3.536 \(\int \frac {(d x)^m}{(a^2+2 a b x^n+b^2 x^{2 n})^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac {(d x)^{m+1} \left (a+b x^n\right ) \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a^3 d (m+1) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

(d*x)^(1+m)*(a+b*x^n)*hypergeom([3, (1+m)/n],[(1+m+n)/n],-b*x^n/a)/a^3/d/(1+m)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/
2)

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Rubi [A]  time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, 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} \[ \frac {(d x)^{m+1} \left (a+b x^n\right ) \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a^3 d (m+1) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]

[Out]

((d*x)^(1 + m)*(a + b*x^n)*Hypergeometric2F1[3, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a^3*d*(1 + m)*Sqrt[a
^2 + 2*a*b*x^n + b^2*x^(2*n)])

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.80 \[ \frac {x (d x)^m \left (a+b x^n\right ) \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {b x^n}{a}\right )}{a^3 (m+1) \sqrt {\left (a+b x^n\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2),x]

[Out]

(x*(d*x)^m*(a + b*x^n)*Hypergeometric2F1[3, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a^3*(1 + m)*Sqrt[(a + b*
x^n)^2])

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}} \left (d x\right )^{m}}{b^{4} x^{4 \, n} + 4 \, a^{2} b^{2} x^{2 \, n} + 4 \, a^{3} b x^{n} + a^{4} + 2 \, {\left (2 \, a b^{3} x^{n} + a^{2} b^{2}\right )} x^{2 \, n}}, x\right ) \]

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))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2)*(d*x)^m/(b^4*x^(4*n) + 4*a^2*b^2*x^(2*n) + 4*a^3*b*x^n + a^4 + 2*
(2*a*b^3*x^n + a^2*b^2)*x^(2*n)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

integrate((d*x)^m/(b^2*x^(2*n) + 2*a*b*x^n + a^2)^(3/2), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{\left (2 a b \,x^{n}+b^{2} x^{2 n}+a^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(2*a*b*x^n+b^2*x^(2*n)+a^2)^(3/2),x)

[Out]

int((d*x)^m/(2*a*b*x^n+b^2*x^(2*n)+a^2)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (m^{2} - m {\left (3 \, n - 2\right )} + 2 \, n^{2} - 3 \, n + 1\right )} d^{m} \int \frac {x^{m}}{2 \, {\left (a^{2} b n^{2} x^{n} + a^{3} n^{2}\right )}}\,{d x} - \frac {a d^{m} {\left (m - 3 \, n + 1\right )} x x^{m} + b d^{m} {\left (m - 2 \, n + 1\right )} x e^{\left (m \log \relax (x) + n \log \relax (x)\right )}}{2 \, {\left (a^{2} b^{2} n^{2} x^{2 \, n} + 2 \, a^{3} b n^{2} x^{n} + a^{4} n^{2}\right )}} \]

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))^(3/2),x, algorithm="maxima")

[Out]

(m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*d^m*integrate(1/2*x^m/(a^2*b*n^2*x^n + a^3*n^2), x) - 1/2*(a*d^m*(m - 3*
n + 1)*x*x^m + b*d^m*(m - 2*n + 1)*x*e^(m*log(x) + n*log(x)))/(a^2*b^2*n^2*x^(2*n) + 2*a^3*b*n^2*x^n + a^4*n^2
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,x\right )}^m}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2),x)

[Out]

int((d*x)^m/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{\left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

Integral((d*x)**m/((a + b*x**n)**2)**(3/2), x)

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