3.529 \(\int \frac{(d x)^m}{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\)

Optimal. Leaf size=76 \[ \frac{(d x)^{m+1} \left (a+b x^n\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a d (m+1) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

((d*x)^(1 + m)*(a + b*x^n)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*d*(1 + m)*Sqrt[a^2
+ 2*a*b*x^n + b^2*x^(2*n)])

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Rubi [A]  time = 0.0389876, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {1355, 364} \[ \frac{(d x)^{m+1} \left (a+b x^n\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a d (m+1) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)],x]

[Out]

((d*x)^(1 + m)*(a + b*x^n)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a*d*(1 + m)*Sqrt[a^2
+ 2*a*b*x^n + b^2*x^(2*n)])

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(d x)^m}{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \, dx &=\frac{\left (a b+b^2 x^n\right ) \int \frac{(d x)^m}{a b+b^2 x^n} \, dx}{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{(d x)^{1+m} \left (a+b x^n\right ) \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{b x^n}{a}\right )}{a d (1+m) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ \end{align*}

Mathematica [A]  time = 0.0224633, size = 62, normalized size = 0.82 \[ \frac{x (d x)^m \left (a+b x^n\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+1}{n}+1;-\frac{b x^n}{a}\right )}{a (m+1) \sqrt{\left (a+b x^n\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m/Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)],x]

[Out]

(x*(d*x)^m*(a + b*x^n)*Hypergeometric2F1[1, (1 + m)/n, 1 + (1 + m)/n, -((b*x^n)/a)])/(a*(1 + m)*Sqrt[(a + b*x^
n)^2])

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx \right ) ^{m}{\frac{1}{\sqrt{{a}^{2}+2\,ab{x}^{n}+{b}^{2}{x}^{2\,n}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{\sqrt{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d x\right )^{m}}{\sqrt{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{\sqrt{\left (a + b x^{n}\right )^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m/sqrt((a + b*x**n)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{\sqrt{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^m/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)