3.4.34 \(\int \sqrt {x^{2 (-1+n)} (a+b x^n)} \, dx\)

Optimal. Leaf size=44 \[ \frac {2 x^{3 (1-n)} \left (a x^{-2 (1-n)}+b x^{3 n-2}\right )^{3/2}}{3 b n} \]

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Rubi [A]  time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1979, 2000} \begin {gather*} \frac {2 x^{3 (1-n)} \left (a x^{-2 (1-n)}+b x^{3 n-2}\right )^{3/2}}{3 b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^(2*(-1 + n))*(a + b*x^n)],x]

[Out]

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

Rule 1979

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedBinomialQ[u, x] &&  !Gene
ralizedBinomialMatchQ[u, x]

Rule 2000

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rubi steps

\begin {align*} \int \sqrt {x^{2 (-1+n)} \left (a+b x^n\right )} \, dx &=\int \sqrt {a x^{2 (-1+n)}+b x^{2 (-1+n)+n}} \, dx\\ &=\frac {2 x^{3 (1-n)} \left (a x^{-2 (1-n)}+b x^{-2+3 n}\right )^{3/2}}{3 b n}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.82 \begin {gather*} \frac {2 x^{3-3 n} \left (x^{2 n-2} \left (a+b x^n\right )\right )^{3/2}}{3 b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x^(2*(-1 + n))*(a + b*x^n)],x]

[Out]

(2*x^(3 - 3*n)*(x^(-2 + 2*n)*(a + b*x^n))^(3/2))/(3*b*n)

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IntegrateAlgebraic [A]  time = 0.07, size = 43, normalized size = 0.98 \begin {gather*} \frac {2 x^{n-1} \left (a+b x^n\right )^2}{3 b n \sqrt {x^{2 n-2} \left (a+b x^n\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x^(2*(-1 + n))*(a + b*x^n)],x]

[Out]

(2*x^(-1 + n)*(a + b*x^n)^2)/(3*b*n*Sqrt[x^(-2 + 2*n)*(a + b*x^n)])

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fricas [A]  time = 0.41, size = 44, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (b x x^{n} + a x\right )} \sqrt {\frac {b x^{3 \, n} + a x^{2 \, n}}{x^{2}}}}{3 \, b n x^{n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x*x^n + a*x)*sqrt((b*x^(3*n) + a*x^(2*n))/x^2)/(b*n*x^n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {{\left (b x^{n} + a\right )} x^{2 \, n - 2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt((b*x^n + a)*x^(2*n - 2)), x)

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maple [A]  time = 0.19, size = 40, normalized size = 0.91 \begin {gather*} \frac {2 \sqrt {\frac {\left (b \,x^{n}+a \right ) x^{2 n}}{x^{2}}}\, \left (b \,x^{n}+a \right ) x \,x^{-n}}{3 b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.55, size = 17, normalized size = 0.39 \begin {gather*} \frac {2 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}}}{3 \, b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x^n + a)^(3/2)/(b*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \sqrt {x^{2\,n-2}\,\left (a+b\,x^n\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**(-2+2*n)*(a+b*x**n))**(1/2),x)

[Out]

Timed out

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