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

Optimal. Leaf size=341 \[ -\frac {2 d \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{f \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}-\frac {2 (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {d \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \sqrt {f} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {d \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \sqrt {f} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 1.04, antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6725, 636, 1018, 1033, 724, 206} \[ -\frac {2 d \left (a \left (2 a c f+b^2 (-f)+2 c^2 d\right )+b c x (c d-a f)\right )}{f \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (b^2 d f-(a f+c d)^2\right )}-\frac {2 (2 a+b x)}{f \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {d \tanh ^{-1}\left (\frac {-2 a \sqrt {f}+x \left (2 c \sqrt {d}-b \sqrt {f}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}\right )}{2 \sqrt {f} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {d \tanh ^{-1}\left (\frac {2 a \sqrt {f}+x \left (b \sqrt {f}+2 c \sqrt {d}\right )+b \sqrt {d}}{2 \sqrt {a+b x+c x^2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}\right )}{2 \sqrt {f} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
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 1018

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^(q + 1)*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*(
2*a*f)) + c*(g*(2*c^2*d + b^2*f - c*(2*a*f)) - h*(b*c*d + a*b*f))*x))/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)
*(p + 1)), x] + Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1)*(d + f
*x^2)^q*Simp[(b*h - 2*g*c)*((c*d - a*f)^2 - (b*d)*(-(b*f)))*(p + 1) + (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*
(c*d - a*f)))*(a*f*(p + 1) - c*d*(p + 2)) - (2*f*((g*c)*(-(b*(c*d + a*f))) + (g*b - a*h)*(2*c^2*d + b^2*f - c*
(2*a*f)))*(p + q + 2) - (b^2*(g*f) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(b*f*(p + 1)))*x - c*f*(b^2*(g*f
) - b*(h*c*d + a*h*f) + 2*(g*c*(c*d - a*f)))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}
, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1
])

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 6725

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

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Mathematica [A]  time = 1.20, size = 414, normalized size = 1.21 \[ \frac {1}{2} \left (\frac {8 a^3 f+4 a^2 (b f x+2 c d)-4 a b d (b-3 c x)-4 b^3 d x}{\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \left (f \left (b^2 d-a^2 f\right )-2 a c d f-c^2 d^2\right )}-\frac {d \log \left (\sqrt {d} \sqrt {f}-f x\right )}{\sqrt {f} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}}-\frac {d \log \left (\sqrt {d} \sqrt {f}+f x\right )}{\sqrt {f} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {d \log \left (\sqrt {d} \left (2 \sqrt {a+x (b+c x)} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}+2 a \sqrt {f}-b \sqrt {d}+b \sqrt {f} x-2 c \sqrt {d} x\right )\right )}{\sqrt {f} \left (a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d\right )^{3/2}}+\frac {d \log \left (\sqrt {d} \left (2 \left (\sqrt {a+x (b+c x)} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}+a \sqrt {f}+c \sqrt {d} x\right )+b \left (\sqrt {d}+\sqrt {f} x\right )\right )\right )}{\sqrt {f} \left (a f+b \sqrt {d} \sqrt {f}+c d\right )^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^3*f - 4*b^3*d*x - 4*a*b*d*(b - 3*c*x) + 4*a^2*(2*c*d + b*f*x))/((b^2 - 4*a*c)*(-(c^2*d^2) - 2*a*c*d*f +
f*(b^2*d - a^2*f))*Sqrt[a + x*(b + c*x)]) - (d*Log[Sqrt[d]*Sqrt[f] - f*x])/(Sqrt[f]*(c*d + b*Sqrt[d]*Sqrt[f] +
a*f)^(3/2)) - (d*Log[Sqrt[d]*Sqrt[f] + f*x])/(Sqrt[f]*(c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)) + (d*Log[Sqrt[d]
*(-(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)])])/(Sqrt[f]*(c*d - b*Sqrt[d]*Sqrt[f] + a*f)^(3/2)) + (d*Log[Sqrt[d]*(b*(Sqrt[d] + Sqrt[f]*x) + 2*(a
*Sqrt[f] + c*Sqrt[d]*x + Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*Sqrt[a + x*(b + c*x)]))])/(Sqrt[f]*(c*d + b*Sqrt[
d]*Sqrt[f] + a*f)^(3/2)))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valueint()  Bad Argument Typ
e

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maple [B]  time = 0.02, size = 1480, normalized size = 4.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f/c/(c*x^2+b*x+a)^(1/2)+2/f*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/f*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-1/
2/f*d/(a*f+c*d-(d*f)^(1/2)*b)/((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)-2/f^2*d/(a*f+c*d-(d*f)^(1/2)*b)/(4*a*c-b^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)*(d*f)^(1/2)*x*c^2+1/f*d/(a*f+c*d-(d*f)^(1/2)*b)/(4*a*c-b^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*b*c-1/f^2*
d/(a*f+c*d-(d*f)^(1/2)*b)/(4*a*c-b^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)*(d*f)^(1/2)*b*c+1/2/f*d/(a*f+c*d-(d*f)^(1/2)*b)/(4*a*c-b^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^2+1/2/f*d/(a*f+c*d-(d*f)^(1/2)*b)/
((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))-1/2/f*d/(a*f+c*d+(d*f)^(1/2)*b)/((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)+2/f^2*d/(a*f+c*d+(d*f)^(1/2)*b)/(4*a*c-b^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)*(d*f)^(1/2)*x*c^
2+1/f*d/(a*f+c*d+(d*f)^(1/2)*b)/(4*a*c-b^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*b*c+1/f^2*d/(a*f+c*d+(d*f)^(1/2)*b)/(4*a*c-b^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)*(d*f)^(1/2)*b*c+1/2/f*d/(a*f+c*d+(d*f)^(
1/2)*b)/(4*a*c-b^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^2+1/2/f*d/(a*f+c*d+(d*f)^(1/2)*b)/((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))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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 zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{3}}{- a d \sqrt {a + b x + c x^{2}} + a f x^{2} \sqrt {a + b x + c x^{2}} - b d x \sqrt {a + b x + c x^{2}} + b f x^{3} \sqrt {a + b x + c x^{2}} - c d x^{2} \sqrt {a + b x + c x^{2}} + c f x^{4} \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**3/(-a*d*sqrt(a + b*x + c*x**2) + a*f*x**2*sqrt(a + b*x + c*x**2) - b*d*x*sqrt(a + b*x + c*x**2) +
b*f*x**3*sqrt(a + b*x + c*x**2) - c*d*x**2*sqrt(a + b*x + c*x**2) + c*f*x**4*sqrt(a + b*x + c*x**2)), x)

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