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

Optimal. Leaf size=82 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{d}-\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} d} \]

[Out]

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

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Rubi [A]  time = 0.0542935, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {402, 217, 206, 377, 208} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{d}-\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/(c + d*x^2),x]

[Out]

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

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

Mathematica [A]  time = 0.0446294, size = 84, normalized size = 1.02 \[ \frac{\sqrt{a d-b c} \tan ^{-1}\left (\frac{x \sqrt{a d-b c}}{\sqrt{c} \sqrt{a+b x^2}}\right )}{\sqrt{c} d}+\frac{\sqrt{b} \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 932, normalized size = 11.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.91823, size = 1288, normalized size = 15.71 \begin{align*} \left [\frac{2 \, \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + \sqrt{\frac{b c - a d}{c}} \log \left (\frac{{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \,{\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} - 4 \,{\left (a c^{2} x +{\left (2 \, b c^{2} - a c d\right )} x^{3}\right )} \sqrt{b x^{2} + a} \sqrt{\frac{b c - a d}{c}}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right )}{4 \, d}, -\frac{4 \, \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - \sqrt{\frac{b c - a d}{c}} \log \left (\frac{{\left (8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \,{\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2} - 4 \,{\left (a c^{2} x +{\left (2 \, b c^{2} - a c d\right )} x^{3}\right )} \sqrt{b x^{2} + a} \sqrt{\frac{b c - a d}{c}}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}\right )}{4 \, d}, \frac{\sqrt{-\frac{b c - a d}{c}} \arctan \left (\frac{{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt{b x^{2} + a} \sqrt{-\frac{b c - a d}{c}}}{2 \,{\left ({\left (b^{2} c - a b d\right )} x^{3} +{\left (a b c - a^{2} d\right )} x\right )}}\right ) + \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right )}{2 \, d}, -\frac{2 \, \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - \sqrt{-\frac{b c - a d}{c}} \arctan \left (\frac{{\left ({\left (2 \, b c - a d\right )} x^{2} + a c\right )} \sqrt{b x^{2} + a} \sqrt{-\frac{b c - a d}{c}}}{2 \,{\left ({\left (b^{2} c - a b d\right )} x^{3} +{\left (a b c - a^{2} d\right )} x\right )}}\right )}{2 \, d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1362, size = 150, normalized size = 1.83 \begin{align*} \frac{{\left (b^{\frac{3}{2}} c - a \sqrt{b} d\right )} \arctan \left (\frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt{-b^{2} c^{2} + a b c d}}\right )}{\sqrt{-b^{2} c^{2} + a b c d} d} - \frac{\sqrt{b} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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