3.2192 \(\int \frac {x^2}{a+b \sqrt {x}} \, dx\)

Optimal. Leaf size=79 \[ -\frac {2 a^5 \log \left (a+b \sqrt {x}\right )}{b^6}+\frac {2 a^4 \sqrt {x}}{b^5}-\frac {a^3 x}{b^4}+\frac {2 a^2 x^{3/2}}{3 b^3}-\frac {a x^2}{2 b^2}+\frac {2 x^{5/2}}{5 b} \]

[Out]

-a^3*x/b^4+2/3*a^2*x^(3/2)/b^3-1/2*a*x^2/b^2+2/5*x^(5/2)/b-2*a^5*ln(a+b*x^(1/2))/b^6+2*a^4*x^(1/2)/b^5

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Rubi [A]  time = 0.05, antiderivative size = 79, 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 = {266, 43} \[ \frac {2 a^2 x^{3/2}}{3 b^3}+\frac {2 a^4 \sqrt {x}}{b^5}-\frac {a^3 x}{b^4}-\frac {2 a^5 \log \left (a+b \sqrt {x}\right )}{b^6}-\frac {a x^2}{2 b^2}+\frac {2 x^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*Sqrt[x]),x]

[Out]

(2*a^4*Sqrt[x])/b^5 - (a^3*x)/b^4 + (2*a^2*x^(3/2))/(3*b^3) - (a*x^2)/(2*b^2) + (2*x^(5/2))/(5*b) - (2*a^5*Log
[a + b*Sqrt[x]])/b^6

Rule 43

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^2}{a+b \sqrt {x}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^5}{a+b x} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {a^4}{b^5}-\frac {a^3 x}{b^4}+\frac {a^2 x^2}{b^3}-\frac {a x^3}{b^2}+\frac {x^4}{b}-\frac {a^5}{b^5 (a+b x)}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2 a^4 \sqrt {x}}{b^5}-\frac {a^3 x}{b^4}+\frac {2 a^2 x^{3/2}}{3 b^3}-\frac {a x^2}{2 b^2}+\frac {2 x^{5/2}}{5 b}-\frac {2 a^5 \log \left (a+b \sqrt {x}\right )}{b^6}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 75, normalized size = 0.95 \[ \frac {-60 a^5 \log \left (a+b \sqrt {x}\right )+60 a^4 b \sqrt {x}-30 a^3 b^2 x+20 a^2 b^3 x^{3/2}-15 a b^4 x^2+12 b^5 x^{5/2}}{30 b^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*Sqrt[x]),x]

[Out]

(60*a^4*b*Sqrt[x] - 30*a^3*b^2*x + 20*a^2*b^3*x^(3/2) - 15*a*b^4*x^2 + 12*b^5*x^(5/2) - 60*a^5*Log[a + b*Sqrt[
x]])/(30*b^6)

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fricas [A]  time = 0.92, size = 66, normalized size = 0.84 \[ -\frac {15 \, a b^{4} x^{2} + 30 \, a^{3} b^{2} x + 60 \, a^{5} \log \left (b \sqrt {x} + a\right ) - 4 \, {\left (3 \, b^{5} x^{2} + 5 \, a^{2} b^{3} x + 15 \, a^{4} b\right )} \sqrt {x}}{30 \, b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*x^(1/2)),x, algorithm="fricas")

[Out]

-1/30*(15*a*b^4*x^2 + 30*a^3*b^2*x + 60*a^5*log(b*sqrt(x) + a) - 4*(3*b^5*x^2 + 5*a^2*b^3*x + 15*a^4*b)*sqrt(x
))/b^6

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giac [A]  time = 0.15, size = 67, normalized size = 0.85 \[ -\frac {2 \, a^{5} \log \left ({\left | b \sqrt {x} + a \right |}\right )}{b^{6}} + \frac {12 \, b^{4} x^{\frac {5}{2}} - 15 \, a b^{3} x^{2} + 20 \, a^{2} b^{2} x^{\frac {3}{2}} - 30 \, a^{3} b x + 60 \, a^{4} \sqrt {x}}{30 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*x^(1/2)),x, algorithm="giac")

[Out]

-2*a^5*log(abs(b*sqrt(x) + a))/b^6 + 1/30*(12*b^4*x^(5/2) - 15*a*b^3*x^2 + 20*a^2*b^2*x^(3/2) - 30*a^3*b*x + 6
0*a^4*sqrt(x))/b^5

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maple [A]  time = 0.00, size = 66, normalized size = 0.84 \[ \frac {2 x^{\frac {5}{2}}}{5 b}-\frac {a \,x^{2}}{2 b^{2}}+\frac {2 a^{2} x^{\frac {3}{2}}}{3 b^{3}}-\frac {2 a^{5} \ln \left (b \sqrt {x}+a \right )}{b^{6}}-\frac {a^{3} x}{b^{4}}+\frac {2 a^{4} \sqrt {x}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^(1/2)+a),x)

[Out]

-a^3*x/b^4+2/3*a^2*x^(3/2)/b^3-1/2*a/b^2*x^2+2/5*x^(5/2)/b-2*a^5*ln(b*x^(1/2)+a)/b^6+2*a^4*x^(1/2)/b^5

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maxima [A]  time = 0.86, size = 95, normalized size = 1.20 \[ -\frac {2 \, a^{5} \log \left (b \sqrt {x} + a\right )}{b^{6}} + \frac {2 \, {\left (b \sqrt {x} + a\right )}^{5}}{5 \, b^{6}} - \frac {5 \, {\left (b \sqrt {x} + a\right )}^{4} a}{2 \, b^{6}} + \frac {20 \, {\left (b \sqrt {x} + a\right )}^{3} a^{2}}{3 \, b^{6}} - \frac {10 \, {\left (b \sqrt {x} + a\right )}^{2} a^{3}}{b^{6}} + \frac {10 \, {\left (b \sqrt {x} + a\right )} a^{4}}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(a+b*x^(1/2)),x, algorithm="maxima")

[Out]

-2*a^5*log(b*sqrt(x) + a)/b^6 + 2/5*(b*sqrt(x) + a)^5/b^6 - 5/2*(b*sqrt(x) + a)^4*a/b^6 + 20/3*(b*sqrt(x) + a)
^3*a^2/b^6 - 10*(b*sqrt(x) + a)^2*a^3/b^6 + 10*(b*sqrt(x) + a)*a^4/b^6

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mupad [B]  time = 0.03, size = 65, normalized size = 0.82 \[ \frac {2\,x^{5/2}}{5\,b}-\frac {a\,x^2}{2\,b^2}-\frac {a^3\,x}{b^4}-\frac {2\,a^5\,\ln \left (a+b\,\sqrt {x}\right )}{b^6}+\frac {2\,a^2\,x^{3/2}}{3\,b^3}+\frac {2\,a^4\,\sqrt {x}}{b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b*x^(1/2)),x)

[Out]

(2*x^(5/2))/(5*b) - (a*x^2)/(2*b^2) - (a^3*x)/b^4 - (2*a^5*log(a + b*x^(1/2)))/b^6 + (2*a^2*x^(3/2))/(3*b^3) +
(2*a^4*x^(1/2))/b^5

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sympy [A]  time = 0.65, size = 82, normalized size = 1.04 \[ \begin {cases} - \frac {2 a^{5} \log {\left (\frac {a}{b} + \sqrt {x} \right )}}{b^{6}} + \frac {2 a^{4} \sqrt {x}}{b^{5}} - \frac {a^{3} x}{b^{4}} + \frac {2 a^{2} x^{\frac {3}{2}}}{3 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 a} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+b*x**(1/2)),x)

[Out]

Piecewise((-2*a**5*log(a/b + sqrt(x))/b**6 + 2*a**4*sqrt(x)/b**5 - a**3*x/b**4 + 2*a**2*x**(3/2)/(3*b**3) - a*
x**2/(2*b**2) + 2*x**(5/2)/(5*b), Ne(b, 0)), (x**3/(3*a), True))

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