\(\int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx\) [723]

Optimal result

Integrand size = 20, antiderivative size = 118 \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {11}{16} \sqrt {1-x} \sqrt {1+x}-\frac {11}{48} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{120} \sqrt {1-x} (1+x)^{5/2} (18+19 x)+\frac {11 \arcsin (x)}{16} \]

[Out]

11/16*arcsin(x)-11/48*(1-x)^(1/2)*(1+x)^(3/2)-1/15*x^2*(1+x)^(5/2)*(1-x)^(1/2)-1/6*x^3*(1+x)^(5/2)*(1-x)^(1/2)
-1/120*(1+x)^(5/2)*(18+19*x)*(1-x)^(1/2)-11/16*(1-x)^(1/2)*(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {102, 158, 152, 52, 41, 222} \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\frac {11 \arcsin (x)}{16}-\frac {1}{6} \sqrt {1-x} (x+1)^{5/2} x^3-\frac {1}{15} \sqrt {1-x} (x+1)^{5/2} x^2-\frac {11}{48} \sqrt {1-x} (x+1)^{3/2}-\frac {1}{120} \sqrt {1-x} (x+1)^{5/2} (19 x+18)-\frac {11}{16} \sqrt {1-x} \sqrt {x+1} \]

[In]

Int[(x^4*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

(-11*Sqrt[1 - x]*Sqrt[1 + x])/16 - (11*Sqrt[1 - x]*(1 + x)^(3/2))/48 - (Sqrt[1 - x]*x^2*(1 + x)^(5/2))/15 - (S
qrt[1 - x]*x^3*(1 + x)^(5/2))/6 - (Sqrt[1 - x]*(1 + x)^(5/2)*(18 + 19*x))/120 + (11*ArcSin[x])/16

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 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 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 158

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{6} \int \frac {(-3-2 x) x^2 (1+x)^{3/2}}{\sqrt {1-x}} \, dx \\ & = -\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}+\frac {1}{30} \int \frac {x (1+x)^{3/2} (4+19 x)}{\sqrt {1-x}} \, dx \\ & = -\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{120} \sqrt {1-x} (1+x)^{5/2} (18+19 x)+\frac {11}{24} \int \frac {(1+x)^{3/2}}{\sqrt {1-x}} \, dx \\ & = -\frac {11}{48} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{120} \sqrt {1-x} (1+x)^{5/2} (18+19 x)+\frac {11}{16} \int \frac {\sqrt {1+x}}{\sqrt {1-x}} \, dx \\ & = -\frac {11}{16} \sqrt {1-x} \sqrt {1+x}-\frac {11}{48} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{120} \sqrt {1-x} (1+x)^{5/2} (18+19 x)+\frac {11}{16} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = -\frac {11}{16} \sqrt {1-x} \sqrt {1+x}-\frac {11}{48} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{120} \sqrt {1-x} (1+x)^{5/2} (18+19 x)+\frac {11}{16} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {11}{16} \sqrt {1-x} \sqrt {1+x}-\frac {11}{48} \sqrt {1-x} (1+x)^{3/2}-\frac {1}{15} \sqrt {1-x} x^2 (1+x)^{5/2}-\frac {1}{6} \sqrt {1-x} x^3 (1+x)^{5/2}-\frac {1}{120} \sqrt {1-x} (1+x)^{5/2} (18+19 x)+\frac {11}{16} \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {\sqrt {1-x} \left (256+421 x+293 x^2+238 x^3+206 x^4+136 x^5+40 x^6\right )}{240 \sqrt {1+x}}-\frac {11}{8} \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]

[In]

Integrate[(x^4*(1 + x)^(3/2))/Sqrt[1 - x],x]

[Out]

-1/240*(Sqrt[1 - x]*(256 + 421*x + 293*x^2 + 238*x^3 + 206*x^4 + 136*x^5 + 40*x^6))/Sqrt[1 + x] - (11*ArcTan[S
qrt[1 - x]/Sqrt[1 + x]])/8

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\left (40 x^{5}+96 x^{4}+110 x^{3}+128 x^{2}+165 x +256\right ) \left (-1+x \right ) \sqrt {1+x}\, \sqrt {\left (1+x \right ) \left (1-x \right )}}{240 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {11 \arcsin \left (x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{16 \sqrt {1-x}\, \sqrt {1+x}}\)	\(92\)
default	\(\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (-40 x^{5} \sqrt {-x^{2}+1}-96 x^{4} \sqrt {-x^{2}+1}-110 x^{3} \sqrt {-x^{2}+1}-128 x^{2} \sqrt {-x^{2}+1}-165 x \sqrt {-x^{2}+1}+165 \arcsin \left (x \right )-256 \sqrt {-x^{2}+1}\right )}{240 \sqrt {-x^{2}+1}}\)	\(108\)

[In]

int(x^4*(1+x)^(3/2)/(1-x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/240*(40*x^5+96*x^4+110*x^3+128*x^2+165*x+256)*(-1+x)*(1+x)^(1/2)/(-(-1+x)*(1+x))^(1/2)*((1+x)*(1-x))^(1/2)/(
1-x)^(1/2)+11/16*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 = 62, normalized size of antiderivative = 0.53 \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{240} \, {\left (40 \, x^{5} + 96 \, x^{4} + 110 \, x^{3} + 128 \, x^{2} + 165 \, x + 256\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {11}{8} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

[In]

integrate(x^4*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="fricas")

[Out]

-1/240*(40*x^5 + 96*x^4 + 110*x^3 + 128*x^2 + 165*x + 256)*sqrt(x + 1)*sqrt(-x + 1) - 11/8*arctan((sqrt(x + 1)
*sqrt(-x + 1) - 1)/x)

Sympy [F]

\[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\int \frac {x^{4} \left (x + 1\right )^{\frac {3}{2}}}{\sqrt {1 - x}}\, dx \]

[In]

integrate(x**4*(1+x)**(3/2)/(1-x)**(1/2),x)

[Out]

Integral(x**4*(x + 1)**(3/2)/sqrt(1 - x), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{6} \, \sqrt {-x^{2} + 1} x^{5} - \frac {2}{5} \, \sqrt {-x^{2} + 1} x^{4} - \frac {11}{24} \, \sqrt {-x^{2} + 1} x^{3} - \frac {8}{15} \, \sqrt {-x^{2} + 1} x^{2} - \frac {11}{16} \, \sqrt {-x^{2} + 1} x - \frac {16}{15} \, \sqrt {-x^{2} + 1} + \frac {11}{16} \, \arcsin \left (x\right ) \]

[In]

integrate(x^4*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(-x^2 + 1)*x^5 - 2/5*sqrt(-x^2 + 1)*x^4 - 11/24*sqrt(-x^2 + 1)*x^3 - 8/15*sqrt(-x^2 + 1)*x^2 - 11/16*
sqrt(-x^2 + 1)*x - 16/15*sqrt(-x^2 + 1) + 11/16*arcsin(x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=-\frac {1}{240} \, {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, x - 8\right )} {\left (x + 1\right )} + 63\right )} {\left (x + 1\right )} - 13\right )} {\left (x + 1\right )} + 55\right )} {\left (x + 1\right )} + 165\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {11}{8} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[In]

integrate(x^4*(1+x)^(3/2)/(1-x)^(1/2),x, algorithm="giac")

[Out]

-1/240*((2*((4*(5*x - 8)*(x + 1) + 63)*(x + 1) - 13)*(x + 1) + 55)*(x + 1) + 165)*sqrt(x + 1)*sqrt(-x + 1) + 1
1/8*arcsin(1/2*sqrt(2)*sqrt(x + 1))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (1+x)^{3/2}}{\sqrt {1-x}} \, dx=\int \frac {x^4\,{\left (x+1\right )}^{3/2}}{\sqrt {1-x}} \,d x \]

[In]

int((x^4*(x + 1)^(3/2))/(1 - x)^(1/2),x)

[Out]

int((x^4*(x + 1)^(3/2))/(1 - x)^(1/2), x)