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

Optimal. Leaf size=140 \[ \frac{5 b^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{7/2}}-\frac{5 b d}{2 \sqrt{c+d x^2} (b c-a d)^3}-\frac{1}{2 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{5 d}{6 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

[Out]

(-5*d)/(6*(b*c - a*d)^2*(c + d*x^2)^(3/2)) - 1/(2*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) - (5*b*d)/(2*(b*c
- a*d)^3*Sqrt[c + d*x^2]) + (5*b^(3/2)*d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*(b*c - a*d)^(
7/2))

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Rubi [A]  time = 0.102821, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {444, 51, 63, 208} \[ \frac{5 b^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{7/2}}-\frac{5 b d}{2 \sqrt{c+d x^2} (b c-a d)^3}-\frac{1}{2 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{5 d}{6 \left (c+d x^2\right )^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d)/(6*(b*c - a*d)^2*(c + d*x^2)^(3/2)) - 1/(2*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) - (5*b*d)/(2*(b*c
- a*d)^3*Sqrt[c + d*x^2]) + (5*b^(3/2)*d*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(2*(b*c - a*d)^(
7/2))

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 )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(a+b x)^2 (c+d x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{(5 d) \operatorname{Subst}\left (\int \frac{1}{(a+b x) (c+d x)^{5/2}} \, dx,x,x^2\right )}{4 (b c-a d)}\\ &=-\frac{5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{(5 b 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)^2}\\ &=-\frac{5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{5 b d}{2 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{\left (5 b^2 d\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{4 (b c-a d)^3}\\ &=-\frac{5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{5 b d}{2 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{\left (5 b^2\right ) \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)^3}\\ &=-\frac{5 d}{6 (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac{1}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{5 b d}{2 (b c-a d)^3 \sqrt{c+d x^2}}+\frac{5 b^{3/2} d \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0237449, size = 54, normalized size = 0.39 \[ -\frac{d \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};-\frac{b \left (d x^2+c\right )}{a d-b c}\right )}{3 \left (c+d x^2\right )^{3/2} (a d-b c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*Hypergeometric2F1[-3/2, 2, -1/2, -((b*(c + d*x^2))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^2*(c + d*x^2)^(3/2)
)

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Maple [B]  time = 0.012, size = 1639, normalized size = 11.7 \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)^(5/2),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)^(3/2)-5/12*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)^(3/2)-5/12*(-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)^(3/2)*x-5/6*(-a*b)^(1/2)/b*d^2/(a*d-b*c)^2/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+5/4*b*d/(a*d-b*c)^3/((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)+5/4*(-a*b)^(1/2)*d^2/(a*d-b*c)^3/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-5/4*b*d/(a*d-b*c)^3/(-(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)))+1/3*(-a*b)^(1/2)/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+2/3*(-a*b)^(1/2)/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-1/4*(-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)^(3/2)-5/12*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)
^(3/2)+5/12*(-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)^(3/2)*x+5/6*(-a*b)^(1/2)/b*d^2/(a*d-b*c)^2/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+5/4*b*d/(a*d-b*c)^3/((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)-5/4*(-a*b)^(1/2)*d^2/(a*d-b*c)^3/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-5/4*b*d/(a*d-b*c)^3/(-(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)))-1/3*(-a*b)^(1/2)/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-2/3*(-a*b)^(1/2)/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

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.37066, size = 1816, normalized size = 12.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(15*(b^2*d^3*x^6 + a*b*c^2*d + (2*b^2*c*d^2 + a*b*d^3)*x^4 + (b^2*c^2*d + 2*a*b*c*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)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)
) + 4*(15*b^2*d^2*x^4 + 3*b^2*c^2 + 14*a*b*c*d - 2*a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a
*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4
- a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5
- a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2), -1/12*(15*(b^2*d^3*x^6 + a*b*c^2*d +
(2*b^2*c*d^2 + a*b*d^3)*x^4 + (b^2*c^2*d + 2*a*b*c*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*(15*b^2*d^2*x^4 + 3*b^2*c^2 + 14*a*b*c*d - 2*
a^2*d^2 + 10*(2*b^2*c*d + a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*
c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 +
3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3
- 2*a^4*c*d^4)*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)**(5/2),x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.14011, size = 301, normalized size = 2.15 \begin{align*} -\frac{1}{6} \,{\left (\frac{15 \, b^{2} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} + \frac{3 \, \sqrt{d x^{2} + c} b^{2}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\left ({\left (d x^{2} + c\right )} b - b c + a d\right )}} + \frac{2 \,{\left (6 \,{\left (d x^{2} + c\right )} b + b c - a d\right )}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\right )} d \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*b^2*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^
3)*sqrt(-b^2*c + a*b*d)) + 3*sqrt(d*x^2 + c)*b^2/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x^2
+ c)*b - b*c + a*d)) + 2*(6*(d*x^2 + c)*b + b*c - a*d)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d
*x^2 + c)^(3/2)))*d