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

Optimal. Leaf size=113 \[ -\frac{3 d}{2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{1}{2 \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}+\frac{3 \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{5/2}} \]

[Out]

(-3*d)/(2*(b*c - a*d)^2*Sqrt[c + d*x^2]) - 1/(2*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) + (3*Sqrt[b]*d*ArcTan
h[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.0852439, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {444, 51, 63, 208} \[ -\frac{3 d}{2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{1}{2 \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}+\frac{3 \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d)/(2*(b*c - a*d)^2*Sqrt[c + d*x^2]) - 1/(2*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) + (3*Sqrt[b]*d*ArcTan
h[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*(b*c - a*d)^(5/2))

Rule 444

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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{x}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt{c+d x^2}}-\frac{(3 d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 (b c-a d)}\\ &=-\frac{3 d}{2 (b c-a d)^2 \sqrt{c+d x^2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt{c+d x^2}}-\frac{(3 b d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{4 (b c-a d)^2}\\ &=-\frac{3 d}{2 (b c-a d)^2 \sqrt{c+d x^2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt{c+d x^2}}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{2 (b c-a d)^2}\\ &=-\frac{3 d}{2 (b c-a d)^2 \sqrt{c+d x^2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \sqrt{c+d x^2}}+\frac{3 \sqrt{b} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0160044, size = 52, normalized size = 0.46 \[ -\frac{d \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{b \left (d x^2+c\right )}{a d-b c}\right )}{\sqrt{c+d x^2} (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Hypergeometric2F1[-1/2, 2, 1/2, -((b*(c + d*x^2))/(-(b*c) + a*d))])/((-(b*c) + a*d)^2*Sqrt[c + d*x^2]))

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Maple [B]  time = 0.011, size = 989, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.04625, size = 1110, normalized size = 9.82 \begin{align*} \left [\frac{3 \,{\left (b d^{2} x^{4} + a c d +{\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \,{\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \,{\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} +{\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c} \sqrt{\frac{b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \,{\left (3 \, b d x^{2} + b c + 2 \, a d\right )} \sqrt{d x^{2} + c}}{8 \,{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}, -\frac{3 \,{\left (b d^{2} x^{4} + a c d +{\left (b c d + a d^{2}\right )} x^{2}\right )} \sqrt{-\frac{b}{b c - a d}} \arctan \left (\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{d x^{2} + c} \sqrt{-\frac{b}{b c - a d}}}{2 \,{\left (b d x^{2} + b c\right )}}\right ) + 2 \,{\left (3 \, b d x^{2} + b c + 2 \, a d\right )} \sqrt{d x^{2} + c}}{4 \,{\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{4} +{\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.12848, size = 203, normalized size = 1.8 \begin{align*} -\frac{1}{2} \, d{\left (\frac{3 \, b \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d}} + \frac{3 \,{\left (d x^{2} + c\right )} b - 2 \, b c + 2 \, a d}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left ({\left (d x^{2} + c\right )}^{\frac{3}{2}} b - \sqrt{d x^{2} + c} b c + \sqrt{d x^{2} + c} a d\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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