\(\int \frac {\tan ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx\) [104]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {5 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{4 \sqrt {2} \sqrt {a} f}-\frac {\sec (e+f x)}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 \sec (e+f x) \sqrt {a+a \sin (e+f x)}}{4 a f} \] Output:

5/8*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2) 
/a^(1/2)/f-1/2*sec(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)+3/4*sec(f*x+e)*(a+a*sin 
(f*x+e))^(1/2)/a/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.10 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\sec (e+f x) \left (-1+(5+5 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-3 \sin (e+f x)\right )}{4 f \sqrt {a (1+\sin (e+f x))}} \] Input:

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

Output:

-1/4*(Sec[e + f*x]*(-1 + (5 + 5*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/ 
4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e 
+ f*x)/2] + Sin[(e + f*x)/2])^2 - 3*Sin[e + f*x]))/(f*Sqrt[a*(1 + Sin[e + 
f*x])])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3191, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3191

\(\displaystyle \frac {\int -\frac {1}{2} \sec ^2(e+f x) (a-4 a \sin (e+f x)) \sqrt {\sin (e+f x) a+a}dx}{2 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \sec ^2(e+f x) (a-4 a \sin (e+f x)) \sqrt {\sin (e+f x) a+a}dx}{4 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {(a-4 a \sin (e+f x)) \sqrt {\sin (e+f x) a+a}}{\cos (e+f x)^2}dx}{4 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3334

\(\displaystyle -\frac {\frac {5}{2} a^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx-\frac {3 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}}{4 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {5}{2} a^2 \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx-\frac {3 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}}{4 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {-\frac {5 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}-\frac {3 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}}{4 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {-\frac {5 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {2} f}-\frac {3 a \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{f}}{4 a^2}-\frac {\sec (e+f x)}{2 f \sqrt {a \sin (e+f x)+a}}\)

Input:

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

Output:

-1/2*Sec[e + f*x]/(f*Sqrt[a + a*Sin[e + f*x]]) - ((-5*a^(3/2)*ArcTanh[(Sqr 
t[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[2]*f) - (3*a 
*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/f)/(4*a^2)
 

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 3191

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^2, 
x_Symbol] :> Simp[b*((a + b*Sin[e + f*x])^m/(a*f*(2*m - 1)*Cos[e + f*x])), 
x] - Simp[1/(a^2*(2*m - 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*((a*m - b*(2 
*m - 1)*Sin[e + f*x])/Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f}, 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 = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {5 \sqrt {2}\, \sqrt {a -\sin \left (f x +e \right ) a}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (f x +e \right ) a +5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sqrt {a -\sin \left (f x +e \right ) a}+6 a^{\frac {3}{2}} \sin \left (f x +e \right )+2 a^{\frac {3}{2}}}{8 a^{\frac {3}{2}} \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(134\)

Input:

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

Output:

1/8*(5*2^(1/2)*(a-sin(f*x+e)*a)^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2 
^(1/2)/a^(1/2))*sin(f*x+e)*a+5*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)* 
2^(1/2)/a^(1/2))*a*(a-sin(f*x+e)*a)^(1/2)+6*a^(3/2)*sin(f*x+e)+2*a^(3/2))/ 
a^(3/2)/cos(f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (88) = 176\).

Time = 0.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.87 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {5 \, \sqrt {2} {\left (\cos \left (f x + e\right ) \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (3 \, \sin \left (f x + e\right ) + 1\right )}}{16 \, {\left (a f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a f \cos \left (f x + e\right )\right )}} \] Input:

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

Output:

1/16*(5*sqrt(2)*(cos(f*x + e)*sin(f*x + e) + cos(f*x + e))*sqrt(a)*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)*(3*sin(f*x + e) + 1))/(a*f*cos(f*x + 
e)*sin(f*x + e) + a*f*cos(f*x + e))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.46 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\frac {5 \, \sqrt {2} \log \left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {5 \, \sqrt {2} \log \left (-\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \sqrt {2} {\left (3 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{16 \, f} \] Input:

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

Output:

1/16*(5*sqrt(2)*log(sin(3/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(a)*sgn(cos(-1 
/4*pi + 1/2*f*x + 1/2*e))) - 5*sqrt(2)*log(-sin(3/4*pi + 1/2*f*x + 1/2*e) 
+ 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*sqrt(2)*(3*sin(3/4* 
pi + 1/2*f*x + 1/2*e)^2 - 2)/((sin(3/4*pi + 1/2*f*x + 1/2*e)^3 - sin(3/4*p 
i + 1/2*f*x + 1/2*e))*sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f
 

Mupad [F(-1)]

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

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

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

Output:

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