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

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

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Rubi [A]  time = 1.02, 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 = {1071, 1070, 1078, 621, 206, 1033, 724} \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]

-((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])/(64*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) + (Sq
rt[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 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 621

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 724

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 1033

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 1070

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 1071

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 1078

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 [A]  time = 1.02, size = 395, normalized size = 0.95 \begin {gather*} \frac {-\left (\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+x (b+c x)}}\right )\right )-2 \sqrt {c} \left (f \sqrt {a+x (b+c x)} \left (4 b c \left (5 a f+20 c d+6 c f x^2\right )+8 c^2 x \left (5 a f+4 c d+2 c f x^2\right )-3 b^3 f+2 b^2 c f x\right )+32 c^2 \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}-b \sqrt {d}+b \sqrt {f} x-2 c \sqrt {d} x}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )+32 c^2 \sqrt {d} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2} \tanh ^{-1}\left (\frac {-2 \left (a \sqrt {f}+c \sqrt {d} x\right )-b \left (\sqrt {d}+\sqrt {f} x\right )}{2 \sqrt {a+x (b+c x)} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )\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]

(-((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 + x*(b + c*x)])]) - 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)) + 32*c^2*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]*x + b*Sqrt[f]*x)/(2*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*
f]*Sqrt[a + x*(b + c*x)])] + 32*c^2*Sqrt[d]*(c*d + b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)*ArcTanh[(-2*(a*Sqrt[f] + c*S
qrt[d]*x) - b*(Sqrt[d] + Sqrt[f]*x))/(2*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + x*(b + c*x)])]))/(128*c^(
5/2)*f^3)

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IntegrateAlgebraic [C]  time = 2.56, size = 620, normalized size = 1.49 \begin {gather*} -\frac {d \text {RootSum}\left [\text {$\#$1}^4 (-f)+2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d-4 \text {$\#$1} b \sqrt {c} d-a^2 f+b^2 d\&,\frac {2 \text {$\#$1}^2 b c d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1}^2 a b f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a^2 b f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a^2 \sqrt {c} f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^3 d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b^2 \sqrt {c} d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} c^{5/2} d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-4 \text {$\#$1} a c^{3/2} d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b c^2 d^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{\text {$\#$1}^3 f-\text {$\#$1} a f-2 \text {$\#$1} c d+b \sqrt {c} d}\&\right ]}{2 f^3}+\frac {\left (48 a^2 c^2 f^2-24 a b^2 c f^2+192 a c^3 d f+3 b^4 f^2+48 b^2 c^2 d f+128 c^4 d^2\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{128 c^{5/2} f^3}+\frac {\sqrt {a+b x+c x^2} \left (-20 a b c f-40 a c^2 f x+3 b^3 f-2 b^2 c f x-80 b c^2 d-24 b c^2 f x^2-32 c^3 d x-16 c^3 f x^3\right )}{64 c^2 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-80*b*c^2*d + 3*b^3*f - 20*a*b*c*f - 32*c^3*d*x - 2*b^2*c*f*x - 40*a*c^2*f*x - 24*b*c^
2*f*x^2 - 16*c^3*f*x^3))/(64*c^2*f^2) + ((128*c^4*d^2 + 48*b^2*c^2*d*f + 192*a*c^3*d*f + 3*b^4*f^2 - 24*a*b^2*
c*f^2 + 48*a^2*c^2*f^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(128*c^(5/2)*f^3) - (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) & ])/(2*f^3)

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

[Out]

Timed out

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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,x):;OUTPUT:Erro
r: Bad Argument Type

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maple [B]  time = 0.02, size = 4900, normalized size = 11.75 \begin {gather*} \text {output too large to display} \end {gather*}

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)

[Out]

-3/8/f/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-3/128/f/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b
*x+a)^(1/2))*b^4-1/6*d/(d*f)^(1/2)/f*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d
+(d*f)^(1/2)*b)/f)^(3/2)-5/8*d/f^2*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(
d*f)^(1/2)*b)/f)^(1/2)*b-1/2*d^2/f^3*ln(((x-(d*f)^(1/2)/f)*c+1/2*(b*f+2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x-(d*f)^(1
/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))*c^(3/2)+1/6*d/(d*f)^(1/
2)/f*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(3/2)-5/8*d/f
^2*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*b-1/2*d^2
/f^3*ln(((x+(d*f)^(1/2)/f)*c+1/2*(b*f-2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)
*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))*c^(3/2)-1/8/f/c*(c*x^2+b*x+a)^(3/2)*b-3/8/f*(c*x^2+b*x+
a)^(1/2)*x*a+3/64/f/c^2*(c*x^2+b*x+a)^(1/2)*b^3-d^2/(d*f)^(1/2)/f^2/((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a
*f+c*d-(d*f)^(1/2)*b)/f+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+2*((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*((x+(d*f
)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))/(x+(d*f)^(1/2)/f))*
a*c+3/8*d/(d*f)^(1/2)/f/c^(1/2)*ln(((x+(d*f)^(1/2)/f)*c+1/2*(b*f-2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x+(d*f)^(1/2)/f
)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))*a*b-3/8*d/(d*f)^(1/2)/f/c^(1
/2)*ln(((x-(d*f)^(1/2)/f)*c+1/2*(b*f+2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*
(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))*a*b+1/2*d/(d*f)^(1/2)/f/((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2
)*ln((2*(a*f+c*d+(d*f)^(1/2)*b)/f+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+2*((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2
)*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))/(x-(d*f)^
(1/2)/f))*a^2+1/2*d^3/(d*f)^(1/2)/f^3/((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a*f+c*d+(d*f)^(1/2)*b)/f+(b*f+2
*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+2*((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1
/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))/(x-(d*f)^(1/2)/f))*c^2-1/8*d/(d*f)^(1/2)/f*((x-(d
*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*x*b-1/16*d/(d*f)^(
1/2)/f/c*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*b^2
+1/32*d/(d*f)^(1/2)/f/c^(3/2)*ln(((x-(d*f)^(1/2)/f)*c+1/2*(b*f+2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x-(d*f)^(1/2)/f)^
2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))*b^3-3/4*d^2/(d*f)^(1/2)/f^2*ln
(((x-(d*f)^(1/2)/f)*c+1/2*(b*f+2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*
f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))*c^(1/2)*b+1/2*d^2/(d*f)^(1/2)/f^2/((a*f+c*d+(d*f)^(1/2)*b)/f)^
(1/2)*ln((2*(a*f+c*d+(d*f)^(1/2)*b)/f+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+2*((a*f+c*d+(d*f)^(1/2)*b)/f)^
(1/2)*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))/(x-(d
*f)^(1/2)/f))*b^2+d/f^2/((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a*f+c*d+(d*f)^(1/2)*b)/f+(b*f+2*(d*f)^(1/2)*c
)*(x-(d*f)^(1/2)/f)/f+2*((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f
)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))/(x-(d*f)^(1/2)/f))*b*a+d^2/f^3/((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2
)*ln((2*(a*f+c*d+(d*f)^(1/2)*b)/f+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+2*((a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2
)*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))/(x-(d*f)^
(1/2)/f))*b*c+d^2/f^3/((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a*f+c*d-(d*f)^(1/2)*b)/f+(b*f-2*(d*f)^(1/2)*c)*
(x+(d*f)^(1/2)/f)/f+2*((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^
(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))/(x+(d*f)^(1/2)/f))*b*c-1/2*d/(d*f)^(1/2)/f/((a*f+c*d-(d*f)^(1/2)*
b)/f)^(1/2)*ln((2*(a*f+c*d-(d*f)^(1/2)*b)/f+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+2*((a*f+c*d-(d*f)^(1/2)*
b)/f)^(1/2)*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))
/(x+(d*f)^(1/2)/f))*a^2-1/2*d^3/(d*f)^(1/2)/f^3/((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a*f+c*d-(d*f)^(1/2)*b
)/f+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+2*((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*((x+(d*f)^(1/2)/f)^2*c+(b*f-
2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))/(x+(d*f)^(1/2)/f))*c^2+1/8*d/(d*f)^(1/2
)/f*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*x*b+1/16
*d/(d*f)^(1/2)/f/c*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)
^(1/2)*b^2-1/32*d/(d*f)^(1/2)/f/c^(3/2)*ln(((x+(d*f)^(1/2)/f)*c+1/2*(b*f-2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x+(d*f)
^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))*b^3+3/4*d^2/(d*f)^(1
/2)/f^2*ln(((x+(d*f)^(1/2)/f)*c+1/2*(b*f-2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)
*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))*c^(1/2)*b-1/2*d^2/(d*f)^(1/2)/f^2/((a*f+c*d-(d*f)^(1
/2)*b)/f)^(1/2)*ln((2*(a*f+c*d-(d*f)^(1/2)*b)/f+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+2*((a*f+c*d-(d*f)^(1
/2)*b)/f)^(1/2)*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1
/2))/(x+(d*f)^(1/2)/f))*b^2+d/f^2/((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a*f+c*d-(d*f)^(1/2)*b)/f+(b*f-2*(d*
f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+2*((a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*
c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))/(x+(d*f)^(1/2)/f))*b*a+d^2/(d*f)^(1/2)/f^2/((a*f+c*d+
(d*f)^(1/2)*b)/f)^(1/2)*ln((2*(a*f+c*d+(d*f)^(1/2)*b)/f+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+2*((a*f+c*d+
(d*f)^(1/2)*b)/f)^(1/2)*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*
b)/f)^(1/2))/(x-(d*f)^(1/2)/f))*a*c-1/4/f*x*(c*x^2+b*x+a)^(3/2)-3/16*d/f^2*ln(((x+(d*f)^(1/2)/f)*c+1/2*(b*f-2*
(d*f)^(1/2)*c)/f)/c^(1/2)+((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2
)*b)/f)^(1/2))/c^(1/2)*b^2+1/2*d/(d*f)^(1/2)/f*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/
f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*a+1/2*d^2/(d*f)^(1/2)/f^2*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(
d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2)*c-1/4*d/f^2*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(
d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*x*c-3/4*d/f^2*ln(((x-(d*f)^(1/2)/f)*c+1/2*(b*f+2*(d*f)^(1/2)*
c)/f)/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2
))*c^(1/2)*a-3/16*d/f^2*ln(((x-(d*f)^(1/2)/f)*c+1/2*(b*f+2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x-(d*f)^(1/2)/f)^2*c+(b
*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2))/c^(1/2)*b^2-1/2*d/(d*f)^(1/2)/f*((x-
(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*a-1/2*d^2/(d*f)^
(1/2)/f^2*((x-(d*f)^(1/2)/f)^2*c+(b*f+2*(d*f)^(1/2)*c)*(x-(d*f)^(1/2)/f)/f+(a*f+c*d+(d*f)^(1/2)*b)/f)^(1/2)*c+
3/32/f/c*(c*x^2+b*x+a)^(1/2)*x*b^2-3/16/f/c*(c*x^2+b*x+a)^(1/2)*b*a+3/16/f/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x
^2+b*x+a)^(1/2))*b^2*a-1/4*d/f^2*((x+(d*f)^(1/2)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*
f)^(1/2)*b)/f)^(1/2)*x*c-3/4*d/f^2*ln(((x+(d*f)^(1/2)/f)*c+1/2*(b*f-2*(d*f)^(1/2)*c)/f)/c^(1/2)+((x+(d*f)^(1/2
)/f)^2*c+(b*f-2*(d*f)^(1/2)*c)*(x+(d*f)^(1/2)/f)/f+(a*f+c*d-(d*f)^(1/2)*b)/f)^(1/2))*c^(1/2)*a

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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?` for more details)Is ((c*sqrt(4*d*f))/(2*f^2)    +b/(2*f))    ^2    -(c*((b*sqrt(4*d*f))
/(2*f)                  +(c*d)/f+a))     /f^2 positive, negative or zero?

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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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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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