3.10.31 \(\int \sqrt {x+x^{3/2}} \, dx\) [931]

Optimal. Leaf size=59 \[ \frac {32 \left (x+x^{3/2}\right )^{3/2}}{105 x^{3/2}}-\frac {16 \left (x+x^{3/2}\right )^{3/2}}{35 x}+\frac {4 \left (x+x^{3/2}\right )^{3/2}}{7 \sqrt {x}} \]

[Out]

32/105*(x+x^(3/2))^(3/2)/x^(3/2)-16/35*(x+x^(3/2))^(3/2)/x+4/7*(x+x^(3/2))^(3/2)/x^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2027, 2041, 2039} \begin {gather*} \frac {4 \left (x^{3/2}+x\right )^{3/2}}{7 \sqrt {x}}-\frac {16 \left (x^{3/2}+x\right )^{3/2}}{35 x}+\frac {32 \left (x^{3/2}+x\right )^{3/2}}{105 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x + x^(3/2)],x]

[Out]

(32*(x + x^(3/2))^(3/2))/(105*x^(3/2)) - (16*(x + x^(3/2))^(3/2))/(35*x) + (4*(x + x^(3/2))^(3/2))/(7*Sqrt[x])

Rule 2027

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[b*((n*p + n - j + 1)/(a*(j*p + 1))), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

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])

Rule 2041

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*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \sqrt {x+x^{3/2}} \, dx &=\frac {4 \left (x+x^{3/2}\right )^{3/2}}{7 \sqrt {x}}-\frac {4}{7} \int \frac {\sqrt {x+x^{3/2}}}{\sqrt {x}} \, dx\\ &=-\frac {16 \left (x+x^{3/2}\right )^{3/2}}{35 x}+\frac {4 \left (x+x^{3/2}\right )^{3/2}}{7 \sqrt {x}}+\frac {8}{35} \int \frac {\sqrt {x+x^{3/2}}}{x} \, dx\\ &=\frac {32 \left (x+x^{3/2}\right )^{3/2}}{105 x^{3/2}}-\frac {16 \left (x+x^{3/2}\right )^{3/2}}{35 x}+\frac {4 \left (x+x^{3/2}\right )^{3/2}}{7 \sqrt {x}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 39, normalized size = 0.66 \begin {gather*} \frac {4 \sqrt {x+x^{3/2}} \left (8-4 \sqrt {x}+3 x+15 x^{3/2}\right )}{105 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x + x^(3/2)],x]

[Out]

(4*Sqrt[x + x^(3/2)]*(8 - 4*Sqrt[x] + 3*x + 15*x^(3/2)))/(105*Sqrt[x])

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Maple [A]
time = 0.22, size = 28, normalized size = 0.47

method	result	size
derivativedivides	\(\frac {4 \sqrt {x +x^{\frac {3}{2}}}\, \left (1+\sqrt {x}\right ) \left (15 x -12 \sqrt {x}+8\right )}{105 \sqrt {x}}\)	\(28\)
default	\(\frac {4 \sqrt {x +x^{\frac {3}{2}}}\, \left (1+\sqrt {x}\right ) \left (15 x -12 \sqrt {x}+8\right )}{105 \sqrt {x}}\)	\(28\)
meijerg	\(-\frac {\frac {32 \sqrt {\pi }}{105}-\frac {4 \sqrt {\pi }\, \left (1+\sqrt {x}\right )^{\frac {3}{2}} \left (15 x -12 \sqrt {x}+8\right )}{105}}{\sqrt {\pi }}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+x^(3/2))^(1/2),x,method=_RETURNVERBOSE)

[Out]

4/105*(x+x^(3/2))^(1/2)*(1+x^(1/2))*(15*x-12*x^(1/2)+8)/x^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+x^(3/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^(3/2) + x), x)

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Fricas [A]
time = 0.36, size = 30, normalized size = 0.51 \begin {gather*} \frac {4 \, {\left (15 \, x^{2} + {\left (3 \, x + 8\right )} \sqrt {x} - 4 \, x\right )} \sqrt {x^{\frac {3}{2}} + x}}{105 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+x^(3/2))^(1/2),x, algorithm="fricas")

[Out]

4/105*(15*x^2 + (3*x + 8)*sqrt(x) - 4*x)*sqrt(x^(3/2) + x)/x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {x^{\frac {3}{2}} + x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+x**(3/2))**(1/2),x)

[Out]

Integral(sqrt(x**(3/2) + x), x)

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Giac [A]
time = 5.35, size = 33, normalized size = 0.56 \begin {gather*} \frac {4}{105} \, {\left (15 \, {\left (\sqrt {x} + 1\right )}^{\frac {7}{2}} - 42 \, {\left (\sqrt {x} + 1\right )}^{\frac {5}{2}} + 35 \, {\left (\sqrt {x} + 1\right )}^{\frac {3}{2}} - 8\right )} \mathrm {sgn}\left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+x^(3/2))^(1/2),x, algorithm="giac")

[Out]

4/105*(15*(sqrt(x) + 1)^(7/2) - 42*(sqrt(x) + 1)^(5/2) + 35*(sqrt(x) + 1)^(3/2) - 8)*sgn(x)

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Mupad [B]
time = 3.54, size = 27, normalized size = 0.46 \begin {gather*} \frac {2\,x\,\sqrt {x+x^{3/2}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},3;\ 4;\ -\sqrt {x}\right )}{3\,\sqrt {\sqrt {x}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^(3/2))^(1/2),x)

[Out]

(2*x*(x + x^(3/2))^(1/2)*hypergeom([-1/2, 3], 4, -x^(1/2)))/(3*(x^(1/2) + 1)^(1/2))

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