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

Optimal. Leaf size=72 \[ \frac{\left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{2 a d}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{2 a^{3/2} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0520451, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {1591, 242, 51, 63, 208} \[ \frac{\left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{2 a d}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b/(c + d*x^2)],x]

[Out]

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

Rule 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{a+\frac{b}{c+d x^2}}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b}{x}}} \, dx,x,c+d x^2\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{c+d x^2}\right )}{2 d}\\ &=\frac{\left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{2 a d}+\frac{b \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{c+d x^2}\right )}{4 a d}\\ &=\frac{\left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{2 a d}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{c+d x^2}}\right )}{2 a d}\\ &=\frac{\left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}}{2 a d}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{2 a^{3/2} d}\\ \end{align*}

Mathematica [A]  time = 0.0774736, size = 70, normalized size = 0.97 \[ \frac{\sqrt{a} \left (c+d x^2\right ) \sqrt{a+\frac{b}{c+d x^2}}-b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{c+d x^2}}}{\sqrt{a}}\right )}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.007, size = 184, normalized size = 2.6 \begin{align*}{\frac{d{x}^{2}+c}{4\,ad}\sqrt{{\frac{ad{x}^{2}+ac+b}{d{x}^{2}+c}}} \left ( -b\ln \left ({\frac{1}{2} \left ( 2\,a{d}^{2}{x}^{2}+2\,acd+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}}+bd \right ){\frac{1}{\sqrt{a{d}^{2}}}}} \right ) d+2\,\sqrt{a{d}^{2}{x}^{4}+2\,acd{x}^{2}+bd{x}^{2}+{c}^{2}a+bc}\sqrt{a{d}^{2}} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( ad{x}^{2}+ac+b \right ) }}}{\frac{1}{\sqrt{a{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-b*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x
^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d
^2)^(1/2))/((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)/a/d/(a*d^2)^(1/2)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.32526, size = 612, normalized size = 8.5 \begin{align*} \left [\frac{\sqrt{a} b \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \,{\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} - 4 \,{\left (2 \, a d^{2} x^{4} +{\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt{a} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \,{\left (a d x^{2} + a c\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, a^{2} d}, \frac{\sqrt{-a} b \arctan \left (\frac{{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt{-a} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{2 \,{\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) + 2 \,{\left (a d x^{2} + a c\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, a^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.42982, size = 190, normalized size = 2.64 \begin{align*} \frac{b \log \left ({\left | -2 \, a^{\frac{3}{2}} c d - 2 \,{\left (\sqrt{a d^{2}} x^{2} - \sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a{\left | d \right |} - \sqrt{a} b d \right |}\right )}{4 \, a^{\frac{3}{2}}{\left | d \right |} \mathrm{sgn}\left (d x^{2} + c\right )} + \frac{\sqrt{a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}}{2 \, a d \mathrm{sgn}\left (d x^{2} + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*log(abs(-2*a^(3/2)*c*d - 2*(sqrt(a*d^2)*x^2 - sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c))*a*a
bs(d) - sqrt(a)*b*d))/(a^(3/2)*abs(d)*sgn(d*x^2 + c)) + 1/2*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b
*c)/(a*d*sgn(d*x^2 + c))