Integrand size = 26, antiderivative size = 57 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{f}-\frac {\sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{f} \] Output:
arctanh(sin(f*x+e))*(a*cos(f*x+e)^2)^(1/2)*sec(f*x+e)/f-(a*cos(f*x+e)^2)^( 1/2)*tan(f*x+e)/f
Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.70 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\sqrt {a \cos ^2(e+f x)} \sec (e+f x) (\text {arctanh}(\sin (e+f x))-\sin (e+f x))}{f} \] Input:
Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^2,x]
Output:
(Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]*(ArcTanh[Sin[e + f*x]] - Sin[e + f*x] ))/f
Time = 0.37 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.70, 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, 3686, 3042, 3072, 262, 219}
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.
\(\displaystyle \int \tan ^2(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^2 \sqrt {a-a \sin (e+f x)^2}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \tan ^2(e+f x) \sqrt {a \cos ^2(e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}{\tan \left (e+f x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \sin (e+f x) \tan (e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \sin (e+f x) \tan (e+f x)dx\) |
\(\Big \downarrow \) 3072 |
\(\displaystyle \frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \frac {\sin ^2(e+f x)}{1-\sin ^2(e+f x)}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \left (\int \frac {1}{1-\sin ^2(e+f x)}d\sin (e+f x)-\sin (e+f x)\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} (\text {arctanh}(\sin (e+f x))-\sin (e+f x))}{f}\) |
Input:
Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^2,x]
Output:
(Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]*(ArcTanh[Sin[e + f*x]] - Sin[e + f*x] ))/f
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ (ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x ]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
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[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.57 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.68
method | result | size |
default | \(-\frac {\cos \left (f x +e \right ) \sqrt {a \sin \left (f x +e \right )^{2}}\, \left (\sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}-\ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}+2 a}{\cos \left (f x +e \right )}\right ) a \right )}{\sqrt {a}\, \sin \left (f x +e \right ) \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\) | \(96\) |
risch | \(\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}-\frac {\ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(238\) |
Input:
int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x,method=_RETURNVERBOSE)
Output:
-cos(f*x+e)*(a*sin(f*x+e)^2)^(1/2)*(a^(1/2)*(a*sin(f*x+e)^2)^(1/2)-ln(2/co s(f*x+e)*(a^(1/2)*(a*sin(f*x+e)^2)^(1/2)+a))*a)/a^(1/2)/sin(f*x+e)/(a*cos( f*x+e)^2)^(1/2)/f
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (\log \left (-\frac {\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, \sin \left (f x + e\right )\right )}}{2 \, f \cos \left (f x + e\right )} \] Input:
integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="fricas")
Output:
-1/2*sqrt(a*cos(f*x + e)^2)*(log(-(sin(f*x + e) - 1)/(sin(f*x + e) + 1)) + 2*sin(f*x + e))/(f*cos(f*x + e))
\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate((a-a*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**2,x)
Output:
Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**2, x )
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\sqrt {a} {\left (\log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 2 \, \sin \left (f x + e\right )\right )}}{2 \, f} \] Input:
integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="maxima")
Output:
1/2*sqrt(a)*(log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - l og(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 2*sin(f*x + e)) /f
Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (53) = 106\).
Time = 0.48 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.23 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {{\left (\log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \frac {4 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}\right )} \sqrt {a}}{2 \, f} \] Input:
integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="giac")
Output:
-1/2*(log(abs(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e) + 2))*sgn(tan( 1/2*f*x + 1/2*e)^4 - 1) - log(abs(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1 /2*e) - 2))*sgn(tan(1/2*f*x + 1/2*e)^4 - 1) - 4*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)/(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e)))*sqrt(a)/f
Timed out. \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \] Input:
int(tan(e + f*x)^2*(a - a*sin(e + f*x)^2)^(1/2),x)
Output:
int(tan(e + f*x)^2*(a - a*sin(e + f*x)^2)^(1/2), x)
\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\sqrt {a}\, \left (\int \sqrt {-\sin \left (f x +e \right )^{2}+1}\, \tan \left (f x +e \right )^{2}d x \right ) \] Input:
int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x)
Output:
sqrt(a)*int(sqrt( - sin(e + f*x)**2 + 1)*tan(e + f*x)**2,x)