac {a^2 x^3}{3}+\frac {1}{2} a b x^4+\frac {b^2 x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^2/x^2,x]

[Out]

(a^2*x^3)/3 + (a*b*x^4)/2 + (b^2*x^5)/5

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Maple [A]
time = 0.33, size = 25, normalized size = 0.83

method	result	size
gosper	\(\frac {x^{3} \left (6 b^{2} x^{2}+15 a b x +10 a^{2}\right )}{30}\)	\(25\)
default	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
risch	\(\frac {1}{3} a^{2} x^{3}+\frac {1}{2} a b \,x^{4}+\frac {1}{5} b^{2} x^{5}\)	\(25\)
norman	\(\frac {\frac {1}{3} a^{2} x^{4}+\frac {1}{5} b^{2} x^{6}+\frac {1}{2} a b \,x^{5}}{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*x^3+1/2*a*b*x^4+1/5*b^2*x^5

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Maxima [A]
time = 0.27, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5 + 1/2*a*b*x^4 + 1/3*a^2*x^3

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Fricas [A]
time = 1.15, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5 + 1/2*a*b*x^4 + 1/3*a^2*x^3

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Sympy [A]
time = 0.01, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^{2} x^{3}}{3} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{5}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**3/3 + a*b*x**4/2 + b**2*x**5/5

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Giac [A]
time = 0.98, size = 24, normalized size = 0.80 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {1}{2} \, a b x^{4} + \frac {1}{3} \, a^{2} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5 + 1/2*a*b*x^4 + 1/3*a^2*x^3

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Mupad [B]
time = 0.03, size = 24, normalized size = 0.80 \begin {gather*} \frac {a^2\,x^3}{3}+\frac {a\,b\,x^4}{2}+\frac {b^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^2/x^2,x)

[Out]

(a^2*x^3)/3 + (b^2*x^5)/5 + (a*b*x^4)/2

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