3.1.77 \(\int \frac {x^3 \sqrt {a+b x+c x^2}}{d-f x^2} \, dx\) [77]

Optimal. Leaf size=392 \[ -\frac {d \sqrt {a+b x+c x^2}}{f^2}+\frac {b (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2 f}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac {b d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}-\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} f}-\frac {d \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^{5/2}}+\frac {d \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.60, antiderivative size = 392, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {6857, 654, 626, 635, 212, 1035, 1092, 1047, 738} \begin {gather*} -\frac {b \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 )}{16 c^{5/2} f}+\frac {b (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2 f}-\frac {d \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c 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 f^{5/2}}+\frac {d \sqrt {a f+b \sqrt {d} \sqrt {f}+c 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 f^{5/2}}-\frac {d \sqrt {a+b x+c x^2}}{f^2}-\frac {b d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Sqrt[a + b*x + c*x^2])/f^2) + (b*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(8*c^2*f) - (a + b*x + c*x^2)^(3/2)/(
3*c*f) - (b*d*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c]*f^2) - (b*(b^2 - 4*a*c)*ArcTa
nh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(5/2)*f) - (d*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]*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*f^(5/2)) + (d*Sqrt[c*d + b*Sqrt[d]*Sqrt[f] + a*f]*ArcTanh[(b*Sqrt[d] + 2*a*Sqrt[f] + (2*c*Sq
rt[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^(5/2))

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 626

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

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 654

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

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 1035

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[h*(a + b*x + c*x^2)^p*((d + f*x^2)^(q + 1)/(2*f*(p + q + 1))), x] - Dist[1/(2*f*(p + q + 1)), Int[(a + b*x +
c*x^2)^(p - 1)*(d + f*x^2)^q*Simp[h*p*(b*d) + a*(-2*g*f)*(p + q + 1) + (2*h*p*(c*d - a*f) + b*(-2*g*f)*(p + q
+ 1))*x + (h*p*((-b)*f) + c*(-2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}, x] && NeQ
[b^2 - 4*a*c, 0] && GtQ[p, 0] && NeQ[p + q + 1, 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 {x^3 \sqrt {a+b x+c x^2}}{d-f x^2} \, dx &=\int \left (-\frac {x \sqrt {a+b x+c x^2}}{f}+\frac {d x \sqrt {a+b x+c x^2}}{f \left (d-f x^2\right )}\right ) \, dx\\ &=-\frac {\int x \sqrt {a+b x+c x^2} \, dx}{f}+\frac {d \int \frac {x \sqrt {a+b x+c x^2}}{d-f x^2} \, dx}{f}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{f^2}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f}+\frac {d \int \frac {\frac {b d}{2}+(c d+a f) x+\frac {1}{2} b f x^2}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f^2}+\frac {b \int \sqrt {a+b x+c x^2} \, dx}{2 c f}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{f^2}+\frac {b (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2 f}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac {d \int \frac {-b d f-f (c d+a f) x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f^3}-\frac {(b d) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 f^2}-\frac {\left (b \left (b^2-4 a c\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^2 f}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{f^2}+\frac {b (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2 f}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac {(b d) \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 )}{f^2}-\frac {\left (b \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 )}{8 c^2 f}+\frac {\left (d \left (c d-b \sqrt {d} \sqrt {f}+a f\right )\right ) \int \frac {1}{\left (-\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f^2}+\frac {\left (d \left (c d+b \sqrt {d} \sqrt {f}+a f\right )\right ) \int \frac {1}{\left (\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f^2}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{f^2}+\frac {b (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2 f}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac {b d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}-\frac {b \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 )}{16 c^{5/2} f}-\frac {\left (d \left (c d-b \sqrt {d} \sqrt {f}+a f\right )\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^2}-\frac {\left (d \left (c d+b \sqrt {d} \sqrt {f}+a f\right )\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^2}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{f^2}+\frac {b (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2 f}-\frac {\left (a+b x+c x^2\right )^{3/2}}{3 c f}-\frac {b d \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}-\frac {b \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 )}{16 c^{5/2} f}-\frac {d \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \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^{5/2}}+\frac {d \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \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^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.75, size = 409, normalized size = 1.04 \begin {gather*} \frac {-2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-3 b^2 f+2 c f (4 a+b x)+8 c^2 \left (3 d+f x^2\right )\right )+3 b \left (8 c^2 d+b^2 f-4 a c f\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )-24 c^{5/2} d \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {b^2 d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-a^2 f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-2 b \sqrt {c} d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c d \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+a f \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{b \sqrt {c} d-2 c d \text {$\#$1}-a f \text {$\#$1}+f \text {$\#$1}^3}\&\right ]}{48 c^{5/2} f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-3*b^2*f + 2*c*f*(4*a + b*x) + 8*c^2*(3*d + f*x^2)) + 3*b*(8*c^2*d + b^2*f
- 4*a*c*f)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]] - 24*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^2*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*c*d*Log
[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*b*Sqrt[
c]*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 + c*d*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1
^2 + a*f*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) & ])/
(48*c^(5/2)*f^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(862\) vs. \(2(306)=612\).
time = 0.18, size = 863, normalized size = 2.20 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(c*x^2+b*x+a)^(1/2)/(-f*x^2+d),x,method=_RETURNVERBOSE)

[Out]

-1/f*(1/3*(c*x^2+b*x+a)^(3/2)/c-1/2*b/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/f^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)+1/2/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/f^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)+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^3*(c*x^2+b*x+a)^(1/2)/(-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^3*(c*x^2+b*x+a)^(1/2)/(-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 {x^{3} \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**3*(c*x**2+b*x+a)**(1/2)/(-f*x**2+d),x)

[Out]

-Integral(x**3*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^3*(c*x^2+b*x+a)^(1/2)/(-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^3\,\sqrt {c\,x^2+b\,x+a}}{d-f\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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