3.9.95 \(\int (1+\frac {2 x}{1+x^2})^{5/2} \, dx\) [895]

Optimal. Leaf size=133 \[ -\frac {4}{3} (1-2 x) (1+x) \sqrt {1+\frac {2 x}{1+x^2}}-\frac {(1-x) (1+x)^3 \sqrt {1+\frac {2 x}{1+x^2}}}{3 \left (1+x^2\right )}-\frac {(4+3 x) \left (1+x^2\right ) \sqrt {1+\frac {2 x}{1+x^2}}}{1+x}+\frac {5 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}} \sinh ^{-1}(x)}{1+x} \]

[Out]

-4/3*(1-2*x)*(1+x)*(1+2*x/(x^2+1))^(1/2)-1/3*(1-x)*(1+x)^3*(1+2*x/(x^2+1))^(1/2)/(x^2+1)-(4+3*x)*(x^2+1)*(1+2*
x/(x^2+1))^(1/2)/(1+x)+5*arcsinh(x)*(x^2+1)^(1/2)*(1+2*x/(x^2+1))^(1/2)/(1+x)

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Rubi [A]
time = 0.05, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6855, 984, 753, 833, 794, 221} \begin {gather*} -\frac {(1-x) \sqrt {\frac {2 x}{x^2+1}+1} (x+1)^3}{3 \left (x^2+1\right )}-\frac {4}{3} (1-2 x) \sqrt {\frac {2 x}{x^2+1}+1} (x+1)-\frac {(3 x+4) \left (x^2+1\right ) \sqrt {\frac {2 x}{x^2+1}+1}}{x+1}+\frac {5 \sqrt {x^2+1} \sqrt {\frac {2 x}{x^2+1}+1} \sinh ^{-1}(x)}{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + (2*x)/(1 + x^2))^(5/2),x]

[Out]

(-4*(1 - 2*x)*(1 + x)*Sqrt[1 + (2*x)/(1 + x^2)])/3 - ((1 - x)*(1 + x)^3*Sqrt[1 + (2*x)/(1 + x^2)])/(3*(1 + x^2
)) - ((4 + 3*x)*(1 + x^2)*Sqrt[1 + (2*x)/(1 + x^2)])/(1 + x) + (5*Sqrt[1 + x^2]*Sqrt[1 + (2*x)/(1 + x^2)]*ArcS
inh[x])/(1 + x)

Rule 221

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

Rule 753

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 984

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 6855

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_)*(x_)^(m_.))^(p_), x_Symbol] :> Dist[(a + b*x^m*v^n)^FracPart[p]/(v^(n*FracP
art[p])*(b*x^m + a/v^n)^FracPart[p]), Int[u*v^(n*p)*(b*x^m + a/v^n)^p, x], x] /; FreeQ[{a, b, m, p}, x] &&  !I
ntegerQ[p] && ILtQ[n, 0] && BinomialQ[v, x]

Rubi steps

\begin {align*} \int \left (1+\frac {2 x}{1+x^2}\right )^{5/2} \, dx &=\frac {\left (\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}\right ) \int \frac {\left (1+2 x+x^2\right )^{5/2}}{\left (1+x^2\right )^{5/2}} \, dx}{\sqrt {1+2 x+x^2}}\\ &=\frac {\left (\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}\right ) \int \frac {(2+2 x)^5}{\left (1+x^2\right )^{5/2}} \, dx}{16 (2+2 x)}\\ &=-\frac {(1-x) (1+x)^3 \sqrt {1+\frac {2 x}{1+x^2}}}{3 \left (1+x^2\right )}+\frac {\left (\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}\right ) \int \frac {(24-8 x) (2+2 x)^3}{\left (1+x^2\right )^{3/2}} \, dx}{48 (2+2 x)}\\ &=-\frac {4}{3} (1-2 x) (1+x) \sqrt {1+\frac {2 x}{1+x^2}}-\frac {(1-x) (1+x)^3 \sqrt {1+\frac {2 x}{1+x^2}}}{3 \left (1+x^2\right )}+\frac {\left (\sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}\right ) \int \frac {(96-288 x) (2+2 x)}{\sqrt {1+x^2}} \, dx}{48 (2+2 x)}\\ &=-\frac {4}{3} (1-2 x) (1+x) \sqrt {1+\frac {2 x}{1+x^2}}-\frac {(1-x) (1+x)^3 \sqrt {1+\frac {2 x}{1+x^2}}}{3 \left (1+x^2\right )}-\frac {(4+3 x) \left (1+x^2\right ) \sqrt {1+\frac {2 x}{1+x^2}}}{1+x}+\frac {\left (10 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}}\right ) \int \frac {1}{\sqrt {1+x^2}} \, dx}{2+2 x}\\ &=-\frac {4}{3} (1-2 x) (1+x) \sqrt {1+\frac {2 x}{1+x^2}}-\frac {(1-x) (1+x)^3 \sqrt {1+\frac {2 x}{1+x^2}}}{3 \left (1+x^2\right )}-\frac {(4+3 x) \left (1+x^2\right ) \sqrt {1+\frac {2 x}{1+x^2}}}{1+x}+\frac {5 \sqrt {1+x^2} \sqrt {1+\frac {2 x}{1+x^2}} \sinh ^{-1}(x)}{1+x}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 74, normalized size = 0.56 \begin {gather*} \frac {(1+x) \left (-17-12 x-18 x^2-8 x^3+3 x^4+15 \left (1+x^2\right )^{3/2} \tanh ^{-1}\left (\frac {x}{\sqrt {1+x^2}}\right )\right )}{3 \sqrt {\frac {(1+x)^2}{1+x^2}} \left (1+x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + (2*x)/(1 + x^2))^(5/2),x]

[Out]

((1 + x)*(-17 - 12*x - 18*x^2 - 8*x^3 + 3*x^4 + 15*(1 + x^2)^(3/2)*ArcTanh[x/Sqrt[1 + x^2]]))/(3*Sqrt[(1 + x)^
2/(1 + x^2)]*(1 + x^2)^2)

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Maple [A]
time = 0.09, size = 62, normalized size = 0.47

method	result	size
default	\(\frac {\left (\frac {x^{2}+2 x +1}{x^{2}+1}\right )^{\frac {5}{2}} \left (x^{2}+1\right ) \left (15 \arcsinh \left (x \right ) \left (x^{2}+1\right )^{\frac {3}{2}}+3 x^{4}-8 x^{3}-18 x^{2}-12 x -17\right )}{3 \left (1+x \right )^{5}}\)	\(62\)
risch	\(\frac {\left (3 x^{4}-8 x^{3}-18 x^{2}-12 x -17\right ) \sqrt {\frac {\left (1+x \right )^{2}}{x^{2}+1}}}{3 \left (x^{2}+1\right ) \left (1+x \right )}+\frac {5 \arcsinh \left (x \right ) \sqrt {x^{2}+1}\, \sqrt {\frac {\left (1+x \right )^{2}}{x^{2}+1}}}{1+x}\)	\(82\)
trager	\(\frac {\left (3 x^{4}-8 x^{3}-18 x^{2}-12 x -17\right ) \sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}}{3 \left (x^{2}+1\right ) \left (1+x \right )}+5 \ln \left (\frac {\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}\, x^{2}+x^{2}+\sqrt {-\frac {-x^{2}-2 x -1}{x^{2}+1}}+x}{1+x}\right )\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x/(x^2+1))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((x^2+2*x+1)/(x^2+1))^(5/2)/(1+x)^5*(x^2+1)*(15*arcsinh(x)*(x^2+1)^(3/2)+3*x^4-8*x^3-18*x^2-12*x-17)

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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+2*x/(x^2+1))^(5/2),x, algorithm="maxima")

[Out]

integrate((2*x/(x^2 + 1) + 1)^(5/2), x)

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Fricas [A]
time = 0.35, size = 117, normalized size = 0.88 \begin {gather*} -\frac {8 \, x^{3} + 8 \, x^{2} + 15 \, {\left (x^{3} + x^{2} + x + 1\right )} \log \left (-\frac {x^{2} - {\left (x^{2} + 1\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + x}{x + 1}\right ) - {\left (3 \, x^{4} - 8 \, x^{3} - 18 \, x^{2} - 12 \, x - 17\right )} \sqrt {\frac {x^{2} + 2 \, x + 1}{x^{2} + 1}} + 8 \, x + 8}{3 \, {\left (x^{3} + x^{2} + x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x^2+1))^(5/2),x, algorithm="fricas")

[Out]

-1/3*(8*x^3 + 8*x^2 + 15*(x^3 + x^2 + x + 1)*log(-(x^2 - (x^2 + 1)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + x)/(x + 1
)) - (3*x^4 - 8*x^3 - 18*x^2 - 12*x - 17)*sqrt((x^2 + 2*x + 1)/(x^2 + 1)) + 8*x + 8)/(x^3 + x^2 + x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (\frac {2 x}{x^{2} + 1} + 1\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x**2+1))**(5/2),x)

[Out]

Integral((2*x/(x**2 + 1) + 1)**(5/2), x)

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Giac [A]
time = 4.03, size = 86, normalized size = 0.65 \begin {gather*} {\left (\sqrt {2} + 5 \, \log \left (\sqrt {2} + 1\right )\right )} \mathrm {sgn}\left (x + 1\right ) - 5 \, \log \left (-x + \sqrt {x^{2} + 1}\right ) \mathrm {sgn}\left (x + 1\right ) + \frac {{\left ({\left ({\left (3 \, x \mathrm {sgn}\left (x + 1\right ) - 8 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 18 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 12 \, \mathrm {sgn}\left (x + 1\right )\right )} x - 17 \, \mathrm {sgn}\left (x + 1\right )}{3 \, {\left (x^{2} + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x/(x^2+1))^(5/2),x, algorithm="giac")

[Out]

(sqrt(2) + 5*log(sqrt(2) + 1))*sgn(x + 1) - 5*log(-x + sqrt(x^2 + 1))*sgn(x + 1) + 1/3*((((3*x*sgn(x + 1) - 8*
sgn(x + 1))*x - 18*sgn(x + 1))*x - 12*sgn(x + 1))*x - 17*sgn(x + 1))/(x^2 + 1)^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {2\,x}{x^2+1}+1\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x)/(x^2 + 1) + 1)^(5/2),x)

[Out]

int(((2*x)/(x^2 + 1) + 1)^(5/2), x)

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