3.3.45 \(\int \frac {(a x^2+b x^3)^{3/2}}{x^3} \, dx\) [245]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2039} \begin {gather*} \frac {2 \left (a x^2+b x^3\right )^{5/2}}{5 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^3,x]

[Out]

(2*(a*x^2 + b*x^3)^(5/2))/(5*b*x^5)

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 23, normalized size = 0.92 \begin {gather*} \frac {2 \left (x^2 (a+b x)\right )^{5/2}}{5 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x^2*(a + b*x))^(5/2))/(5*b*x^5)

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Maple [A]
time = 0.36, size = 27, normalized size = 1.08

method	result	size
gosper	\(\frac {2 \left (b x +a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{5 b \,x^{3}}\)	\(27\)
default	\(\frac {2 \left (b x +a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{5 b \,x^{3}}\)	\(27\)
risch	\(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (b^{2} x^{2}+2 a b x +a^{2}\right )}{5 x b}\)	\(36\)
trager	\(\frac {2 \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{5 b x}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 28, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x + a}}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)/b

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Fricas [A]
time = 2.05, size = 37, normalized size = 1.48 \begin {gather*} \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b x^{3} + a x^{2}}}{5 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (21) = 42\).
time = 1.56, size = 89, normalized size = 3.56 \begin {gather*} -\frac {2 \, a^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{5 \, b} + \frac {2 \, {\left (15 \, \sqrt {b x + a} a^{2} \mathrm {sgn}\left (x\right ) + 10 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a \mathrm {sgn}\left (x\right ) + {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} \mathrm {sgn}\left (x\right )\right )}}{15 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*a^(5/2)*sgn(x)/b + 2/15*(15*sqrt(b*x + a)*a^2*sgn(x) + 10*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a*sgn(x)
+ (3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*sgn(x))/b

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Mupad [B]
time = 5.62, size = 28, normalized size = 1.12 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2}{5\,b\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(3/2)/x^3,x)

[Out]

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

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