3.1.64 \(\int \frac {x^3}{\sqrt {a+c x^2} (d+e x+f x^2)} \, dx\) [64]

Optimal. Leaf size=380 \[ \frac {\sqrt {a+c x^2}}{c f}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f^2}-\frac {\left (2 d e f-\left (e^2-d f\right ) \left (e-\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\left (2 d e f-\left (e^2-d f\right ) \left (e+\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}} \]

[Out]

-e*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/f^2/c^(1/2)+(c*x^2+a)^(1/2)/c/f-1/2*arctanh(1/2*(2*a*f-c*x*(e-(-4*d*f+e^
2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2))*(2*d*e*f-(-d*f+e^2)*(e-
(-4*d*f+e^2)^(1/2)))/f^2*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)+1/2*arc
tanh(1/2*(2*a*f-c*x*(e+(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2)
))^(1/2))*(2*d*e*f-(-d*f+e^2)*(e+(-4*d*f+e^2)^(1/2)))/f^2*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(
-4*d*f+e^2)^(1/2)))^(1/2)

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Rubi [A]
time = 0.71, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6860, 223, 212, 267, 1048, 739} \begin {gather*} -\frac {\left (2 d e f-\left (e^2-d f\right ) \left (e-\sqrt {e^2-4 d f}\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (2 d e f-\left (e^2-d f\right ) \left (\sqrt {e^2-4 d f}+e\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{\sqrt {c} f^2}+\frac {\sqrt {a+c x^2}}{c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a + c*x^2]/(c*f) - (e*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(Sqrt[c]*f^2) - ((2*d*e*f - (e^2 - d*f)*(e -
Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*S
qrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*f^2*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^
2 - 4*d*f])]) + ((2*d*e*f - (e^2 - d*f)*(e + Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)
/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*f^2*Sqrt[e^2 - 4*d
*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 1048

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.38, size = 312, normalized size = 0.82 \begin {gather*} \frac {f \sqrt {a+c x^2}+\sqrt {c} e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )-c \text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-a d f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 \sqrt {c} d e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+d f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{c f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f*Sqrt[a + c*x^2] + Sqrt[c]*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]] - c*RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*
d*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (a*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - a*d*f*Lo
g[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*Sqrt[c]*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - e^2*Log[-
(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + d*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4
*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ])/(c*f^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(712\) vs. \(2(335)=670\).
time = 0.14, size = 713, normalized size = 1.88

method	result	size
default	\(\frac {\sqrt {c \,x^{2}+a}}{c f}-\frac {e \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{f^{2} \sqrt {c}}-\frac {\left (e^{3}-3 d e f +e^{2} \sqrt {-4 d f +e^{2}}-d f \sqrt {-4 d f +e^{2}}\right ) \sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{3} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}-\frac {\left (-d f \sqrt {-4 d f +e^{2}}+e^{2} \sqrt {-4 d f +e^{2}}+3 d e f -e^{3}\right ) \sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{2 f^{3} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}\)	\(713\)
risch	\(\text {Expression too large to display}\)	\(2397\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(f*x^2+e*x+d)/(c*x^2+a)^(1/2),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(4*d*f-%e^2>0)', see `assume?`
for more det

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(f*x**2+e*x+d)/(c*x**2+a)**(1/2),x)

[Out]

Integral(x**3/(sqrt(a + c*x**2)*(d + e*x + f*x**2)), x)

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Giac [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/(f*x^2+e*x+d)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

Timed out

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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+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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