\(\int (a-a \sin ^2(x))^{5/2} \, dx\) [116]

Optimal result

Integrand size = 13, antiderivative size = 53 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {8}{15} a^2 \sqrt {a \cos ^2(x)} \tan (x)+\frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x) \]

[Out]

4/15*a*(a*cos(x)^2)^(3/2)*tan(x)+1/5*(a*cos(x)^2)^(5/2)*tan(x)+8/15*a^2*(a*cos(x)^2)^(1/2)*tan(x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3255, 3282, 3286, 2717} \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {8}{15} a^2 \tan (x) \sqrt {a \cos ^2(x)}+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}+\frac {4}{15} a \tan (x) \left (a \cos ^2(x)\right )^{3/2} \]

[In]

Int[(a - a*Sin[x]^2)^(5/2),x]

[Out]

(8*a^2*Sqrt[a*Cos[x]^2]*Tan[x])/15 + (4*a*(a*Cos[x]^2)^(3/2)*Tan[x])/15 + ((a*Cos[x]^2)^(5/2)*Tan[x])/5

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3255

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3282

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-Cot[e + f*x])*((b*Sin[e + f*x]^2)^p/(2*f*p)),
x] + Dist[b*((2*p - 1)/(2*p)), Int[(b*Sin[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] &&  !IntegerQ[p]
&& GtQ[p, 1]

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps \begin{align*} \text {integral}& = \int \left (a \cos ^2(x)\right )^{5/2} \, dx \\ & = \frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)+\frac {1}{5} (4 a) \int \left (a \cos ^2(x)\right )^{3/2} \, dx \\ & = \frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)+\frac {1}{15} \left (8 a^2\right ) \int \sqrt {a \cos ^2(x)} \, dx \\ & = \frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x)+\frac {1}{15} \left (8 a^2 \sqrt {a \cos ^2(x)} \sec (x)\right ) \int \cos (x) \, dx \\ & = \frac {8}{15} a^2 \sqrt {a \cos ^2(x)} \tan (x)+\frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.62 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {1}{15} a^2 \sqrt {a \cos ^2(x)} \left (15-10 \sin ^2(x)+3 \sin ^4(x)\right ) \tan (x) \]

[In]

Integrate[(a - a*Sin[x]^2)^(5/2),x]

[Out]

(a^2*Sqrt[a*Cos[x]^2]*(15 - 10*Sin[x]^2 + 3*Sin[x]^4)*Tan[x])/15

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {\cos \left (x \right ) a^{3} \sin \left (x \right ) \left (3 \left (\cos ^{4}\left (x \right )\right )+4 \left (\cos ^{2}\left (x \right )\right )+8\right )}{15 \sqrt {a \left (\cos ^{2}\left (x \right )\right )}}\)	\(32\)
risch	\(-\frac {i a^{2} {\mathrm e}^{6 i x} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{160 \left ({\mathrm e}^{2 i x}+1\right )}-\frac {5 i a^{2} {\mathrm e}^{2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}+1\right )}+\frac {5 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}+1\right )}+\frac {5 i a^{2} {\mathrm e}^{-2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{96 \left ({\mathrm e}^{2 i x}+1\right )}-\frac {11 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}\, \cos \left (4 x \right )}{240 \left ({\mathrm e}^{2 i x}+1\right )}+\frac {7 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}\, \sin \left (4 x \right )}{120 \left ({\mathrm e}^{2 i x}+1\right )}\)	\(222\)

[In]

int((a-a*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15*cos(x)*a^3*sin(x)*(3*cos(x)^4+4*cos(x)^2+8)/(a*cos(x)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.75 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {{\left (3 \, a^{2} \cos \left (x\right )^{4} + 4 \, a^{2} \cos \left (x\right )^{2} + 8 \, a^{2}\right )} \sqrt {a \cos \left (x\right )^{2}} \sin \left (x\right )}{15 \, \cos \left (x\right )} \]

[In]

integrate((a-a*sin(x)^2)^(5/2),x, algorithm="fricas")

[Out]

1/15*(3*a^2*cos(x)^4 + 4*a^2*cos(x)^2 + 8*a^2)*sqrt(a*cos(x)^2)*sin(x)/cos(x)

Sympy [F(-1)]

Timed out. \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((a-a*sin(x)**2)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {1}{240} \, {\left (3 \, a^{2} \sin \left (5 \, x\right ) + 25 \, a^{2} \sin \left (3 \, x\right ) + 150 \, a^{2} \sin \left (x\right )\right )} \sqrt {a} \]

[In]

integrate((a-a*sin(x)^2)^(5/2),x, algorithm="maxima")

[Out]

1/240*(3*a^2*sin(5*x) + 25*a^2*sin(3*x) + 150*a^2*sin(x))*sqrt(a)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (41) = 82\).

Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.58 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=-\frac {2 \, {\left (15 \, a^{\frac {5}{2}} {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right ) - 40 \, a^{\frac {5}{2}} {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right ) + 48 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right )\right )}}{15 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{5}} \]

[In]

integrate((a-a*sin(x)^2)^(5/2),x, algorithm="giac")

[Out]

-2/15*(15*a^(5/2)*(1/tan(1/2*x) + tan(1/2*x))^4*sgn(tan(1/2*x)^4 - 1) - 40*a^(5/2)*(1/tan(1/2*x) + tan(1/2*x))
^2*sgn(tan(1/2*x)^4 - 1) + 48*a^(5/2)*sgn(tan(1/2*x)^4 - 1))/(1/tan(1/2*x) + tan(1/2*x))^5

Mupad [F(-1)]

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

[In]

int((a - a*sin(x)^2)^(5/2),x)

[Out]

int((a - a*sin(x)^2)^(5/2), x)