3.682 \(\int \frac{\sqrt{c+d x^2}}{x^2 (a+b x^2)} \, dx\)

Optimal. Leaf size=70 \[ -\frac{\sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2}}-\frac{\sqrt{c+d x^2}}{a x} \]

[Out]

-(Sqrt[c + d*x^2]/(a*x)) - (Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/a^(3/2)

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Rubi [A]  time = 0.0506841, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {475, 12, 377, 205} \[ -\frac{\sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2}}-\frac{\sqrt{c+d x^2}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d*x^2]/(a*x)) - (Sqrt[b*c - a*d]*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/a^(3/2)

Rule 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.018657, size = 51, normalized size = 0.73 \[ -\frac{\sqrt{c+d x^2} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{(a d-b c) x^2}{a \left (d x^2+c\right )}\right )}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.017, size = 1017, normalized size = 14.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b/a/(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/2/
a*d^(1/2)*ln((-d*(-a*b)^(1/2)/b+(x+1/b*(-a*b)^(1/2))*d)/d^(1/2)+((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(
x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+1/2/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1
/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(
1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*d-1/2*b/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b
-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(
x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))*c-1/2*b/a/(-a*b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d
+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-1/2/a*d^(1/2)*ln((d*(-a*b)^(1/2)/b+(x-1/b*(-a*b)^(
1/2))*d)/d^(1/2)+((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))-1/2/(-a
*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/
2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))
*d+1/2*b/a/(-a*b)^(1/2)/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a
*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*
(-a*b)^(1/2)))*c-1/a/c/x*(d*x^2+c)^(3/2)+1/a*d/c*x*(d*x^2+c)^(1/2)+1/a*d^(1/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.39687, size = 570, normalized size = 8.14 \begin{align*} \left [\frac{x \sqrt{-\frac{b c - a d}{a}} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left (a^{2} c x -{\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, \sqrt{d x^{2} + c}}{4 \, a x}, -\frac{x \sqrt{\frac{b c - a d}{a}} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{a}}}{2 \,{\left ({\left (b c d - a d^{2}\right )} x^{3} +{\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, \sqrt{d x^{2} + c}}{2 \, a x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(x*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*
x^2 + 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)) -
4*sqrt(d*x^2 + c))/(a*x), -1/2*(x*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sq
rt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)) + 2*sqrt(d*x^2 + c))/(a*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.90558, size = 158, normalized size = 2.26 \begin{align*} \frac{{\left (b c \sqrt{d} - a d^{\frac{3}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{\sqrt{a b c d - a^{2} d^{2}} a} + \frac{2 \, c \sqrt{d}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c*sqrt(d) - a*d^(3/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2)
)/(sqrt(a*b*c*d - a^2*d^2)*a) + 2*c*sqrt(d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)*a)