\(\int \sqrt {x \sqrt {x^{3/2}}} \, dx\) [219]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8}{15} x \sqrt {x \sqrt {x^{3/2}}} \]

[Out]

8/15*x*(x*(x^(3/2))^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6851, 15, 30} \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8}{15} x \sqrt {x \sqrt {x^{3/2}}} \]

[In]

Int[Sqrt[x*Sqrt[x^(3/2)]],x]

[Out]

(8*x*Sqrt[x*Sqrt[x^(3/2)]])/15

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6851

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n)^FracPart[p]/(v^(m*Fr
acPart[p])*w^(n*FracPart[p]))), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x \sqrt {x^2 \sqrt {x^3}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {\left (2 \sqrt {x \sqrt {x^{3/2}}}\right ) \text {Subst}\left (\int x^2 \sqrt [4]{x^3} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{x^{3/2}}} \\ & = \frac {\left (2 \sqrt {x \sqrt {x^{3/2}}}\right ) \text {Subst}\left (\int x^{11/4} \, dx,x,\sqrt {x}\right )}{x^{7/8}} \\ & = \frac {8}{15} x \sqrt {x \sqrt {x^{3/2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8}{15} x \sqrt {x \sqrt {x^{3/2}}} \]

[In]

Integrate[Sqrt[x*Sqrt[x^(3/2)]],x]

[Out]

(8*x*Sqrt[x*Sqrt[x^(3/2)]])/15

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {8 x \sqrt {x \sqrt {x^{\frac {3}{2}}}}}{15}\)	\(13\)
derivativedivides	\(\frac {8 x \sqrt {x \sqrt {x^{\frac {3}{2}}}}}{15}\)	\(13\)
default	\(\frac {8 x \sqrt {x \sqrt {x^{\frac {3}{2}}}}}{15}\)	\(13\)

[In]

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

[Out]

8/15*x*(x*(x^(3/2))^(1/2))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8}{15} \, \sqrt {\sqrt {x^{\frac {3}{2}}} x} x \]

[In]

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

[Out]

8/15*sqrt(sqrt(x^(3/2))*x)*x

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8 x \sqrt {x \sqrt {x^{\frac {3}{2}}}}}{15} \]

[In]

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

[Out]

8*x*sqrt(x*sqrt(x**(3/2)))/15

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.25 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8}{15} \, x^{\frac {15}{8}} \]

[In]

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

[Out]

8/15*x^(15/8)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8}{15} \, x^{\frac {15}{8}} \mathrm {sgn}\left (x^{\frac {7}{4}} + 4 \, x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x + 4\right ) \]

[In]

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

[Out]

8/15*x^(15/8)*sgn(x^(7/4) + 4*x^(3/4))*sgn(x + 4)

Mupad [B] (verification not implemented)

Time = 15.40 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \sqrt {x \sqrt {x^{3/2}}} \, dx=\frac {8\,x\,\sqrt {x\,\sqrt {x^{3/2}}}}{15} \]

[In]

int((x*(x^(3/2))^(1/2))^(1/2),x)

[Out]

(8*x*(x*(x^(3/2))^(1/2))^(1/2))/15