\(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx\) [458]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {a}{f \sqrt {a \cos ^2(e+f x)}}+\frac {\sqrt {a \cos ^2(e+f x)}}{f} \]

[Out]

a/f/(a*cos(f*x+e)^2)^(1/2)+(a*cos(f*x+e)^2)^(1/2)/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3255, 3284, 16, 45} \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {a}{f \sqrt {a \cos ^2(e+f x)}}+\frac {\sqrt {a \cos ^2(e+f x)}}{f} \]

[In]

Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^3,x]

[Out]

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

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 45

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])

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 3284

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
]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {a \left (1+\cos ^2(e+f x)\right )}{f \sqrt {a \cos ^2(e+f x)}} \]

[In]

Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {a \left (\cos ^{2}\left (f x +e \right )+1\right )}{\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\)	\(28\)
risch	\(\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{4 i \left (f x +e \right )}+6 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}\)	\(67\)

[In]

int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x,method=_RETURNVERBOSE)

[Out]

a*(cos(f*x+e)^2+1)/(a*cos(f*x+e)^2)^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (\cos \left (f x + e\right )^{2} + 1\right )}}{f \cos \left (f x + e\right )^{2}} \]

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="fricas")

[Out]

sqrt(a*cos(f*x + e)^2)*(cos(f*x + e)^2 + 1)/(f*cos(f*x + e)^2)

Sympy [F]

\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{3}{\left (e + f x \right )}\, dx \]

[In]

integrate((a-a*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**3,x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a} a^{2} + \frac {a^{3}}{\sqrt {-a \sin \left (f x + e\right )^{2} + a}}}{a^{2} f} \]

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="maxima")

[Out]

(sqrt(-a*sin(f*x + e)^2 + a)*a^2 + a^3/sqrt(-a*sin(f*x + e)^2 + a))/(a^2*f)

Giac [A] (verification not implemented)

none

Time = 0.96 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {4 \, \sqrt {a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} f} \]

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.82 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^3(e+f x) \, dx=\frac {\sqrt {2}\,\sqrt {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )}\,\left (8\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )+7\right )}{2\,f\,\left (4\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )+3\right )} \]

[In]

int(tan(e + f*x)^3*(a - a*sin(e + f*x)^2)^(1/2),x)

[Out]

(2^(1/2)*(a*(cos(2*e + 2*f*x) + 1))^(1/2)*(8*cos(2*e + 2*f*x) + cos(4*e + 4*f*x) + 7))/(2*f*(4*cos(2*e + 2*f*x
) + cos(4*e + 4*f*x) + 3))