3.8.24 \(\int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx\) [724]

Optimal. Leaf size=110 \[ -\frac {3}{4} \sqrt {1-x} \sqrt {1+x}-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3}{4} \sin ^{-1}(x) \]

[Out]

3/4*arcsin(x)-1/4*(1-x)^(1/2)*(1+x)^(3/2)-1/10*(1-x)^(1/2)*(1+x)^(5/2)-1/5*x^2*(1+x)^(5/2)*(1-x)^(1/2)-1/10*(1
+x)^(7/2)*(1-x)^(1/2)-3/4*(1-x)^(1/2)*(1+x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {102, 21, 81, 52, 41, 222} \begin {gather*} \frac {3 \text {ArcSin}(x)}{4}-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}-\frac {1}{10} \sqrt {1-x} (x+1)^{7/2}-\frac {1}{10} \sqrt {1-x} (x+1)^{5/2}-\frac {1}{4} \sqrt {1-x} (x+1)^{3/2}-\frac {3}{4} \sqrt {1-x} \sqrt {x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

(-3*Sqrt[1 - x]*Sqrt[1 + x])/4 - (Sqrt[1 - x]*(1 + x)^(3/2))/4 - (Sqrt[1 - x]*(1 + x)^(5/2))/10 - (Sqrt[1 - x]
*x^2*(1 + x)^(5/2))/5 - (Sqrt[1 - x]*(1 + x)^(7/2))/10 + (3*ArcSin[x])/4

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 222

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 \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx &=-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{5} \int \frac {(-2-2 x) x (1+x)^{3/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}+\frac {2}{5} \int \frac {x (1+x)^{5/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3}{10} \int \frac {(1+x)^{5/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {1}{2} \int \frac {(1+x)^{3/2}}{\sqrt {1-x}} \, dx\\ &=-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3}{4} \int \frac {\sqrt {1+x}}{\sqrt {1-x}} \, dx\\ &=-\frac {3}{4} \sqrt {1-x} \sqrt {1+x}-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3}{4} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx\\ &=-\frac {3}{4} \sqrt {1-x} \sqrt {1+x}-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3}{4} \int \frac {1}{\sqrt {1-x^2}} \, dx\\ &=-\frac {3}{4} \sqrt {1-x} \sqrt {1+x}-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3}{4} \sin ^{-1}(x)\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 68, normalized size = 0.62 \begin {gather*} -\frac {\sqrt {1-x} \left (24+39 x+27 x^2+22 x^3+14 x^4+4 x^5\right )}{20 \sqrt {1+x}}-\frac {3}{2} \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

-1/20*(Sqrt[1 - x]*(24 + 39*x + 27*x^2 + 22*x^3 + 14*x^4 + 4*x^5))/Sqrt[1 + x] - (3*ArcTan[Sqrt[1 - x]/Sqrt[1
+ x]])/2

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Maple [A]
time = 0.09, size = 94, normalized size = 0.85

method	result	size
risch	\(\frac {\left (4 x^{4}+10 x^{3}+12 x^{2}+15 x +24\right ) \sqrt {1+x}\, \left (-1+x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{20 \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}}+\frac {3 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{4 \sqrt {1-x}\, \sqrt {1+x}}\)	\(87\)
default	\(\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (-4 x^{4} \sqrt {-x^{2}+1}-10 x^{3} \sqrt {-x^{2}+1}-12 x^{2} \sqrt {-x^{2}+1}-15 x \sqrt {-x^{2}+1}+15 \arcsin \left (x \right )-24 \sqrt {-x^{2}+1}\right )}{20 \sqrt {-x^{2}+1}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(1+x)^(3/2)/(1-x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/20*(1+x)^(1/2)*(1-x)^(1/2)*(-4*x^4*(-x^2+1)^(1/2)-10*x^3*(-x^2+1)^(1/2)-12*x^2*(-x^2+1)^(1/2)-15*x*(-x^2+1)^
(1/2)+15*arcsin(x)-24*(-x^2+1)^(1/2))/(-x^2+1)^(1/2)

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Maxima [A]
time = 0.52, size = 70, normalized size = 0.64 \begin {gather*} -\frac {1}{5} \, \sqrt {-x^{2} + 1} x^{4} - \frac {1}{2} \, \sqrt {-x^{2} + 1} x^{3} - \frac {3}{5} \, \sqrt {-x^{2} + 1} x^{2} - \frac {3}{4} \, \sqrt {-x^{2} + 1} x - \frac {6}{5} \, \sqrt {-x^{2} + 1} + \frac {3}{4} \, \arcsin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="maxima")

[Out]

-1/5*sqrt(-x^2 + 1)*x^4 - 1/2*sqrt(-x^2 + 1)*x^3 - 3/5*sqrt(-x^2 + 1)*x^2 - 3/4*sqrt(-x^2 + 1)*x - 6/5*sqrt(-x
^2 + 1) + 3/4*arcsin(x)

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Fricas [A]
time = 1.08, size = 57, normalized size = 0.52 \begin {gather*} -\frac {1}{20} \, {\left (4 \, x^{4} + 10 \, x^{3} + 12 \, x^{2} + 15 \, x + 24\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {3}{2} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="fricas")

[Out]

-1/20*(4*x^4 + 10*x^3 + 12*x^2 + 15*x + 24)*sqrt(x + 1)*sqrt(-x + 1) - 3/2*arctan((sqrt(x + 1)*sqrt(-x + 1) -
1)/x)

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Sympy [A]
time = 120.13, size = 434, normalized size = 3.95 \begin {gather*} - 2 \left (\begin {cases} - \frac {x \sqrt {1 - x} \sqrt {x + 1}}{4} - \sqrt {1 - x} \sqrt {x + 1} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )}}{2} & \text {for}\: \sqrt {x + 1} > - \sqrt {2} \wedge \sqrt {x + 1} < \sqrt {2} \end {cases}\right ) + 6 \left (\begin {cases} - \frac {3 x \sqrt {1 - x} \sqrt {x + 1}}{4} + \frac {\left (1 - x\right )^{\frac {3}{2}} \left (x + 1\right )^{\frac {3}{2}}}{6} - 2 \sqrt {1 - x} \sqrt {x + 1} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )}}{2} & \text {for}\: \sqrt {x + 1} > - \sqrt {2} \wedge \sqrt {x + 1} < \sqrt {2} \end {cases}\right ) - 6 \left (\begin {cases} - \frac {7 x \sqrt {1 - x} \sqrt {x + 1}}{4} + \frac {2 \left (1 - x\right )^{\frac {3}{2}} \left (x + 1\right )^{\frac {3}{2}}}{3} + \frac {\sqrt {1 - x} \sqrt {x + 1} \left (- 5 x - 2 \left (x + 1\right )^{3} + 6 \left (x + 1\right )^{2} - 4\right )}{16} - 4 \sqrt {1 - x} \sqrt {x + 1} + \frac {35 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )}}{8} & \text {for}\: \sqrt {x + 1} > - \sqrt {2} \wedge \sqrt {x + 1} < \sqrt {2} \end {cases}\right ) + 2 \left (\begin {cases} - \frac {15 x \sqrt {1 - x} \sqrt {x + 1}}{4} - \frac {\left (1 - x\right )^{\frac {5}{2}} \left (x + 1\right )^{\frac {5}{2}}}{10} + 2 \left (1 - x\right )^{\frac {3}{2}} \left (x + 1\right )^{\frac {3}{2}} + \frac {5 \sqrt {1 - x} \sqrt {x + 1} \left (- 5 x - 2 \left (x + 1\right )^{3} + 6 \left (x + 1\right )^{2} - 4\right )}{16} - 8 \sqrt {1 - x} \sqrt {x + 1} + \frac {63 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {x + 1}}{2} \right )}}{8} & \text {for}\: \sqrt {x + 1} > - \sqrt {2} \wedge \sqrt {x + 1} < \sqrt {2} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(1+x)**(3/2)/(1-x)**(1/2),x)

[Out]

-2*Piecewise((-x*sqrt(1 - x)*sqrt(x + 1)/4 - sqrt(1 - x)*sqrt(x + 1) + 3*asin(sqrt(2)*sqrt(x + 1)/2)/2, (sqrt(
x + 1) < sqrt(2)) & (sqrt(x + 1) > -sqrt(2)))) + 6*Piecewise((-3*x*sqrt(1 - x)*sqrt(x + 1)/4 + (1 - x)**(3/2)*
(x + 1)**(3/2)/6 - 2*sqrt(1 - x)*sqrt(x + 1) + 5*asin(sqrt(2)*sqrt(x + 1)/2)/2, (sqrt(x + 1) < sqrt(2)) & (sqr
t(x + 1) > -sqrt(2)))) - 6*Piecewise((-7*x*sqrt(1 - x)*sqrt(x + 1)/4 + 2*(1 - x)**(3/2)*(x + 1)**(3/2)/3 + sqr
t(1 - x)*sqrt(x + 1)*(-5*x - 2*(x + 1)**3 + 6*(x + 1)**2 - 4)/16 - 4*sqrt(1 - x)*sqrt(x + 1) + 35*asin(sqrt(2)
*sqrt(x + 1)/2)/8, (sqrt(x + 1) < sqrt(2)) & (sqrt(x + 1) > -sqrt(2)))) + 2*Piecewise((-15*x*sqrt(1 - x)*sqrt(
x + 1)/4 - (1 - x)**(5/2)*(x + 1)**(5/2)/10 + 2*(1 - x)**(3/2)*(x + 1)**(3/2) + 5*sqrt(1 - x)*sqrt(x + 1)*(-5*
x - 2*(x + 1)**3 + 6*(x + 1)**2 - 4)/16 - 8*sqrt(1 - x)*sqrt(x + 1) + 63*asin(sqrt(2)*sqrt(x + 1)/2)/8, (sqrt(
x + 1) < sqrt(2)) & (sqrt(x + 1) > -sqrt(2))))

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Giac [A]
time = 0.82, size = 52, normalized size = 0.47 \begin {gather*} -\frac {1}{20} \, {\left ({\left (2 \, {\left ({\left (2 \, x - 1\right )} {\left (x + 1\right )} + 3\right )} {\left (x + 1\right )} + 5\right )} {\left (x + 1\right )} + 15\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {3}{2} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="giac")

[Out]

-1/20*((2*((2*x - 1)*(x + 1) + 3)*(x + 1) + 5)*(x + 1) + 15)*sqrt(x + 1)*sqrt(-x + 1) + 3/2*arcsin(1/2*sqrt(2)
*sqrt(x + 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(x + 1)^(3/2))/(1 - x)^(1/2),x)

[Out]

int((x^3*(x + 1)^(3/2))/(1 - x)^(1/2), x)

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