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

Optimal. Leaf size=53 \[ \frac{8}{15} a^2 \tan (x) \sqrt{a \cos ^2(x)}+\frac{1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}+\frac{4}{15} a \tan (x) \left (a \cos ^2(x)\right )^{3/2} \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.0540351, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3176, 3203, 3207, 2637} \[ \frac{8}{15} a^2 \tan (x) \sqrt{a \cos ^2(x)}+\frac{1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}+\frac{4}{15} a \tan (x) \left (a \cos ^2(x)\right )^{3/2} \]

Antiderivative was successfully verified.

[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 3176

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 3203

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 3207

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
]])

Rule 2637

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

Rubi steps

\begin{align*} \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx &=\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]  time = 0.0215734, size = 36, normalized size = 0.68 \[ \frac{1}{240} a^2 (150 \sin (x)+25 \sin (3 x)+3 \sin (5 x)) \sec (x) \sqrt{a \cos ^2(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[x]^2)^(5/2),x]

[Out]

(a^2*Sqrt[a*Cos[x]^2]*Sec[x]*(150*Sin[x] + 25*Sin[3*x] + 3*Sin[5*x]))/240

________________________________________________________________________________________

Maple [A]  time = 0.677, size = 32, normalized size = 0.6 \begin{align*}{\frac{{a}^{3}\cos \left ( x \right ) \sin \left ( x \right ) \left ( 3\, \left ( \cos \left ( x \right ) \right ) ^{4}+4\, \left ( \cos \left ( x \right ) \right ) ^{2}+8 \right ) }{15}{\frac{1}{\sqrt{a \left ( \cos \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.6172, size = 42, normalized size = 0.79 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 1.67462, size = 107, normalized size = 2.02 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.26189, size = 113, normalized size = 2.13 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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