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

Optimal. Leaf size=112 \[ -\frac{\sqrt{a} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^2}+\frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2 \sqrt{d}}+\frac{x \sqrt{c+d x^2}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0962023, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {478, 523, 217, 206, 377, 205} \[ -\frac{\sqrt{a} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{b^2}+\frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 b^2 \sqrt{d}}+\frac{x \sqrt{c+d x^2}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 478

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

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

Mathematica [A]  time = 0.123648, size = 108, normalized size = 0.96 \[ \frac{\frac{(b c-2 a d) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{\sqrt{d}}-2 \sqrt{a} \sqrt{b c-a d} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )+b x \sqrt{c+d x^2}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 1010, normalized size = 9. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98577, size = 1544, normalized size = 13.79 \begin{align*} \left [\frac{2 \, \sqrt{d x^{2} + c} b d x -{\left (b c - 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + \sqrt{-a b c + a^{2} d} d \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 ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b^{2} d}, \frac{2 \, \sqrt{d x^{2} + c} b d x - 2 \,{\left (b c - 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) + \sqrt{-a b c + a^{2} d} d \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 ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{4 \, b^{2} d}, \frac{2 \, \sqrt{d x^{2} + c} b d x - 2 \, \sqrt{a b c - a^{2} d} d \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) -{\left (b c - 2 \, a d\right )} \sqrt{d} \log \left (-2 \, d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right )}{4 \, b^{2} d}, \frac{\sqrt{d x^{2} + c} b d x - \sqrt{a b c - a^{2} d} d \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) -{\left (b c - 2 \, a d\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right )}{2 \, b^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15268, size = 185, normalized size = 1.65 \begin{align*} \frac{\sqrt{d x^{2} + c} x}{2 \, b} + \frac{{\left (a b c \sqrt{d} - a^{2} 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}} b^{2}} - \frac{{\left (b c - 2 \, a d\right )} \log \left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2}\right )}{4 \, b^{2} \sqrt{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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