\(\int \frac {\tan ^3(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [479]

Optimal result

Integrand size = 26, antiderivative size = 44 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[Out]

1/5*a/f/(a*cos(f*x+e)^2)^(5/2)-1/3/f/(a*cos(f*x+e)^2)^(3/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 44, 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} \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a \left (3-5 \cos ^2(e+f x)\right )}{15 f \left (a \cos ^2(e+f x)\right )^{5/2}} \]

[In]

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

[Out]

(a*(3 - 5*Cos[e + f*x]^2))/(15*f*(a*Cos[e + f*x]^2)^(5/2))

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (5 \left (\cos ^{2}\left (f x +e \right )\right )-3\right )}{15 a^{2} \cos \left (f x +e \right )^{6} f}\)	\(41\)
risch	\(-\frac {8 \left (5 \,{\mathrm e}^{6 i \left (f x +e \right )}-2 \,{\mathrm e}^{4 i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4} a}\)	\(82\)

[In]

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

[Out]

-1/15/a^2/cos(f*x+e)^6*(a*cos(f*x+e)^2)^(1/2)*(5*cos(f*x+e)^2-3)/f

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (5 \, \cos \left (f x + e\right )^{2} - 3\right )}}{15 \, a^{2} f \cos \left (f x + e\right )^{6}} \]

[In]

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

[Out]

-1/15*sqrt(a*cos(f*x + e)^2)*(5*cos(f*x + e)^2 - 3)/(a^2*f*cos(f*x + e)^6)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {5 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{2} + 3 \, a^{3}}{15 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} a^{2} f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (36) = 72\).

Time = 0.97 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.82 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} a^{\frac {3}{2}} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \]

[In]

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

[Out]

4/15*(15*tan(1/2*f*x + 1/2*e)^6 + 5*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*a^(3/2)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))

Mupad [B] (verification not implemented)

Time = 19.30 (sec) , antiderivative size = 389, normalized size of antiderivative = 8.84 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {16\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {272\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{15\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {128\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {64\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \]

[In]

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

[Out]

(272*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(15*a^2*f*(
exp(e*2i + f*x*2i) + 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (16*exp(e*3i + f*x*3i)*(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*(exp(e*2i + f*x*2i) + 1)^2*(exp(e*1i + f*x
*1i) + exp(e*3i + f*x*3i))) - (128*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i
)*1i)/2)^2)^(1/2))/(5*a^2*f*(exp(e*2i + f*x*2i) + 1)^4*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (64*exp(e*
3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(5*a^2*f*(exp(e*2i + f
*x*2i) + 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))