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

Optimal. Leaf size=69 \[ \frac {\left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{2 d}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 \sqrt {a} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, 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, 47, 63, 208} \[ \frac {\left (c+d x^2\right ) \sqrt {a+\frac {b}{c+d x^2}}}{2 d}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 \sqrt {a} d} \]

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

Rule 47

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

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 77, normalized size = 1.12 \[ \frac {\sqrt {a} \left (c+d x^2\right ) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}+b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{c+d x^2}}}{\sqrt {a}}\right )}{2 \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 267, normalized size = 3.87 \[ \left [\frac {\sqrt {a} b \log \left (8 \, a^{2} d^{2} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (2 \, a^{2} c + a b\right )} d x^{2} + 8 \, a b c + b^{2} + 4 \, {\left (2 \, a d^{2} x^{4} + {\left (4 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + b c\right )} \sqrt {a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right ) + 4 \, {\left (a d x^{2} + a c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{8 \, a d}, -\frac {\sqrt {-a} b \arctan \left (\frac {{\left (2 \, a d x^{2} + 2 \, a c + b\right )} \sqrt {-a} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} d x^{2} + a^{2} c + a b\right )}}\right ) - 2 \, {\left (a d x^{2} + a c\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{4 \, a d}\right ] \]

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*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*d)]

________________________________________________________________________________________

giac [B]  time = 0.51, size = 127, normalized size = 1.84 \[ -\frac {1}{4} \, {\left (\frac {b \log \left ({\left | -8 \, a^{\frac {3}{2}} c d - 8 \, {\left (\sqrt {a d^{2}} x^{2} - \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}\right )} a {\left | d \right |} - 4 \, \sqrt {a} b d \right |}\right )}{\sqrt {a} {\left | d \right |}} - \frac {2 \, \sqrt {a d^{2} x^{4} + 2 \, a c d x^{2} + b d x^{2} + a c^{2} + b c}}{d}\right )} \mathrm {sgn}\left (d x^{2} + c\right ) \]

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(-8*a^(3/2)*c*d - 8*(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
*abs(d) - 4*sqrt(a)*b*d))/(sqrt(a)*abs(d)) - 2*sqrt(a*d^2*x^4 + 2*a*c*d*x^2 + b*d*x^2 + a*c^2 + b*c)/d)*sgn(d*
x^2 + c)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 180, normalized size = 2.61 \[ \frac {\sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (b d \ln \left (\frac {2 a \,d^{2} x^{2}+2 a c d +b d +2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}}{2 \sqrt {a \,d^{2}}}\right )+2 \sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}\, \sqrt {a \,d^{2}}\right )}{4 \sqrt {\left (d \,x^{2}+c \right ) \left (a d \,x^{2}+a c +b \right )}\, \sqrt {a \,d^{2}}\, d} \]

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

________________________________________________________________________________________

maxima [B]  time = 1.64, size = 126, normalized size = 1.83 \[ -\frac {b \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a d - \frac {{\left (a d x^{2} + a c + b\right )} d}{d x^{2} + c}\right )}} - \frac {b \log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{4 \, \sqrt {a} d} \]

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]

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

________________________________________________________________________________________

mupad [B]  time = 3.01, size = 120, normalized size = 1.74 \[ \frac {\sqrt {\frac {b\,\left (d\,x^2+c\right )+a\,{\left (d\,x^2+c\right )}^2}{{\left (d\,x^2+c\right )}^2}}\,\left (d\,x^2+c\right )\,\left (\frac {b\,\ln \left (\frac {\frac {b}{2}+a\,\left (d\,x^2+c\right )+\sqrt {a}\,\sqrt {b\,\left (d\,x^2+c\right )+a\,{\left (d\,x^2+c\right )}^2}}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b\,\left (d\,x^2+c\right )+a\,{\left (d\,x^2+c\right )}^2}}+2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/(c + d*x^2))^(1/2),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {\frac {a c + a d x^{2} + b}{c + d x^{2}}}\, dx \]

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)

________________________________________________________________________________________