Integrand size = 26, antiderivative size = 42 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \] Output:
1/3*a/f/(a*cos(f*x+e)^2)^(3/2)-1/f/(a*cos(f*x+e)^2)^(1/2)
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {-3+\sec ^2(e+f x)}{3 f \sqrt {a \cos ^2(e+f x)}} \] Input:
Integrate[Tan[e + f*x]^3/Sqrt[a - a*Sin[e + f*x]^2],x]
Output:
(-3 + Sec[e + f*x]^2)/(3*f*Sqrt[a*Cos[e + f*x]^2])
Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21, 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.
\(\displaystyle \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (e+f x)^3}{\sqrt {a-a \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {\tan ^3(e+f x)}{\sqrt {a \cos ^2(e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\tan \left (e+f x+\frac {\pi }{2}\right )^3 \sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\sqrt {a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2} \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^3}dx\) |
\(\Big \downarrow \) 3684 |
\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right ) \sec ^4(e+f x)}{\sqrt {a \cos ^2(e+f x)}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {a^2 \int \frac {1-\cos ^2(e+f x)}{\left (a \cos ^2(e+f x)\right )^{5/2}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {a^2 \int \left (\frac {1}{\left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {1}{a \left (a \cos ^2(e+f x)\right )^{3/2}}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \left (\frac {2}{a^2 \sqrt {a \cos ^2(e+f x)}}-\frac {2}{3 a \left (a \cos ^2(e+f x)\right )^{3/2}}\right )}{2 f}\) |
Input:
Int[Tan[e + f*x]^3/Sqrt[a - a*Sin[e + f*x]^2],x]
Output:
-1/2*(a^2*(-2/(3*a*(a*Cos[e + f*x]^2)^(3/2)) + 2/(a^2*Sqrt[a*Cos[e + f*x]^ 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 = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {3 \cos \left (f x +e \right )^{2}-1}{3 \cos \left (f x +e \right )^{2} \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\) | \(38\) |
risch | \(-\frac {2 \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )}+3\right )}{3 \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} f}\) | \(69\) |
Input:
int(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3/cos(f*x+e)^2*(3*cos(f*x+e)^2-1)/(a*cos(f*x+e)^2)^(1/2)/f
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (3 \, \cos \left (f x + e\right )^{2} - 1\right )}}{3 \, a f \cos \left (f x + e\right )^{4}} \] Input:
integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
-1/3*sqrt(a*cos(f*x + e)^2)*(3*cos(f*x + e)^2 - 1)/(a*f*cos(f*x + e)^4)
\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \] Input:
integrate(tan(f*x+e)**3/(a-a*sin(f*x+e)**2)**(1/2),x)
Output:
Integral(tan(e + f*x)**3/sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1)), x )
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {3 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{2} + a^{3}}{3 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} f} \] Input:
integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
1/3*(3*(a*sin(f*x + e)^2 - a)*a^2 + a^3)/((-a*sin(f*x + e)^2 + a)^(3/2)*a^ 2*f)
Time = 0.51 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {4 \, {\left (3 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} \sqrt {a} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \] Input:
integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
4/3*(3*tan(1/2*f*x + 1/2*e)^2 - 1)/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*sqrt(a) *f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))
Time = 39.83 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.38 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {4\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,\left (2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+3\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+3\right )}{3\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \] Input:
int(tan(e + f*x)^3/(a - a*sin(e + f*x)^2)^(1/2),x)
Output:
-(4*exp(e*2i + f*x*2i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f *x*1i)*1i)/2)^2)^(1/2)*(2*exp(e*2i + f*x*2i) + 3*exp(e*4i + f*x*4i) + 3))/ (3*a*f*(exp(e*2i + f*x*2i) + 1)^4)
\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )^{2}+1}\, \tan \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{2}-1}d x \right )}{a} \] Input:
int(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x)
Output:
( - sqrt(a)*int((sqrt( - sin(e + f*x)**2 + 1)*tan(e + f*x)**3)/(sin(e + f* x)**2 - 1),x))/a