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

Optimal. Leaf size=113 \[ \frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 \sqrt{c}}-\frac{\sqrt{b} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2}-\frac{\sqrt{c+d x^2}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.120464, antiderivative size = 113, 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, 99, 156, 63, 208} \[ \frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^2 \sqrt{c}}-\frac{\sqrt{b} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{a^2}-\frac{\sqrt{c+d x^2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[c + d*x^2]/(2*a*x^2) + ((2*b*c - a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*a^2*Sqrt[c]) - (Sqrt[b]*Sqrt[
b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/a^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 99

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

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

Mathematica [A]  time = 0.122045, size = 107, normalized size = 0.95 \[ \frac{\frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{\sqrt{c}}-2 \sqrt{b} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )-\frac{a \sqrt{c+d x^2}}{x^2}}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 1054, normalized size = 9.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.95258, size = 1578, normalized size = 13.96 \begin{align*} \left [\frac{\sqrt{b^{2} c - a b d} c x^{2} \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 (b d x^{2} + 2 \, b c - a d\right )} \sqrt{b^{2} c - a b d} \sqrt{d x^{2} + c}}{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 x^{2}}, -\frac{2 \,{\left (2 \, b c - a d\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) - \sqrt{b^{2} c - a b d} c x^{2} \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 (b d x^{2} + 2 \, b c - a d\right )} \sqrt{b^{2} c - a b d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, \sqrt{d x^{2} + c} a c}{4 \, a^{2} c x^{2}}, -\frac{2 \, \sqrt{-b^{2} c + a b d} c x^{2} \arctan \left (-\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{-b^{2} c + a b d} \sqrt{d x^{2} + c}}{2 \,{\left (b^{2} c^{2} - a b c d +{\left (b^{2} c d - a b d^{2}\right )} x^{2}\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 x^{2}}, -\frac{\sqrt{-b^{2} c + a b d} c x^{2} \arctan \left (-\frac{{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt{-b^{2} c + a b d} \sqrt{d x^{2} + c}}{2 \,{\left (b^{2} c^{2} - a b c d +{\left (b^{2} c d - a b d^{2}\right )} x^{2}\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 x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12783, size = 163, normalized size = 1.44 \begin{align*} \frac{1}{2} \, d^{2}{\left (\frac{2 \,{\left (b^{2} c - a b d\right )} \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} d^{2}} - \frac{\sqrt{d x^{2} + c}}{a d^{2} x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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