3.6.8 \(\int \frac {1+x}{\sqrt {16+18 x+13 x^2+4 x^3+x^4}} \, dx\) [508]
Optimal. Leaf size=39 \[ \frac {1}{2} \log \left (9+4 x+2 x^2+2 \sqrt {16+18 x+13 x^2+4 x^3+x^4}\right ) \]
[Out]
1/2*ln(9+4*x+2*x^2+2*(x^4+4*x^3+13*x^2+18*x+16)^(1/2))
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Rubi [A]
time = 0.03, antiderivative size = 36, normalized size of antiderivative = 0.92, number of steps
used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1694, 1121,
635, 212} \begin {gather*} \frac {1}{2} \tanh ^{-1}\left (\frac {2 (x+1)^2+7}{2 \sqrt {(x+1)^4+7 (x+1)^2+8}}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(1 + x)/Sqrt[16 + 18*x + 13*x^2 + 4*x^3 + x^4],x]
[Out]
ArcTanh[(7 + 2*(1 + x)^2)/(2*Sqrt[8 + 7*(1 + x)^2 + (1 + x)^4])]/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 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 1121
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Rule 1694
Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] && !IGtQ[p, 0]
Rubi steps
\begin {align*} \int \frac {1+x}{\sqrt {16+18 x+13 x^2+4 x^3+x^4}} \, dx &=\text {Subst}\left (\int \frac {x}{\sqrt {8+7 x^2+x^4}} \, dx,x,1+x\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {8+7 x+x^2}} \, dx,x,(1+x)^2\right )\\ &=\text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {7+2 (1+x)^2}{\sqrt {8+7 (1+x)^2+(1+x)^4}}\right )\\ &=\frac {1}{2} \tanh ^{-1}\left (\frac {7+2 (1+x)^2}{2 \sqrt {8+7 (1+x)^2+(1+x)^4}}\right )\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 39, normalized size = 1.00 \begin {gather*} \frac {1}{2} \log \left (9+4 x+2 x^2+2 \sqrt {16+18 x+13 x^2+4 x^3+x^4}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(1 + x)/Sqrt[16 + 18*x + 13*x^2 + 4*x^3 + x^4],x]
[Out]
Log[9 + 4*x + 2*x^2 + 2*Sqrt[16 + 18*x + 13*x^2 + 4*x^3 + x^4]]/2
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 1.26, size = 1422, normalized size = 36.46
| | |
| method |
result |
size |
| | |
| trager |
\(\frac {\ln \left (9+4 x +2 x^{2}+2 \sqrt {x^{4}+4 x^{3}+13 x^{2}+18 x +16}\right )}{2}\) |
\(36\) |
| default |
\(\text {Expression too large to display}\) |
\(1422\) |
| elliptic |
\(\text {Expression too large to display}\) |
\(1422\) |
| | |
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|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1+x)/(x^4+4*x^3+13*x^2+18*x+16)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-2*I*(-1/2*I*(14+2*17^(1/2))^(1/2)-1/2*I*(14-2*17^(1/2))^(1/2))*((1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(
1/2))^(1/2))*(x+1+1/2*I*(14+2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))/(x+1-
1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2)*(x+1-1/2*I*(14+2*17^(1/2))^(1/2))^2*(I*(14+2*17^(1/2))^(1/2)*(x+1+1/2*I*(1
4-2*17^(1/2))^(1/2))/(-1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))/(x+1-1/2*I*(14+2*17^(1/2))^(1/
2)))^(1/2)*(I*(14+2*17^(1/2))^(1/2)*(x+1-1/2*I*(14-2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2
*17^(1/2))^(1/2))/(x+1-1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2)/(1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^
(1/2))/(14+2*17^(1/2))^(1/2)/((x+1+1/2*I*(14+2*17^(1/2))^(1/2))*(x+1-1/2*I*(14+2*17^(1/2))^(1/2))*(x+1+1/2*I*(
14-2*17^(1/2))^(1/2))*(x+1-1/2*I*(14-2*17^(1/2))^(1/2)))^(1/2)*EllipticF(((1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(
14+2*17^(1/2))^(1/2))*(x+1+1/2*I*(14+2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/
2))/(x+1-1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2),((1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))*(-1/2*
I*(14+2*17^(1/2))^(1/2)-1/2*I*(14-2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^(1/2))
/(-1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2))-2*I*(-1/2*I*(14+2*17^(1/2))^(1/2)-1/2*I*(1
4-2*17^(1/2))^(1/2))*((1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^(1/2))*(x+1+1/2*I*(14+2*17^(1/2))^(1/
2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))/(x+1-1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2)*(x+1-1/
2*I*(14+2*17^(1/2))^(1/2))^2*(I*(14+2*17^(1/2))^(1/2)*(x+1+1/2*I*(14-2*17^(1/2))^(1/2))/(-1/2*I*(14-2*17^(1/2)
)^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))/(x+1-1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2)*(I*(14+2*17^(1/2))^(1/2)*(x+1-1/
2*I*(14-2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))/(x+1-1/2*I*(14+2*17^(1/2)
)^(1/2)))^(1/2)/(1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^(1/2))/(14+2*17^(1/2))^(1/2)/((x+1+1/2*I*(1
4+2*17^(1/2))^(1/2))*(x+1-1/2*I*(14+2*17^(1/2))^(1/2))*(x+1+1/2*I*(14-2*17^(1/2))^(1/2))*(x+1-1/2*I*(14-2*17^(
1/2))^(1/2)))^(1/2)*((-1+1/2*I*(14+2*17^(1/2))^(1/2))*EllipticF(((1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(
1/2))^(1/2))*(x+1+1/2*I*(14+2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))/(x+1-
1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2),((1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2))*(-1/2*I*(14+2*1
7^(1/2))^(1/2)-1/2*I*(14-2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^(1/2))/(-1/2*I*
(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2))-I*(14+2*17^(1/2))^(1/2)*EllipticPi(((1/2*I*(14-2*17
^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^(1/2))*(x+1+1/2*I*(14+2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*
I*(14+2*17^(1/2))^(1/2))/(x+1-1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2),(1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^
(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)-1/2*I*(14+2*17^(1/2))^(1/2)),((1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(1
4+2*17^(1/2))^(1/2))*(-1/2*I*(14+2*17^(1/2))^(1/2)-1/2*I*(14-2*17^(1/2))^(1/2))/(1/2*I*(14-2*17^(1/2))^(1/2)-1
/2*I*(14+2*17^(1/2))^(1/2))/(-1/2*I*(14-2*17^(1/2))^(1/2)+1/2*I*(14+2*17^(1/2))^(1/2)))^(1/2)))
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(x^4+4*x^3+13*x^2+18*x+16)^(1/2),x, algorithm="maxima")
[Out]
integrate((x + 1)/sqrt(x^4 + 4*x^3 + 13*x^2 + 18*x + 16), x)
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Fricas [A]
time = 0.40, size = 35, normalized size = 0.90 \begin {gather*} \frac {1}{2} \, \log \left (2 \, x^{2} + 4 \, x + 2 \, \sqrt {x^{4} + 4 \, x^{3} + 13 \, x^{2} + 18 \, x + 16} + 9\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(x^4+4*x^3+13*x^2+18*x+16)^(1/2),x, algorithm="fricas")
[Out]
1/2*log(2*x^2 + 4*x + 2*sqrt(x^4 + 4*x^3 + 13*x^2 + 18*x + 16) + 9)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x + 1}{\sqrt {x^{4} + 4 x^{3} + 13 x^{2} + 18 x + 16}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(x**4+4*x**3+13*x**2+18*x+16)**(1/2),x)
[Out]
Integral((x + 1)/sqrt(x**4 + 4*x**3 + 13*x**2 + 18*x + 16), x)
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Giac [A]
time = 0.40, size = 46, normalized size = 1.18 \begin {gather*} -\frac {1}{2} \, \log \left (\sqrt {2} {\left (\sqrt {2} {\left (x^{2} + 2 \, x\right )} - 2 \, \sqrt {\frac {1}{2} \, {\left (x^{2} + 2 \, x\right )}^{2} + \frac {9}{2} \, x^{2} + 9 \, x + 8}\right )} + 9\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((1+x)/(x^4+4*x^3+13*x^2+18*x+16)^(1/2),x, algorithm="giac")
[Out]
-1/2*log(sqrt(2)*(sqrt(2)*(x^2 + 2*x) - 2*sqrt(1/2*(x^2 + 2*x)^2 + 9/2*x^2 + 9*x + 8)) + 9)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x+1}{\sqrt {x^4+4\,x^3+13\,x^2+18\,x+16}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x + 1)/(18*x + 13*x^2 + 4*x^3 + x^4 + 16)^(1/2),x)
[Out]
int((x + 1)/(18*x + 13*x^2 + 4*x^3 + x^4 + 16)^(1/2), x)
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