\(\int (c x)^m \sqrt {b x^2} \, dx\) [63]

Optimal result

Integrand size = 15, antiderivative size = 28 \[ \int (c x)^m \sqrt {b x^2} \, dx=\frac {(c x)^{2+m} \sqrt {b x^2}}{c^2 (2+m) x} \]

[Out]

(c*x)^(2+m)*(b*x^2)^(1/2)/c^2/(2+m)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {15, 16, 32} \[ \int (c x)^m \sqrt {b x^2} \, dx=\frac {\sqrt {b x^2} (c x)^{m+2}}{c^2 (m+2) x} \]

[In]

Int[(c*x)^m*Sqrt[b*x^2],x]

[Out]

((c*x)^(2 + m)*Sqrt[b*x^2])/(c^2*(2 + m)*x)

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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {b x^2} \int x (c x)^m \, dx}{x} \\ & = \frac {\sqrt {b x^2} \int (c x)^{1+m} \, dx}{c x} \\ & = \frac {(c x)^{2+m} \sqrt {b x^2}}{c^2 (2+m) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int (c x)^m \sqrt {b x^2} \, dx=\frac {x (c x)^m \sqrt {b x^2}}{2+m} \]

[In]

Integrate[(c*x)^m*Sqrt[b*x^2],x]

[Out]

(x*(c*x)^m*Sqrt[b*x^2])/(2 + m)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71

method	result	size
gosper	\(\frac {x \left (c x \right )^{m} \sqrt {b \,x^{2}}}{2+m}\)	\(20\)
risch	\(\frac {x \left (c x \right )^{m} \sqrt {b \,x^{2}}}{2+m}\)	\(20\)

[In]

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

[Out]

x/(2+m)*(c*x)^m*(b*x^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int (c x)^m \sqrt {b x^2} \, dx=\frac {\sqrt {b x^{2}} \left (c x\right )^{m} x}{m + 2} \]

[In]

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

[Out]

sqrt(b*x^2)*(c*x)^m*x/(m + 2)

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int (c x)^m \sqrt {b x^2} \, dx=\begin {cases} \frac {x \sqrt {b x^{2}} \left (c x\right )^{m}}{m + 2} & \text {for}\: m \neq -2 \\\frac {\sqrt {b x^{2}} \log {\left (x \right )}}{c^{2} x} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x)**m*(b*x**2)**(1/2),x)

[Out]

Piecewise((x*sqrt(b*x**2)*(c*x)**m/(m + 2), Ne(m, -2)), (sqrt(b*x**2)*log(x)/(c**2*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int (c x)^m \sqrt {b x^2} \, dx=\frac {\sqrt {b} c^{m} x^{2} x^{m}}{m + 2} \]

[In]

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

[Out]

sqrt(b)*c^m*x^2*x^m/(m + 2)

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 5.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int (c x)^m \sqrt {b x^2} \, dx=\frac {\sqrt {b}\,x\,{\left (c\,x\right )}^m\,\sqrt {x^2}}{m+2} \]

[In]

int((c*x)^m*(b*x^2)^(1/2),x)

[Out]

(b^(1/2)*x*(c*x)^m*(x^2)^(1/2))/(m + 2)