3.720 \(\int \frac{1}{x^3 (a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=156 \[ -\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 (b c-a d)^{3/2}}+\frac{(3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{5/2}}-\frac{d (b c-3 a d)}{2 a c^2 \sqrt{c+d x^2} (b c-a d)}-\frac{1}{2 a c x^2 \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.216236, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 103, 152, 156, 63, 208} \[ -\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 (b c-a d)^{3/2}}+\frac{(3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{5/2}}-\frac{d (b c-3 a d)}{2 a c^2 \sqrt{c+d x^2} (b c-a d)}-\frac{1}{2 a c x^2 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(b*c - 3*a*d))/(2*a*c^2*(b*c - a*d)*Sqrt[c + d*x^2]) - 1/(2*a*c*x^2*Sqrt[c + d*x^2]) + ((2*b*c + 3*a*d)*Ar
cTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*a^2*c^(5/2)) - (b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])
/(a^2*(b*c - a*d)^(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 103

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

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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{1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{1}{2 a c x^2 \sqrt{c+d x^2}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (2 b c+3 a d)+\frac{3 b d x}{2}}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^2\right )}{2 a c}\\ &=-\frac{d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt{c+d x^2}}-\frac{1}{2 a c x^2 \sqrt{c+d x^2}}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{4} (b c-a d) (2 b c+3 a d)-\frac{1}{4} b d (b c-3 a d) x}{x (a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{a c^2 (b c-a d)}\\ &=-\frac{d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt{c+d x^2}}-\frac{1}{2 a c x^2 \sqrt{c+d x^2}}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^2\right )}{2 a^2 (b c-a d)}-\frac{(2 b c+3 a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{4 a^2 c^2}\\ &=-\frac{d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt{c+d x^2}}-\frac{1}{2 a c x^2 \sqrt{c+d x^2}}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{a^2 d (b c-a d)}-\frac{(2 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 a^2 c^2 d}\\ &=-\frac{d (b c-3 a d)}{2 a c^2 (b c-a d) \sqrt{c+d x^2}}-\frac{1}{2 a c x^2 \sqrt{c+d x^2}}+\frac{(2 b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{5/2}}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.046871, size = 117, normalized size = 0.75 \[ \frac{2 b^2 c^2 x^2 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )+(a d-b c) \left (x^2 (3 a d+2 b c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^2}{c}+1\right )+a c\right )}{2 a^2 c^2 x^2 \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 763, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b/a^2/c/(d*x^2+c)^(1/2)+b/a^2/c^(3/2)*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x)-1/2*b^2/a^2/(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)^(1/2)-1/2*b/a^2*(-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)^(1/2)*x*d+1/2*b^2/a^2/(a*d-b*c)/
(-(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)))-1/2*b^2/a
^2/(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)^(1/2)+1/2*b/a^2*(-
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)^(1/2)*x*
d+1/2*b^2/a^2/(a*d-b*c)/(-(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)))-1/2/a/c/x^2/(d*x^2+c)^(1/2)-3/2/a*d/c^2/(d*x^2+c)^(1/2)+3/2/a*d/c^(5/2)*ln((2*c+2*c^(1/2)*(d*x^
2+c)^(1/2))/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 5.52174, size = 2649, normalized size = 16.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((b^2*c^3*d*x^4 + b^2*c^4*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)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^4 + (2*b^2*c^3 +
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*b*c^3 - a^2*c
^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^4 + (a^2*b*c^5 - a^3*c^4
*d)*x^2), -1/4*(2*((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^4 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^2)*sqrt
(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (b^2*c^3*d*x^4 + b^2*c^4*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)) + 2*(a*b*c^3 - a^2*c^2*d +
(a*b*c^2*d - 3*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^4 + (a^2*b*c^5 - a^3*c^4*d)*x^2
), 1/4*(2*(b^2*c^3*d*x^4 + b^2*c^4*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*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^4 + (2*b^2*c^3 + 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*b*c^3 - a^2*c^2*d + (a*
b*c^2*d - 3*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^4 + (a^2*b*c^5 - a^3*c^4*d)*x^2),
1/2*((b^2*c^3*d*x^4 + b^2*c^4*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)*sqr
t(-b/(b*c - a*d))/(b*d*x^2 + b*c)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^4 + (2*b^2*c^3 + a*b*c^2*d - 3*a
^2*c*d^2)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - (a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^
2)*sqrt(d*x^2 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^4 + (a^2*b*c^5 - a^3*c^4*d)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15079, size = 248, normalized size = 1.59 \begin{align*} \frac{1}{2} \,{\left (\frac{2 \, b^{3} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b^{2} c + a b d}} - \frac{{\left (d x^{2} + c\right )} b c - 3 \,{\left (d x^{2} + c\right )} a d + 2 \, a c d}{{\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )}{\left ({\left (d x^{2} + c\right )}^{\frac{3}{2}} - \sqrt{d x^{2} + c} c\right )}} - \frac{{\left (2 \, b c + 3 \, a d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c^{2} d^{2}}\right )} d^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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