3.1.95 \(\int \frac {x^3}{\sqrt {a+b x+c x^2} (d-f x^2)} \, dx\) [95]

Optimal. Leaf size=287 \[ -\frac {\sqrt {a+b x+c x^2}}{c f}+\frac {b \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} f}-\frac {d \tanh ^{-1}\left (\frac {b \sqrt {d}-2 a \sqrt {f}+\left (2 c \sqrt {d}-b \sqrt {f}\right ) x}{2 \sqrt {c d-b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^{3/2} \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\frac {d \tanh ^{-1}\left (\frac {b \sqrt {d}+2 a \sqrt {f}+\left (2 c \sqrt {d}+b \sqrt {f}\right ) x}{2 \sqrt {c d+b \sqrt {d} \sqrt {f}+a f} \sqrt {a+b x+c x^2}}\right )}{2 f^{3/2} \sqrt {c d+b \sqrt {d} \sqrt {f}+a f}} \]

[Out]

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

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Rubi [A]
time = 0.41, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6857, 654, 635, 212, 1047, 738} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} f}-\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 f^{3/2} \sqrt {a f+b \left (-\sqrt {d}\right ) \sqrt {f}+c d}}+\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 f^{3/2} \sqrt {a f+b \sqrt {d} \sqrt {f}+c d}}-\frac {\sqrt {a+b x+c x^2}}{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]

-(Sqrt[a + b*x + c*x^2]/(c*f)) + (b*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*c^(3/2)*f) - (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*f^(3/2)*Sqrt[c*d - b*Sqrt[d]*Sqrt[f] + a*f]) + (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*f^(3/2)*Sqrt[c*d
+ b*Sqrt[d]*Sqrt[f] + a*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 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 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 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} \left (d-f x^2\right )} \, dx &=\int \left (-\frac {x}{f \sqrt {a+b x+c x^2}}+\frac {d x}{f \sqrt {a+b x+c x^2} \left (d-f x^2\right )}\right ) \, dx\\ &=-\frac {\int \frac {x}{\sqrt {a+b x+c x^2}} \, dx}{f}+\frac {d \int \frac {x}{\sqrt {a+b x+c x^2} \left (d-f x^2\right )} \, dx}{f}\\ &=-\frac {\sqrt {a+b x+c x^2}}{c f}+\frac {b \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c f}+\frac {d \int \frac {1}{\left (-\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f}+\frac {d \int \frac {1}{\left (\sqrt {d} \sqrt {f}-f x\right ) \sqrt {a+b x+c x^2}} \, dx}{2 f}\\ &=-\frac {\sqrt {a+b x+c x^2}}{c f}+\frac {b \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 )}{c f}-\frac {d \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}-\frac {d \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}\\ &=-\frac {\sqrt {a+b x+c x^2}}{c f}+\frac {b \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} f}-\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 f^{3/2} \sqrt {c d-b \sqrt {d} \sqrt {f}+a f}}+\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 f^{3/2} \sqrt {c d+b \sqrt {d} \sqrt {f}+a 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 = 0.39, size = 210, normalized size = 0.73 \begin {gather*} -\frac {\frac {2 \sqrt {a+x (b+c x)}}{c}+\frac {b \log \left (c f \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{c^{3/2}}+d \text {RootSum}\left [b^2 d-a^2 f-4 b \sqrt {c} d \text {$\#$1}+4 c d \text {$\#$1}^2+2 a f \text {$\#$1}^2-f \text {$\#$1}^4\&,\frac {a \log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right )-\log \left (-\sqrt {c} x+\sqrt {a+b x+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{-b \sqrt {c} d+2 c d \text {$\#$1}+a f \text {$\#$1}-f \text {$\#$1}^3}\&\right ]}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*Sqrt[a + x*(b + c*x)])/c + (b*Log[c*f*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(3/2) + d*Roo
tSum[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 & , (a*Log[-(Sqrt[c]*x) + Sqrt[a + b*
x + c*x^2] - #1] - 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) & ])/f

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Maple [A]
time = 0.13, size = 409, normalized size = 1.43

method	result	size
default	\(-\frac {\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}}{f}+\frac {d \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}+\frac {d \ln \left (\frac {\frac {2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}\)	\(409\)
risch	\(-\frac {\sqrt {c \,x^{2}+b x +a}}{c f}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}} f}+\frac {d \ln \left (\frac {\frac {2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x -\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (2 c \sqrt {d f}+b f \right ) \left (x -\frac {\sqrt {d f}}{f}\right )}{f}+\frac {b \sqrt {d f}+f a +c d}{f}}}{x -\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {\frac {b \sqrt {d f}+f a +c d}{f}}}+\frac {d \ln \left (\frac {\frac {-2 b \sqrt {d f}+2 f a +2 c d}{f}+\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+2 \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}\, \sqrt {\left (x +\frac {\sqrt {d f}}{f}\right )^{2} c +\frac {\left (-2 c \sqrt {d f}+b f \right ) \left (x +\frac {\sqrt {d f}}{f}\right )}{f}+\frac {-b \sqrt {d f}+f a +c d}{f}}}{x +\frac {\sqrt {d f}}{f}}\right )}{2 f^{2} \sqrt {\frac {-b \sqrt {d f}+f a +c d}{f}}}\)	\(410\)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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