3.2.65 \(\int \frac {(c x)^m}{\sqrt {b x^n}} \, dx\) [165]

Optimal. Leaf size=27 \[ \frac {2 x (c x)^m}{(2+2 m-n) \sqrt {b x^n}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 27, 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, 20, 30} \begin {gather*} \frac {2 x (c x)^m}{(2 m-n+2) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(c*x)^m)/((2 + 2*m - n)*Sqrt[b*x^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 20

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

Rule 30

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

Rubi steps

\begin {align*} \int \frac {(c x)^m}{\sqrt {b x^n}} \, dx &=\frac {x^{n/2} \int x^{-n/2} (c x)^m \, dx}{\sqrt {b x^n}}\\ &=\frac {\left (x^{-m+\frac {n}{2}} (c x)^m\right ) \int x^{m-\frac {n}{2}} \, dx}{\sqrt {b x^n}}\\ &=\frac {2 x (c x)^m}{(2+2 m-n) \sqrt {b x^n}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 26, normalized size = 0.96 \begin {gather*} \frac {x (c x)^m}{\left (1+m-\frac {n}{2}\right ) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*x)^m)/((1 + m - n/2)*Sqrt[b*x^n])

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Maple [A]
time = 0.03, size = 26, normalized size = 0.96

method	result	size
gosper	\(\frac {2 x \left (c x \right )^{m}}{\left (2+2 m -n \right ) \sqrt {b \,x^{n}}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 27, normalized size = 1.00 \begin {gather*} \frac {2 \, c^{m} x x^{m}}{\sqrt {b} {\left (2 \, m - n + 2\right )} \sqrt {x^{n}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c^m*x*x^m/(sqrt(b)*(2*m - n + 2)*sqrt(x^n))

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Fricas [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^n)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {2 x \left (c x\right )^{m}}{2 m \sqrt {b x^{n}} - n \sqrt {b x^{n}} + 2 \sqrt {b x^{n}}} & \text {for}\: m \neq \frac {n}{2} - 1 \\\int \frac {\left (c x\right )^{\frac {n}{2} - 1}}{\sqrt {b x^{n}}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*x*(c*x)**m/(2*m*sqrt(b*x**n) - n*sqrt(b*x**n) + 2*sqrt(b*x**n)), Ne(m, n/2 - 1)), (Integral((c*x)
**(n/2 - 1)/sqrt(b*x**n), x), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^m/sqrt(b*x^n), x)

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Mupad [B]
time = 1.03, size = 34, normalized size = 1.26 \begin {gather*} \frac {2\,x^{1-n}\,\sqrt {b\,x^n}\,{\left (c\,x\right )}^m}{b\,\left (2\,m-n+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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