\(\int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx\) [92]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {2} f}+\frac {5 \sec (e+f x) \sqrt {a+a \sin (e+f x)}}{f}-\frac {2 \sec (e+f x) (a+a \sin (e+f x))^{3/2}}{a f} \] Output:

-1/2*a^(1/2)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2) 
)*2^(1/2)/f+5*sec(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f-2*sec(f*x+e)*(a+a*sin(f* 
x+e))^(3/2)/a/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.89 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13 \[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\frac {\sec (e+f x) \left (3+(1-i) \sqrt [4]{-1} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {f x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 e+f x)\right )-\sin \left (\frac {1}{4} (2 e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-2 \sin (e+f x)\right ) \sqrt {a (1+\sin (e+f x))}}{f} \] Input:

Integrate[Sqrt[a + a*Sin[e + f*x]]*Tan[e + f*x]^2,x]
 

Output:

(Sec[e + f*x]*(3 + (1 - I)*(-1)^(1/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[( 
f*x)/4]*(Cos[(2*e + f*x)/4] - Sin[(2*e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin 
[(e + f*x)/2]) - 2*Sin[e + f*x])*Sqrt[a*(1 + Sin[e + f*x])])/f
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3192, 27, 3042, 3334, 3042, 3128, 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.

\(\displaystyle \int \tan ^2(e+f x) \sqrt {a \sin (e+f x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^2 \sqrt {a \sin (e+f x)+a}dx\)

\(\Big \downarrow \) 3192

\(\displaystyle \frac {2 \int \frac {1}{2} \sec ^2(e+f x) \sqrt {\sin (e+f x) a+a} (2 \sin (e+f x) a+3 a)dx}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec ^2(e+f x) \sqrt {\sin (e+f x) a+a} (2 \sin (e+f x) a+3 a)dx}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {\sin (e+f x) a+a} (2 \sin (e+f x) a+3 a)}{\cos (e+f x)^2}dx}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {1}{2} a^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {5 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} a^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {5 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {5 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}-\frac {a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {5 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {2} f}}{a}-\frac {2 \sec (e+f x) (a \sin (e+f x)+a)^{3/2}}{a f}\)

Input:

Int[Sqrt[a + a*Sin[e + f*x]]*Tan[e + f*x]^2,x]
 

Output:

(-2*Sec[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(a*f) + (-((a^(3/2)*ArcTanh[( 
Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*f)) + 
(5*a*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/f)/a
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3192

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, 
x_Symbol] :> Simp[-(a + b*Sin[e + f*x])^(m + 1)/(b*f*m*Cos[e + f*x]), x] + 
Simp[1/(b*m)   Int[(a + b*Sin[e + f*x])^m*((b*(m + 1) + a*Sin[e + f*x])/Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && 
!IntegerQ[m] &&  !LtQ[m, 0]
 

rule 3334

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* 
c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) 
, x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1)))   Int[(g*Cos[e + 
f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, 
 f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \left (\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {a -\sin \left (f x +e \right ) a}+4 \sin \left (f x +e \right ) a -6 a \right )}{2 \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(89\)

Input:

int((a+sin(f*x+e)*a)^(1/2)*tan(f*x+e)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2*(1+sin(f*x+e))*(a^(1/2)*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^ 
(1/2)/a^(1/2))*(a-sin(f*x+e)*a)^(1/2)+4*sin(f*x+e)*a-6*a)/cos(f*x+e)/(a+si 
n(f*x+e)*a)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.67 \[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\frac {\sqrt {2} \sqrt {a} \cos \left (f x + e\right ) \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (2 \, \sin \left (f x + e\right ) - 3\right )}}{4 \, f \cos \left (f x + e\right )} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^2,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/4*(sqrt(2)*sqrt(a)*cos(f*x + e)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt( 
a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f* 
x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos 
(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*sqrt(a*sin(f*x + e) + 
 a)*(2*sin(f*x + e) - 3))/(f*cos(f*x + e))
 

Sympy [F]

\[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate((a+a*sin(f*x+e))**(1/2)*tan(f*x+e)**2,x)
 

Output:

Integral(sqrt(a*(sin(e + f*x) + 1))*tan(e + f*x)**2, x)
 

Maxima [F]

\[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} \tan \left (f x + e\right )^{2} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(a*sin(f*x + e) + a)*tan(f*x + e)^2, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (88) = 176\).

Time = 0.32 (sec) , antiderivative size = 428, normalized size of antiderivative = 4.24 \[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx =\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^2,x, algorithm="giac")
 

Output:

1/4*sqrt(2)*(log(2*(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 2*tan(-1/8*pi + 1/4 
*f*x + 1/4*e) + 1)/(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos(-1/4*pi 
 + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^3 - log(2*(tan(-1/8*pi 
 + 1/4*f*x + 1/4*e)^2 - 2*tan(-1/8*pi + 1/4*f*x + 1/4*e) + 1)/(tan(-1/8*pi 
 + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*p 
i + 1/4*f*x + 1/4*e)^3 - sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 
 1/4*f*x + 1/4*e)^4 + log(2*(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 2*tan(-1/8 
*pi + 1/4*f*x + 1/4*e) + 1)/(tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e) - log(2*(tan( 
-1/8*pi + 1/4*f*x + 1/4*e)^2 - 2*tan(-1/8*pi + 1/4*f*x + 1/4*e) + 1)/(tan( 
-1/8*pi + 1/4*f*x + 1/4*e)^2 + 1))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan 
(-1/8*pi + 1/4*f*x + 1/4*e) - 18*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(- 
1/8*pi + 1/4*f*x + 1/4*e)^2 - sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a) 
/((tan(-1/8*pi + 1/4*f*x + 1/4*e)^3 + tan(-1/8*pi + 1/4*f*x + 1/4*e))*f)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,\sqrt {a+a\,\sin \left (e+f\,x\right )} \,d x \] Input:

int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(1/2),x)
 

Output:

int(tan(e + f*x)^2*(a + a*sin(e + f*x))^(1/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \sqrt {a+a \sin (e+f x)} \tan ^2(e+f x) \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}d x \right ) \] Input:

int((a+a*sin(f*x+e))^(1/2)*tan(f*x+e)^2,x)
 

Output:

sqrt(a)*int(sqrt(sin(e + f*x) + 1)*tan(e + f*x)**2,x)