3.457 \(\int \sqrt{a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=64 \[ \frac{a^2}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac{2 a}{f \sqrt{a \cos ^2(e+f x)}}-\frac{\sqrt{a \cos ^2(e+f x)}}{f} \]

[Out]

a^2/(3*f*(a*Cos[e + f*x]^2)^(3/2)) - (2*a)/(f*Sqrt[a*Cos[e + f*x]^2]) - Sqrt[a*Cos[e + f*x]^2]/f

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Rubi [A]  time = 0.109609, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3205, 16, 43} \[ \frac{a^2}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac{2 a}{f \sqrt{a \cos ^2(e+f x)}}-\frac{\sqrt{a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]

[Out]

a^2/(3*f*(a*Cos[e + f*x]^2)^(3/2)) - (2*a)/(f*Sqrt[a*Cos[e + f*x]^2]) - Sqrt[a*Cos[e + f*x]^2]/f

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 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sqrt{a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx &=\int \sqrt{a \cos ^2(e+f x)} \tan ^5(e+f x) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2 \sqrt{a x}}{x^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{(1-x)^2}{(a x)^{5/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{1}{(a x)^{5/2}}-\frac{2}{a (a x)^{3/2}}+\frac{1}{a^2 \sqrt{a x}}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{a^2}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac{2 a}{f \sqrt{a \cos ^2(e+f x)}}-\frac{\sqrt{a \cos ^2(e+f x)}}{f}\\ \end{align*}

Mathematica [A]  time = 0.0937716, size = 51, normalized size = 0.8 \[ -\frac{\left (3 \cos ^4(e+f x)+6 \cos ^2(e+f x)-1\right ) \sec ^4(e+f x) \sqrt{a \cos ^2(e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]

[Out]

-(Sqrt[a*Cos[e + f*x]^2]*(-1 + 6*Cos[e + f*x]^2 + 3*Cos[e + f*x]^4)*Sec[e + f*x]^4)/(3*f)

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Maple [A]  time = 2.165, size = 48, normalized size = 0.8 \begin{align*} -{\frac{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+6\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-1}{3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}f}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x)

[Out]

-1/3/cos(f*x+e)^4*(a*cos(f*x+e)^2)^(1/2)*(3*cos(f*x+e)^4+6*cos(f*x+e)^2-1)/f

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Maxima [A]  time = 1.01499, size = 93, normalized size = 1.45 \begin{align*} -\frac{3 \, \sqrt{-a \sin \left (f x + e\right )^{2} + a} a^{3} - \frac{6 \,{\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + a^{5}}{{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}}{3 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

-1/3*(3*sqrt(-a*sin(f*x + e)^2 + a)*a^3 - (6*(a*sin(f*x + e)^2 - a)*a^4 + a^5)/(-a*sin(f*x + e)^2 + a)^(3/2))/
(a^3*f)

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Fricas [A]  time = 1.65234, size = 122, normalized size = 1.91 \begin{align*} -\frac{{\left (3 \, \cos \left (f x + e\right )^{4} + 6 \, \cos \left (f x + e\right )^{2} - 1\right )} \sqrt{a \cos \left (f x + e\right )^{2}}}{3 \, f \cos \left (f x + e\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

-1/3*(3*cos(f*x + e)^4 + 6*cos(f*x + e)^2 - 1)*sqrt(a*cos(f*x + e)^2)/(f*cos(f*x + e)^4)

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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(f*x+e)**2)**(1/2)*tan(f*x+e)**5,x)

[Out]

Timed out

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Giac [B]  time = 2.74912, size = 184, normalized size = 2.88 \begin{align*} \frac{2 \, \sqrt{a}{\left (\frac{3 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1} - \frac{3 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 12 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{3}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="giac")

[Out]

2/3*sqrt(a)*(3*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)/(tan(1/2*f*x + 1/2*e)^2 + 1) - (3*sgn(tan(1/2*f*x + 1/2*e)^4 -
1)*tan(1/2*f*x + 1/2*e)^4 - 12*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^2 + 5*sgn(tan(1/2*f*x + 1/
2*e)^4 - 1))/(tan(1/2*f*x + 1/2*e)^2 - 1)^3)/f