\(\int \frac {\tan ^2(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [414]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}}-\frac {\tan (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{4 a f \sqrt {a \cos ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.56 \[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\text {arctanh}(\sin (e+f x)) \cos (e+f x)+\left (-1+2 \sec ^2(e+f x)\right ) \tan (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}} \] Input:

Integrate[Tan[e + f*x]^2/(a - a*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(-(ArcTanh[Sin[e + f*x]]*Cos[e + f*x]) + (-1 + 2*Sec[e + f*x]^2)*Tan[e + f 
*x])/(8*a*f*Sqrt[a*Cos[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3655, 3042, 3686, 3042, 3091, 3042, 4255, 3042, 4257}

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.

\(\displaystyle \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3655

\(\displaystyle \int \frac {\tan ^2(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan \left (e+f x+\frac {\pi }{2}\right )^2 \left (a \sin \left (e+f x+\frac {\pi }{2}\right )^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3686

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3091

\(\displaystyle \frac {\cos (e+f x) \left (\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {1}{4} \int \sec ^3(e+f x)dx\right )}{a \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos (e+f x) \left (\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}-\frac {1}{4} \int \csc \left (e+f x+\frac {\pi }{2}\right )^3dx\right )}{a \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {\cos (e+f x) \left (\frac {1}{4} \left (-\frac {1}{2} \int \sec (e+f x)dx-\frac {\tan (e+f x) \sec (e+f x)}{2 f}\right )+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )}{a \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos (e+f x) \left (\frac {1}{4} \left (-\frac {1}{2} \int \csc \left (e+f x+\frac {\pi }{2}\right )dx-\frac {\tan (e+f x) \sec (e+f x)}{2 f}\right )+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )}{a \sqrt {a \cos ^2(e+f x)}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\cos (e+f x) \left (\frac {1}{4} \left (-\frac {\text {arctanh}(\sin (e+f x))}{2 f}-\frac {\tan (e+f x) \sec (e+f x)}{2 f}\right )+\frac {\tan (e+f x) \sec ^3(e+f x)}{4 f}\right )}{a \sqrt {a \cos ^2(e+f x)}}\)

Input:

Int[Tan[e + f*x]^2/(a - a*Sin[e + f*x]^2)^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3091

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3655

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]
 

rule 3686

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]])
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\left (\ln \left (1+\sin \left (f x +e \right )\right )-\ln \left (\sin \left (f x +e \right )-1\right )\right ) \cos \left (f x +e \right )^{4}+2 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )-4 \sin \left (f x +e \right )}{16 a \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\)	\(104\)
risch	\(\frac {i \left ({\mathrm e}^{6 i \left (f x +e \right )}-7 \,{\mathrm e}^{4 i \left (f x +e \right )}+7 \,{\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{4 a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f}+\frac {\ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}-\frac {\ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}\)	\(195\)

Input:

int(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/a*((ln(1+sin(f*x+e))-ln(sin(f*x+e)-1))*cos(f*x+e)^4+2*cos(f*x+e)^2*si 
n(f*x+e)-4*sin(f*x+e))/(1+sin(f*x+e))/(sin(f*x+e)-1)/cos(f*x+e)/(a*cos(f*x 
+e)^2)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (\cos \left (f x + e\right )^{4} \log \left (-\frac {\sin \left (f x + e\right ) + 1}{\sin \left (f x + e\right ) - 1}\right ) + 2 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{16 \, a^{2} f \cos \left (f x + e\right )^{5}} \] Input:

integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(f*x+e)**2/(a-a*sin(f*x+e)**2)**(3/2),x)
 

Output:

Integral(tan(e + f*x)**2/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2) 
, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1532 vs. \(2 (94) = 188\).

Time = 0.32 (sec) , antiderivative size = 1532, normalized size of antiderivative = 14.45 \[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

-1/16*(4*(sin(7*f*x + 7*e) - 7*sin(5*f*x + 5*e) + 7*sin(3*f*x + 3*e) - sin 
(f*x + e))*cos(8*f*x + 8*e) - 8*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 
 2*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) - 16*(7*sin(5*f*x + 5*e) - 7*sin(3*f 
*x + 3*e) + sin(f*x + e))*cos(6*f*x + 6*e) + 56*(3*sin(4*f*x + 4*e) + 2*si 
n(2*f*x + 2*e))*cos(5*f*x + 5*e) + 24*(7*sin(3*f*x + 3*e) - sin(f*x + e))* 
cos(4*f*x + 4*e) + (2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos(2*f 
*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4* 
e) + 4*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12 
*(4*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 + 16*co 
s(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f* 
x + 2*e))*sin(8*f*x + 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 
 2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + 16*sin(6*f*x + 6*e)^2 + 36*sin(4*f 
*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x + 2*e)^2 
 + 8*cos(2*f*x + 2*e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x 
 + e) + 1) - (2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2 
*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4*e) + 4 
*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*co 
s(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 + 16*cos(2*f* 
x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2* 
e))*sin(8*f*x + 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*...
 

Giac [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} + \frac {4}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + 4 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{4 \, {\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} - 4\right )}^{2} a^{\frac {3}{2}} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \] Input:

integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

-1/4*((1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))^3 + 4/tan(1/2*f*x + 
1/2*e) + 4*tan(1/2*f*x + 1/2*e))/(((1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 
 1/2*e))^2 - 4)^2*a^(3/2)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a-a\,{\sin \left (e+f\,x\right )}^2\right )}^{3/2}} \,d x \] Input:

int(tan(e + f*x)^2/(a - a*sin(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^2/(a - a*sin(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )^{2}+1}\, \tan \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{2}+1}d x \right )}{a^{2}} \] Input:

int(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x)
 

Output:

(sqrt(a)*int((sqrt( - sin(e + f*x)**2 + 1)*tan(e + f*x)**2)/(sin(e + f*x)* 
*4 - 2*sin(e + f*x)**2 + 1),x))/a**2