3.1065 \(\int (1-x)^{5/2} \sqrt {1+x} \, dx\)

Optimal. Leaf size=68 \[ \frac {1}{4} (x+1)^{3/2} (1-x)^{5/2}+\frac {5}{12} (x+1)^{3/2} (1-x)^{3/2}+\frac {5}{8} x \sqrt {x+1} \sqrt {1-x}+\frac {5}{8} \sin ^{-1}(x) \]

[Out]

5/12*(1-x)^(3/2)*(1+x)^(3/2)+1/4*(1-x)^(5/2)*(1+x)^(3/2)+5/8*arcsin(x)+5/8*x*(1-x)^(1/2)*(1+x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {49, 38, 41, 216} \[ \frac {1}{4} (x+1)^{3/2} (1-x)^{5/2}+\frac {5}{12} (x+1)^{3/2} (1-x)^{3/2}+\frac {5}{8} x \sqrt {x+1} \sqrt {1-x}+\frac {5}{8} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^(5/2)*Sqrt[1 + x],x]

[Out]

(5*Sqrt[1 - x]*x*Sqrt[1 + x])/8 + (5*(1 - x)^(3/2)*(1 + x)^(3/2))/12 + ((1 - x)^(5/2)*(1 + x)^(3/2))/4 + (5*Ar
cSin[x])/8

Rule 38

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 49

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[(2*c*n)/(m + n + 1), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]

Rule 216

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

Rubi steps

\begin {align*} \int (1-x)^{5/2} \sqrt {1+x} \, dx &=\frac {1}{4} (1-x)^{5/2} (1+x)^{3/2}+\frac {5}{4} \int (1-x)^{3/2} \sqrt {1+x} \, dx\\ &=\frac {5}{12} (1-x)^{3/2} (1+x)^{3/2}+\frac {1}{4} (1-x)^{5/2} (1+x)^{3/2}+\frac {5}{4} \int \sqrt {1-x} \sqrt {1+x} \, dx\\ &=\frac {5}{8} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{12} (1-x)^{3/2} (1+x)^{3/2}+\frac {1}{4} (1-x)^{5/2} (1+x)^{3/2}+\frac {5}{8} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx\\ &=\frac {5}{8} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{12} (1-x)^{3/2} (1+x)^{3/2}+\frac {1}{4} (1-x)^{5/2} (1+x)^{3/2}+\frac {5}{8} \int \frac {1}{\sqrt {1-x^2}} \, dx\\ &=\frac {5}{8} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{12} (1-x)^{3/2} (1+x)^{3/2}+\frac {1}{4} (1-x)^{5/2} (1+x)^{3/2}+\frac {5}{8} \sin ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 50, normalized size = 0.74 \[ \frac {1}{24} \left (\sqrt {1-x^2} \left (6 x^3-16 x^2+9 x+16\right )-30 \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - x^2]*(16 + 9*x - 16*x^2 + 6*x^3) - 30*ArcSin[Sqrt[1 - x]/Sqrt[2]])/24

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fricas [A]  time = 0.45, size = 52, normalized size = 0.76 \[ \frac {1}{24} \, {\left (6 \, x^{3} - 16 \, x^{2} + 9 \, x + 16\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {5}{4} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*x^3 - 16*x^2 + 9*x + 16)*sqrt(x + 1)*sqrt(-x + 1) - 5/4*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)

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giac [B]  time = 1.29, size = 101, normalized size = 1.49 \[ \frac {1}{24} \, {\left ({\left (2 \, {\left (3 \, x - 10\right )} {\left (x + 1\right )} + 43\right )} {\left (x + 1\right )} - 39\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {1}{6} \, {\left ({\left (2 \, x - 5\right )} {\left (x + 1\right )} + 9\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {1}{2} \, \sqrt {x + 1} {\left (x - 2\right )} \sqrt {-x + 1} + \sqrt {x + 1} \sqrt {-x + 1} + \frac {5}{4} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((2*(3*x - 10)*(x + 1) + 43)*(x + 1) - 39)*sqrt(x + 1)*sqrt(-x + 1) - 1/6*((2*x - 5)*(x + 1) + 9)*sqrt(x
+ 1)*sqrt(-x + 1) - 1/2*sqrt(x + 1)*(x - 2)*sqrt(-x + 1) + sqrt(x + 1)*sqrt(-x + 1) + 5/4*arcsin(1/2*sqrt(2)*s
qrt(x + 1))

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maple [A]  time = 0.01, size = 85, normalized size = 1.25 \[ \frac {5 \sqrt {\left (x +1\right ) \left (-x +1\right )}\, \arcsin \relax (x )}{8 \sqrt {x +1}\, \sqrt {-x +1}}+\frac {\left (-x +1\right )^{\frac {5}{2}} \left (x +1\right )^{\frac {3}{2}}}{4}+\frac {5 \left (-x +1\right )^{\frac {3}{2}} \left (x +1\right )^{\frac {3}{2}}}{12}+\frac {5 \sqrt {-x +1}\, \left (x +1\right )^{\frac {3}{2}}}{8}-\frac {5 \sqrt {-x +1}\, \sqrt {x +1}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x+1)^(5/2)*(x+1)^(1/2),x)

[Out]

1/4*(-x+1)^(5/2)*(x+1)^(3/2)+5/12*(-x+1)^(3/2)*(x+1)^(3/2)+5/8*(-x+1)^(1/2)*(x+1)^(3/2)-5/8*(-x+1)^(1/2)*(x+1)
^(1/2)+5/8*((x+1)*(-x+1))^(1/2)/(x+1)^(1/2)/(-x+1)^(1/2)*arcsin(x)

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maxima [A]  time = 2.95, size = 40, normalized size = 0.59 \[ -\frac {1}{4} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x + \frac {2}{3} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} + \frac {5}{8} \, \sqrt {-x^{2} + 1} x + \frac {5}{8} \, \arcsin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-x^2 + 1)^(3/2)*x + 2/3*(-x^2 + 1)^(3/2) + 5/8*sqrt(-x^2 + 1)*x + 5/8*arcsin(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x)^(5/2)*(x + 1)^(1/2),x)

[Out]

int((1 - x)^(5/2)*(x + 1)^(1/2), x)

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sympy [A]  time = 9.03, size = 218, normalized size = 3.21 \[ \begin {cases} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )}}{4} + \frac {i \left (x + 1\right )^{\frac {9}{2}}}{4 \sqrt {x - 1}} - \frac {23 i \left (x + 1\right )^{\frac {7}{2}}}{12 \sqrt {x - 1}} + \frac {127 i \left (x + 1\right )^{\frac {5}{2}}}{24 \sqrt {x - 1}} - \frac {133 i \left (x + 1\right )^{\frac {3}{2}}}{24 \sqrt {x - 1}} + \frac {5 i \sqrt {x + 1}}{4 \sqrt {x - 1}} & \text {for}\: \frac {\left |{x + 1}\right |}{2} > 1 \\\frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )}}{4} - \frac {\left (x + 1\right )^{\frac {9}{2}}}{4 \sqrt {1 - x}} + \frac {23 \left (x + 1\right )^{\frac {7}{2}}}{12 \sqrt {1 - x}} - \frac {127 \left (x + 1\right )^{\frac {5}{2}}}{24 \sqrt {1 - x}} + \frac {133 \left (x + 1\right )^{\frac {3}{2}}}{24 \sqrt {1 - x}} - \frac {5 \sqrt {x + 1}}{4 \sqrt {1 - x}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-5*I*acosh(sqrt(2)*sqrt(x + 1)/2)/4 + I*(x + 1)**(9/2)/(4*sqrt(x - 1)) - 23*I*(x + 1)**(7/2)/(12*sq
rt(x - 1)) + 127*I*(x + 1)**(5/2)/(24*sqrt(x - 1)) - 133*I*(x + 1)**(3/2)/(24*sqrt(x - 1)) + 5*I*sqrt(x + 1)/(
4*sqrt(x - 1)), Abs(x + 1)/2 > 1), (5*asin(sqrt(2)*sqrt(x + 1)/2)/4 - (x + 1)**(9/2)/(4*sqrt(1 - x)) + 23*(x +
1)**(7/2)/(12*sqrt(1 - x)) - 127*(x + 1)**(5/2)/(24*sqrt(1 - x)) + 133*(x + 1)**(3/2)/(24*sqrt(1 - x)) - 5*sq
rt(x + 1)/(4*sqrt(1 - x)), True))

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