\(\int \frac {\tan (e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [480]

Optimal result

Integrand size = 24, antiderivative size = 21 \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3255, 3284, 16, 32} \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

1/(3*f*(a*Cos[e + f*x]^2)^(3/2))

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

1/(3*f*(a*Cos[e + f*x]^2)^(3/2))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {1}{3 {\left (a -a \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} f}\)	\(21\)
default	\(\frac {1}{3 {\left (a -a \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}} f}\)	\(21\)
risch	\(\frac {8 \,{\mathrm e}^{2 i \left (f x +e \right )}}{3 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {a \cos \left (f x + e\right )^{2}}}{3 \, a^{2} f \cos \left (f x + e\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (17) = 34\).

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 4.52 \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right ) + \sqrt {-a \sin \left (f x + e\right )^{2} + a} a} - \frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right ) - \sqrt {-a \sin \left (f x + e\right )^{2} + a} a}}{6 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (17) = 34\).

Time = 0.63 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.57 \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 1\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a^{\frac {3}{2}} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 17.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.43 \[ \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {16\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \]

[In]

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

[Out]

(16*exp(e*4i + f*x*4i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(3*a^2*f*(ex
p(e*2i + f*x*2i) + 1)^4)