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

Optimal. Leaf size=44 \[ \frac {(d x)^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a^2 d (m+1)} \]

[Out]

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

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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 = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {28, 364} \[ \frac {(d x)^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a^2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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}{a^2+2 a b x^2+b^2 x^4} \, dx &=b^2 \int \frac {(d x)^m}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=\frac {(d x)^{1+m} \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a^2 d (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 0.95 \[ \frac {x (d x)^m \, _2F_1\left (2,\frac {m+1}{2};\frac {m+1}{2}+1;-\frac {b x^2}{a}\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d x\right )^{m}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

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),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\,{d x} \]

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),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{b^{2} x^{4}+2 a b \,x^{2}+a^{2}}\, dx \]

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),x)

[Out]

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

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

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),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (d\,x\right )}^m}{a^2+2\,a\,b\,x^2+b^2\,x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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),x)

[Out]

Integral((d*x)**m/(a + b*x**2)**2, x)

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