3.12.20 \(\int \frac {x \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx\)

Optimal. Leaf size=83 \[ \frac {1}{96} \sqrt {x+\sqrt {x+1}} (8 x+27)+\frac {1}{48} \sqrt {x+1} (24 x-41) \sqrt {x+\sqrt {x+1}}-\frac {35}{64} \log \left (2 \sqrt {x+1}-2 \sqrt {x+\sqrt {x+1}}+1\right ) \]

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Rubi [A]  time = 0.21, antiderivative size = 103, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1661, 640, 612, 621, 206} \begin {gather*} \frac {1}{2} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {5}{12} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {7}{32} \left (2 \sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}+\frac {35}{64} \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*Sqrt[x + Sqrt[1 + x]])/Sqrt[1 + x],x]

[Out]

(-5*(x + Sqrt[1 + x])^(3/2))/12 + (Sqrt[1 + x]*(x + Sqrt[1 + x])^(3/2))/2 - (7*Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sq
rt[1 + x]))/32 + (35*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/64

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 \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx &=2 \operatorname {Subst}\left (\int \left (-1+x^2\right ) \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \operatorname {Subst}\left (\int \left (-3-\frac {5 x}{2}\right ) \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {7}{8} \operatorname {Subst}\left (\int \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {7}{32} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {35}{64} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {7}{32} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {35}{32} \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}-\frac {7}{32} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {35}{64} \tanh ^{-1}\left (\frac {1+2 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 76, normalized size = 0.92 \begin {gather*} \frac {1}{96} \sqrt {x+\sqrt {x+1}} \left (8 x \left (6 \sqrt {x+1}+1\right )-82 \sqrt {x+1}+27\right )+\frac {35}{64} \tanh ^{-1}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x + Sqrt[1 + x]]*(27 - 82*Sqrt[1 + x] + 8*x*(1 + 6*Sqrt[1 + x])))/96 + (35*ArcTanh[(1 + 2*Sqrt[1 + x])/(
2*Sqrt[x + Sqrt[1 + x]])])/64

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IntegrateAlgebraic [A]  time = 0.18, size = 74, normalized size = 0.89 \begin {gather*} \frac {1}{96} \sqrt {x+\sqrt {1+x}} \left (19-130 \sqrt {1+x}+8 (1+x)+48 (1+x)^{3/2}\right )-\frac {35}{64} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x + Sqrt[1 + x]]*(19 - 130*Sqrt[1 + x] + 8*(1 + x) + 48*(1 + x)^(3/2)))/96 - (35*Log[-1 - 2*Sqrt[1 + x]
+ 2*Sqrt[x + Sqrt[1 + x]]])/64

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fricas [A]  time = 1.19, size = 64, normalized size = 0.77 \begin {gather*} \frac {1}{96} \, {\left (2 \, {\left (24 \, x - 41\right )} \sqrt {x + 1} + 8 \, x + 27\right )} \sqrt {x + \sqrt {x + 1}} + \frac {35}{128} \, \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*(x+(1+x)^(1/2))^(1/2)/(1+x)^(1/2),x, algorithm="fricas")

[Out]

1/96*(2*(24*x - 41)*sqrt(x + 1) + 8*x + 27)*sqrt(x + sqrt(x + 1)) + 35/128*log(4*sqrt(x + sqrt(x + 1))*(2*sqrt
(x + 1) + 1) + 8*x + 8*sqrt(x + 1) + 5)

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giac [A]  time = 0.15, size = 62, normalized size = 0.75 \begin {gather*} \frac {1}{96} \, {\left (2 \, {\left (4 \, \sqrt {x + 1} {\left (6 \, \sqrt {x + 1} + 1\right )} - 65\right )} \sqrt {x + 1} + 19\right )} \sqrt {x + \sqrt {x + 1}} - \frac {35}{64} \, \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*(x+(1+x)^(1/2))^(1/2)/(1+x)^(1/2),x, algorithm="giac")

[Out]

1/96*(2*(4*sqrt(x + 1)*(6*sqrt(x + 1) + 1) - 65)*sqrt(x + 1) + 19)*sqrt(x + sqrt(x + 1)) - 35/64*log(-2*sqrt(x
+ sqrt(x + 1)) + 2*sqrt(x + 1) + 1)

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maple [A]  time = 0.04, size = 68, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {\sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{2}-\frac {5 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{12}-\frac {7 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{32}+\frac {35 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{64}\)	\(68\)
default	\(\frac {\sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{2}-\frac {5 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{12}-\frac {7 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{32}+\frac {35 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{64}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1+x)^(1/2)*(x+(1+x)^(1/2))^(3/2)-5/12*(x+(1+x)^(1/2))^(3/2)-7/32*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)+
35/64*ln(1/2+(1+x)^(1/2)+(x+(1+x)^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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