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

Optimal. Leaf size=214 \[ 2 \sqrt{x+1}+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{\sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{\sqrt{2 \left (1+\sqrt{2}\right )}} \]

[Out]

2*Sqrt[1 + x] + ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(1 + Sqrt[2])] -
ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(1 + Sqrt[2])] + (Sqrt[(1 + Sqr
t[2])/2]*Log[1 + Sqrt[2] + x - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]])/2 - (Sqrt[(1 + Sqrt[2])/2]*Log[1 + Sqrt[2]
+ x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]])/2

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Rubi [A]  time = 0.246522, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {825, 827, 1169, 634, 618, 204, 628} \[ 2 \sqrt{x+1}+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{x+1}+\sqrt{2}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{x+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{\sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 \sqrt{x+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{\sqrt{2 \left (1+\sqrt{2}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[1 + x])/(1 + x^2),x]

[Out]

2*Sqrt[1 + x] + ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(1 + Sqrt[2])] -
ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + x])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(1 + Sqrt[2])] + (Sqrt[(1 + Sqr
t[2])/2]*Log[1 + Sqrt[2] + x - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]])/2 - (Sqrt[(1 + Sqrt[2])/2]*Log[1 + Sqrt[2]
+ x + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + x]])/2

Rule 825

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g*(d + e*x)^m)/
(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x])/(a + c*x^2), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x \sqrt{1+x}}{1+x^2} \, dx &=2 \sqrt{1+x}+\int \frac{-1+x}{\sqrt{1+x} \left (1+x^2\right )} \, dx\\ &=2 \sqrt{1+x}+2 \operatorname{Subst}\left (\int \frac{-2+x^2}{2-2 x^2+x^4} \, dx,x,\sqrt{1+x}\right )\\ &=2 \sqrt{1+x}+\frac{\operatorname{Subst}\left (\int \frac{-2 \sqrt{2 \left (1+\sqrt{2}\right )}-\left (-2-\sqrt{2}\right ) x}{\sqrt{2}-\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )}{2 \sqrt{1+\sqrt{2}}}+\frac{\operatorname{Subst}\left (\int \frac{-2 \sqrt{2 \left (1+\sqrt{2}\right )}+\left (-2-\sqrt{2}\right ) x}{\sqrt{2}+\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )}{2 \sqrt{1+\sqrt{2}}}\\ &=2 \sqrt{1+x}-\frac{1}{2} \sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )-\frac{1}{2} \sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}+\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{-\sqrt{2 \left (1+\sqrt{2}\right )}+2 x}{\sqrt{2}-\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{\sqrt{2 \left (1+\sqrt{2}\right )}+2 x}{\sqrt{2}+\sqrt{2 \left (1+\sqrt{2}\right )} x+x^2} \, dx,x,\sqrt{1+x}\right )\\ &=2 \sqrt{1+x}+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (1+\sqrt{2}+x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (1+\sqrt{2}+x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )+\sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{2 \left (1-\sqrt{2}\right )-x^2} \, dx,x,-\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}\right )+\sqrt{3-2 \sqrt{2}} \operatorname{Subst}\left (\int \frac{1}{2 \left (1-\sqrt{2}\right )-x^2} \, dx,x,\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}\right )\\ &=2 \sqrt{1+x}+\sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{1+x}}{\sqrt{2 \left (-1+\sqrt{2}\right )}}\right )-\sqrt{\frac{1}{2} \left (-1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}+2 \sqrt{1+x}}{\sqrt{2 \left (-1+\sqrt{2}\right )}}\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (1+\sqrt{2}+x-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \log \left (1+\sqrt{2}+x+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{1+x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0433481, size = 60, normalized size = 0.28 \[ 2 \sqrt{x+1}-\sqrt{1-i} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1-i}}\right )-\sqrt{1+i} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{\sqrt{1+i}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[1 + x] - Sqrt[1 - I]*ArcTanh[Sqrt[1 + x]/Sqrt[1 - I]] - Sqrt[1 + I]*ArcTanh[Sqrt[1 + x]/Sqrt[1 + I]]

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Maple [A]  time = 0.077, size = 240, normalized size = 1.1 \begin{align*} 2\,\sqrt{1+x}-{\frac{\sqrt{2+2\,\sqrt{2}}}{4}\ln \left ( 1+x+\sqrt{2}+\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }+{\frac{1}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }-{\frac{\sqrt{2}}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}+\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}}{4}\ln \left ( 1+x+\sqrt{2}-\sqrt{1+x}\sqrt{2+2\,\sqrt{2}} \right ) }-{\frac{\sqrt{2}}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}} \right ) } \right ) }+{\frac{1}{\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+x}-\sqrt{2+2\,\sqrt{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1+x)^(1/2)-1/4*ln(1+x+2^(1/2)+(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(1/2)+1/(-2+2*2^(1/2))^(1/2)*a
rctan((2*(1+x)^(1/2)+(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))-1/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+x)^(1/2)+(
2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+1/4*ln(1+x+2^(1/2)-(1+x)^(1/2)*(2+2*2^(1/2))^(1/2))*(2+2*2^(
1/2))^(1/2)-1/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+x)^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+1/
(-2+2*2^(1/2))^(1/2)*arctan((2*(1+x)^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + 1)*x/(x^2 + 1), x)

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Fricas [A]  time = 1.67516, size = 1048, normalized size = 4.9 \begin{align*} -\frac{1}{8} \cdot 2^{\frac{1}{4}}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} \log \left (\frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{x + 1}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} + x + \sqrt{2} + 1\right ) + \frac{1}{8} \cdot 2^{\frac{1}{4}}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} \log \left (-\frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{x + 1}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} + x + \sqrt{2} + 1\right ) + \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{-2 \, \sqrt{2} + 4} \arctan \left (\frac{1}{4} \cdot 2^{\frac{3}{4}} \sqrt{2^{\frac{1}{4}} \sqrt{x + 1}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} + 2 \, x + 2 \, \sqrt{2} + 2}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} - \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1}{\left (\sqrt{2} + 1\right )} \sqrt{-2 \, \sqrt{2} + 4} - \sqrt{2} - 1\right ) + \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{-2 \, \sqrt{2} + 4} \arctan \left (\frac{1}{4} \cdot 2^{\frac{3}{4}} \sqrt{-2^{\frac{1}{4}} \sqrt{x + 1}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} + 2 \, x + 2 \, \sqrt{2} + 2}{\left (\sqrt{2} + 2\right )} \sqrt{-2 \, \sqrt{2} + 4} - \frac{1}{2} \cdot 2^{\frac{3}{4}} \sqrt{x + 1}{\left (\sqrt{2} + 1\right )} \sqrt{-2 \, \sqrt{2} + 4} + \sqrt{2} + 1\right ) + 2 \, \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*2^(1/4)*(sqrt(2) + 2)*sqrt(-2*sqrt(2) + 4)*log(1/2*2^(1/4)*sqrt(x + 1)*(sqrt(2) + 2)*sqrt(-2*sqrt(2) + 4)
+ x + sqrt(2) + 1) + 1/8*2^(1/4)*(sqrt(2) + 2)*sqrt(-2*sqrt(2) + 4)*log(-1/2*2^(1/4)*sqrt(x + 1)*(sqrt(2) + 2
)*sqrt(-2*sqrt(2) + 4) + x + sqrt(2) + 1) + 1/2*2^(3/4)*sqrt(-2*sqrt(2) + 4)*arctan(1/4*2^(3/4)*sqrt(2^(1/4)*s
qrt(x + 1)*(sqrt(2) + 2)*sqrt(-2*sqrt(2) + 4) + 2*x + 2*sqrt(2) + 2)*(sqrt(2) + 2)*sqrt(-2*sqrt(2) + 4) - 1/2*
2^(3/4)*sqrt(x + 1)*(sqrt(2) + 1)*sqrt(-2*sqrt(2) + 4) - sqrt(2) - 1) + 1/2*2^(3/4)*sqrt(-2*sqrt(2) + 4)*arcta
n(1/4*2^(3/4)*sqrt(-2^(1/4)*sqrt(x + 1)*(sqrt(2) + 2)*sqrt(-2*sqrt(2) + 4) + 2*x + 2*sqrt(2) + 2)*(sqrt(2) + 2
)*sqrt(-2*sqrt(2) + 4) - 1/2*2^(3/4)*sqrt(x + 1)*(sqrt(2) + 1)*sqrt(-2*sqrt(2) + 4) + sqrt(2) + 1) + 2*sqrt(x
+ 1)

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Sympy [A]  time = 8.44238, size = 68, normalized size = 0.32 \begin{align*} 2 \sqrt{x + 1} - 4 \operatorname{RootSum}{\left (512 t^{4} + 32 t^{2} + 1, \left ( t \mapsto t \log{\left (- 128 t^{3} + \sqrt{x + 1} \right )} \right )\right )} + 2 \operatorname{RootSum}{\left (128 t^{4} + 16 t^{2} + 1, \left ( t \mapsto t \log{\left (64 t^{3} + 4 t + \sqrt{x + 1} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x + 1) - 4*RootSum(512*_t**4 + 32*_t**2 + 1, Lambda(_t, _t*log(-128*_t**3 + sqrt(x + 1)))) + 2*RootSum(
128*_t**4 + 16*_t**2 + 1, Lambda(_t, _t*log(64*_t**3 + 4*_t + sqrt(x + 1))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + 1)*x/(x^2 + 1), x)