\(\int (1-x)^{7/2} (1+x)^{5/2} \, dx\) [1090]

Optimal result

Integrand size = 17, antiderivative size = 90 \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=\frac {5}{16} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{24} (1-x)^{3/2} x (1+x)^{3/2}+\frac {1}{6} (1-x)^{5/2} x (1+x)^{5/2}+\frac {1}{7} (1-x)^{7/2} (1+x)^{7/2}+\frac {5 \arcsin (x)}{16} \]

[Out]

5/24*(1-x)^(3/2)*x*(1+x)^(3/2)+1/6*(1-x)^(5/2)*x*(1+x)^(5/2)+1/7*(1-x)^(7/2)*(1+x)^(7/2)+5/16*arcsin(x)+5/16*x
*(1-x)^(1/2)*(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {51, 38, 41, 222} \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=\frac {5 \arcsin (x)}{16}+\frac {1}{7} (1-x)^{7/2} (x+1)^{7/2}+\frac {1}{6} (1-x)^{5/2} x (x+1)^{5/2}+\frac {5}{24} (1-x)^{3/2} x (x+1)^{3/2}+\frac {5}{16} \sqrt {1-x} x \sqrt {x+1} \]

[In]

Int[(1 - x)^(7/2)*(1 + x)^(5/2),x]

[Out]

(5*Sqrt[1 - x]*x*Sqrt[1 + x])/16 + (5*(1 - x)^(3/2)*x*(1 + x)^(3/2))/24 + ((1 - x)^(5/2)*x*(1 + x)^(5/2))/6 +
((1 - x)^(7/2)*(1 + x)^(7/2))/7 + (5*ArcSin[x])/16

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 51

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 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}{7} (1-x)^{7/2} (1+x)^{7/2}+\int (1-x)^{5/2} (1+x)^{5/2} \, dx \\ & = \frac {1}{6} (1-x)^{5/2} x (1+x)^{5/2}+\frac {1}{7} (1-x)^{7/2} (1+x)^{7/2}+\frac {5}{6} \int (1-x)^{3/2} (1+x)^{3/2} \, dx \\ & = \frac {5}{24} (1-x)^{3/2} x (1+x)^{3/2}+\frac {1}{6} (1-x)^{5/2} x (1+x)^{5/2}+\frac {1}{7} (1-x)^{7/2} (1+x)^{7/2}+\frac {5}{8} \int \sqrt {1-x} \sqrt {1+x} \, dx \\ & = \frac {5}{16} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{24} (1-x)^{3/2} x (1+x)^{3/2}+\frac {1}{6} (1-x)^{5/2} x (1+x)^{5/2}+\frac {1}{7} (1-x)^{7/2} (1+x)^{7/2}+\frac {5}{16} \int \frac {1}{\sqrt {1-x} \sqrt {1+x}} \, dx \\ & = \frac {5}{16} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{24} (1-x)^{3/2} x (1+x)^{3/2}+\frac {1}{6} (1-x)^{5/2} x (1+x)^{5/2}+\frac {1}{7} (1-x)^{7/2} (1+x)^{7/2}+\frac {5}{16} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = \frac {5}{16} \sqrt {1-x} x \sqrt {1+x}+\frac {5}{24} (1-x)^{3/2} x (1+x)^{3/2}+\frac {1}{6} (1-x)^{5/2} x (1+x)^{5/2}+\frac {1}{7} (1-x)^{7/2} (1+x)^{7/2}+\frac {5}{16} \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.76 \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=\frac {1}{336} \sqrt {1-x^2} \left (48+231 x-144 x^2-182 x^3+144 x^4+56 x^5-48 x^6\right )-\frac {5}{8} \arctan \left (\frac {\sqrt {1-x^2}}{-1+x}\right ) \]

[In]

Integrate[(1 - x)^(7/2)*(1 + x)^(5/2),x]

[Out]

(Sqrt[1 - x^2]*(48 + 231*x - 144*x^2 - 182*x^3 + 144*x^4 + 56*x^5 - 48*x^6))/336 - (5*ArcTan[Sqrt[1 - x^2]/(-1
+ x)])/8

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {\left (48 x^{6}-56 x^{5}-144 x^{4}+182 x^{3}+144 x^{2}-231 x -48\right ) \left (-1+x \right ) \sqrt {1+x}\, \sqrt {\left (1+x \right ) \left (1-x \right )}}{336 \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {5 \sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{16 \sqrt {1+x}\, \sqrt {1-x}}\)	\(97\)
default	\(\frac {\left (1-x \right )^{\frac {7}{2}} \left (1+x \right )^{\frac {7}{2}}}{7}+\frac {\left (1-x \right )^{\frac {5}{2}} \left (1+x \right )^{\frac {7}{2}}}{6}+\frac {\left (1-x \right )^{\frac {3}{2}} \left (1+x \right )^{\frac {7}{2}}}{6}+\frac {\sqrt {1-x}\, \left (1+x \right )^{\frac {7}{2}}}{8}-\frac {\sqrt {1-x}\, \left (1+x \right )^{\frac {5}{2}}}{24}-\frac {5 \sqrt {1-x}\, \left (1+x \right )^{\frac {3}{2}}}{48}-\frac {5 \sqrt {1-x}\, \sqrt {1+x}}{16}+\frac {5 \sqrt {\left (1+x \right ) \left (1-x \right )}\, \arcsin \left (x \right )}{16 \sqrt {1+x}\, \sqrt {1-x}}\)	\(127\)

[In]

int((1-x)^(7/2)*(1+x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/336*(48*x^6-56*x^5-144*x^4+182*x^3+144*x^2-231*x-48)*(-1+x)*(1+x)^(1/2)/(-(-1+x)*(1+x))^(1/2)*((1+x)*(1-x))^
(1/2)/(1-x)^(1/2)+5/16*((1+x)*(1-x))^(1/2)/(1+x)^(1/2)/(1-x)^(1/2)*arcsin(x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=-\frac {1}{336} \, {\left (48 \, x^{6} - 56 \, x^{5} - 144 \, x^{4} + 182 \, x^{3} + 144 \, x^{2} - 231 \, x - 48\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {5}{8} \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

[In]

integrate((1-x)^(7/2)*(1+x)^(5/2),x, algorithm="fricas")

[Out]

-1/336*(48*x^6 - 56*x^5 - 144*x^4 + 182*x^3 + 144*x^2 - 231*x - 48)*sqrt(x + 1)*sqrt(-x + 1) - 5/8*arctan((sqr
t(x + 1)*sqrt(-x + 1) - 1)/x)

Sympy [F(-1)]

Timed out. \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((1-x)**(7/2)*(1+x)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58 \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=\frac {1}{7} \, {\left (-x^{2} + 1\right )}^{\frac {7}{2}} + \frac {1}{6} \, {\left (-x^{2} + 1\right )}^{\frac {5}{2}} x + \frac {5}{24} \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x + \frac {5}{16} \, \sqrt {-x^{2} + 1} x + \frac {5}{16} \, \arcsin \left (x\right ) \]

[In]

integrate((1-x)^(7/2)*(1+x)^(5/2),x, algorithm="maxima")

[Out]

1/7*(-x^2 + 1)^(7/2) + 1/6*(-x^2 + 1)^(5/2)*x + 5/24*(-x^2 + 1)^(3/2)*x + 5/16*sqrt(-x^2 + 1)*x + 5/16*arcsin(
x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (64) = 128\).

Time = 0.33 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.59 \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=-\frac {1}{1680} \, {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, x - 37\right )} {\left (x + 1\right )} + 661\right )} {\left (x + 1\right )} - 4551\right )} {\left (x + 1\right )} + 4781\right )} {\left (x + 1\right )} - 6335\right )} {\left (x + 1\right )} + 2835\right )} \sqrt {x + 1} \sqrt {-x + 1} + \frac {1}{40} \, {\left ({\left (2 \, {\left (3 \, {\left (4 \, x - 17\right )} {\left (x + 1\right )} + 133\right )} {\left (x + 1\right )} - 295\right )} {\left (x + 1\right )} + 195\right )} \sqrt {x + 1} \sqrt {-x + 1} - \frac {1}{2} \, {\left ({\left (2 \, x - 5\right )} {\left (x + 1\right )} + 9\right )} \sqrt {x + 1} \sqrt {-x + 1} + \sqrt {x + 1} \sqrt {-x + 1} + \frac {5}{8} \, \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {x + 1}\right ) \]

[In]

integrate((1-x)^(7/2)*(1+x)^(5/2),x, algorithm="giac")

[Out]

-1/1680*((2*((4*(5*(6*x - 37)*(x + 1) + 661)*(x + 1) - 4551)*(x + 1) + 4781)*(x + 1) - 6335)*(x + 1) + 2835)*s
qrt(x + 1)*sqrt(-x + 1) + 1/40*((2*(3*(4*x - 17)*(x + 1) + 133)*(x + 1) - 295)*(x + 1) + 195)*sqrt(x + 1)*sqrt
(-x + 1) - 1/2*((2*x - 5)*(x + 1) + 9)*sqrt(x + 1)*sqrt(-x + 1) + sqrt(x + 1)*sqrt(-x + 1) + 5/8*arcsin(1/2*sq
rt(2)*sqrt(x + 1))

Mupad [F(-1)]

Timed out. \[ \int (1-x)^{7/2} (1+x)^{5/2} \, dx=\int {\left (1-x\right )}^{7/2}\,{\left (x+1\right )}^{5/2} \,d x \]

[In]

int((1 - x)^(7/2)*(x + 1)^(5/2),x)

[Out]

int((1 - x)^(7/2)*(x + 1)^(5/2), x)