3.3.20 \(\int \frac {\sqrt {x}}{(1+x)^{7/2}} \, dx\) [220]

Optimal. Leaf size=33 \[ \frac {2 x^{3/2}}{5 (1+x)^{5/2}}+\frac {4 x^{3/2}}{15 (1+x)^{3/2}} \]

[Out]

2/5*x^(3/2)/(1+x)^(5/2)+4/15*x^(3/2)/(1+x)^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.00, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {47, 37} \begin {gather*} \frac {4 x^{3/2}}{15 (x+1)^{3/2}}+\frac {2 x^{3/2}}{5 (x+1)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(1 + x)^(7/2),x]

[Out]

(2*x^(3/2))/(5*(1 + x)^(5/2)) + (4*x^(3/2))/(15*(1 + x)^(3/2))

Rule 37

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps

\begin {align*} \int \frac {\sqrt {x}}{(1+x)^{7/2}} \, dx &=\frac {2 x^{3/2}}{5 (1+x)^{5/2}}+\frac {2}{5} \int \frac {\sqrt {x}}{(1+x)^{5/2}} \, dx\\ &=\frac {2 x^{3/2}}{5 (1+x)^{5/2}}+\frac {4 x^{3/2}}{15 (1+x)^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 21, normalized size = 0.64 \begin {gather*} \frac {2 x^{3/2} (5+2 x)}{15 (1+x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(1 + x)^(7/2),x]

[Out]

(2*x^(3/2)*(5 + 2*x))/(15*(1 + x)^(5/2))

________________________________________________________________________________________

Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 2 in optimal.
time = 5.34, size = 89, normalized size = 2.70 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {2 I x \left (5+2 x\right ) \sqrt {-\frac {x}{1+x}}}{15 \left (1+2 x+x^2\right )},\frac {1}{\text {Abs}\left [1+x\right ]}>1\right \}\right \},\frac {-2 \sqrt {1-\frac {1}{1+x}}}{5 \left (1+x\right )^2}+\frac {2 \sqrt {1-\frac {1}{1+x}}}{15 \left (1+x\right )}+\frac {4 \sqrt {1-\frac {1}{1+x}}}{15}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x^(1/2)*(x+1)^(-7/2),x]')

[Out]

Piecewise[{{2 I / 15 x (5 + 2 x) Sqrt[-x / (1 + x)] / (1 + 2 x + x ^ 2), 1 / Abs[1 + x] > 1}}, -2 Sqrt[1 - 1 /
(1 + x)] / (5 (1 + x) ^ 2) + 2 Sqrt[1 - 1 / (1 + x)] / (15 (1 + x)) + 4 Sqrt[1 - 1 / (1 + x)] / 15]

________________________________________________________________________________________

Maple [A]
time = 0.03, size = 32, normalized size = 0.97

method	result	size
gosper	\(\frac {2 x^{\frac {3}{2}} \left (5+2 x \right )}{15 \left (1+x \right )^{\frac {5}{2}}}\)	\(16\)
meijerg	\(\frac {2 x^{\frac {3}{2}} \left (5+2 x \right )}{15 \left (1+x \right )^{\frac {5}{2}}}\)	\(16\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (5+2 x \right )}{15 \left (1+x \right )^{\frac {5}{2}}}\)	\(16\)
default	\(-\frac {2 \sqrt {x}}{5 \left (1+x \right )^{\frac {5}{2}}}+\frac {2 \sqrt {x}}{15 \left (1+x \right )^{\frac {3}{2}}}+\frac {4 \sqrt {x}}{15 \sqrt {1+x}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(1+x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/(1+x)^(5/2)*x^(1/2)+2/15/(1+x)^(3/2)*x^(1/2)+4/15/(1+x)^(1/2)*x^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.25, size = 20, normalized size = 0.61 \begin {gather*} \frac {2 \, x^{\frac {5}{2}} {\left (\frac {5 \, {\left (x + 1\right )}}{x} - 3\right )}}{15 \, {\left (x + 1\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(1+x)^(7/2),x, algorithm="maxima")

[Out]

2/15*x^(5/2)*(5*(x + 1)/x - 3)/(x + 1)^(5/2)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (21) = 42\).
time = 0.30, size = 50, normalized size = 1.52 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} + {\left (2 \, x^{2} + 5 \, x\right )} \sqrt {x + 1} \sqrt {x} + 6 \, x^{2} + 6 \, x + 2\right )}}{15 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(1+x)^(7/2),x, algorithm="fricas")

[Out]

2/15*(2*x^3 + (2*x^2 + 5*x)*sqrt(x + 1)*sqrt(x) + 6*x^2 + 6*x + 2)/(x^3 + 3*x^2 + 3*x + 1)

________________________________________________________________________________________

Sympy [A]
time = 2.90, size = 167, normalized size = 5.06 \begin {gather*} \begin {cases} \frac {4 i \sqrt {-1 + \frac {1}{x + 1}} \left (x + 1\right )^{2}}{- 15 x + 15 \left (x + 1\right )^{2} - 15} - \frac {2 i \sqrt {-1 + \frac {1}{x + 1}} \left (x + 1\right )}{- 15 x + 15 \left (x + 1\right )^{2} - 15} - \frac {8 i \sqrt {-1 + \frac {1}{x + 1}}}{- 15 x + 15 \left (x + 1\right )^{2} - 15} + \frac {6 i \sqrt {-1 + \frac {1}{x + 1}}}{\left (x + 1\right ) \left (- 15 x + 15 \left (x + 1\right )^{2} - 15\right )} & \text {for}\: \frac {1}{\left |{x + 1}\right |} > 1 \\\frac {4 \sqrt {1 - \frac {1}{x + 1}}}{15} + \frac {2 \sqrt {1 - \frac {1}{x + 1}}}{15 \left (x + 1\right )} - \frac {2 \sqrt {1 - \frac {1}{x + 1}}}{5 \left (x + 1\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(1+x)**(7/2),x)

[Out]

Piecewise((4*I*sqrt(-1 + 1/(x + 1))*(x + 1)**2/(-15*x + 15*(x + 1)**2 - 15) - 2*I*sqrt(-1 + 1/(x + 1))*(x + 1)
/(-15*x + 15*(x + 1)**2 - 15) - 8*I*sqrt(-1 + 1/(x + 1))/(-15*x + 15*(x + 1)**2 - 15) + 6*I*sqrt(-1 + 1/(x + 1
))/((x + 1)*(-15*x + 15*(x + 1)**2 - 15)), 1/Abs(x + 1) > 1), (4*sqrt(1 - 1/(x + 1))/15 + 2*sqrt(1 - 1/(x + 1)
)/(15*(x + 1)) - 2*sqrt(1 - 1/(x + 1))/(5*(x + 1)**2), True))

________________________________________________________________________________________

Giac [A]
time = 0.00, size = 46, normalized size = 1.39 \begin {gather*} \frac {4 \left (\frac {1}{15} \sqrt {x} \sqrt {x}+\frac 1{6}\right ) \sqrt {x} \sqrt {x} \sqrt {x} \sqrt {x+1}}{\left (x+1\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(1+x)^(7/2),x)

[Out]

2/15*(2*x + 5)*x^(3/2)/(x + 1)^(5/2)

________________________________________________________________________________________

Mupad [B]
time = 0.30, size = 15, normalized size = 0.45 \begin {gather*} \frac {2\,x^{3/2}\,\left (2\,x+5\right )}{15\,{\left (x+1\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x + 1)^(7/2),x)

[Out]

(2*x^(3/2)*(2*x + 5))/(15*(x + 1)^(5/2))

________________________________________________________________________________________