Integrand size = 23, antiderivative size = 58 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=\frac {\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f}-\frac {\sqrt {a+b \sin ^2(e+f x)}}{f} \] Output:
(a+b)^(1/2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/f-(a+b*sin(f*x+e )^2)^(1/2)/f
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=\frac {\sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b-b \cos ^2(e+f x)}}{\sqrt {a+b}}\right )-\sqrt {a+b-b \cos ^2(e+f x)}}{f} \] Input:
Integrate[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x],x]
Output:
(Sqrt[a + b]*ArcTanh[Sqrt[a + b - b*Cos[e + f*x]^2]/Sqrt[a + b]] - Sqrt[a + b - b*Cos[e + f*x]^2])/f
Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3673, 60, 73, 221}
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.
\(\displaystyle \int \tan (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x) \sqrt {a+b \sin (e+f x)^2}dx\) |
\(\Big \downarrow \) 3673 |
\(\displaystyle \frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{1-\sin ^2(e+f x)}d\sin ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(a+b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)-2 \sqrt {a+b \sin ^2(e+f x)}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 (a+b) \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b}-2 \sqrt {a+b \sin ^2(e+f x)}}{2 f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {a+b} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )-2 \sqrt {a+b \sin ^2(e+f x)}}{2 f}\) |
Input:
Int[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x],x]
Output:
(2*Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - 2*Sqrt[a + b*Sin[e + f*x]^2])/(2*f)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(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)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(128\) vs. \(2(50)=100\).
Time = 0.47 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.22
method | result | size |
default | \(\frac {\frac {\sqrt {a +b}\, \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{2}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{2}-\sqrt {a +b -b \cos \left (f x +e \right )^{2}}}{f}\) | \(129\) |
Input:
int((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e),x,method=_RETURNVERBOSE)
Output:
(1/2*(a+b)^(1/2)*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/ 2)+b*sin(f*x+e)+a))+1/2*(a+b)^(1/2)*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b- b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))-(a+b-b*cos(f*x+e)^2)^(1/2))/f
Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.76 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=\left [\frac {\sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, f}, \frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) - \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{f}\right ] \] Input:
integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e),x, algorithm="fricas")
Output:
[1/2*(sqrt(a + b)*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b )*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*sqrt(-b*cos(f*x + e)^2 + a + b))/f, (sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b) /(b*cos(f*x + e)^2 - a - b)) - sqrt(-b*cos(f*x + e)^2 + a + b))/f]
\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sin(f*x+e)**2)**(1/2)*tan(f*x+e),x)
Output:
Integral(sqrt(a + b*sin(e + f*x)**2)*tan(e + f*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (50) = 100\).
Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.10 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=-\frac {\sqrt {a + b} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right ) - \sqrt {a + b} \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right ) + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{2 \, f} \] Input:
integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e),x, algorithm="maxima")
Output:
-1/2*(sqrt(a + b)*arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1))) - sqrt(a + b)*arcsinh(-b*sin(f*x + e)/(s qrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1))) + 2*sqrt( b*sin(f*x + e)^2 + a))/f
Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (50) = 100\).
Time = 0.45 (sec) , antiderivative size = 313, normalized size of antiderivative = 5.40 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=-\frac {2 \, {\left (\frac {{\left (a + b\right )} \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} - \sqrt {a}}{2 \, \sqrt {-a - b}}\right )}{\sqrt {-a - b}} - \frac {2 \, {\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} b - \sqrt {a} b\right )}}{{\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} \sqrt {a} + a + 4 \, b}\right )}}{f} \] Input:
integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e),x, algorithm="giac")
Output:
-2*((a + b)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f *x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/sqrt(-a - b) - 2*((sqrt(a)*tan(1/2*f*x + 1/2*e )^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan (1/2*f*x + 1/2*e)^2 + a))*b - sqrt(a)*b)/((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2 *f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1 /2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^ 2 + a))*sqrt(a) + a + 4*b))/f
Timed out. \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=\int \mathrm {tan}\left (e+f\,x\right )\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \] Input:
int(tan(e + f*x)*(a + b*sin(e + f*x)^2)^(1/2),x)
Output:
int(tan(e + f*x)*(a + b*sin(e + f*x)^2)^(1/2), x)
\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x) \, dx=\int \sqrt {\sin \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )d x \] Input:
int((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e),x)
Output:
int(sqrt(sin(e + f*x)**2*b + a)*tan(e + f*x),x)