3.1.89 \(\int \frac {(a+b x+c x^2)^{3/2}}{x^2 (d-f x^2)} \, dx\) [89]

Optimal. Leaf size=463 \[ \frac {3 (3 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}-\frac {3 \sqrt {a} b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 d}+\frac {3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d}-\frac {\left (8 c^2 d+3 b^2 f+12 a c f\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d f}+\frac {\left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^{3/2} f}+\frac {\left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^{3/2} f} \]

[Out]

-(c*x^2+b*x+a)^(3/2)/d/x-3/2*b*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))*a^(1/2)/d+3/8*(4*a*c+b^2)*ar
ctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/d/c^(1/2)-1/8*(12*a*c*f+3*b^2*f+8*c^2*d)*arctanh(1/2*(2*c*x+b
)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/d/f/c^(1/2)+1/2*arctanh(1/2*(b*d^(1/2)-2*a*f^(1/2)+x*(2*c*d^(1/2)-b*f^(1/2)))/(
c*x^2+b*x+a)^(1/2)/(c*d+a*f-b*d^(1/2)*f^(1/2))^(1/2))*(c*d+a*f-b*d^(1/2)*f^(1/2))^(3/2)/d^(3/2)/f+1/2*arctanh(
1/2*(b*d^(1/2)+2*a*f^(1/2)+x*(2*c*d^(1/2)+b*f^(1/2)))/(c*x^2+b*x+a)^(1/2)/(c*d+a*f+b*d^(1/2)*f^(1/2))^(1/2))*(
c*d+a*f+b*d^(1/2)*f^(1/2))^(3/2)/d^(3/2)/f+3/4*(2*c*x+3*b)*(c*x^2+b*x+a)^(1/2)/d-1/4*(2*c*x+5*b)*(c*x^2+b*x+a)
^(1/2)/d

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Rubi [A]
time = 0.74, antiderivative size = 463, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6857, 746, 828, 857, 635, 212, 738, 992, 1092, 1047} \begin {gather*} -\frac {\left (12 a c f+3 b^2 f+8 c^2 d\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d f}+\frac {3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d}+\frac {\left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2} \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 d^{3/2} f}+\frac {\left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2} \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 d^{3/2} f}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}+\frac {3 (3 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {3 \sqrt {a} b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(3*b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*d) - ((5*b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(4*d) - (a + b*x + c*x^2
)^(3/2)/(d*x) - (3*Sqrt[a]*b*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(2*d) + (3*(b^2 + 4*a*c)*
ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c]*d) - ((8*c^2*d + 3*b^2*f + 12*a*c*f)*ArcTan
h[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*Sqrt[c]*d*f) + ((c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)*Arc
Tanh[(b*Sqrt[d] - 2*a*Sqrt[f] + (2*c*Sqrt[d] - b*Sqrt[f])*x)/(2*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b
*x + c*x^2])])/(2*d^(3/2)*f) + ((c*d + b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*
Sqrt[d] + b*Sqrt[f])*x)/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + b*x + c*x^2])])/(2*d^(3/2)*f)

Rule 212

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

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 746

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 1))), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 992

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b*(3*p + 2*q) + 2
*c*(p + q)*x)*(a + b*x + c*x^2)^(p - 1)*((d + f*x^2)^(q + 1)/(2*f*(p + q)*(2*p + 2*q + 1))), x] - Dist[1/(2*f*
(p + q)*(2*p + 2*q + 1)), Int[(a + b*x + c*x^2)^(p - 2)*(d + f*x^2)^q*Simp[b^2*d*(p - 1)*(2*p + q) - (p + q)*(
b^2*d*(1 - p) - 2*a*(c*d - a*f*(2*p + 2*q + 1))) - (2*b*(c*d - a*f)*(1 - p)*(2*p + q) - 2*(p + q)*b*(2*c*d*(2*
p + q) - (c*d + a*f)*(2*p + 2*q + 1)))*x + (b^2*f*p*(1 - p) + 2*c*(p + q)*(c*d*(2*p - 1) - a*f*(4*p + 2*q - 1)
))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[p, 1] && NeQ[p + q, 0] && NeQ
[2*p + 2*q + 1, 0] &&  !IGtQ[p, 0] &&  !IGtQ[q, 0]

Rule 1047

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

Rule 1092

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

Rule 6857

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^2 \left (d-f x^2\right )} \, dx &=\int \left (\frac {\left (a+b x+c x^2\right )^{3/2}}{d x^2}+\frac {f \left (a+b x+c x^2\right )^{3/2}}{d \left (d-f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b x+c x^2\right )^{3/2}}{x^2} \, dx}{d}+\frac {f \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d-f x^2} \, dx}{d}\\ &=-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}+\frac {\int \frac {\frac {1}{4} \left (5 b^2 d+4 a (c d+2 a f)\right )+4 b (c d+a f) x+\frac {1}{4} \left (8 c^2 d+3 b^2 f+12 a c f\right ) x^2}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{2 d}+\frac {3 \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{x} \, dx}{2 d}\\ &=\frac {3 (3 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}-\frac {3 \int \frac {-4 a b c-c \left (b^2+4 a c\right ) x}{x \sqrt {a+b x+c x^2}} \, dx}{8 c d}-\frac {\int \frac {-\frac {1}{4} d \left (8 c^2 d+3 b^2 f+12 a c f\right )-\frac {1}{4} f \left (5 b^2 d+4 a (c d+2 a f)\right )-4 b f (c d+a f) x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{2 d f}-\frac {\left (8 c^2 d+3 b^2 f+12 a c f\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 d f}\\ &=\frac {3 (3 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}+\frac {(3 a b) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{2 d}+\frac {\left (3 \left (b^2+4 a c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 d}-\frac {\left (c d-b \sqrt {d} \sqrt {f}+a f\right )^2 \int \frac {1}{\left (-\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 d^{3/2} \sqrt {f}}+\frac {\left (c d+b \sqrt {d} \sqrt {f}+a f\right )^2 \int \frac {1}{\left (\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 d^{3/2} \sqrt {f}}-\frac {\left (8 c^2 d+3 b^2 f+12 a c f\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{4 d f}\\ &=\frac {3 (3 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}-\frac {\left (8 c^2 d+3 b^2 f+12 a c f\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d f}-\frac {(3 a b) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x}{\sqrt {a+b x+c x^2}}\right )}{d}+\frac {\left (3 \left (b^2+4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{4 d}+\frac {\left (c d-b \sqrt {d} \sqrt {f}+a f\right )^2 \text {Subst}\left (\int \frac {1}{4 c d f-4 b \sqrt {d} f^{3/2}+4 a f^2-x^2} \, dx,x,\frac {b \sqrt {d} \sqrt {f}-2 a f-\left (-2 c \sqrt {d} \sqrt {f}+b f\right ) x}{\sqrt {a+b x+c x^2}}\right )}{d^{3/2} \sqrt {f}}-\frac {\left (c d+b \sqrt {d} \sqrt {f}+a f\right )^2 \text {Subst}\left (\int \frac {1}{4 c d f+4 b \sqrt {d} f^{3/2}+4 a f^2-x^2} \, dx,x,\frac {-b \sqrt {d} \sqrt {f}-2 a f-\left (2 c \sqrt {d} \sqrt {f}+b f\right ) x}{\sqrt {a+b x+c x^2}}\right )}{d^{3/2} \sqrt {f}}\\ &=\frac {3 (3 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {(5 b+2 c x) \sqrt {a+b x+c x^2}}{4 d}-\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}-\frac {3 \sqrt {a} b \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 d}+\frac {3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d}-\frac {\left (8 c^2 d+3 b^2 f+12 a c f\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} d f}+\frac {\left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^{3/2} f}+\frac {\left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 d^{3/2} f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.69, size = 541, normalized size = 1.17 \begin {gather*} \frac {-2 a f \sqrt {a+x (b+c x)}+6 \sqrt {a} b f x \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )+2 c^{3/2} d x \log \left (f \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )-x \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b c^2 d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )+b^3 d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a^2 b f^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 c^{5/2} d^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 b^2 \sqrt {c} d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-4 a c^{3/2} d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a^2 \sqrt {c} f^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 b c d f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 a b f^2 \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{2 d f x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*f*Sqrt[a + x*(b + c*x)] + 6*Sqrt[a]*b*f*x*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]] + 2*c^(3/
2)*d*x*Log[f*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - x*RootSum[b^2*d - a^2*f - 4*b*Sqrt[c]*d*#1 + 4*c
*d*#1^2 + 2*a*f*#1^2 - f*#1^4 & , (b*c^2*d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + b^3*d*f*Log[-(Sq
rt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^2*b*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*c^(5/2)*
d^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*b^2*Sqrt[c]*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*
x^2] - #1]*#1 - 4*a*c^(3/2)*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 2*a^2*Sqrt[c]*f^2*Log[-(Sq
rt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + 2*b*c*d*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 + 2*
a*b*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2)/(b*Sqrt[c]*d - 2*c*d*#1 - a*f*#1 + f*#1^3) & ])/(
2*d*f*x)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1786\) vs. \(2(367)=734\).
time = 0.15, size = 1787, normalized size = 3.86

method	result	size
default	\(\text {Expression too large to display}\)	\(1787\)
risch	\(\text {Expression too large to display}\)	\(2283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/x^2/(-f*x^2+d),x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/a/x*(c*x^2+b*x+a)^(5/2)+3/2*b/a*(1/3*(c*x^2+b*x+a)^(3/2)+1/2*b*(1/4*(2*c*x+b)/c*(c*x^2+b*x+a)^(1/2)+1/
8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))+a*((c*x^2+b*x+a)^(1/2)+1/2*b*ln((1/2*b+c*x)
/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)))+4*c/a*(1/8*(2*c*
x+b)/c*(c*x^2+b*x+a)^(3/2)+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)/c*(c*x^2+b*x+a)^(1/2)+1/8*(4*a*c-b^2)/c^(3/2)*ln(
(1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))))+1/2*f/d/(d*f)^(1/2)*(1/3*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1
/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(3/2)+1/2/f*(-2*c*(d*f)^(1/2)+b*f)*(1/4*(2*c*(x+(d*f)
^(1/2)/f)+1/f*(-2*c*(d*f)^(1/2)+b*f))/c*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/
f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2)+1/8*(4*c/f*(-b*(d*f)^(1/2)+f*a+c*d)-1/f^2*(-2*c*(d*f)^(1/2)+b*f)^2)/c^(3/2)*
ln((1/2/f*(-2*c*(d*f)^(1/2)+b*f)+c*(x+(d*f)^(1/2)/f))/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f
)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2)))+1/f*(-b*(d*f)^(1/2)+f*a+c*d)*(((x+(d*f)^(1/2)/f)^2*c
+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2)+1/2/f*(-2*c*(d*f)^(1/2)+b*f)
*ln((1/2/f*(-2*c*(d*f)^(1/2)+b*f)+c*(x+(d*f)^(1/2)/f))/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*
f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)^(1/2)+f*a+c*d))^(1/2))/c^(1/2)-1/f*(-b*(d*f)^(1/2)+f*a+c*d)/(1/f*(-b*(d*f)^
(1/2)+f*a+c*d))^(1/2)*ln((2/f*(-b*(d*f)^(1/2)+f*a+c*d)+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+2*(1/f*(-b
*(d*f)^(1/2)+f*a+c*d))^(1/2)*((x+(d*f)^(1/2)/f)^2*c+1/f*(-2*c*(d*f)^(1/2)+b*f)*(x+(d*f)^(1/2)/f)+1/f*(-b*(d*f)
^(1/2)+f*a+c*d))^(1/2))/(x+(d*f)^(1/2)/f))))-1/2*f/d/(d*f)^(1/2)*(1/3*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+
b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+f*a+c*d)/f)^(3/2)+1/2*(2*c*(d*f)^(1/2)+b*f)/f*(1/4*(2*c*(x-(d*f)^(1/2)
/f)+(2*c*(d*f)^(1/2)+b*f)/f)/c*((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)
+f*a+c*d)/f)^(1/2)+1/8*(4*c*(b*(d*f)^(1/2)+f*a+c*d)/f-(2*c*(d*f)^(1/2)+b*f)^2/f^2)/c^(3/2)*ln((1/2*(2*c*(d*f)^
(1/2)+b*f)/f+c*(x-(d*f)^(1/2)/f))/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*
(d*f)^(1/2)+f*a+c*d)/f)^(1/2)))+(b*(d*f)^(1/2)+f*a+c*d)/f*(((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(
d*f)^(1/2)/f)+(b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2)+1/2*(2*c*(d*f)^(1/2)+b*f)/f*ln((1/2*(2*c*(d*f)^(1/2)+b*f)/f+c*(
x-(d*f)^(1/2)/f))/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+f*a+
c*d)/f)^(1/2))/c^(1/2)-(b*(d*f)^(1/2)+f*a+c*d)/f/((b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2)*ln((2*(b*(d*f)^(1/2)+f*a+c*
d)/f+(2*c*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+2*((b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(2*c
*(d*f)^(1/2)+b*f)/f*(x-(d*f)^(1/2)/f)+(b*(d*f)^(1/2)+f*a+c*d)/f)^(1/2))/(x-(d*f)^(1/2)/f))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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+b*x+a)^(3/2)/x^2/(-f*x^2+d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a \sqrt {a + b x + c x^{2}}}{- d x^{2} + f x^{4}}\, dx - \int \frac {b x \sqrt {a + b x + c x^{2}}}{- d x^{2} + f x^{4}}\, dx - \int \frac {c x^{2} \sqrt {a + b x + c x^{2}}}{- d x^{2} + f x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*sqrt(a + b*x + c*x**2)/(-d*x**2 + f*x**4), x) - Integral(b*x*sqrt(a + b*x + c*x**2)/(-d*x**2 + f*x
**4), x) - Integral(c*x**2*sqrt(a + b*x + c*x**2)/(-d*x**2 + f*x**4), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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