3.8.94 \(\int \frac {(d x)^m}{\sqrt {a^2+2 a b x^2+b^2 x^4}} \, dx\) [794]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 = {1126, 371} \begin {gather*} \frac {\left (a+b x^2\right ) (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a d (m+1) \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 1126

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 62, normalized size = 0.85 \begin {gather*} \frac {x (d x)^m \left (a+b x^2\right ) \, _2F_1\left (1,\frac {1+m}{2};1+\frac {1+m}{2};-\frac {b x^2}{a}\right )}{a (1+m) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d x \right )^{m}}{\sqrt {b^{2} x^{4}+2 a b \,x^{2}+a^{2}}}\, dx\]

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)^(1/2),x)

[Out]

int((d*x)^m/(b^2*x^4+2*a*b*x^2+a^2)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(1/2),x, algorithm="fricas")

[Out]

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

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

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)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(1/2),x, algorithm="giac")

[Out]

integrate((d*x)^m/sqrt(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 = -0.01 \begin {gather*} \int \frac {{\left (d\,x\right )}^m}{\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}} \,d x \end {gather*}

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)^(1/2),x)

[Out]

int((d*x)^m/(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2), x)

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