3.1.85 \(\int \frac {x^2 (a+b x+c x^2)^{3/2}}{d-f x^2} \, dx\) [85]

Optimal. Leaf size=417 \[ -\frac {\left (b \left (80 c^2 d-3 b^2 f+12 a c f\right )+2 c \left (16 c^2 d-3 b^2 f+12 a c f\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 f^2}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}-\frac {\left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} f^3}+\frac {\sqrt {d} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^3}+\frac {\sqrt {d} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^{3/2} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^3} \]

[Out]

-1/8*(2*c*x+b)*(c*x^2+b*x+a)^(3/2)/c/f-1/128*(128*c^4*d^2+192*a*c^3*d*f+3*b^4*f^2-24*a*b^2*c*f^2+48*c^2*f*(a^2
*f+b^2*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)/f^3+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))*d^(1/2)*(c*d+a*f-b*d^(1/2
)*f^(1/2))^(3/2)/f^3+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))*d^(1/2)*(c*d+a*f+b*d^(1/2)*f^(1/2))^(3/2)/f^3-1/64*(b*(12*a*c*f-3*b^2*f+80*c^2
*d)+2*c*(12*a*c*f-3*b^2*f+16*c^2*d)*x)*(c*x^2+b*x+a)^(1/2)/c^2/f^2

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Rubi [A]
time = 0.63, antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1085, 1084, 1092, 635, 212, 1047, 738} \begin {gather*} -\frac {\left (48 c^2 f \left (a^2 f+b^2 d\right )-24 a b^2 c f^2+192 a c^3 d f+3 b^4 f^2+128 c^4 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} f^3}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (12 a c f-3 b^2 f+16 c^2 d\right )+b \left (12 a c f-3 b^2 f+80 c^2 d\right )\right )}{64 c^2 f^2}+\frac {\sqrt {d} \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 f^3}+\frac {\sqrt {d} \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 f^3}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*((b*(80*c^2*d - 3*b^2*f + 12*a*c*f) + 2*c*(16*c^2*d - 3*b^2*f + 12*a*c*f)*x)*Sqrt[a + b*x + c*x^2])/(c^2
*f^2) - ((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(8*c*f) - ((128*c^4*d^2 + 192*a*c^3*d*f + 3*b^4*f^2 - 24*a*b^2*c
*f^2 + 48*c^2*f*(b^2*d + a^2*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(5/2)*f^3) + (
Sqrt[d]*(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*f^3) + (Sqrt[d]*(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*f^3)

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 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 1084

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

Rule 1085

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

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]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b x+c x^2\right )^{3/2}}{d-f x^2} \, dx &=-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}-\frac {\int \frac {\sqrt {a+b x+c x^2} \left (-\frac {3}{4} \left (3 b^2+4 a c\right ) d f-12 b c d f x-\frac {3}{4} f \left (16 c^2 d-3 \left (b^2-4 a c\right ) f\right ) x^2\right )}{d-f x^2} \, dx}{12 c f^2}\\ &=-\frac {\left (b \left (80 c^2 d-3 b^2 f+12 a c f\right )+2 c \left (16 c^2 d-3 b^2 f+12 a c f\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 f^2}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}+\frac {\int \frac {-\frac {3}{16} d f^2 \left (3 b^4 f-8 b^2 c (10 c d+3 a f)-16 a c^2 (4 c d+5 a f)\right )+48 b c^2 d f^2 (c d+a f) x+\frac {3}{16} f^2 \left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right ) x^2}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{24 c^2 f^4}\\ &=-\frac {\left (b \left (80 c^2 d-3 b^2 f+12 a c f\right )+2 c \left (16 c^2 d-3 b^2 f+12 a c f\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 f^2}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}-\frac {\int \frac {\frac {3}{16} d f^3 \left (3 b^4 f-8 b^2 c (10 c d+3 a f)-16 a c^2 (4 c d+5 a f)\right )-\frac {3}{16} d f^2 \left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right )-48 b c^2 d f^3 (c d+a f) x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{24 c^2 f^5}-\frac {\left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^2 f^3}\\ &=-\frac {\left (b \left (80 c^2 d-3 b^2 f+12 a c f\right )+2 c \left (16 c^2 d-3 b^2 f+12 a c f\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 f^2}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}-\frac {\left (\sqrt {d} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^2\right ) \int \frac {1}{\left (-\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f^{5/2}}+\frac {\left (\sqrt {d} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^2\right ) \int \frac {1}{\left (\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f^{5/2}}-\frac {\left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\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 )}{64 c^2 f^3}\\ &=-\frac {\left (b \left (80 c^2 d-3 b^2 f+12 a c f\right )+2 c \left (16 c^2 d-3 b^2 f+12 a c f\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 f^2}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}-\frac {\left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} f^3}+\frac {\left (\sqrt {d} \left (c d-b \sqrt {d} \sqrt {f}+a f\right )^2\right ) \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 )}{f^{5/2}}-\frac {\left (\sqrt {d} \left (c d+b \sqrt {d} \sqrt {f}+a f\right )^2\right ) \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 )}{f^{5/2}}\\ &=-\frac {\left (b \left (80 c^2 d-3 b^2 f+12 a c f\right )+2 c \left (16 c^2 d-3 b^2 f+12 a c f\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2 f^2}-\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c f}-\frac {\left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2} f^3}+\frac {\sqrt {d} \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 f^3}+\frac {\sqrt {d} \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 f^3}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 1.88, size = 608, normalized size = 1.46 \begin {gather*} \frac {-2 \sqrt {c} f \sqrt {a+x (b+c x)} \left (-3 b^3 f+2 b^2 c f x+8 c^2 x \left (4 c d+5 a f+2 c f x^2\right )+4 b c \left (20 c d+5 a f+6 c f x^2\right )\right )+\left (128 c^4 d^2+192 a c^3 d f+3 b^4 f^2-24 a b^2 c f^2+48 c^2 f \left (b^2 d+a^2 f\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )-64 c^{5/2} d \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 ]}{128 c^{5/2} f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c]*f*Sqrt[a + x*(b + c*x)]*(-3*b^3*f + 2*b^2*c*f*x + 8*c^2*x*(4*c*d + 5*a*f + 2*c*f*x^2) + 4*b*c*(20*
c*d + 5*a*f + 6*c*f*x^2)) + (128*c^4*d^2 + 192*a*c^3*d*f + 3*b^4*f^2 - 24*a*b^2*c*f^2 + 48*c^2*f*(b^2*d + a^2*
f))*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]] - 64*c^(5/2)*d*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[
-(Sqrt[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[
-(Sqrt[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) &
])/(128*c^(5/2)*f^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1592\) vs. \(2(343)=686\).
time = 0.15, size = 1593, normalized size = 3.82

method	result	size
default	\(\text {Expression too large to display}\)	\(1593\)
risch	\(\text {Expression too large to display}\)	\(2506\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f*(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*d/(d*f)^(1/2)/f*(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*d/(d*f)^(1/2)/f*(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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((c*sqrt(4*d*f))/(2*f^2)>0)',
see `assume?

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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(x^2*(c*x^2+b*x+a)^(3/2)/(-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 x^{2} \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx - \int \frac {b x^{3} \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx - \int \frac {c x^{4} \sqrt {a + b x + c x^{2}}}{- d + f x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*x**2*sqrt(a + b*x + c*x**2)/(-d + f*x**2), x) - Integral(b*x**3*sqrt(a + b*x + c*x**2)/(-d + f*x**
2), x) - Integral(c*x**4*sqrt(a + b*x + c*x**2)/(-d + f*x**2), 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(x^2*(c*x^2+b*x+a)^(3/2)/(-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 {x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{d-f\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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