\(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx\) [394]

Optimal result

Integrand size = 26, antiderivative size = 94 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=-\frac {3 \text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{2 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{f}+\frac {\sqrt {a \cos ^2(e+f x)} \sec ^2(e+f x) \tan (e+f x)}{2 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.59 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\frac {a (-3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)+(2+\cos (2 (e+f x))) \tan (e+f x))}{2 f \sqrt {a \cos ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3655, 3042, 3686, 3042, 3072, 252, 262, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3655

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3686

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3072

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 3042

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

rule 3072

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f   Subst[Int[ 
(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, a*(Sin[e + f*x]/ff)], x 
]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
 

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]])
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {\sqrt {a \sin \left (f x +e \right )^{2}}\, \left (2 \sqrt {a \sin \left (f x +e \right )^{2}}\, \cos \left (f x +e \right )^{2} \sqrt {a}-3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}+2 a}{\cos \left (f x +e \right )}\right ) \cos \left (f x +e \right )^{2} a +\sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}\right )}{2 \cos \left (f x +e \right ) \sqrt {a}\, \sin \left (f x +e \right ) \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\)	\(131\)
risch	\(-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (i {\mathrm e}^{6 i \left (f x +e \right )}+3 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) {\mathrm e}^{5 i \left (f x +e \right )}-3 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) {\mathrm e}^{5 i \left (f x +e \right )}+3 i {\mathrm e}^{4 i \left (f x +e \right )}+6 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) {\mathrm e}^{3 i \left (f x +e \right )}-6 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) {\mathrm e}^{3 i \left (f x +e \right )}-3 i {\mathrm e}^{2 i \left (f x +e \right )}+3 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) {\mathrm e}^{i \left (f x +e \right )}-3 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) {\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\)	\(246\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (3 \, \cos \left (f x + e\right )^{2} \log \left (-\frac {\sin \left (f x + e\right ) + 1}{\sin \left (f x + e\right ) - 1}\right ) - 2 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right )\right )}}{4 \, f \cos \left (f x + e\right )^{3}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 827 vs. \(2 (84) = 168\).

Time = 0.22 (sec) , antiderivative size = 827, normalized size of antiderivative = 8.80 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(2*(sin(5*f*x + 5*e) + 2*sin(3*f*x + 3*e) + sin(f*x + e))*cos(6*f*x + 
 6*e) - 6*(sin(4*f*x + 4*e) - sin(2*f*x + 2*e))*cos(5*f*x + 5*e) + 6*(2*si 
n(3*f*x + 3*e) + sin(f*x + e))*cos(4*f*x + 4*e) + 3*(2*(2*cos(3*f*x + 3*e) 
 + cos(f*x + e))*cos(5*f*x + 5*e) + cos(5*f*x + 5*e)^2 + 4*cos(3*f*x + 3*e 
)^2 + 4*cos(3*f*x + 3*e)*cos(f*x + e) + cos(f*x + e)^2 + 2*(2*sin(3*f*x + 
3*e) + sin(f*x + e))*sin(5*f*x + 5*e) + sin(5*f*x + 5*e)^2 + 4*sin(3*f*x + 
 3*e)^2 + 4*sin(3*f*x + 3*e)*sin(f*x + e) + sin(f*x + e)^2)*log(cos(f*x + 
e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 3*(2*(2*cos(3*f*x + 3*e) + c 
os(f*x + e))*cos(5*f*x + 5*e) + cos(5*f*x + 5*e)^2 + 4*cos(3*f*x + 3*e)^2 
+ 4*cos(3*f*x + 3*e)*cos(f*x + e) + cos(f*x + e)^2 + 2*(2*sin(3*f*x + 3*e) 
 + sin(f*x + e))*sin(5*f*x + 5*e) + sin(5*f*x + 5*e)^2 + 4*sin(3*f*x + 3*e 
)^2 + 4*sin(3*f*x + 3*e)*sin(f*x + e) + sin(f*x + e)^2)*log(cos(f*x + e)^2 
 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 2*(cos(5*f*x + 5*e) + 2*cos(3*f* 
x + 3*e) + cos(f*x + e))*sin(6*f*x + 6*e) + 2*(3*cos(4*f*x + 4*e) - 3*cos( 
2*f*x + 2*e) - 1)*sin(5*f*x + 5*e) - 6*(2*cos(3*f*x + 3*e) + cos(f*x + e)) 
*sin(4*f*x + 4*e) - 4*(3*cos(2*f*x + 2*e) + 1)*sin(3*f*x + 3*e) + 12*cos(3 
*f*x + 3*e)*sin(2*f*x + 2*e) + 6*cos(f*x + e)*sin(2*f*x + 2*e) - 6*cos(2*f 
*x + 2*e)*sin(f*x + e) - 2*sin(f*x + e))*sqrt(a)/((2*(2*cos(3*f*x + 3*e) + 
 cos(f*x + e))*cos(5*f*x + 5*e) + cos(5*f*x + 5*e)^2 + 4*cos(3*f*x + 3*e)^ 
2 + 4*cos(3*f*x + 3*e)*cos(f*x + e) + cos(f*x + e)^2 + 2*(2*sin(3*f*x +...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (84) = 168\).

Time = 0.60 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.10 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\frac {{\left (3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \frac {4 \, {\left (3 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )\right )}}{{\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} - \frac {4}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - 4 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}\right )} \sqrt {a}}{4 \, f} \] Input:

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

Output:

1/4*(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) - 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))^2*sgn(tan(1/2*f*x + 1/2*e)^4 - 1) - 8*sgn(tan( 
1/2*f*x + 1/2*e)^4 - 1))/((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)))*sqrt(a)/f
 

Mupad [F(-1)]

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

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

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

Output:

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