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

Optimal. Leaf size=89 \[ \frac{c \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 \sqrt{a} (b c-a d)^{3/2}}-\frac{x \sqrt{c+d x^2}}{2 \left (a+b x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0533037, antiderivative size = 89, 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} \[ \frac{c \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 \sqrt{a} (b c-a d)^{3/2}}-\frac{x \sqrt{c+d x^2}}{2 \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.424221, size = 124, normalized size = 1.39 \[ \frac{\sqrt{c+d x^2} \left (-\frac{x^2 (b c-a d)}{a+b x^2}-\frac{c \sqrt{x^2 \left (\frac{d}{c}-\frac{b}{a}\right )} \tanh ^{-1}\left (\frac{\sqrt{x^2 \left (\frac{d}{c}-\frac{b}{a}\right )}}{\sqrt{\frac{d x^2}{c}+1}}\right )}{\sqrt{\frac{d x^2}{c}+1}}\right )}{2 x (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 817, normalized size = 9.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/b/(a*d-b*c)/(x+1/b*(-a*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)+1/4/b^2*d*(-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)))+1/4/(-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)))-1/4/(-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)))+1/4/b/(a*d-b*c)/(x-1/b*(-a*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)-1/4/b^2*d*(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (b x^{2} + a\right )}^{2} \sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.92807, size = 871, normalized size = 9.79 \begin{align*} \left [-\frac{4 \,{\left (a b c - a^{2} d\right )} \sqrt{d x^{2} + c} x -{\left (b c x^{2} + a c\right )} \sqrt{-a b c + a^{2} 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 )}{8 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} +{\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}}, -\frac{2 \,{\left (a b c - a^{2} d\right )} \sqrt{d x^{2} + c} x -{\left (b c x^{2} + a c\right )} \sqrt{a b c - a^{2} 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 )}{4 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} +{\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b 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)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*(4*(a*b*c - a^2*d)*sqrt(d*x^2 + c)*x - (b*c*x^2 + a*c)*sqrt(-a*b*c + a^2*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)))/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (a*b^3*c^2 - 2*a^2*b^2*c
*d + a^3*b*d^2)*x^2), -1/4*(2*(a*b*c - a^2*d)*sqrt(d*x^2 + c)*x - (b*c*x^2 + a*c)*sqrt(a*b*c - a^2*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)))/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^2)]

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Sympy [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)**2/(d*x**2+c)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [B]  time = 3.1892, size = 312, normalized size = 3.51 \begin{align*} \frac{c \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 )}{2 \, \sqrt{a b c d - a^{2} d^{2}}{\left (b c - a d\right )}} + \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c \sqrt{d} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d^{\frac{3}{2}} - b c^{2} \sqrt{d}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}{\left (b^{2} c - a b d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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