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

Optimal. Leaf size=103 \[ \frac{-\frac{3 a^2 d}{c}+4 a b-\frac{2 b^2 c}{d}}{2 c \sqrt{c+d x^2}}-\frac{a^2}{2 c x^2 \sqrt{c+d x^2}}-\frac{a (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{5/2}} \]

[Out]

(4*a*b - (2*b^2*c)/d - (3*a^2*d)/c)/(2*c*Sqrt[c + d*x^2]) - a^2/(2*c*x^2*Sqrt[c + d*x^2]) - (a*(4*b*c - 3*a*d)
*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*c^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0976093, antiderivative size = 103, 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, 78, 63, 208} \[ \frac{-\frac{3 a^2 d}{c}+4 a b-\frac{2 b^2 c}{d}}{2 c \sqrt{c+d x^2}}-\frac{a^2}{2 c x^2 \sqrt{c+d x^2}}-\frac{a (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*b - (2*b^2*c)/d - (3*a^2*d)/c)/(2*c*Sqrt[c + d*x^2]) - a^2/(2*c*x^2*Sqrt[c + d*x^2]) - (a*(4*b*c - 3*a*d)
*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*c^(5/2))

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 78

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

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

Mathematica [A]  time = 0.0826468, size = 96, normalized size = 0.93 \[ \frac{a (3 a d-4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 c^{5/2}}-\frac{a^2 d \left (c+3 d x^2\right )-4 a b c d x^2+2 b^2 c^2 x^2}{2 c^2 d x^2 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*b^2*c^2*x^2 - 4*a*b*c*d*x^2 + a^2*d*(c + 3*d*x^2))/(2*c^2*d*x^2*Sqrt[c + d*x^2]) + (a*(-4*b*c + 3*a*d)*Arc
Tanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*c^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 135, normalized size = 1.3 \begin{align*} -{\frac{{b}^{2}}{d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+2\,{\frac{ab}{c\sqrt{d{x}^{2}+c}}}-2\,{\frac{ab}{{c}^{3/2}}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c}}{x}} \right ) }-{\frac{{a}^{2}}{2\,c{x}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{3\,{a}^{2}d}{2\,{c}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{3\,{a}^{2}d}{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{5}{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)^(3/2),x)

[Out]

-b^2/d/(d*x^2+c)^(1/2)+2*a*b/c/(d*x^2+c)^(1/2)-2*a*b/c^(3/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)-1/2*a^2/c/x
^2/(d*x^2+c)^(1/2)-3/2*a^2*d/c^2/(d*x^2+c)^(1/2)+3/2*a^2*d/c^(5/2)*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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.40858, size = 620, normalized size = 6.02 \begin{align*} \left [-\frac{{\left ({\left (4 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{4} +{\left (4 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt{c} \log \left (-\frac{d x^{2} + 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (a^{2} c^{2} d +{\left (2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{4 \,{\left (c^{3} d^{2} x^{4} + c^{4} d x^{2}\right )}}, \frac{{\left ({\left (4 \, a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{4} +{\left (4 \, a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) -{\left (a^{2} c^{2} d +{\left (2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{2 \,{\left (c^{3} d^{2} x^{4} + c^{4} d x^{2}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{x^{3} \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.15972, size = 189, normalized size = 1.83 \begin{align*} \frac{{\left (4 \, a b c - 3 \, a^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{2 \, \sqrt{-c} c^{2}} - \frac{2 \,{\left (d x^{2} + c\right )} b^{2} c^{2} - 2 \, b^{2} c^{3} - 4 \,{\left (d x^{2} + c\right )} a b c d + 4 \, a b c^{2} d + 3 \,{\left (d x^{2} + c\right )} a^{2} d^{2} - 2 \, a^{2} c d^{2}}{2 \,{\left ({\left (d x^{2} + c\right )}^{\frac{3}{2}} - \sqrt{d x^{2} + c} c\right )} 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)^(3/2),x, algorithm="giac")

[Out]

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