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

Optimal. Leaf size=65 \[ \frac{a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac{2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

a^2/(5*f*(a*Cos[e + f*x]^2)^(5/2)) - (2*a)/(3*f*(a*Cos[e + f*x]^2)^(3/2)) + 1/(f*Sqrt[a*Cos[e + f*x]^2])

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Rubi [A]  time = 0.11512, antiderivative size = 65, 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} \[ \frac{a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac{2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac{1}{f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2/(5*f*(a*Cos[e + f*x]^2)^(5/2)) - (2*a)/(3*f*(a*Cos[e + f*x]^2)^(3/2)) + 1/(f*Sqrt[a*Cos[e + f*x]^2])

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 \frac{\tan ^5(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx &=\int \frac{\tan ^5(e+f x)}{\sqrt{a \cos ^2(e+f x)}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^3 \sqrt{a x}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \frac{(1-x)^2}{(a x)^{7/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \left (\frac{1}{(a x)^{7/2}}-\frac{2}{a (a x)^{5/2}}+\frac{1}{a^2 (a x)^{3/2}}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac{2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac{1}{f \sqrt{a \cos ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0811033, size = 43, normalized size = 0.66 \[ \frac{3 \sec ^4(e+f x)-10 \sec ^2(e+f x)+15}{15 f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15 - 10*Sec[e + f*x]^2 + 3*Sec[e + f*x]^4)/(15*f*Sqrt[a*Cos[e + f*x]^2])

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Maple [A]  time = 1.846, size = 51, normalized size = 0.8 \begin{align*}{\frac{15\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+3}{15\,a \left ( \cos \left ( fx+e \right ) \right ) ^{6}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(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

1/15/a/cos(f*x+e)^6*(a*cos(f*x+e)^2)^(1/2)*(15*cos(f*x+e)^4-10*cos(f*x+e)^2+3)/f

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Maxima [A]  time = 1.03155, size = 93, normalized size = 1.43 \begin{align*} \frac{15 \,{\left (a \sin \left (f x + e\right )^{2} - a\right )}^{2} a^{3} + 10 \,{\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + 3 \, a^{5}}{15 \,{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}} a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(a*sin(f*x + e)^2 - a)^2*a^3 + 10*(a*sin(f*x + e)^2 - a)*a^4 + 3*a^5)/((-a*sin(f*x + e)^2 + a)^(5/2)*
a^3*f)

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Fricas [A]  time = 1.61127, size = 127, normalized size = 1.95 \begin{align*} \frac{{\left (15 \, \cos \left (f x + e\right )^{4} - 10 \, \cos \left (f x + e\right )^{2} + 3\right )} \sqrt{a \cos \left (f x + e\right )^{2}}}{15 \, a f \cos \left (f x + e\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*cos(f*x + e)^4 - 10*cos(f*x + e)^2 + 3)*sqrt(a*cos(f*x + e)^2)/(a*f*cos(f*x + e)^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (e + f x \right )}}{\sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right ) \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(e + f*x)**5/sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1)), x)

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Giac [A]  time = 2.91545, size = 96, normalized size = 1.48 \begin{align*} \frac{16 \,{\left (10 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 5 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}}{15 \,{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{5} \sqrt{a} f \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/15*(10*tan(1/2*f*x + 1/2*e)^4 - 5*tan(1/2*f*x + 1/2*e)^2 + 1)/((tan(1/2*f*x + 1/2*e)^2 - 1)^5*sqrt(a)*f*sgn
(tan(1/2*f*x + 1/2*e)^4 - 1))