\(\int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx\) [1115]

Optimal result

Integrand size = 17, antiderivative size = 101 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=\frac {\sqrt {1+x}}{9 (1-x)^{9/2}}+\frac {4 \sqrt {1+x}}{63 (1-x)^{7/2}}+\frac {4 \sqrt {1+x}}{105 (1-x)^{5/2}}+\frac {8 \sqrt {1+x}}{315 (1-x)^{3/2}}+\frac {8 \sqrt {1+x}}{315 \sqrt {1-x}} \]

[Out]

1/9*(1+x)^(1/2)/(1-x)^(9/2)+4/63*(1+x)^(1/2)/(1-x)^(7/2)+4/105*(1+x)^(1/2)/(1-x)^(5/2)+8/315*(1+x)^(1/2)/(1-x)
^(3/2)+8/315*(1+x)^(1/2)/(1-x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {47, 37} \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=\frac {8 \sqrt {x+1}}{315 \sqrt {1-x}}+\frac {8 \sqrt {x+1}}{315 (1-x)^{3/2}}+\frac {4 \sqrt {x+1}}{105 (1-x)^{5/2}}+\frac {4 \sqrt {x+1}}{63 (1-x)^{7/2}}+\frac {\sqrt {x+1}}{9 (1-x)^{9/2}} \]

[In]

Int[1/((1 - x)^(11/2)*Sqrt[1 + x]),x]

[Out]

Sqrt[1 + x]/(9*(1 - x)^(9/2)) + (4*Sqrt[1 + x])/(63*(1 - x)^(7/2)) + (4*Sqrt[1 + x])/(105*(1 - x)^(5/2)) + (8*
Sqrt[1 + x])/(315*(1 - x)^(3/2)) + (8*Sqrt[1 + x])/(315*Sqrt[1 - x])

Rule 37

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

Rule 47

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+x}}{9 (1-x)^{9/2}}+\frac {4}{9} \int \frac {1}{(1-x)^{9/2} \sqrt {1+x}} \, dx \\ & = \frac {\sqrt {1+x}}{9 (1-x)^{9/2}}+\frac {4 \sqrt {1+x}}{63 (1-x)^{7/2}}+\frac {4}{21} \int \frac {1}{(1-x)^{7/2} \sqrt {1+x}} \, dx \\ & = \frac {\sqrt {1+x}}{9 (1-x)^{9/2}}+\frac {4 \sqrt {1+x}}{63 (1-x)^{7/2}}+\frac {4 \sqrt {1+x}}{105 (1-x)^{5/2}}+\frac {8}{105} \int \frac {1}{(1-x)^{5/2} \sqrt {1+x}} \, dx \\ & = \frac {\sqrt {1+x}}{9 (1-x)^{9/2}}+\frac {4 \sqrt {1+x}}{63 (1-x)^{7/2}}+\frac {4 \sqrt {1+x}}{105 (1-x)^{5/2}}+\frac {8 \sqrt {1+x}}{315 (1-x)^{3/2}}+\frac {8}{315} \int \frac {1}{(1-x)^{3/2} \sqrt {1+x}} \, dx \\ & = \frac {\sqrt {1+x}}{9 (1-x)^{9/2}}+\frac {4 \sqrt {1+x}}{63 (1-x)^{7/2}}+\frac {4 \sqrt {1+x}}{105 (1-x)^{5/2}}+\frac {8 \sqrt {1+x}}{315 (1-x)^{3/2}}+\frac {8 \sqrt {1+x}}{315 \sqrt {1-x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.40 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=\frac {\sqrt {1+x} \left (83-100 x+84 x^2-40 x^3+8 x^4\right )}{315 (1-x)^{9/2}} \]

[In]

Integrate[1/((1 - x)^(11/2)*Sqrt[1 + x]),x]

[Out]

(Sqrt[1 + x]*(83 - 100*x + 84*x^2 - 40*x^3 + 8*x^4))/(315*(1 - x)^(9/2))

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.35

method	result	size
gosper	\(\frac {\sqrt {1+x}\, \left (8 x^{4}-40 x^{3}+84 x^{2}-100 x +83\right )}{315 \left (1-x \right )^{\frac {9}{2}}}\)	\(35\)
risch	\(\frac {\sqrt {\left (1+x \right ) \left (1-x \right )}\, \left (8 x^{5}-32 x^{4}+44 x^{3}-16 x^{2}-17 x +83\right )}{315 \sqrt {1-x}\, \sqrt {1+x}\, \left (-1+x \right )^{4} \sqrt {-\left (-1+x \right ) \left (1+x \right )}}\)	\(66\)
default	\(\frac {\sqrt {1+x}}{9 \left (1-x \right )^{\frac {9}{2}}}+\frac {4 \sqrt {1+x}}{63 \left (1-x \right )^{\frac {7}{2}}}+\frac {4 \sqrt {1+x}}{105 \left (1-x \right )^{\frac {5}{2}}}+\frac {8 \sqrt {1+x}}{315 \left (1-x \right )^{\frac {3}{2}}}+\frac {8 \sqrt {1+x}}{315 \sqrt {1-x}}\)	\(72\)

[In]

int(1/(1-x)^(11/2)/(1+x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/315*(1+x)^(1/2)/(1-x)^(9/2)*(8*x^4-40*x^3+84*x^2-100*x+83)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=\frac {83 \, x^{5} - 415 \, x^{4} + 830 \, x^{3} - 830 \, x^{2} - {\left (8 \, x^{4} - 40 \, x^{3} + 84 \, x^{2} - 100 \, x + 83\right )} \sqrt {x + 1} \sqrt {-x + 1} + 415 \, x - 83}{315 \, {\left (x^{5} - 5 \, x^{4} + 10 \, x^{3} - 10 \, x^{2} + 5 \, x - 1\right )}} \]

[In]

integrate(1/(1-x)^(11/2)/(1+x)^(1/2),x, algorithm="fricas")

[Out]

1/315*(83*x^5 - 415*x^4 + 830*x^3 - 830*x^2 - (8*x^4 - 40*x^3 + 84*x^2 - 100*x + 83)*sqrt(x + 1)*sqrt(-x + 1)
+ 415*x - 83)/(x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 55.84 (sec) , antiderivative size = 850, normalized size of antiderivative = 8.42 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(1-x)**(11/2)/(1+x)**(1/2),x)

[Out]

Piecewise((8*(x + 1)**4/(315*sqrt(-1 + 2/(x + 1))*(x + 1)**4 - 2520*sqrt(-1 + 2/(x + 1))*(x + 1)**3 + 7560*sqr
t(-1 + 2/(x + 1))*(x + 1)**2 - 10080*sqrt(-1 + 2/(x + 1))*(x + 1) + 5040*sqrt(-1 + 2/(x + 1))) - 72*(x + 1)**3
/(315*sqrt(-1 + 2/(x + 1))*(x + 1)**4 - 2520*sqrt(-1 + 2/(x + 1))*(x + 1)**3 + 7560*sqrt(-1 + 2/(x + 1))*(x +
1)**2 - 10080*sqrt(-1 + 2/(x + 1))*(x + 1) + 5040*sqrt(-1 + 2/(x + 1))) + 252*(x + 1)**2/(315*sqrt(-1 + 2/(x +
1))*(x + 1)**4 - 2520*sqrt(-1 + 2/(x + 1))*(x + 1)**3 + 7560*sqrt(-1 + 2/(x + 1))*(x + 1)**2 - 10080*sqrt(-1
+ 2/(x + 1))*(x + 1) + 5040*sqrt(-1 + 2/(x + 1))) - 420*(x + 1)/(315*sqrt(-1 + 2/(x + 1))*(x + 1)**4 - 2520*sq
rt(-1 + 2/(x + 1))*(x + 1)**3 + 7560*sqrt(-1 + 2/(x + 1))*(x + 1)**2 - 10080*sqrt(-1 + 2/(x + 1))*(x + 1) + 50
40*sqrt(-1 + 2/(x + 1))) + 315/(315*sqrt(-1 + 2/(x + 1))*(x + 1)**4 - 2520*sqrt(-1 + 2/(x + 1))*(x + 1)**3 + 7
560*sqrt(-1 + 2/(x + 1))*(x + 1)**2 - 10080*sqrt(-1 + 2/(x + 1))*(x + 1) + 5040*sqrt(-1 + 2/(x + 1))), 1/Abs(x
+ 1) > 1/2), (-8*I*(x + 1)**4/(315*sqrt(1 - 2/(x + 1))*(x + 1)**4 - 2520*sqrt(1 - 2/(x + 1))*(x + 1)**3 + 756
0*sqrt(1 - 2/(x + 1))*(x + 1)**2 - 10080*sqrt(1 - 2/(x + 1))*(x + 1) + 5040*sqrt(1 - 2/(x + 1))) + 72*I*(x + 1
)**3/(315*sqrt(1 - 2/(x + 1))*(x + 1)**4 - 2520*sqrt(1 - 2/(x + 1))*(x + 1)**3 + 7560*sqrt(1 - 2/(x + 1))*(x +
1)**2 - 10080*sqrt(1 - 2/(x + 1))*(x + 1) + 5040*sqrt(1 - 2/(x + 1))) - 252*I*(x + 1)**2/(315*sqrt(1 - 2/(x +
1))*(x + 1)**4 - 2520*sqrt(1 - 2/(x + 1))*(x + 1)**3 + 7560*sqrt(1 - 2/(x + 1))*(x + 1)**2 - 10080*sqrt(1 - 2
/(x + 1))*(x + 1) + 5040*sqrt(1 - 2/(x + 1))) + 420*I*(x + 1)/(315*sqrt(1 - 2/(x + 1))*(x + 1)**4 - 2520*sqrt(
1 - 2/(x + 1))*(x + 1)**3 + 7560*sqrt(1 - 2/(x + 1))*(x + 1)**2 - 10080*sqrt(1 - 2/(x + 1))*(x + 1) + 5040*sqr
t(1 - 2/(x + 1))) - 315*I/(315*sqrt(1 - 2/(x + 1))*(x + 1)**4 - 2520*sqrt(1 - 2/(x + 1))*(x + 1)**3 + 7560*sqr
t(1 - 2/(x + 1))*(x + 1)**2 - 10080*sqrt(1 - 2/(x + 1))*(x + 1) + 5040*sqrt(1 - 2/(x + 1))), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=-\frac {\sqrt {-x^{2} + 1}}{9 \, {\left (x^{5} - 5 \, x^{4} + 10 \, x^{3} - 10 \, x^{2} + 5 \, x - 1\right )}} + \frac {4 \, \sqrt {-x^{2} + 1}}{63 \, {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1\right )}} - \frac {4 \, \sqrt {-x^{2} + 1}}{105 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}} + \frac {8 \, \sqrt {-x^{2} + 1}}{315 \, {\left (x^{2} - 2 \, x + 1\right )}} - \frac {8 \, \sqrt {-x^{2} + 1}}{315 \, {\left (x - 1\right )}} \]

[In]

integrate(1/(1-x)^(11/2)/(1+x)^(1/2),x, algorithm="maxima")

[Out]

-1/9*sqrt(-x^2 + 1)/(x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 1) + 4/63*sqrt(-x^2 + 1)/(x^4 - 4*x^3 + 6*x^2 - 4*x
+ 1) - 4/105*sqrt(-x^2 + 1)/(x^3 - 3*x^2 + 3*x - 1) + 8/315*sqrt(-x^2 + 1)/(x^2 - 2*x + 1) - 8/315*sqrt(-x^2
+ 1)/(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.42 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=-\frac {{\left (4 \, {\left ({\left (2 \, {\left (x + 1\right )} {\left (x - 8\right )} + 63\right )} {\left (x + 1\right )} - 105\right )} {\left (x + 1\right )} + 315\right )} \sqrt {x + 1} \sqrt {-x + 1}}{315 \, {\left (x - 1\right )}^{5}} \]

[In]

integrate(1/(1-x)^(11/2)/(1+x)^(1/2),x, algorithm="giac")

[Out]

-1/315*(4*((2*(x + 1)*(x - 8) + 63)*(x + 1) - 105)*(x + 1) + 315)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^5

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-x)^{11/2} \sqrt {1+x}} \, dx=\frac {17\,x\,\sqrt {1-x}-83\,\sqrt {1-x}+16\,x^2\,\sqrt {1-x}-44\,x^3\,\sqrt {1-x}+32\,x^4\,\sqrt {1-x}-8\,x^5\,\sqrt {1-x}}{315\,{\left (x-1\right )}^5\,\sqrt {x+1}} \]

[In]

int(1/((1 - x)^(11/2)*(x + 1)^(1/2)),x)

[Out]

(17*x*(1 - x)^(1/2) - 83*(1 - x)^(1/2) + 16*x^2*(1 - x)^(1/2) - 44*x^3*(1 - x)^(1/2) + 32*x^4*(1 - x)^(1/2) -
8*x^5*(1 - x)^(1/2))/(315*(x - 1)^5*(x + 1)^(1/2))