3.1.38 \(\int \sqrt {x} (1+x)^{5/2} \, dx\) [38]

Optimal. Leaf size=75 \[ \frac {5}{64} \sqrt {x} \sqrt {1+x}+\frac {5}{32} x^{3/2} \sqrt {1+x}+\frac {5}{24} x^{3/2} (1+x)^{3/2}+\frac {1}{4} x^{3/2} (1+x)^{5/2}-\frac {5}{64} \sinh ^{-1}\left (\sqrt {x}\right ) \]

[Out]

5/24*x^(3/2)*(1+x)^(3/2)+1/4*x^(3/2)*(1+x)^(5/2)-5/64*arcsinh(x^(1/2))+5/32*x^(3/2)*(1+x)^(1/2)+5/64*x^(1/2)*(
1+x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {52, 56, 221} \begin {gather*} \frac {1}{4} x^{3/2} (x+1)^{5/2}+\frac {5}{24} x^{3/2} (x+1)^{3/2}+\frac {5}{32} x^{3/2} \sqrt {x+1}+\frac {5}{64} \sqrt {x} \sqrt {x+1}-\frac {5}{64} \sinh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(1 + x)^(5/2),x]

[Out]

(5*Sqrt[x]*Sqrt[1 + x])/64 + (5*x^(3/2)*Sqrt[1 + x])/32 + (5*x^(3/2)*(1 + x)^(3/2))/24 + (x^(3/2)*(1 + x)^(5/2
))/4 - (5*ArcSinh[Sqrt[x]])/64

Rule 52

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

Rule 56

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

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]

Rubi steps

\begin {align*} \int \sqrt {x} (1+x)^{5/2} \, dx &=\frac {1}{4} x^{3/2} (1+x)^{5/2}+\frac {5}{8} \int \sqrt {x} (1+x)^{3/2} \, dx\\ &=\frac {5}{24} x^{3/2} (1+x)^{3/2}+\frac {1}{4} x^{3/2} (1+x)^{5/2}+\frac {5}{16} \int \sqrt {x} \sqrt {1+x} \, dx\\ &=\frac {5}{32} x^{3/2} \sqrt {1+x}+\frac {5}{24} x^{3/2} (1+x)^{3/2}+\frac {1}{4} x^{3/2} (1+x)^{5/2}+\frac {5}{64} \int \frac {\sqrt {x}}{\sqrt {1+x}} \, dx\\ &=\frac {5}{64} \sqrt {x} \sqrt {1+x}+\frac {5}{32} x^{3/2} \sqrt {1+x}+\frac {5}{24} x^{3/2} (1+x)^{3/2}+\frac {1}{4} x^{3/2} (1+x)^{5/2}-\frac {5}{128} \int \frac {1}{\sqrt {x} \sqrt {1+x}} \, dx\\ &=\frac {5}{64} \sqrt {x} \sqrt {1+x}+\frac {5}{32} x^{3/2} \sqrt {1+x}+\frac {5}{24} x^{3/2} (1+x)^{3/2}+\frac {1}{4} x^{3/2} (1+x)^{5/2}-\frac {5}{64} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {5}{64} \sqrt {x} \sqrt {1+x}+\frac {5}{32} x^{3/2} \sqrt {1+x}+\frac {5}{24} x^{3/2} (1+x)^{3/2}+\frac {1}{4} x^{3/2} (1+x)^{5/2}-\frac {5}{64} \sinh ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 47, normalized size = 0.63 \begin {gather*} \frac {1}{192} \left (\sqrt {x} \sqrt {1+x} \left (15+118 x+136 x^2+48 x^3\right )-15 \tanh ^{-1}\left (\sqrt {\frac {x}{1+x}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(1 + x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[1 + x]*(15 + 118*x + 136*x^2 + 48*x^3) - 15*ArcTanh[Sqrt[x/(1 + x)]])/192

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 22.16, size = 130, normalized size = 1.73 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {-15 \sqrt {x} \text {ArcCosh}\left [\sqrt {1+x}\right ]-56 \left (1+x\right )^{\frac {7}{2}}-5 \left (1+x\right )^{\frac {3}{2}}-2 \left (1+x\right )^{\frac {5}{2}}+15 \sqrt {1+x}+48 \left (1+x\right )^{\frac {9}{2}}}{192 \sqrt {x}},\text {Abs}\left [1+x\right ]>1\right \}\right \},-\frac {I \left (1+x\right )^{\frac {9}{2}}}{4 \sqrt {-x}}-\frac {5 I \sqrt {1+x}}{64 \sqrt {-x}}+\frac {I \left (1+x\right )^{\frac {5}{2}}}{96 \sqrt {-x}}+\frac {I 5 \left (1+x\right )^{\frac {3}{2}}}{192 \sqrt {-x}}+\frac {I 5 \text {ArcSin}\left [\sqrt {1+x}\right ]}{64}+\frac {I 7 \left (1+x\right )^{\frac {7}{2}}}{24 \sqrt {-x}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Piecewise[{{(-15 Sqrt[x] ArcCosh[Sqrt[1 + x]] - 56 (1 + x) ^ (7 / 2) - 5 (1 + x) ^ (3 / 2) - 2 (1 + x) ^ (5 /
2) + 15 Sqrt[1 + x] + 48 (1 + x) ^ (9 / 2)) / (192 Sqrt[x]), Abs[1 + x] > 1}}, -I (1 + x) ^ (9 / 2) / (4 Sqrt[
-x]) - 5 I Sqrt[1 + x] / (64 Sqrt[-x]) + I (1 + x) ^ (5 / 2) / (96 Sqrt[-x]) + I 5 (1 + x) ^ (3 / 2) / (192 Sq
rt[-x]) + I 5 ArcSin[Sqrt[1 + x]] / 64 + I 7 (1 + x) ^ (7 / 2) / (24 Sqrt[-x])]

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Maple [A]
time = 0.04, size = 70, normalized size = 0.93

method result size
meijerg \(-\frac {15 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \left (48 x^{3}+136 x^{2}+118 x +15\right ) \sqrt {1+x}}{360}+\frac {\sqrt {\pi }\, \arcsinh \left (\sqrt {x}\right )}{24}\right )}{8 \sqrt {\pi }}\) \(44\)
risch \(\frac {\left (48 x^{3}+136 x^{2}+118 x +15\right ) \sqrt {x}\, \sqrt {1+x}}{192}-\frac {5 \sqrt {x \left (1+x \right )}\, \ln \left (x +\frac {1}{2}+\sqrt {x^{2}+x}\right )}{128 \sqrt {1+x}\, \sqrt {x}}\) \(55\)
default \(\frac {\sqrt {x}\, \left (1+x \right )^{\frac {7}{2}}}{4}-\frac {\sqrt {x}\, \left (1+x \right )^{\frac {5}{2}}}{24}-\frac {5 \sqrt {x}\, \left (1+x \right )^{\frac {3}{2}}}{96}-\frac {5 \sqrt {x}\, \sqrt {1+x}}{64}-\frac {5 \sqrt {x \left (1+x \right )}\, \ln \left (x +\frac {1}{2}+\sqrt {x^{2}+x}\right )}{128 \sqrt {1+x}\, \sqrt {x}}\) \(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^(1/2)*(1+x)^(7/2)-1/24*x^(1/2)*(1+x)^(5/2)-5/96*x^(1/2)*(1+x)^(3/2)-5/64*x^(1/2)*(1+x)^(1/2)-5/128*(x*(1
+x))^(1/2)/(1+x)^(1/2)/x^(1/2)*ln(x+1/2+(x^2+x)^(1/2))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (47) = 94\).
time = 0.25, size = 113, normalized size = 1.51 \begin {gather*} \frac {\frac {15 \, {\left (x + 1\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}} + \frac {73 \, {\left (x + 1\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}} - \frac {55 \, {\left (x + 1\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}} + \frac {15 \, \sqrt {x + 1}}{\sqrt {x}}}{192 \, {\left (\frac {{\left (x + 1\right )}^{4}}{x^{4}} - \frac {4 \, {\left (x + 1\right )}^{3}}{x^{3}} + \frac {6 \, {\left (x + 1\right )}^{2}}{x^{2}} - \frac {4 \, {\left (x + 1\right )}}{x} + 1\right )}} - \frac {5}{128} \, \log \left (\frac {\sqrt {x + 1}}{\sqrt {x}} + 1\right ) + \frac {5}{128} \, \log \left (\frac {\sqrt {x + 1}}{\sqrt {x}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(15*(x + 1)^(7/2)/x^(7/2) + 73*(x + 1)^(5/2)/x^(5/2) - 55*(x + 1)^(3/2)/x^(3/2) + 15*sqrt(x + 1)/sqrt(x)
)/((x + 1)^4/x^4 - 4*(x + 1)^3/x^3 + 6*(x + 1)^2/x^2 - 4*(x + 1)/x + 1) - 5/128*log(sqrt(x + 1)/sqrt(x) + 1) +
 5/128*log(sqrt(x + 1)/sqrt(x) - 1)

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Fricas [A]
time = 0.31, size = 44, normalized size = 0.59 \begin {gather*} \frac {1}{192} \, {\left (48 \, x^{3} + 136 \, x^{2} + 118 \, x + 15\right )} \sqrt {x + 1} \sqrt {x} + \frac {5}{128} \, \log \left (2 \, \sqrt {x + 1} \sqrt {x} - 2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(48*x^3 + 136*x^2 + 118*x + 15)*sqrt(x + 1)*sqrt(x) + 5/128*log(2*sqrt(x + 1)*sqrt(x) - 2*x - 1)

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Sympy [A]
time = 20.00, size = 190, normalized size = 2.53 \begin {gather*} \begin {cases} - \frac {5 \operatorname {acosh}{\left (\sqrt {x + 1} \right )}}{64} + \frac {\left (x + 1\right )^{\frac {9}{2}}}{4 \sqrt {x}} - \frac {7 \left (x + 1\right )^{\frac {7}{2}}}{24 \sqrt {x}} - \frac {\left (x + 1\right )^{\frac {5}{2}}}{96 \sqrt {x}} - \frac {5 \left (x + 1\right )^{\frac {3}{2}}}{192 \sqrt {x}} + \frac {5 \sqrt {x + 1}}{64 \sqrt {x}} & \text {for}\: \left |{x + 1}\right | > 1 \\\frac {5 i \operatorname {asin}{\left (\sqrt {x + 1} \right )}}{64} - \frac {i \left (x + 1\right )^{\frac {9}{2}}}{4 \sqrt {- x}} + \frac {7 i \left (x + 1\right )^{\frac {7}{2}}}{24 \sqrt {- x}} + \frac {i \left (x + 1\right )^{\frac {5}{2}}}{96 \sqrt {- x}} + \frac {5 i \left (x + 1\right )^{\frac {3}{2}}}{192 \sqrt {- x}} - \frac {5 i \sqrt {x + 1}}{64 \sqrt {- x}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-5*acosh(sqrt(x + 1))/64 + (x + 1)**(9/2)/(4*sqrt(x)) - 7*(x + 1)**(7/2)/(24*sqrt(x)) - (x + 1)**(5
/2)/(96*sqrt(x)) - 5*(x + 1)**(3/2)/(192*sqrt(x)) + 5*sqrt(x + 1)/(64*sqrt(x)), Abs(x + 1) > 1), (5*I*asin(sqr
t(x + 1))/64 - I*(x + 1)**(9/2)/(4*sqrt(-x)) + 7*I*(x + 1)**(7/2)/(24*sqrt(-x)) + I*(x + 1)**(5/2)/(96*sqrt(-x
)) + 5*I*(x + 1)**(3/2)/(192*sqrt(-x)) - 5*I*sqrt(x + 1)/(64*sqrt(-x)), True))

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Giac [A]
time = 0.01, size = 203, normalized size = 2.71 \begin {gather*} 2 \left (2 \left (\left (\left (\frac {1}{16} \sqrt {x} \sqrt {x}+\frac 1{96}\right ) \sqrt {x} \sqrt {x}-\frac {5}{384}\right ) \sqrt {x} \sqrt {x}+\frac {5}{256}\right ) \sqrt {x} \sqrt {x+1}+\frac {5}{128} \ln \left (\sqrt {x+1}-\sqrt {x}\right )\right )+4 \left (2 \left (\left (\frac {1}{12} \sqrt {x} \sqrt {x}+\frac 1{48}\right ) \sqrt {x} \sqrt {x}-\frac 1{32}\right ) \sqrt {x} \sqrt {x+1}-\frac {\ln \left (\sqrt {x+1}-\sqrt {x}\right )}{16}\right )+2 \left (2 \left (\frac {1}{8} \sqrt {x} \sqrt {x}+\frac 1{16}\right ) \sqrt {x} \sqrt {x+1}+\frac {\ln \left (\sqrt {x+1}-\sqrt {x}\right )}{8}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(2*(4*(6*x + 1)*x - 5)*x + 15)*sqrt(x + 1)*sqrt(x) + 1/12*(2*(4*x + 1)*x - 3)*sqrt(x + 1)*sqrt(x) + 1/4*
(2*x + 1)*sqrt(x + 1)*sqrt(x) + 5/64*log(sqrt(x + 1) - sqrt(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(x + 1)^(5/2),x)

[Out]

int(x^(1/2)*(x + 1)^(5/2), x)

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