\(\int x^2 \sqrt {b x^n} \, dx\) [131]

Optimal result

Integrand size = 13, antiderivative size = 19 \[ \int x^2 \sqrt {b x^n} \, dx=\frac {2 x^3 \sqrt {b x^n}}{6+n} \]

[Out]

2*x^3*(b*x^n)^(1/2)/(6+n)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {15, 30} \[ \int x^2 \sqrt {b x^n} \, dx=\frac {2 x^3 \sqrt {b x^n}}{n+6} \]

[In]

Int[x^2*Sqrt[b*x^n],x]

[Out]

(2*x^3*Sqrt[b*x^n])/(6 + n)

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]

Rubi steps \begin{align*} \text {integral}& = \left (x^{-n/2} \sqrt {b x^n}\right ) \int x^{2+\frac {n}{2}} \, dx \\ & = \frac {2 x^3 \sqrt {b x^n}}{6+n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^2 \sqrt {b x^n} \, dx=\frac {2 x^3 \sqrt {b x^n}}{6+n} \]

[In]

Integrate[x^2*Sqrt[b*x^n],x]

[Out]

(2*x^3*Sqrt[b*x^n])/(6 + n)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {2 x^{3} \sqrt {b \,x^{n}}}{6+n}\)	\(18\)
risch	\(\frac {2 b \,x^{3} x^{n}}{\left (6+n \right ) \sqrt {b \,x^{n}}}\)	\(22\)

[In]

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

[Out]

2*x^3*(b*x^n)^(1/2)/(6+n)

Fricas [F(-2)]

Exception generated. \[ \int x^2 \sqrt {b x^n} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (15) = 30\).

Time = 0.50 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int x^2 \sqrt {b x^n} \, dx=\begin {cases} \frac {2 x^{3} \sqrt {b x^{n}}}{n + 6} & \text {for}\: n \neq -6 \\x^{3} \sqrt {\frac {b}{x^{6}}} \log {\left (x \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(b*x**n)**(1/2),x)

[Out]

Piecewise((2*x**3*sqrt(b*x**n)/(n + 6), Ne(n, -6)), (x**3*sqrt(b/x**6)*log(x), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int x^2 \sqrt {b x^n} \, dx=\frac {2 \, \sqrt {b x^{n}} x^{3}}{n + 6} \]

[In]

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

[Out]

2*sqrt(b*x^n)*x^3/(n + 6)

Giac [F]

\[ \int x^2 \sqrt {b x^n} \, dx=\int { \sqrt {b x^{n}} x^{2} \,d x } \]

[In]

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

[Out]

integrate(sqrt(b*x^n)*x^2, x)

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int x^2 \sqrt {b x^n} \, dx=\frac {2\,x^3\,\sqrt {b\,x^n}}{n+6} \]

[In]

int(x^2*(b*x^n)^(1/2),x)

[Out]

(2*x^3*(b*x^n)^(1/2))/(n + 6)