3.61 \(\int \frac{(a+c x^2)^{3/2}}{x (d+e x+f x^2)} \, dx\)

Optimal. Leaf size=496 \[ -\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (\sqrt{e^2-4 d f}+e\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d}-\frac{c^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{\sqrt{a+c x^2} (c d-a f)}{d f}+\frac{a \sqrt{a+c x^2}}{d} \]

[Out]

(a*Sqrt[a + c*x^2])/d + ((c*d - a*f)*Sqrt[a + c*x^2])/(d*f) - (c^(3/2)*e*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])
/f^2 - ((2*e*f*(c^2*d^2 - a^2*f^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*(e - Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c
*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/
(Sqrt[2]*d*f^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + ((2*e*f*(c^2*d^2 - a
^2*f^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*(e + Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)
/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d*f^2*Sqrt[e^2 - 4
*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]) - (a^(3/2)*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/d

________________________________________________________________________________________

Rubi [A]  time = 2.56885, antiderivative size = 496, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 11, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.407, Rules used = {6728, 266, 50, 63, 208, 1020, 1080, 217, 206, 1034, 725} \[ -\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (e-\sqrt{e^2-4 d f}\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (\sqrt{e^2-4 d f}+e\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} d f^2 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d}-\frac{c^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{\sqrt{a+c x^2} (c d-a f)}{d f}+\frac{a \sqrt{a+c x^2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a + c*x^2])/d + ((c*d - a*f)*Sqrt[a + c*x^2])/(d*f) - (c^(3/2)*e*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])
/f^2 - ((2*e*f*(c^2*d^2 - a^2*f^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*(e - Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c
*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/
(Sqrt[2]*d*f^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + ((2*e*f*(c^2*d^2 - a
^2*f^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*(e + Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)
/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d*f^2*Sqrt[e^2 - 4
*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]) - (a^(3/2)*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/d

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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]

Rule 1020

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(h*(a + c*x^2)^p*(d + e*x + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] + Dist[1/(2*f*(p + q + 1)), Int[(a + c*x^2)
^(p - 1)*(d + e*x + f*x^2)^q*Simp[a*h*e*p - a*(h*e - 2*g*f)*(p + q + 1) - 2*h*p*(c*d - a*f)*x - (h*c*e*p + c*(
h*e - 2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h, q}, x] && NeQ[e^2 - 4*d*f, 0] && GtQ[
p, 0] && NeQ[p + q + 1, 0]

Rule 1080

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + (B*c - b*C)*x)/((a + b*x + c*x^2)
*Sqrt[d + f*x^2]), x], x] /; FreeQ[{a, b, c, d, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 725

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

Rubi steps

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

Mathematica [A]  time = 1.62658, size = 746, normalized size = 1.5 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{d}-\frac{c^{3/2} e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{f^2}+\frac{a \sqrt{a f^2+\frac{1}{2} c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{2 d f}-\frac{a e \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{4 d f \sqrt{e^2-4 d f}}-\frac{\left (c d \left (\sqrt{e^2-4 d f}-e\right )-a f \left (\sqrt{e^2-4 d f}+e\right )\right ) \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )} \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )}{4 d f^2 \sqrt{e^2-4 d f}}-\frac{c \sqrt{a f^2+\frac{1}{2} c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{2 f^2}-\frac{c e \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{4 f^2 \sqrt{e^2-4 d f}}+\frac{c \sqrt{a+c x^2}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[a + c*x^2])/f - (c^(3/2)*e*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/f^2 - ((c*d*(-e + Sqrt[e^2 - 4*d*f])
- a*f*(e + Sqrt[e^2 - 4*d*f]))*Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*ArcTanh[(2*a*f + c*(-e
+ Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(4*d*f^2
*Sqrt[e^2 - 4*d*f]) - (c*Sqrt[a*f^2 + (c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]))/2]*ArcTanh[(2*a*f - c*(e + Sqrt[
e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(2*f^2) + (a*Sqrt
[a*f^2 + (c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]))/2]*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^
2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(2*d*f) - (c*e*Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f
+ e*Sqrt[e^2 - 4*d*f])]*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sq
rt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(4*f^2*Sqrt[e^2 - 4*d*f]) - (a*e*Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt
[e^2 - 4*d*f])]*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 -
4*d*f])]*Sqrt[a + c*x^2])])/(4*d*f*Sqrt[e^2 - 4*d*f]) - (a^(3/2)*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/d

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Maple [B]  time = 0.262, size = 9728, normalized size = 19.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^(3/2)/x/(f*x^2+e*x+d),x)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^(3/2)/((f*x^2 + e*x + d)*x), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(3/2)/x/(f*x**2+e*x+d),x)

[Out]

Integral((a + c*x**2)**(3/2)/(x*(d + e*x + f*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x