3.796 \(\int \frac{(d x)^m}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{\left (a+b x^2\right ) (d x)^{m+1} \, _2F_1\left (5,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^5 d (m+1) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

((d*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[5, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^5*d*(1 + m)*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0327895, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {1112, 364} \[ \frac{\left (a+b x^2\right ) (d x)^{m+1} \, _2F_1\left (5,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^5 d (m+1) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

((d*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[5, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^5*d*(1 + m)*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && 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}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x^2\right )\right ) \int \frac{(d x)^m}{\left (a b+b^2 x^2\right )^5} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(d x)^{1+m} \left (a+b x^2\right ) \, _2F_1\left (5,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{a^5 d (1+m) \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0189352, size = 60, normalized size = 0.82 \[ \frac{x \left (a+b x^2\right ) (d x)^m \, _2F_1\left (5,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^5 (m+1) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(x*(d*x)^m*(a + b*x^2)*Hypergeometric2F1[5, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^5*(1 + m)*Sqrt[(a + b*x^2)
^2])

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Maple [F]  time = 0.232, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx \right ) ^{m} \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

int((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

integrate((d*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}} \left (d x\right )^{m}}{b^{6} x^{12} + 6 \, a b^{5} x^{10} + 15 \, a^{2} b^{4} x^{8} + 20 \, a^{3} b^{3} x^{6} + 15 \, a^{4} b^{2} x^{4} + 6 \, a^{5} b x^{2} + a^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)*(d*x)^m/(b^6*x^12 + 6*a*b^5*x^10 + 15*a^2*b^4*x^8 + 20*a^3*b^3*x^6 +
15*a^4*b^2*x^4 + 6*a^5*b*x^2 + a^6), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral((d*x)**m/((a + b*x**2)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

integrate((d*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2)^(5/2), x)