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

Optimal. Leaf size=98 \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac{b}{\sqrt{c+d x^2} (b c-a d)^2}+\frac{1}{3 \left (c+d x^2\right )^{3/2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0848251, antiderivative size = 98, 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} \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}+\frac{b}{\sqrt{c+d x^2} (b c-a d)^2}+\frac{1}{3 \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.021596, size = 52, normalized size = 0.53 \[ \frac{\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )}{3 \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.009, size = 1086, normalized size = 11.1 \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)/(d*x^2+c)^(5/2),x)

[Out]

-1/6/(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/b*d*(-
a*b)^(1/2)/(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/b*d*(-a*b)^(1/2)/(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*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*d-b*c)^2*(-a*b)^(1/2)/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*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*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/b*d*(-a*b)^(1/2)/(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/b*d*(-a*b)^(1/2)/(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*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*d-b*c)^2*(-a*b)^(1/2)/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*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/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.86772, size = 1061, normalized size = 10.83 \begin{align*} \left [\frac{3 \,{\left (b d^{2} x^{4} + 2 \, b c d x^{2} + b c^{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} + 4 \, b c - a d\right )} \sqrt{d x^{2} + c}}{12 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c d^{3}\right )} x^{2}\right )}}, \frac{3 \,{\left (b d^{2} x^{4} + 2 \, b c d x^{2} + b c^{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} + 4 \, b c - a d\right )} \sqrt{d x^{2} + c}}{6 \,{\left (b^{2} c^{4} - 2 \, a b c^{3} d + a^{2} c^{2} d^{2} +{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} x^{4} + 2 \,{\left (b^{2} c^{3} d - 2 \, a b c^{2} d^{2} + a^{2} c 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)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b*d^2*x^4 + 2*b*c*d*x^2 + b*c^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*(3*b*d*x^2 + 4*b*c - a*d)*sqrt(d*x^2 + c))/(b^2*c^
4 - 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2
*c*d^3)*x^2), 1/6*(3*(b*d^2*x^4 + 2*b*c*d*x^2 + b*c^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 + 4*b*c - a*d)*sqrt(d*x^2 + c))/(b^2*c^4
- 2*a*b*c^3*d + a^2*c^2*d^2 + (b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*x^4 + 2*(b^2*c^3*d - 2*a*b*c^2*d^2 + a^2*
c*d^3)*x^2)]

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Sympy [A]  time = 13.8985, size = 85, normalized size = 0.87 \begin{align*} \frac{b}{\sqrt{c + d x^{2}} \left (a d - b c\right )^{2}} + \frac{b \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{\frac{a d - b c}{b}}} \right )}}{\sqrt{\frac{a d - b c}{b}} \left (a d - b c\right )^{2}} - \frac{1}{3 \left (c + d x^{2}\right )^{\frac{3}{2}} \left (a d - b c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10842, size = 159, normalized size = 1.62 \begin{align*} \frac{b^{2} \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 + b c - a d}{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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