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

Optimal. Leaf size=151 \[ -\frac{b d \left (3 a c^2-d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2} \left (a c^2-d^2\right )^2}-\frac{a c-d \sqrt{a+b x^2}}{2 a x^2 \left (a c^2-d^2\right )}+\frac{b c^3 \log \left (c \sqrt{a+b x^2}+d\right )}{\left (a c^2-d^2\right )^2}-\frac{b c^3 \log (x)}{\left (a c^2-d^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.352349, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2155, 741, 801, 635, 206, 260} \[ -\frac{b d \left (3 a c^2-d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2} \left (a c^2-d^2\right )^2}-\frac{a c-d \sqrt{a+b x^2}}{2 a x^2 \left (a c^2-d^2\right )}+\frac{b c^3 \log \left (c \sqrt{a+b x^2}+d\right )}{\left (a c^2-d^2\right )^2}-\frac{b c^3 \log (x)}{\left (a c^2-d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2155

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/n, Subst[Int
[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c
- a*d, 0] && IntegerQ[(m + 1)/n]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 1.05364, size = 291, normalized size = 1.93 \[ \frac{\frac{\sqrt{a} \left (-a^2 c^3 \sqrt{a+b x^2}+a^2 c^2 d+a b c^3 x^2 \sqrt{a+b x^2} \log \left (a c^2+b c^2 x^2-d^2\right )+b d x^2 \sqrt{\frac{b x^2}{a}+1} \left (a c^2-d^2\right ) \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )+a b c^2 d x^2+2 a b c^3 x^2 \sqrt{a+b x^2} \tanh ^{-1}\left (\frac{c \sqrt{a+b x^2}}{d}\right )-2 a b c^3 x^2 \log (x) \sqrt{a+b x^2}+a c d^2 \sqrt{a+b x^2}-a d^3-b d^3 x^2\right )}{x^2 \sqrt{a+b x^2}}+2 b d \left (d^2-2 a c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2} \left (d^2-a c^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.047, size = 2459, normalized size = 16.3 \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/(a*c+b*c*x^2+d*(b*x^2+a)^(1/2)),x)

[Out]

1/2*a*c^5*b/(a*c^2-d^2)^2/d^2*ln(b*c^2*x^2+a*c^2-d^2)-1/2*c/(a*c^2-d^2)/x^2-2*b*c^3*ln(x)/(a*c^2-d^2)^2+1/a*c*
b/(a*c^2-d^2)^2*ln(x)*d^2-1/2*b*c^3/(a*c^2-d^2)/d^2*ln(b*c^2*x^2+a*c^2-d^2)+b*c/a/(a*c^2-d^2)*ln(x)-1/2*d*c^2*
b^2/a^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*((x+(-a*b)
^(1/2)/b)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2)+1/2*d*c^2*b^(3/2)/a^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-
d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*(-a*b)^(1/2)*ln((b*(x+(-a*b)^(1/2)/b)-(-a*b)^(1/2))
/b^(1/2)+((x+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2))-2*d*b/a^(1/2)/(a*c^2-d^2)^2*ln((2*a
+2*a^(1/2)*(b*x^2+a)^(1/2))/x)*c^2+2*d*b/a/(a*c^2-d^2)^2*(b*x^2+a)^(1/2)*c^2+b/a^(3/2)/(a*c^2-d^2)^2*ln((2*a+2
*a^(1/2)*(b*x^2+a)^(1/2))/x)*d^3-b/a^2/(a*c^2-d^2)^2*(b*x^2+a)^(1/2)*d^3+1/2*d*c^6*b^2/(a*c^2-d^2)^2/((-a*b)^(
1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*((x-(-c^2*b*(a*c^2-d^2))^(1
/2)/c^2/b)^2*b+2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2)+1/2*d*c^
4*b^(3/2)/(a*c^2-d^2)^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(
1/2))*(-c^2*b*(a*c^2-d^2))^(1/2)*ln(((-c^2*b*(a*c^2-d^2))^(1/2)/c^2+b*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))/b^
(1/2)+((x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b+2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^2*b*(a*c^2-d^2))^(1/2)
/c^2/b)+1/c^2*d^2)^(1/2))-1/2*c^4*b^2/(a*c^2-d^2)^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2
)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*d^3/(1/c^2*d^2)^(1/2)*ln((2/c^2*d^2+2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x-(-c^
2*b*(a*c^2-d^2))^(1/2)/c^2/b)+2*(1/c^2*d^2)^(1/2)*((x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b+2*(-c^2*b*(a*c^2-d
^2))^(1/2)/c^2*(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2))/(x-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))+1/
2*d*c^6*b^2/(a*c^2-d^2)^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))
^(1/2))*((x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b-2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(-c^2*b*(a*c^2-d^2))^(1/
2)/c^2/b)+1/c^2*d^2)^(1/2)-1/2*d*c^4*b^(3/2)/(a*c^2-d^2)^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*
b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*(-c^2*b*(a*c^2-d^2))^(1/2)*ln((-(-c^2*b*(a*c^2-d^2))^(1/2)/c^2+b*(x+(
-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))/b^(1/2)+((x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)^2*b-2*(-c^2*b*(a*c^2-d^2))^(1/
2)/c^2*(x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2))-1/2*c^4*b^2/(a*c^2-d^2)^2/((-a*b)^(1/2)*c^2+(-c^
2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*d^3/(1/c^2*d^2)^(1/2)*ln((2/c^2*d^2-2*(-
c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+2*(1/c^2*d^2)^(1/2)*((x+(-c^2*b*(a*c^2-d^2))
^(1/2)/c^2/b)^2*b-2*(-c^2*b*(a*c^2-d^2))^(1/2)/c^2*(x+(-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b)+1/c^2*d^2)^(1/2))/(x+(
-c^2*b*(a*c^2-d^2))^(1/2)/c^2/b))-1/2*d*c^2*b^2/a^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2
)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b))^(1/2)-1/2*d*c^2*b
^(3/2)/a^2/((-a*b)^(1/2)*c^2+(-c^2*b*(a*c^2-d^2))^(1/2))/((-a*b)^(1/2)*c^2-(-c^2*b*(a*c^2-d^2))^(1/2))*(-a*b)^
(1/2)*ln((b*(x-(-a*b)^(1/2)/b)+(-a*b)^(1/2))/b^(1/2)+((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)
)^(1/2))+1/2*d/a^2/(a*c^2-d^2)/x^2*(b*x^2+a)^(3/2)+1/2*d/a^(3/2)/(a*c^2-d^2)*b*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/
2))/x)-1/2*d/a^2/(a*c^2-d^2)*b*(b*x^2+a)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 5.32345, size = 1131, normalized size = 7.49 \begin{align*} \left [\frac{2 \, a^{2} b c^{3} x^{2} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a^{2} b c^{3} x^{2} \log \left (x\right ) + a^{2} b c^{3} x^{2} \log \left (-\frac{b c^{2} x^{2} + a c^{2} + 2 \, \sqrt{b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a^{2} b c^{3} x^{2} \log \left (-\frac{b c^{2} x^{2} + a c^{2} - 2 \, \sqrt{b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} -{\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt{a} x^{2} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt{b x^{2} + a}}{4 \,{\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{2}}, \frac{2 \, a^{2} b c^{3} x^{2} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a^{2} b c^{3} x^{2} \log \left (x\right ) + a^{2} b c^{3} x^{2} \log \left (-\frac{b c^{2} x^{2} + a c^{2} + 2 \, \sqrt{b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a^{2} b c^{3} x^{2} \log \left (-\frac{b c^{2} x^{2} + a c^{2} - 2 \, \sqrt{b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} + 2 \,{\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + 2 \,{\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt{b x^{2} + a}}{4 \,{\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.18803, size = 275, normalized size = 1.82 \begin{align*} \frac{1}{2} \,{\left (\frac{2 \, c^{4} \log \left ({\left | \sqrt{b x^{2} + a} c + d \right |}\right )}{a^{2} c^{5} - 2 \, a c^{3} d^{2} + c d^{4}} - \frac{c^{3} \log \left (b x^{2}\right )}{a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}} + \frac{{\left (3 \, a c^{2} d - d^{3}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{{\left (a^{3} c^{4} - 2 \, a^{2} c^{2} d^{2} + a d^{4}\right )} \sqrt{-a}} - \frac{a^{2} c^{3} - a c d^{2} -{\left (a c^{2} d - d^{3}\right )} \sqrt{b x^{2} + a}}{{\left (a c^{2} - d^{2}\right )}^{2} a b x^{2}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*c^4*log(abs(sqrt(b*x^2 + a)*c + d))/(a^2*c^5 - 2*a*c^3*d^2 + c*d^4) - c^3*log(b*x^2)/(a^2*c^4 - 2*a*c^2
*d^2 + d^4) + (3*a*c^2*d - d^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/((a^3*c^4 - 2*a^2*c^2*d^2 + a*d^4)*sqrt(-a))
- (a^2*c^3 - a*c*d^2 - (a*c^2*d - d^3)*sqrt(b*x^2 + a))/((a*c^2 - d^2)^2*a*b*x^2))*b