3.1.10 \(\int \frac {(a x+b x^3)^2}{x^2} \, dx\) [10]

Optimal. Leaf size=25 \[ a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1598, 200} \begin {gather*} a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^3)^2/x^2,x]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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+b x^3\right )^2}{x^2} \, dx &=\int \left (a+b x^2\right )^2 \, dx\\ &=\int \left (a^2+2 a b x^2+b^2 x^4\right ) \, dx\\ &=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 25, normalized size = 1.00 \begin {gather*} a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^3)^2/x^2,x]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

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Maple [A]
time = 0.33, size = 22, normalized size = 0.88

method	result	size
default	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
risch	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
gosper	\(\frac {x \left (3 b^{2} x^{4}+10 a b \,x^{2}+15 a^{2}\right )}{15}\)	\(25\)
norman	\(\frac {a^{2} x^{2}+\frac {1}{5} b^{2} x^{6}+\frac {2}{3} a b \,x^{4}}{x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

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Fricas [A]
time = 1.29, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

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Sympy [A]
time = 0.01, size = 22, normalized size = 0.88 \begin {gather*} a^{2} x + \frac {2 a b x^{3}}{3} + \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/x**2,x)

[Out]

a**2*x + 2*a*b*x**3/3 + b**2*x**5/5

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Giac [A]
time = 1.32, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

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Mupad [B]
time = 0.03, size = 21, normalized size = 0.84 \begin {gather*} a^2\,x+\frac {2\,a\,b\,x^3}{3}+\frac {b^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^3)^2/x^2,x)

[Out]

a^2*x + (b^2*x^5)/5 + (2*a*b*x^3)/3

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