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

Optimal. Leaf size=115 \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{3/2}}-\frac{\sqrt{c+d x^2}}{2 a c x^2} \]

[Out]

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

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Rubi [A]  time = 0.115032, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 103, 156, 63, 208} \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 c^{3/2}}-\frac{\sqrt{c+d x^2}}{2 a c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.274271, size = 109, normalized size = 0.95 \[ \frac{-\frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{c^{3/2}}-\frac{a \sqrt{c+d x^2}}{c x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 385, normalized size = 3.4 \begin{align*}{\frac{b}{{a}^{2}}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){\frac{1}{\sqrt{c}}}}-{\frac{b}{2\,{a}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{b}{2\,{a}^{2}}\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{1}{2\,ac{x}^{2}}\sqrt{d{x}^{2}+c}}+{\frac{d}{2\,a}\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(1/x^3/(b*x^2+a)/(d*x^2+c)^(1/2),x)

[Out]

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

[Out]

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

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Fricas [A]  time = 2.22528, size = 1620, normalized size = 14.09 \begin{align*} \left [\frac{b c^{2} x^{2} \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 ) +{\left (2 \, b c + a d\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 \, \sqrt{d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}, \frac{b c^{2} x^{2} \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 ) - 2 \,{\left (2 \, b c + a d\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) - 2 \, \sqrt{d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}, \frac{2 \, b c^{2} x^{2} \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 ) +{\left (2 \, b c + a d\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 \, \sqrt{d x^{2} + c} a c}{4 \, a^{2} c^{2} x^{2}}, \frac{b c^{2} x^{2} \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 ) -{\left (2 \, b c + a d\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) - \sqrt{d x^{2} + c} a c}{2 \, a^{2} c^{2} x^{2}}\right ] \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)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [A]  time = 1.12779, size = 159, normalized size = 1.38 \begin{align*} \frac{1}{2} \, d^{2}{\left (\frac{2 \, b^{2} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{2} d^{2}} - \frac{{\left (2 \, b c + a d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c d^{2}} - \frac{\sqrt{d x^{2} + c}}{a c d^{2} x^{2}}\right )} \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)^(1/2),x, algorithm="giac")

[Out]

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