3.724 \(\int \frac{x^2}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=115 \[ \frac{x (a d+2 b c)}{3 c \sqrt{c+d x^2} (b c-a d)^2}+\frac{x}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{\sqrt{a} b \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{5/2}} \]

[Out]

x/(3*(b*c - a*d)*(c + d*x^2)^(3/2)) + ((2*b*c + a*d)*x)/(3*c*(b*c - a*d)^2*Sqrt[c + d*x^2]) - (Sqrt[a]*b*ArcTa
n[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(b*c - a*d)^(5/2)

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Rubi [A]  time = 0.0875847, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {471, 527, 12, 377, 205} \[ \frac{x (a d+2 b c)}{3 c \sqrt{c+d x^2} (b c-a d)^2}+\frac{x}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{\sqrt{a} b \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

x/(3*(b*c - a*d)*(c + d*x^2)^(3/2)) + ((2*b*c + a*d)*x)/(3*c*(b*c - a*d)^2*Sqrt[c + d*x^2]) - (Sqrt[a]*b*ArcTa
n[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(b*c - a*d)^(5/2)

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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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 )^{5/2}} \, dx &=\frac{x}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{a-2 b x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{3 (b c-a d)}\\ &=\frac{x}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{(2 b c+a d) x}{3 c (b c-a d)^2 \sqrt{c+d x^2}}-\frac{\int \frac{3 a b c}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{3 c (b c-a d)^2}\\ &=\frac{x}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{(2 b c+a d) x}{3 c (b c-a d)^2 \sqrt{c+d x^2}}-\frac{(a b) \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{(b c-a d)^2}\\ &=\frac{x}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{(2 b c+a d) x}{3 c (b c-a d)^2 \sqrt{c+d x^2}}-\frac{(a b) \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)^2}\\ &=\frac{x}{3 (b c-a d) \left (c+d x^2\right )^{3/2}}+\frac{(2 b c+a d) x}{3 c (b c-a d)^2 \sqrt{c+d x^2}}-\frac{\sqrt{a} b \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{(b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 2.83197, size = 257, normalized size = 2.23 \[ \frac{12 x^6 \left (c+d x^2\right ) (b c-a d)^3 \, _2F_1\left (2,2;\frac{9}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )-\frac{35 c \left (a+b x^2\right ) \left (5 c+2 d x^2\right ) \left (-3 a^2 \left (c+d x^2\right )^2 \sin ^{-1}\left (\sqrt{\frac{x^2 (b c-a d)}{c \left (a+b x^2\right )}}\right )-c \left (a+b x^2\right ) \sqrt{\frac{a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}} \left (-3 a c-4 a d x^2+b c x^2\right )\right )}{\sqrt{\frac{a x^2 \left (c+d x^2\right ) (b c-a d)}{c^2 \left (a+b x^2\right )^2}}}}{315 c^3 x \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/((a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

((-35*c*(a + b*x^2)*(5*c + 2*d*x^2)*(-(c*(a + b*x^2)*Sqrt[(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)]
*(-3*a*c + b*c*x^2 - 4*a*d*x^2)) - 3*a^2*(c + d*x^2)^2*ArcSin[Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]]))/Sqrt[
(a*(b*c - a*d)*x^2*(c + d*x^2))/(c^2*(a + b*x^2)^2)] + 12*(b*c - a*d)^3*x^6*(c + d*x^2)*Hypergeometric2F1[2, 2
, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(315*c^3*(b*c - a*d)^2*x*(a + b*x^2)^2*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.011, size = 1134, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

1/3/b*x/c/(d*x^2+c)^(3/2)+2/3/b/c^2*x/(d*x^2+c)^(1/2)-1/6*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)^(3/2)-1/6*a/b*d/(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)^(3/2)*x-1/3*a/b*d/(a*d-b*c)/c^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+1/2*a/(-a*b)^(1/2)*b/(a*d-b*c)^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/(a*d-b*c)^2/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)*b/(a*d-b*c)^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)))+1/6*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)^(3/2)-1/6*a/b*d/(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)^(3/2)*x-1/3*a/b*d/(a*d-b*c
)/c^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-1/2*a/(-a*b)^(1/2
)*b/(a*d-b*c)^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/(a*
d-b*c)^2/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)*b/(a*d-b*c)^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)))

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.68625, size = 1130, normalized size = 9.83 \begin{align*} \left [\frac{3 \,{\left (b c d^{2} x^{4} + 2 \, b c^{2} d x^{2} + b c^{3}\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 \,{\left (3 \, b c^{2} x +{\left (2 \, b c d + a d^{2}\right )} x^{3}\right )} \sqrt{d x^{2} + c}}{12 \,{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} +{\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \,{\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} x^{2}\right )}}, \frac{3 \,{\left (b c d^{2} x^{4} + 2 \, b c^{2} d x^{2} + b c^{3}\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 \,{\left (3 \, b c^{2} x +{\left (2 \, b c d + a d^{2}\right )} x^{3}\right )} \sqrt{d x^{2} + c}}{6 \,{\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2} +{\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{4} + 2 \,{\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b*c*d^2*x^4 + 2*b*c^2*d*x^2 + b*c^3)*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*(3*b*c^2*x + (2*b*c*d + a*d^2)*x^3)*sq
rt(d*x^2 + c))/(b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^4 + 2*(b^2*c
^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x^2), 1/6*(3*(b*c*d^2*x^4 + 2*b*c^2*d*x^2 + b*c^3)*sqrt(a/(b*c - a*d))*arc
tan(-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*(3*b*c^2*x + (2*
b*c*d + a*d^2)*x^3)*sqrt(d*x^2 + c))/(b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^2 + (b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2
*c*d^4)*x^4 + 2*(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*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{5}{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)**(5/2),x)

[Out]

Integral(x**2/((a + b*x**2)*(c + d*x**2)**(5/2)), x)

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Giac [B]  time = 1.14976, size = 393, normalized size = 3.42 \begin{align*} \frac{a b \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 )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a b c d - a^{2} d^{2}}} + \frac{{\left (\frac{{\left (2 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} + a^{3} d^{5}\right )} x^{2}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}} + \frac{3 \,{\left (b^{3} c^{4} d - 2 \, a b^{2} c^{3} d^{2} + a^{2} b c^{2} d^{3}\right )}}{b^{4} c^{5} d - 4 \, a b^{3} c^{4} d^{2} + 6 \, a^{2} b^{2} c^{3} d^{3} - 4 \, a^{3} b c^{2} d^{4} + a^{4} c d^{5}}\right )} x}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

a*b*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))/((b^2*c^2 -
2*a*b*c*d + a^2*d^2)*sqrt(a*b*c*d - a^2*d^2)) + 1/3*((2*b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 + a^3*d^5)*x^2/(b^4*c^5*
d - 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5) + 3*(b^3*c^4*d - 2*a*b^2*c^3*d^2 + a^2*
b*c^2*d^3)/(b^4*c^5*d - 4*a*b^3*c^4*d^2 + 6*a^2*b^2*c^3*d^3 - 4*a^3*b*c^2*d^4 + a^4*c*d^5))*x/(d*x^2 + c)^(3/2
)