Integrand size = 26, antiderivative size = 44 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \] Output:
1/5*a/f/(a*cos(f*x+e)^2)^(5/2)-1/3/f/(a*cos(f*x+e)^2)^(3/2)
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a \left (3-5 \cos ^2(e+f x)\right )}{15 f \left (a \cos ^2(e+f x)\right )^{5/2}} \] Input:
Integrate[Tan[e + f*x]^3/(a - a*Sin[e + f*x]^2)^(3/2),x]
Output:
(a*(3 - 5*Cos[e + f*x]^2))/(15*f*(a*Cos[e + f*x]^2)^(5/2))
Time = 0.36 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3655, 3042, 25, 3684, 8, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^3}{\left (a-a \sin (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {\tan ^3(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\tan \left (e+f x+\frac {\pi }{2}\right )^3 \left (a \sin \left (e+f x+\frac {\pi }{2}\right )^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\left (a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2\right )^{3/2} \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^3}dx\) |
\(\Big \downarrow \) 3684 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right ) \sec ^4(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {a^2 \int \frac {1-\cos ^2(e+f x)}{\left (a \cos ^2(e+f x)\right )^{7/2}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {a^2 \int \left (\frac {1}{\left (a \cos ^2(e+f x)\right )^{7/2}}-\frac {1}{a \left (a \cos ^2(e+f x)\right )^{5/2}}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \left (\frac {2}{3 a^2 \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {2}{5 a \left (a \cos ^2(e+f x)\right )^{5/2}}\right )}{2 f}\) |
Input:
Int[Tan[e + f*x]^3/(a - a*Sin[e + f*x]^2)^(3/2),x]
Output:
-1/2*(a^2*(-2/(5*a*(a*Cos[e + f*x]^2)^(5/2)) + 2/(3*a^2*(a*Cos[e + f*x]^2) ^(3/2))))/f
Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m Int[u*(a* x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
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] && 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])
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_. ), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[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] && Inte gerQ[(m - 1)/2] && IntegerQ[n/2]
Time = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {\sqrt {a \cos \left (f x +e \right )^{2}}\, \left (5 \cos \left (f x +e \right )^{2}-3\right )}{15 a^{2} \cos \left (f x +e \right )^{6} f}\) | \(41\) |
risch | \(-\frac {8 \left (5 \,{\mathrm e}^{6 i \left (f x +e \right )}-2 \,{\mathrm e}^{4 i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4} a}\) | \(82\) |
Input:
int(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-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
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (5 \, \cos \left (f x + e\right )^{2} - 3\right )}}{15 \, a^{2} f \cos \left (f x + e\right )^{6}} \] Input:
integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
Output:
-1/15*sqrt(a*cos(f*x + e)^2)*(5*cos(f*x + e)^2 - 3)/(a^2*f*cos(f*x + e)^6)
\[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(tan(f*x+e)**3/(a-a*sin(f*x+e)**2)**(3/2),x)
Output:
Integral(tan(e + f*x)**3/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2) , x)
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {5 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{2} + 3 \, a^{3}}{15 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} a^{2} f} \] Input:
integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (36) = 72\).
Time = 0.70 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.82 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} a^{\frac {3}{2}} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \] Input:
integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
Output:
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))
Time = 41.06 (sec) , antiderivative size = 389, normalized size of antiderivative = 8.84 \[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {16\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {272\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{15\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {128\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {64\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{5\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \] Input:
int(tan(e + f*x)^3/(a - a*sin(e + f*x)^2)^(3/2),x)
Output:
(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*1 i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(3*a^2*f*(exp(e*2 i + 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) + e xp(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)))
\[ \int \frac {\tan ^3(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )^{2}+1}\, \tan \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{2}+1}d x \right )}{a^{2}} \] Input:
int(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(3/2),x)
Output:
(sqrt(a)*int((sqrt( - sin(e + f*x)**2 + 1)*tan(e + f*x)**3)/(sin(e + f*x)* *4 - 2*sin(e + f*x)**2 + 1),x))/a**2