3.2.64 \(\int \frac {\text {ArcTan}(\sqrt {x})}{x^{5/2}} \, dx\) [164]

Optimal. Leaf size=37 \[ -\frac {1}{3 x}-\frac {2 \text {ArcTan}\left (\sqrt {x}\right )}{3 x^{3/2}}-\frac {\log (x)}{3}+\frac {1}{3} \log (1+x) \]

[Out]

-1/3/x-2/3*arctan(x^(1/2))/x^(3/2)-1/3*ln(x)+1/3*ln(1+x)

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4946, 46} \begin {gather*} -\frac {2 \text {ArcTan}\left (\sqrt {x}\right )}{3 x^{3/2}}-\frac {1}{3 x}-\frac {\log (x)}{3}+\frac {1}{3} \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[Sqrt[x]]/x^(5/2),x]

[Out]

-1/3*1/x - (2*ArcTan[Sqrt[x]])/(3*x^(3/2)) - Log[x]/3 + Log[1 + x]/3

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}\left (\sqrt {x}\right )}{x^{5/2}} \, dx &=-\frac {2 \tan ^{-1}\left (\sqrt {x}\right )}{3 x^{3/2}}+\frac {1}{3} \int \frac {1}{x^2 (1+x)} \, dx\\ &=-\frac {2 \tan ^{-1}\left (\sqrt {x}\right )}{3 x^{3/2}}+\frac {1}{3} \int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx\\ &=-\frac {1}{3 x}-\frac {2 \tan ^{-1}\left (\sqrt {x}\right )}{3 x^{3/2}}-\frac {\log (x)}{3}+\frac {1}{3} \log (1+x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 31, normalized size = 0.84 \begin {gather*} \frac {1}{3} \left (-\frac {1}{x}-\frac {2 \text {ArcTan}\left (\sqrt {x}\right )}{x^{3/2}}-\log (x)+\log (1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Sqrt[x]]/x^(5/2),x]

[Out]

(-x^(-1) - (2*ArcTan[Sqrt[x]])/x^(3/2) - Log[x] + Log[1 + x])/3

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Maple [A]
time = 0.03, size = 26, normalized size = 0.70

method	result	size
derivativedivides	\(-\frac {1}{3 x}-\frac {2 \arctan \left (\sqrt {x}\right )}{3 x^{\frac {3}{2}}}-\frac {\ln \left (x \right )}{3}+\frac {\ln \left (1+x \right )}{3}\)	\(26\)
default	\(-\frac {1}{3 x}-\frac {2 \arctan \left (\sqrt {x}\right )}{3 x^{\frac {3}{2}}}-\frac {\ln \left (x \right )}{3}+\frac {\ln \left (1+x \right )}{3}\)	\(26\)
meijerg	\(\frac {-10 x +30}{45 x}-\frac {2 \arctan \left (\sqrt {x}\right )}{3 x^{\frac {3}{2}}}+\frac {\ln \left (1+x \right )}{3}+\frac {2}{9}-\frac {\ln \left (x \right )}{3}-\frac {1}{x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/x-2/3*arctan(x^(1/2))/x^(3/2)-1/3*ln(x)+1/3*ln(1+x)

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Maxima [A]
time = 0.26, size = 25, normalized size = 0.68 \begin {gather*} -\frac {2 \, \arctan \left (\sqrt {x}\right )}{3 \, x^{\frac {3}{2}}} - \frac {1}{3 \, x} + \frac {1}{3} \, \log \left (x + 1\right ) - \frac {1}{3} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x^(1/2))/x^(5/2),x, algorithm="maxima")

[Out]

-2/3*arctan(sqrt(x))/x^(3/2) - 1/3/x + 1/3*log(x + 1) - 1/3*log(x)

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Fricas [A]
time = 3.13, size = 33, normalized size = 0.89 \begin {gather*} \frac {x^{2} \log \left (x + 1\right ) - x^{2} \log \left (x\right ) - 2 \, \sqrt {x} \arctan \left (\sqrt {x}\right ) - x}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^2*log(x + 1) - x^2*log(x) - 2*sqrt(x)*arctan(sqrt(x)) - x)/x^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (31) = 62\).
time = 1.18, size = 143, normalized size = 3.86 \begin {gather*} - \frac {2 x^{\frac {3}{2}} \operatorname {atan}{\left (\sqrt {x} \right )}}{3 x^{3} + 3 x^{2}} - \frac {2 \sqrt {x} \operatorname {atan}{\left (\sqrt {x} \right )}}{3 x^{3} + 3 x^{2}} - \frac {x^{3} \log {\left (x \right )}}{3 x^{3} + 3 x^{2}} + \frac {x^{3} \log {\left (x + 1 \right )}}{3 x^{3} + 3 x^{2}} - \frac {x^{2} \log {\left (x \right )}}{3 x^{3} + 3 x^{2}} + \frac {x^{2} \log {\left (x + 1 \right )}}{3 x^{3} + 3 x^{2}} - \frac {x^{2}}{3 x^{3} + 3 x^{2}} - \frac {x}{3 x^{3} + 3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x**(3/2)*atan(sqrt(x))/(3*x**3 + 3*x**2) - 2*sqrt(x)*atan(sqrt(x))/(3*x**3 + 3*x**2) - x**3*log(x)/(3*x**3
+ 3*x**2) + x**3*log(x + 1)/(3*x**3 + 3*x**2) - x**2*log(x)/(3*x**3 + 3*x**2) + x**2*log(x + 1)/(3*x**3 + 3*x*
*2) - x**2/(3*x**3 + 3*x**2) - x/(3*x**3 + 3*x**2)

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Giac [A]
time = 0.41, size = 28, normalized size = 0.76 \begin {gather*} \frac {x - 1}{3 \, x} - \frac {2 \, \arctan \left (\sqrt {x}\right )}{3 \, x^{\frac {3}{2}}} + \frac {1}{3} \, \log \left (x + 1\right ) - \frac {1}{3} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x - 1)/x - 2/3*arctan(sqrt(x))/x^(3/2) + 1/3*log(x + 1) - 1/3*log(x)

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Mupad [B]
time = 0.35, size = 27, normalized size = 0.73 \begin {gather*} \frac {\ln \left (x+1\right )}{3}-\frac {2\,\ln \left (\sqrt {x}\right )}{3}-\frac {2\,\mathrm {atan}\left (\sqrt {x}\right )}{3\,x^{3/2}}-\frac {1}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x^(1/2))/x^(5/2),x)

[Out]

log(x + 1)/3 - (2*log(x^(1/2)))/3 - (2*atan(x^(1/2)))/(3*x^(3/2)) - 1/(3*x)

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