\(\int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) [404]

Optimal result

Integrand size = 26, antiderivative size = 91 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}-\frac {3 \tan (e+f x)}{8 f \sqrt {a \cos ^2(e+f x)}}+\frac {\tan ^3(e+f x)}{4 f \sqrt {a \cos ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)+\tan (e+f x) \left (3-6 \sec ^2(e+f x)+8 \tan ^2(e+f x)\right )}{8 f \sqrt {a \cos ^2(e+f x)}} \] Input:

Integrate[Tan[e + f*x]^4/Sqrt[a - a*Sin[e + f*x]^2],x]
 

Output:

(3*ArcTanh[Sin[e + f*x]]*Cos[e + f*x] + Tan[e + f*x]*(3 - 6*Sec[e + f*x]^2 
 + 8*Tan[e + f*x]^2))/(8*f*Sqrt[a*Cos[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89, 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, 3091, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3655

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan \left (e+f x+\frac {\pi }{2}\right )^4 \sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}dx\)

\(\Big \downarrow \) 3686

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3091

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3091

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

Input:

Int[Tan[e + f*x]^4/Sqrt[a - a*Sin[e + f*x]^2],x]
 

Output:

(Cos[e + f*x]*((Sec[e + f*x]*Tan[e + f*x]^3)/(4*f) - (3*(-1/2*ArcTanh[Sin[ 
e + f*x]]/f + (Sec[e + f*x]*Tan[e + f*x])/(2*f)))/4))/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 4257

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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {\sqrt {a \sin \left (f x +e \right )^{2}}\, \left (-3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}+2 a}{\cos \left (f x +e \right )}\right ) a \cos \left (f x +e \right )^{4}+5 \sqrt {a \sin \left (f x +e \right )^{2}}\, \cos \left (f x +e \right )^{2} \sqrt {a}-2 \sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}\right )}{8 \cos \left (f x +e \right )^{3} a^{\frac {3}{2}} \sin \left (f x +e \right ) \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\)	\(132\)
risch	\(\frac {i \left (5 \,{\mathrm e}^{6 i \left (f x +e \right )}-3 \,{\mathrm e}^{4 i \left (f x +e \right )}+3 \,{\mathrm e}^{2 i \left (f x +e \right )}-5\right )}{4 \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} f}-\frac {3 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}+\frac {3 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}\)	\(188\)

Input:

int(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8/cos(f*x+e)^3*(a*sin(f*x+e)^2)^(1/2)*(-3*ln(2/cos(f*x+e)*(a^(1/2)*(a*s 
in(f*x+e)^2)^(1/2)+a))*a*cos(f*x+e)^4+5*(a*sin(f*x+e)^2)^(1/2)*cos(f*x+e)^ 
2*a^(1/2)-2*a^(1/2)*(a*sin(f*x+e)^2)^(1/2))/a^(3/2)/sin(f*x+e)/(a*cos(f*x+ 
e)^2)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {{\left (3 \, \cos \left (f x + e\right )^{4} \log \left (-\frac {\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, {\left (5 \, \cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{16 \, a f \cos \left (f x + e\right )^{5}} \] Input:

integrate(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1518 vs. \(2 (79) = 158\).

Time = 0.32 (sec) , antiderivative size = 1518, normalized size of antiderivative = 16.68 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
 

Output:

-1/16*(4*(5*sin(7*f*x + 7*e) - 3*sin(5*f*x + 5*e) + 3*sin(3*f*x + 3*e) - 5 
*sin(f*x + e))*cos(8*f*x + 8*e) - 40*(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*(3*sin(5*f*x + 5*e) - 3*si 
n(3*f*x + 3*e) + 5*sin(f*x + e))*cos(6*f*x + 6*e) + 24*(3*sin(4*f*x + 4*e) 
 + 2*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) + 24*(3*sin(3*f*x + 3*e) - 5*sin(f 
*x + e))*cos(4*f*x + 4*e) - 3*(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*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*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) + 3*(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*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*s 
in(2*f*x + 2*e))*sin(8*f*x + 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (79) = 158\).

Time = 0.53 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.07 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\frac {3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} - \frac {3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} - \frac {4 \, {\left (3 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} - \frac {20}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - 20 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} - 4\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}}{16 \, \sqrt {a} f} \] Input:

integrate(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
 

Output:

-1/16*(3*log(abs(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e) + 2))/sgn(t 
an(1/2*f*x + 1/2*e)^4 - 1) - 3*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*(3*(1/tan(1/2*f*x + 1 
/2*e) + tan(1/2*f*x + 1/2*e))^3 - 20/tan(1/2*f*x + 1/2*e) - 20*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*sgn( 
tan(1/2*f*x + 1/2*e)^4 - 1)))/(sqrt(a)*f)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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