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

Optimal result

Integrand size = 20, antiderivative size = 110 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {3}{4} \sqrt {1-x} \sqrt {1+x}-\frac {1}{4} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{5/2}-\frac {1}{5} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{10} \sqrt {1-x} (1+x)^{7/2}+\frac {3 \arcsin (x)}{4} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {102, 21, 81, 52, 41, 222} \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\frac {3 \arcsin (x)}{4}-\frac {1}{5} \sqrt {1-x} x^2 (x+1)^{5/2}-\frac {1}{10} \sqrt {1-x} (x+1)^{7/2}-\frac {1}{10} \sqrt {1-x} (x+1)^{5/2}-\frac {1}{4} \sqrt {1-x} (x+1)^{3/2}-\frac {3}{4} \sqrt {1-x} \sqrt {x+1} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {\sqrt {1-x} \left (24+39 x+27 x^2+22 x^3+14 x^4+4 x^5\right )}{20 \sqrt {1+x}}-\frac {3}{2} \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]

[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

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.52 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{20} \, {\left (4 \, x^{4} + 10 \, x^{3} + 12 \, x^{2} + 15 \, x + 24\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {3}{2} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

[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)

Sympy [F]

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

[In]

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

[Out]

Integral(x**3*(x + 1)**(3/2)/sqrt(1 - x), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{5} \, \sqrt {-x^{2} + 1} x^{4} - \frac {1}{2} \, \sqrt {-x^{2} + 1} x^{3} - \frac {3}{5} \, \sqrt {-x^{2} + 1} x^{2} - \frac {3}{4} \, \sqrt {-x^{2} + 1} x - \frac {6}{5} \, \sqrt {-x^{2} + 1} + \frac {3}{4} \, \arcsin \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.47 \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, {\left ({\left (2 \, x - 1\right )} {\left (x + 1\right )} + 3\right )} {\left (x + 1\right )} + 5\right )} {\left (x + 1\right )} + 15\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {3}{2} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\int \frac {x^3\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x}} \,d x \]

[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)