3.13.93 \(\int \frac {x^2 \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx\)

Optimal. Leaf size=93 \[ \frac {\sqrt {x+\sqrt {x+1}} \left (128 x^2+328 x+563\right )}{3840}+\frac {\sqrt {x+1} \left (640 x^2-872 x+975\right ) \sqrt {x+\sqrt {x+1}}}{1920}+\frac {385}{512} \log \left (2 \sqrt {x+1}-2 \sqrt {x+\sqrt {x+1}}+1\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 147, normalized size of antiderivative = 1.58, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1661, 640, 612, 621, 206} \begin {gather*} \frac {1}{3} (x+1)^{3/2} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {3}{10} (x+1) \left (x+\sqrt {x+1}\right )^{3/2}-\frac {39}{80} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}+\frac {33}{160} \left (x+\sqrt {x+1}\right )^{3/2}+\frac {77}{256} \left (2 \sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}-\frac {385}{512} \tanh ^{-1}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*Sqrt[x + Sqrt[1 + x]])/Sqrt[1 + x],x]

[Out]

(33*(x + Sqrt[1 + x])^(3/2))/160 - (39*Sqrt[1 + x]*(x + Sqrt[1 + x])^(3/2))/80 - (3*(1 + x)*(x + Sqrt[1 + x])^
(3/2))/10 + ((1 + x)^(3/2)*(x + Sqrt[1 + x])^(3/2))/3 + (77*Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sqrt[1 + x]))/256 - (
385*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/512

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 612

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 621

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 640

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 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx &=2 \operatorname {Subst}\left (\int \left (-1+x^2\right )^2 \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} \operatorname {Subst}\left (\int \sqrt {-1+x+x^2} \left (6-9 x^2-\frac {9 x^3}{2}\right ) \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {3}{10} (1+x) \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{15} \operatorname {Subst}\left (\int \left (30-9 x-\frac {117 x^2}{4}\right ) \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {39}{80} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {3}{10} (1+x) \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{60} \operatorname {Subst}\left (\int \left (\frac {363}{4}+\frac {297 x}{8}\right ) \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {33}{160} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {39}{80} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {3}{10} (1+x) \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {77}{64} \operatorname {Subst}\left (\int \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {33}{160} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {39}{80} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {3}{10} (1+x) \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {77}{256} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {385}{512} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {33}{160} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {39}{80} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {3}{10} (1+x) \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {77}{256} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {385}{256} \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )\\ &=\frac {33}{160} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {39}{80} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {3}{10} (1+x) \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{3} (1+x)^{3/2} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {77}{256} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {385}{512} \tanh ^{-1}\left (\frac {1+2 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 92, normalized size = 0.99 \begin {gather*} \frac {2 \sqrt {x+\sqrt {x+1}} \left (128 \left (10 \sqrt {x+1}+1\right ) x^2-8 \left (218 \sqrt {x+1}-41\right ) x+1950 \sqrt {x+1}+563\right )-5775 \tanh ^{-1}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )}{7680} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sqrt[x + Sqrt[1 + x]])/Sqrt[1 + x],x]

[Out]

(2*Sqrt[x + Sqrt[1 + x]]*(563 + 1950*Sqrt[1 + x] + 128*x^2*(1 + 10*Sqrt[1 + x]) - 8*x*(-41 + 218*Sqrt[1 + x]))
- 5775*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/7680

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.23, size = 90, normalized size = 0.97 \begin {gather*} \frac {\sqrt {x+\sqrt {1+x}} \left (363+4974 \sqrt {1+x}+72 (1+x)-4304 (1+x)^{3/2}+128 (1+x)^2+1280 (1+x)^{5/2}\right )}{3840}+\frac {385}{512} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*Sqrt[x + Sqrt[1 + x]])/Sqrt[1 + x],x]

[Out]

(Sqrt[x + Sqrt[1 + x]]*(363 + 4974*Sqrt[1 + x] + 72*(1 + x) - 4304*(1 + x)^(3/2) + 128*(1 + x)^2 + 1280*(1 + x
)^(5/2)))/3840 + (385*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]])/512

________________________________________________________________________________________

fricas [A]  time = 1.13, size = 74, normalized size = 0.80 \begin {gather*} \frac {1}{3840} \, {\left (128 \, x^{2} + 2 \, {\left (640 \, x^{2} - 872 \, x + 975\right )} \sqrt {x + 1} + 328 \, x + 563\right )} \sqrt {x + \sqrt {x + 1}} + \frac {385}{1024} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt {x + 1} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x+(1+x)^(1/2))^(1/2)/(1+x)^(1/2),x, algorithm="fricas")

[Out]

1/3840*(128*x^2 + 2*(640*x^2 - 872*x + 975)*sqrt(x + 1) + 328*x + 563)*sqrt(x + sqrt(x + 1)) + 385/1024*log(4*
sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) - 8*x - 8*sqrt(x + 1) - 5)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 80, normalized size = 0.86 \begin {gather*} \frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, \sqrt {x + 1} {\left (10 \, \sqrt {x + 1} + 1\right )} - 269\right )} \sqrt {x + 1} + 9\right )} \sqrt {x + 1} + 2487\right )} \sqrt {x + 1} + 363\right )} \sqrt {x + \sqrt {x + 1}} + \frac {385}{512} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x+(1+x)^(1/2))^(1/2)/(1+x)^(1/2),x, algorithm="giac")

[Out]

1/3840*(2*(4*(2*(8*sqrt(x + 1)*(10*sqrt(x + 1) + 1) - 269)*sqrt(x + 1) + 9)*sqrt(x + 1) + 2487)*sqrt(x + 1) +
363)*sqrt(x + sqrt(x + 1)) + 385/512*log(-2*sqrt(x + sqrt(x + 1)) + 2*sqrt(x + 1) + 1)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 98, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {\left (1+x \right )^{\frac {3}{2}} \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {3 \left (1+x \right ) \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{10}-\frac {39 \sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{80}+\frac {33 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{160}+\frac {77 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{256}-\frac {385 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{512}\)	\(98\)
default	\(\frac {\left (1+x \right )^{\frac {3}{2}} \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{3}-\frac {3 \left (1+x \right ) \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{10}-\frac {39 \sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{80}+\frac {33 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{160}+\frac {77 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{256}-\frac {385 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{512}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x+(1+x)^(1/2))^(1/2)/(1+x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(1+x)^(3/2)*(x+(1+x)^(1/2))^(3/2)-3/10*(1+x)*(x+(1+x)^(1/2))^(3/2)-39/80*(1+x)^(1/2)*(x+(1+x)^(1/2))^(3/2)
+33/160*(x+(1+x)^(1/2))^(3/2)+77/256*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)-385/512*ln(1/2+(1+x)^(1/2)+(x+(1+
x)^(1/2))^(1/2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x + \sqrt {x + 1}} x^{2}}{\sqrt {x + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(x+(1+x)^(1/2))^(1/2)/(1+x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x + sqrt(x + 1))*x^2/sqrt(x + 1), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\sqrt {x+\sqrt {x+1}}}{\sqrt {x+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(x + (x + 1)^(1/2))^(1/2))/(x + 1)^(1/2),x)

[Out]

int((x^2*(x + (x + 1)^(1/2))^(1/2))/(x + 1)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \sqrt {x + \sqrt {x + 1}}}{\sqrt {x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(x+(1+x)**(1/2))**(1/2)/(1+x)**(1/2),x)

[Out]

Integral(x**2*sqrt(x + sqrt(x + 1))/sqrt(x + 1), x)

________________________________________________________________________________________