3.1.64 \(\int \frac {(c x)^m}{\sqrt {b x^2}} \, dx\)

Optimal. Leaf size=19 \[ \frac {x (c x)^m}{m \sqrt {b x^2}} \]

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Rubi [A]  time = 0.00, antiderivative size = 19, 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 {x (c x)^m}{m \sqrt {b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 \frac {(c x)^m}{\sqrt {b x^2}} \, dx &=\frac {x \int \frac {(c x)^m}{x} \, dx}{\sqrt {b x^2}}\\ &=\frac {(c x) \int (c x)^{-1+m} \, dx}{\sqrt {b x^2}}\\ &=\frac {x (c x)^m}{m \sqrt {b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} \frac {x (c x)^m}{m \sqrt {b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 24, normalized size = 1.26 \begin {gather*} \frac {\sqrt {b x^2} (c x)^m}{b m x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 22, normalized size = 1.16 \begin {gather*} \frac {\sqrt {b x^{2}} \left (c x\right )^{m}}{b m x} \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/(b*m*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x\right )^{m}}{\sqrt {b x^{2}}}\,{d x} \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]

integrate((c*x)^m/sqrt(b*x^2), x)

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maple [A]  time = 0.00, size = 18, normalized size = 0.95 \begin {gather*} \frac {x \left (c x \right )^{m}}{\sqrt {b \,x^{2}}\, m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.42, size = 13, normalized size = 0.68 \begin {gather*} \frac {c^{m} x^{m}}{\sqrt {b} m} \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]

c^m*x^m/(sqrt(b)*m)

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mupad [B]  time = 0.96, size = 18, normalized size = 0.95 \begin {gather*} \frac {x\,{\left (c\,x\right )}^m}{\sqrt {b}\,m\,\sqrt {x^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]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {c^{m} x x^{m}}{\sqrt {b} m \sqrt {x^{2}}} & \text {for}\: m \neq 0 \\\int \frac {1}{\sqrt {b x^{2}}}\, dx & \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((c**m*x*x**m/(sqrt(b)*m*sqrt(x**2)), Ne(m, 0)), (Integral(1/sqrt(b*x**2), x), True))

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