Integrand size = 26, antiderivative size = 64 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {a^2}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {2 a}{f \sqrt {a \cos ^2(e+f x)}}-\frac {\sqrt {a \cos ^2(e+f x)}}{f} \] Output:
1/3*a^2/f/(a*cos(f*x+e)^2)^(3/2)-2*a/f/(a*cos(f*x+e)^2)^(1/2)-(a*cos(f*x+e )^2)^(1/2)/f
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.66 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {a \left (-6-3 \cos ^2(e+f x)+\sec ^2(e+f x)\right )}{3 f \sqrt {a \cos ^2(e+f x)}} \] Input:
Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]
Output:
(a*(-6 - 3*Cos[e + f*x]^2 + Sec[e + f*x]^2))/(3*f*Sqrt[a*Cos[e + f*x]^2])
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09, 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.
\(\displaystyle \int \tan ^5(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^5 \sqrt {a-a \sin (e+f x)^2}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \tan ^5(e+f x) \sqrt {a \cos ^2(e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}{\tan \left (e+f x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\sqrt {a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2}}{\tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^5}dx\) |
\(\Big \downarrow \) 3684 |
\(\displaystyle -\frac {\int \sqrt {a \cos ^2(e+f x)} \left (1-\cos ^2(e+f x)\right )^2 \sec ^6(e+f x)d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {a^3 \int \frac {\left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)\right )^{5/2}}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {a^3 \int \left (\frac {1}{a^2 \sqrt {a \cos ^2(e+f x)}}-\frac {2}{a \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac {1}{\left (a \cos ^2(e+f x)\right )^{5/2}}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 \left (\frac {2 \sqrt {a \cos ^2(e+f x)}}{a^3}+\frac {4}{a^2 \sqrt {a \cos ^2(e+f x)}}-\frac {2}{3 a \left (a \cos ^2(e+f x)\right )^{3/2}}\right )}{2 f}\) |
Input:
Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^5,x]
Output:
-1/2*(a^3*(-2/(3*a*(a*Cos[e + f*x]^2)^(3/2)) + 4/(a^2*Sqrt[a*Cos[e + f*x]^ 2]) + (2*Sqrt[a*Cos[e + f*x]^2])/a^3))/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 = 0.71 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {a \left (3 \cos \left (f x +e \right )^{4}+6 \cos \left (f x +e \right )^{2}-1\right )}{3 \cos \left (f x +e \right )^{2} \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\) | \(49\) |
risch | \(-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (3 \,{\mathrm e}^{8 i \left (f x +e \right )}+36 \,{\mathrm e}^{6 i \left (f x +e \right )}+50 \,{\mathrm e}^{4 i \left (f x +e \right )}+36 \,{\mathrm e}^{2 i \left (f x +e \right )}+3\right )}{6 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}\) | \(91\) |
Input:
int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)
Output:
-1/3/cos(f*x+e)^2*a*(3*cos(f*x+e)^4+6*cos(f*x+e)^2-1)/(a*cos(f*x+e)^2)^(1/ 2)/f
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=-\frac {{\left (3 \, \cos \left (f x + e\right )^{4} + 6 \, \cos \left (f x + e\right )^{2} - 1\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{3 \, f \cos \left (f x + e\right )^{4}} \] Input:
integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="fricas")
Output:
-1/3*(3*cos(f*x + e)^4 + 6*cos(f*x + e)^2 - 1)*sqrt(a*cos(f*x + e)^2)/(f*c os(f*x + e)^4)
\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate((a-a*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**5,x)
Output:
Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**5, x )
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.08 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=-\frac {3 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} a^{3} - \frac {6 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + a^{5}}{{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}}{3 \, a^{3} f} \] Input:
integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="maxima")
Output:
-1/3*(3*sqrt(-a*sin(f*x + e)^2 + a)*a^3 - (6*(a*sin(f*x + e)^2 - a)*a^4 + a^5)/(-a*sin(f*x + e)^2 + a)^(3/2))/(a^3*f)
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (56) = 112\).
Time = 0.83 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.00 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\frac {2 \, \sqrt {a} {\left (\frac {3 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1} - \frac {3 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 12 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3}}\right )}}{3 \, f} \] Input:
integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="giac")
Output:
2/3*sqrt(a)*(3*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)/(tan(1/2*f*x + 1/2*e)^2 + 1 ) - (3*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^4 - 12*sgn(tan (1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^2 + 5*sgn(tan(1/2*f*x + 1/2* e)^4 - 1))/(tan(1/2*f*x + 1/2*e)^2 - 1)^3)/f
Time = 40.34 (sec) , antiderivative size = 326, normalized size of antiderivative = 5.09 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=-\frac {\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{f}-\frac {8\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {16\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {16\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \] Input:
int(tan(e + f*x)^5*(a - a*sin(e + f*x)^2)^(1/2),x)
Output:
(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*f*(exp(e*2i + f*x*2i) + 1)^2*(exp(e*1i + f*x*1i ) + exp(e*3i + f*x*3i))) - (8*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))/(f*(exp(e*2i + f*x*2i) + 1)*(exp(e*1i + f*x*1i) + 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)/f - (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*f*(exp(e*2i + f*x*2i) + 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i )))
\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx=\sqrt {a}\, \left (\int \sqrt {-\sin \left (f x +e \right )^{2}+1}\, \tan \left (f x +e \right )^{5}d x \right ) \] Input:
int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x)
Output:
sqrt(a)*int(sqrt( - sin(e + f*x)**2 + 1)*tan(e + f*x)**5,x)