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

Optimal. Leaf size=80 \[ -\frac{a^2 \sqrt{c+d x^2}}{2 c x^2}-\frac{a (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{3/2}}+\frac{b^2 \sqrt{c+d x^2}}{d} \]

[Out]

(b^2*Sqrt[c + d*x^2])/d - (a^2*Sqrt[c + d*x^2])/(2*c*x^2) - (a*(4*b*c - a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])
/(2*c^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0676449, antiderivative size = 80, 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 = {446, 89, 80, 63, 208} \[ -\frac{a^2 \sqrt{c+d x^2}}{2 c x^2}-\frac{a (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{3/2}}+\frac{b^2 \sqrt{c+d x^2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*Sqrt[c + d*x^2])/d - (a^2*Sqrt[c + d*x^2])/(2*c*x^2) - (a*(4*b*c - a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])
/(2*c^(3/2))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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

Mathematica [A]  time = 0.0532708, size = 77, normalized size = 0.96 \[ \frac{\frac{\sqrt{c} \sqrt{c+d x^2} \left (2 b^2 c x^2-a^2 d\right )}{d x^2}+a (a d-4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[c]*(-(a^2*d) + 2*b^2*c*x^2)*Sqrt[c + d*x^2])/(d*x^2) + a*(-4*b*c + a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]
])/(2*c^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 100, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}}{d}\sqrt{d{x}^{2}+c}}-2\,{\frac{ab}{\sqrt{c}}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c}}{x}} \right ) }-{\frac{{a}^{2}}{2\,c{x}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{{a}^{2}d}{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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((b*x^2+a)^2/x^3/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.46884, size = 387, normalized size = 4.84 \begin{align*} \left [-\frac{{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt{c} x^{2} \log \left (-\frac{d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) - 2 \,{\left (2 \, b^{2} c^{2} x^{2} - a^{2} c d\right )} \sqrt{d x^{2} + c}}{4 \, c^{2} d x^{2}}, \frac{{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (2 \, b^{2} c^{2} x^{2} - a^{2} c d\right )} \sqrt{d x^{2} + c}}{2 \, c^{2} d x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^3/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 43.0845, size = 99, normalized size = 1.24 \begin{align*} - \frac{a^{2} \sqrt{d} \sqrt{\frac{c}{d x^{2}} + 1}}{2 c x} + \frac{a^{2} d \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{2 c^{\frac{3}{2}}} - \frac{2 a b \operatorname{asinh}{\left (\frac{\sqrt{c}}{\sqrt{d} x} \right )}}{\sqrt{c}} + b^{2} \left (\begin{cases} \frac{x^{2}}{2 \sqrt{c}} & \text{for}\: d = 0 \\\frac{\sqrt{c + d x^{2}}}{d} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**3/(d*x**2+c)**(1/2),x)

[Out]

-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*c*x) + a**2*d*asinh(sqrt(c)/(sqrt(d)*x))/(2*c**(3/2)) - 2*a*b*asinh(sqrt
(c)/(sqrt(d)*x))/sqrt(c) + b**2*Piecewise((x**2/(2*sqrt(c)), Eq(d, 0)), (sqrt(c + d*x**2)/d, True))

________________________________________________________________________________________

Giac [A]  time = 1.12102, size = 109, normalized size = 1.36 \begin{align*} \frac{2 \, \sqrt{d x^{2} + c} b^{2} - \frac{\sqrt{d x^{2} + c} a^{2} d}{c x^{2}} + \frac{{\left (4 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^3/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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