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

Optimal. Leaf size=65 \[ \frac{\sqrt{c+d x^2}}{b}-\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0544701, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {444, 50, 63, 208} \[ \frac{\sqrt{c+d x^2}}{b}-\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 444

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0255953, size = 65, normalized size = 1. \[ \frac{\sqrt{c+d x^2}}{b}-\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 936, normalized size = 14.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*((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/b^2*d^(1/2)*(-a
*b)^(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/b^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)))*a*d-1/2/b/(-(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*((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/b^2*d^(1/2)*(-a*b)^(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/b^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)))*a*d-1/2/b/(-(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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.43189, size = 539, normalized size = 8.29 \begin{align*} \left [\frac{\sqrt{\frac{b c - a d}{b}} \log \left (\frac{b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \,{\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \,{\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, \sqrt{d x^{2} + c}}{4 \, b}, -\frac{\sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b c - a d}{b}}}{2 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, \sqrt{d x^{2} + c}}{2 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.42645, size = 61, normalized size = 0.94 \begin{align*} \frac{2 \left (\frac{d \sqrt{c + d x^{2}}}{2 b} - \frac{d \left (a d - b c\right ) \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{2 b^{2} \sqrt{\frac{a d - b c}{b}}}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*sqrt(c + d*x**2)/(2*b) - d*(a*d - b*c)*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*b**2*sqrt((a*d - b*c
)/b)))/d

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Giac [A]  time = 1.11184, size = 86, normalized size = 1.32 \begin{align*} \frac{{\left (b c - a d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b} + \frac{\sqrt{d x^{2} + c}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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