3.657 \(\int \frac{(a+b x^2)^2}{x^5 (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=145 \[ -\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}+\frac{8 b^2 c^2-3 a d (8 b c-5 a d)}{8 c^3 \sqrt{c+d x^2}}-\frac{\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{8 c^{7/2}}-\frac{a (8 b c-5 a d)}{8 c^2 x^2 \sqrt{c+d x^2}} \]

[Out]

(8*b^2*c^2 - 3*a*d*(8*b*c - 5*a*d))/(8*c^3*Sqrt[c + d*x^2]) - a^2/(4*c*x^4*Sqrt[c + d*x^2]) - (a*(8*b*c - 5*a*
d))/(8*c^2*x^2*Sqrt[c + d*x^2]) - ((8*b^2*c^2 - 3*a*d*(8*b*c - 5*a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^
(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.167187, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 89, 78, 51, 63, 208} \[ -\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}+\frac{8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt{c+d x^2}}-\frac{\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{8 c^{7/2}}-\frac{a (8 b c-5 a d)}{8 c^2 x^2 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]

[Out]

(8*b^2 - (3*a*d*(8*b*c - 5*a*d))/c^2)/(8*c*Sqrt[c + d*x^2]) - a^2/(4*c*x^4*Sqrt[c + d*x^2]) - (a*(8*b*c - 5*a*
d))/(8*c^2*x^2*Sqrt[c + d*x^2]) - ((8*b^2*c^2 - 3*a*d*(8*b*c - 5*a*d))*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^
(7/2))

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 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{\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x^3 (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} a (8 b c-5 a d)+2 b^2 c x}{x^2 (c+d x)^{3/2}} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}-\frac{a (8 b c-5 a d)}{8 c^2 x^2 \sqrt{c+d x^2}}+\frac{1}{16} \left (8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac{8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt{c+d x^2}}-\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}-\frac{a (8 b c-5 a d)}{8 c^2 x^2 \sqrt{c+d x^2}}+\frac{\left (8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{16 c}\\ &=\frac{8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt{c+d x^2}}-\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}-\frac{a (8 b c-5 a d)}{8 c^2 x^2 \sqrt{c+d x^2}}+\frac{\left (8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{8 c d}\\ &=\frac{8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}}{8 c \sqrt{c+d x^2}}-\frac{a^2}{4 c x^4 \sqrt{c+d x^2}}-\frac{a (8 b c-5 a d)}{8 c^2 x^2 \sqrt{c+d x^2}}-\frac{\left (8 b^2-\frac{3 a d (8 b c-5 a d)}{c^2}\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{8 c^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0341562, size = 89, normalized size = 0.61 \[ \frac{x^4 \left (15 a^2 d^2-24 a b c d+8 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^2}{c}+1\right )+a c \left (-2 a c+5 a d x^2-8 b c x^2\right )}{8 c^3 x^4 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]

[Out]

(a*c*(-2*a*c - 8*b*c*x^2 + 5*a*d*x^2) + (8*b^2*c^2 - 24*a*b*c*d + 15*a^2*d^2)*x^4*Hypergeometric2F1[-1/2, 1, 1
/2, 1 + (d*x^2)/c])/(8*c^3*x^4*Sqrt[c + d*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 211, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}}{c}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{{b}^{2}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{3}{2}}}}-{\frac{{a}^{2}}{4\,c{x}^{4}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{5\,{a}^{2}d}{8\,{c}^{2}{x}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{15\,{a}^{2}{d}^{2}}{8\,{c}^{3}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{15\,{a}^{2}{d}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){c}^{-{\frac{7}{2}}}}-{\frac{ab}{c{x}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-3\,{\frac{abd}{{c}^{2}\sqrt{d{x}^{2}+c}}}+3\,{\frac{abd}{{c}^{5/2}}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c}}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x)

[Out]

b^2/c/(d*x^2+c)^(1/2)-b^2/c^(3/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)-1/4*a^2/c/x^4/(d*x^2+c)^(1/2)+5/8*a^2*
d/c^2/x^2/(d*x^2+c)^(1/2)+15/8*a^2*d^2/c^3/(d*x^2+c)^(1/2)-15/8*a^2*d^2/c^(7/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1
/2))/x)-a*b/c/x^2/(d*x^2+c)^(1/2)-3*a*b*d/c^2/(d*x^2+c)^(1/2)+3*a*b*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^5/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.41592, size = 782, normalized size = 5.39 \begin{align*} \left [\frac{{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} +{\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt{c} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) - 2 \,{\left (2 \, a^{2} c^{3} -{\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} +{\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{16 \,{\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}, \frac{{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} +{\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) -{\left (2 \, a^{2} c^{3} -{\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} +{\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{8 \,{\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(((8*b^2*c^2*d - 24*a*b*c*d^2 + 15*a^2*d^3)*x^6 + (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4)*sqrt(c)
*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(2*a^2*c^3 - (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2
)*x^4 + (8*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c^4*d*x^6 + c^5*x^4), 1/8*(((8*b^2*c^2*d - 24*a*b*c*d
^2 + 15*a^2*d^3)*x^6 + (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)
) - (2*a^2*c^3 - (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4 + (8*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c
))/(c^4*d*x^6 + c^5*x^4)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{x^{5} \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**5/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/(x**5*(c + d*x**2)**(3/2)), x)

________________________________________________________________________________________

Giac [A]  time = 1.17151, size = 220, normalized size = 1.52 \begin{align*} \frac{{\left (8 \, b^{2} c^{2} - 24 \, a b c d + 15 \, a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{8 \, \sqrt{-c} c^{3}} + \frac{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{\sqrt{d x^{2} + c} c^{3}} - \frac{8 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b c d - 8 \, \sqrt{d x^{2} + c} a b c^{2} d - 7 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a^{2} d^{2} + 9 \, \sqrt{d x^{2} + c} a^{2} c d^{2}}{8 \, c^{3} d^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

1/8*(8*b^2*c^2 - 24*a*b*c*d + 15*a^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) + (b^2*c^2 - 2*a*b*c
*d + a^2*d^2)/(sqrt(d*x^2 + c)*c^3) - 1/8*(8*(d*x^2 + c)^(3/2)*a*b*c*d - 8*sqrt(d*x^2 + c)*a*b*c^2*d - 7*(d*x^
2 + c)^(3/2)*a^2*d^2 + 9*sqrt(d*x^2 + c)*a^2*c*d^2)/(c^3*d^2*x^4)