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

Optimal. Leaf size=604 \[ \frac {a^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2}-\frac {\left (a^2 f^2 \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )+4 a c d^2 f^2+c^2 d^2 \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 f \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (a^2 f^2 \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )+4 a c d^2 f^2+c^2 d^2 \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 f \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {a e \sqrt {a+c x^2}}{d^2}+\frac {\sqrt {a+c x^2} (2 a e-c d x)}{2 d^2}+\frac {\sqrt {c} (2 c d-3 a f) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d f}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {3 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d} \]

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Rubi [A]  time = 2.81, antiderivative size = 604, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 14, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.518, Rules used = {6728, 277, 195, 217, 206, 266, 50, 63, 208, 1020, 1068, 1080, 1034, 725} \begin {gather*} -\frac {\left (a^2 f^2 \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )+4 a c d^2 f^2+c^2 d^2 \left (-e \sqrt {e^2-4 d f}-2 d f+e^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^2 f \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 (a^2 f^2 \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )+4 a c d^2 f^2+c^2 d^2 \left (e \sqrt {e^2-4 d f}-2 d f+e^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^2 f \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} e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2}-\frac {a e \sqrt {a+c x^2}}{d^2}+\frac {\sqrt {a+c x^2} (2 a e-c d x)}{2 d^2}+\frac {\sqrt {c} (2 c d-3 a f) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d f}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {3 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 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 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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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]

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 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 1068

Int[((a_) + (c_.)*(x_)^2)^(p_)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_
Symbol] :> Simp[((B*c*f*(2*p + 2*q + 3) + C*(-(c*e*(2*p + q + 2))) + 2*c*C*f*(p + q + 1)*x)*(a + c*x^2)^p*(d +
e*x + f*x^2)^(q + 1))/(2*c*f^2*(p + q + 1)*(2*p + 2*q + 3)), x] - Dist[1/(2*c*f^2*(p + q + 1)*(2*p + 2*q + 3)
), Int[(a + c*x^2)^(p - 1)*(d + e*x + f*x^2)^q*Simp[p*(-(a*e))*(C*(c*e)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3
)) + (p + q + 1)*(a*c*(C*(2*d*f - e^2*(2*p + q + 2)) + f*(B*e - 2*A*f)*(2*p + 2*q + 3))) + (2*p*(c*d - a*f)*(C
*(c*e)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)*(C*e*f*p*(-4*a*c)))*x + (p*(c*e)*(C*(c*e)*(q + 1
) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)*(C*f^2*p*(-4*a*c) - c^2*(C*(e^2 - 4*d*f)*(2*p + q + 2) + f*(2
*C*d - B*e + 2*A*f)*(2*p + 2*q + 3))))*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, A, B, C, q}, x] && NeQ[e^2 - 4
*d*f, 0] && GtQ[p, 0] && NeQ[p + q + 1, 0] && NeQ[2*p + 2*q + 3, 0] &&  !IGtQ[p, 0] &&  !IGtQ[q, 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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^2\right )^{3/2}}{x^2 \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac {\left (a+c x^2\right )^{3/2}}{d x^2}-\frac {e \left (a+c x^2\right )^{3/2}}{d^2 x}+\frac {\left (e^2-d f+e f x\right ) \left (a+c x^2\right )^{3/2}}{d^2 \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {\left (e^2-d f+e f x\right ) \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx}{d^2}+\frac {\int \frac {\left (a+c x^2\right )^{3/2}}{x^2} \, dx}{d}-\frac {e \int \frac {\left (a+c x^2\right )^{3/2}}{x} \, dx}{d^2}\\ &=\frac {e \left (a+c x^2\right )^{3/2}}{3 d^2}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {(3 c) \int \sqrt {a+c x^2} \, dx}{d}-\frac {e \operatorname {Subst}\left (\int \frac {(a+c x)^{3/2}}{x} \, dx,x,x^2\right )}{2 d^2}+\frac {\int \frac {\sqrt {a+c x^2} \left (3 a f \left (e^2-d f\right )-3 e f (c d-a f) x-3 c d f^2 x^2\right )}{d+e x+f x^2} \, dx}{3 d^2 f}\\ &=\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {(2 a e-c d x) \sqrt {a+c x^2}}{2 d^2}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {(3 a c) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 d}-\frac {(a e) \operatorname {Subst}\left (\int \frac {\sqrt {a+c x}}{x} \, dx,x,x^2\right )}{2 d^2}-\frac {\int \frac {-3 a c f^3 \left (c d^2+2 a e^2-2 a d f\right )+3 a c e f^3 (3 c d-2 a f) x-3 c^2 d f^3 (2 c d-3 a f) x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{6 c d^2 f^3}\\ &=-\frac {a e \sqrt {a+c x^2}}{d^2}+\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {(2 a e-c d x) \sqrt {a+c x^2}}{2 d^2}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {(3 a c) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 d}-\frac {\left (a^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^2}-\frac {\int \frac {3 c^2 d^2 f^3 (2 c d-3 a f)-3 a c f^4 \left (c d^2+2 a e^2-2 a d f\right )+\left (3 c^2 d e f^3 (2 c d-3 a f)+3 a c e f^4 (3 c d-2 a f)\right ) x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{6 c d^2 f^4}+\frac {(c (2 c d-3 a f)) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 d f}\\ &=-\frac {a e \sqrt {a+c x^2}}{d^2}+\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {(2 a e-c d x) \sqrt {a+c x^2}}{2 d^2}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {3 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d}-\frac {\left (a^2 e\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{c d^2}+\frac {(c (2 c d-3 a f)) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 d f}-\frac {\left (4 a c d^2 f^2+a^2 f^2 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+c^2 d^2 \left (e^2-2 d f+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^2 f \sqrt {e^2-4 d f}}+\frac {\left (4 a c d^2 f^2+c^2 d^2 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+a^2 f^2 \left (e^2-2 d f+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^2 f \sqrt {e^2-4 d f}}\\ &=-\frac {a e \sqrt {a+c x^2}}{d^2}+\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {(2 a e-c d x) \sqrt {a+c x^2}}{2 d^2}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {3 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d}+\frac {\sqrt {c} (2 c d-3 a f) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d f}+\frac {a^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2}+\frac {\left (4 a c d^2 f^2+a^2 f^2 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+c^2 d^2 \left (e^2-2 d f+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^2 f \sqrt {e^2-4 d f}}-\frac {\left (4 a c d^2 f^2+c^2 d^2 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+a^2 f^2 \left (e^2-2 d f+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^2 f \sqrt {e^2-4 d f}}\\ &=-\frac {a e \sqrt {a+c x^2}}{d^2}+\frac {3 c x \sqrt {a+c x^2}}{2 d}+\frac {(2 a e-c d x) \sqrt {a+c x^2}}{2 d^2}-\frac {\left (a+c x^2\right )^{3/2}}{d x}+\frac {3 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d}+\frac {\sqrt {c} (2 c d-3 a f) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 d f}-\frac {\left (4 a c d^2 f^2+c^2 d^2 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+a^2 f^2 \left (e^2-2 d f+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^2 f \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 (4 a c d^2 f^2+a^2 f^2 \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )+c^2 d^2 \left (e^2-2 d f+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^2 f \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} e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2}\\ \end {align*}

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Mathematica [C]  time = 4.16, size = 885, normalized size = 1.47 \begin {gather*} \frac {-x \left (2 \sqrt {a} \sqrt {c} d f \sqrt {e^2-4 d f} \sqrt {c x^2+a} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )+\sqrt {\frac {c x^2}{a}+1} \left (-2 c \sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right ) d^2-4 \sqrt {c} (c d-a f) \sqrt {e^2-4 d f} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {c x^2+a}}\right ) d+2 a f \sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right ) d+2 c f \sqrt {e^2-4 d f} x \sqrt {c x^2+a} d+\left (2 c d^2+a \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )\right ) \sqrt {4 a f^2-2 c \left (-e^2+\sqrt {e^2-4 d f} e+2 d f\right )} \tanh ^{-1}\left (\frac {2 a f+c \left (\sqrt {e^2-4 d f}-e\right ) x}{\sqrt {4 a f^2-2 c \left (-e^2+\sqrt {e^2-4 d f} e+2 d f\right )} \sqrt {c x^2+a}}\right )+\sqrt {2} a e \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right )-a e^2 \sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {4 a f^2+2 c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )} \sqrt {c x^2+a}}\right )-4 a^{3/2} e f \sqrt {e^2-4 d f} \tanh ^{-1}\left (\frac {\sqrt {c x^2+a}}{\sqrt {a}}\right )\right )\right )-4 a d f \sqrt {e^2-4 d f} \sqrt {c x^2+a} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {c x^2}{a}\right )}{4 d^2 f \sqrt {e^2-4 d f} x \sqrt {\frac {c x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(x*(2*Sqrt[a]*Sqrt[c]*d*f*Sqrt[e^2 - 4*d*f]*Sqrt[a + c*x^2]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]] + Sqrt[1 + (c*x^2)/
a]*(2*c*d*f*Sqrt[e^2 - 4*d*f]*x*Sqrt[a + c*x^2] - 4*Sqrt[c]*d*(c*d - a*f)*Sqrt[e^2 - 4*d*f]*ArcTanh[(Sqrt[c]*x
)/Sqrt[a + c*x^2]] + (2*c*d^2 + a*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]))*Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sq
rt[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])] + Sqrt[2]*a*e*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^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])] - 2*c*d^2*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 + Sq
rt[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])] - a*e^2*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])] + 2*a*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*a^(3/2)*e*f*Sqrt[e^2 - 4*d*f]*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]]))) - 4*a*d*f*Sqr
t[e^2 - 4*d*f]*Sqrt[a + c*x^2]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((c*x^2)/a)])/(4*d^2*f*Sqrt[e^2 - 4*d*f]*x*
Sqrt[1 + (c*x^2)/a])

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IntegrateAlgebraic [C]  time = 0.74, size = 502, normalized size = 0.83 \begin {gather*} -\frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e+a^2 f\&,\frac {-\text {$\#$1}^2 a^2 e f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 c^2 d^2 e \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+a^3 e f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a^2 \sqrt {c} d f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} a^2 \sqrt {c} e^2 f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} c^{5/2} d^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+4 \text {$\#$1} a c^{3/2} d^2 f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-a c^2 d^2 e \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 \text {$\#$1}^3 f-3 \text {$\#$1}^2 \sqrt {c} e-2 \text {$\#$1} a f+4 \text {$\#$1} c d+a \sqrt {c} e}\&\right ]}{d^2 f}-\frac {2 a^{3/2} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d^2}-\frac {c^{3/2} \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{f}-\frac {a \sqrt {a+c x^2}}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Sqrt[a + c*x^2])/(d*x)) - (2*a^(3/2)*e*ArcTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/d^2 - (c^(
3/2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/f - RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2*S
qrt[c]*e*#1^3 + f*#1^4 & , (-(a*c^2*d^2*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]) + a^3*e*f^2*Log[-(Sqrt[c]*
x) + Sqrt[a + c*x^2] - #1] - 2*c^(5/2)*d^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 4*a*c^(3/2)*d^2*f*Log
[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 2*a^2*Sqrt[c]*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - 2
*a^2*Sqrt[c]*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + c^2*d^2*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]
- #1]*#1^2 - a^2*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*S
qrt[c]*e*#1^2 + 2*f*#1^3) & ]/(d^2*f)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2

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maple [B]  time = 0.02, size = 9912, normalized size = 16.41 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}}}{{\left (f x^{2} + e x + d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}}{x^2\,\left (f\,x^2+e\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{x^{2} \left (d + e x + f x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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