3.715 \(\int \frac{x^2}{(a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=74 \[ \frac{x}{\sqrt{c+d x^2} (b c-a d)}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.049263, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {471, 12, 377, 205} \[ \frac{x}{\sqrt{c+d x^2} (b c-a d)}-\frac{\sqrt{a} \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

Rule 471

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 + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.320419, size = 111, normalized size = 1.5 \[ \frac{x^2 (b c-a d)+a c \sqrt{\frac{d x^2}{c}+1} \sqrt{x^2 \left (\frac{d}{c}-\frac{b}{a}\right )} \tanh ^{-1}\left (\frac{\sqrt{x^2 \left (\frac{d}{c}-\frac{b}{a}\right )}}{\sqrt{\frac{d x^2}{c}+1}}\right )}{x \sqrt{c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/((a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

((b*c - a*d)*x^2 + a*c*Sqrt[(-(b/a) + d/c)*x^2]*Sqrt[1 + (d*x^2)/c]*ArcTanh[Sqrt[(-(b/a) + d/c)*x^2]/Sqrt[1 +
(d*x^2)/c]])/((b*c - a*d)^2*x*Sqrt[c + d*x^2])

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Maple [B]  time = 0.011, size = 653, normalized size = 8.8 \begin{align*}{\frac{x}{bc}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{a}{2\,ad-2\,bc}{\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{adx}{2\, \left ( ad-bc \right ) bc}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}+{\frac{a}{2\,ad-2\,bc}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}+{\frac{a}{2\,ad-2\,bc}{\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{adx}{2\, \left ( ad-bc \right ) bc}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{a}{2\,ad-2\,bc}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2+a)/(d*x^2+c)^(3/2),x)

[Out]

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

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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/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88482, size = 698, normalized size = 9.43 \begin{align*} \left [-\frac{{\left (d x^{2} + c\right )} \sqrt{-\frac{a}{b c - a 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^{2} c^{2} - 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{3} -{\left (a b c^{2} - a^{2} c d\right )} x\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{a}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, \sqrt{d x^{2} + c} x}{4 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}}, \frac{{\left (d x^{2} + c\right )} \sqrt{\frac{a}{b c - a d}} \arctan \left (-\frac{{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c} \sqrt{\frac{a}{b c - a d}}}{2 \,{\left (a d x^{3} + a c x\right )}}\right ) + 2 \, \sqrt{d x^{2} + c} x}{2 \,{\left (b c^{2} - a c d +{\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**2+a)/(d*x**2+c)**(3/2),x)

[Out]

Integral(x**2/((a + b*x**2)*(c + d*x**2)**(3/2)), x)

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Giac [A]  time = 1.18859, size = 139, normalized size = 1.88 \begin{align*} -\frac{a \sqrt{d} \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}}{\left (b c - a d\right )}} + \frac{x}{\sqrt{d x^{2} + c}{\left (b c - a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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