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

Optimal. Leaf size=28 \[ \frac {(c x)^{2+m} \sqrt {b x^2}}{c^2 (2+m) x} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {15, 16, 32} \begin {gather*} \frac {\sqrt {b x^2} (c x)^{m+2}}{c^2 (m+2) x} \end {gather*}

Antiderivative was successfully verified.

[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*} \int (c x)^m \sqrt {b x^2} \, dx &=\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*}

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Mathematica [A]
time = 0.00, size = 21, normalized size = 0.75 \begin {gather*} \frac {x (c x)^m \sqrt {b x^2}}{2+m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*x)^m*Sqrt[b*x^2])/(2 + m)

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Maple [A]
time = 0.01, size = 20, normalized size = 0.71

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.30, size = 18, normalized size = 0.64 \begin {gather*} \frac {\sqrt {b} c^{m} x^{2} x^{m}}{m + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b)*c^m*x^2*x^m/(m + 2)

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Fricas [A]
time = 0.36, size = 19, normalized size = 0.68 \begin {gather*} \frac {\sqrt {b x^{2}} \left (c x\right )^{m} x}{m + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b*x^2)*(c*x)^m*x/(m + 2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {x \sqrt {b x^{2}} \left (c x\right )^{m}}{m + 2} & \text {for}\: m \neq -2 \\\frac {\int \frac {\sqrt {b x^{2}}}{x^{2}}\, dx}{c^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)), (Integral(sqrt(b*x**2)/x**2, x)/c**2, True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 0.94, size = 20, normalized size = 0.71 \begin {gather*} \frac {\sqrt {b}\,x\,{\left (c\,x\right )}^m\,\sqrt {x^2}}{m+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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