\(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx\) [393]

Optimal result

Integrand size = 26, antiderivative size = 130 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\frac {15 \text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{8 f}-\frac {\sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{f}-\frac {9 \sqrt {a \cos ^2(e+f x)} \sec ^2(e+f x) \tan (e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \sec ^4(e+f x) \tan (e+f x)}{4 f} \] Output:

15/8*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-9/8*(a*cos(f*x+e)^2)^(1/2)*sec(f*x+e)^2*tan(f*x+e)/ 
f+1/4*(a*cos(f*x+e)^2)^(1/2)*sec(f*x+e)^4*tan(f*x+e)/f
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.58 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\frac {\sqrt {a \cos ^2(e+f x)} \sec ^5(e+f x) \left (60 \text {arctanh}(\sin (e+f x)) \cos ^4(e+f x)-5 \sin (e+f x)-15 \sin (3 (e+f x))-2 \sin (5 (e+f x))\right )}{32 f} \] Input:

Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^6,x]
 

Output:

(Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]^5*(60*ArcTanh[Sin[e + f*x]]*Cos[e + f 
*x]^4 - 5*Sin[e + f*x] - 15*Sin[3*(e + f*x)] - 2*Sin[5*(e + f*x)]))/(32*f)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.78, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3655, 3042, 3686, 3042, 3072, 252, 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.

\(\displaystyle \int \tan ^6(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^6 \sqrt {a-a \sin (e+f x)^2}dx\)

\(\Big \downarrow \) 3655

\(\displaystyle \int \tan ^6(e+f x) \sqrt {a \cos ^2(e+f x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}{\tan \left (e+f x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 3686

\(\displaystyle \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \sin (e+f x) \tan ^5(e+f x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \sin (e+f x) \tan (e+f x)^5dx\)

\(\Big \downarrow \) 3072

\(\displaystyle \frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \frac {\sin ^6(e+f x)}{\left (1-\sin ^2(e+f x)\right )^3}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \left (\frac {\sin ^5(e+f x)}{4 \left (1-\sin ^2(e+f x)\right )^2}-\frac {5}{4} \left (\frac {\sin ^3(e+f x)}{2 \left (1-\sin ^2(e+f x)\right )}-\frac {3}{2} (\text {arctanh}(\sin (e+f x))-\sin (e+f x))\right )\right )}{f}\)

Input:

Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^6,x]
 

Output:

(Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]*(Sin[e + f*x]^5/(4*(1 - Sin[e + f*x]^ 
2)^2) - (5*((-3*(ArcTanh[Sin[e + f*x]] - Sin[e + f*x]))/2 + Sin[e + f*x]^3 
/(2*(1 - Sin[e + f*x]^2))))/4))/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 = 0.66 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {\sqrt {a \sin \left (f x +e \right )^{2}}\, \left (8 \sqrt {a \sin \left (f x +e \right )^{2}}\, \cos \left (f x +e \right )^{4} \sqrt {a}-15 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}+2 a}{\cos \left (f x +e \right )}\right ) a \cos \left (f x +e \right )^{4}+9 \sqrt {a \sin \left (f x +e \right )^{2}}\, \cos \left (f x +e \right )^{2} \sqrt {a}-2 \sqrt {a}\, \sqrt {a \sin \left (f x +e \right )^{2}}\right )}{8 \cos \left (f x +e \right )^{3} \sqrt {a}\, \sin \left (f x +e \right ) \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\)	\(157\)
risch	\(\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}+\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (9 \,{\mathrm e}^{8 i \left (f x +e \right )}+{\mathrm e}^{6 i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{4 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}+\frac {15 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{8 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {15 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{8 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(327\)

Input:

int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x,method=_RETURNVERBOSE)
 

Output:

-1/8/cos(f*x+e)^3*(a*sin(f*x+e)^2)^(1/2)*(8*(a*sin(f*x+e)^2)^(1/2)*cos(f*x 
+e)^4*a^(1/2)-15*ln(2/cos(f*x+e)*(a^(1/2)*(a*sin(f*x+e)^2)^(1/2)+a))*a*cos 
(f*x+e)^4+9*(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^(1/2)/sin(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.67 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=-\frac {{\left (15 \, \cos \left (f x + e\right )^{4} \log \left (-\frac {\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, {\left (8 \, \cos \left (f x + e\right )^{4} + 9 \, \cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{16 \, f \cos \left (f x + e\right )^{5}} \] Input:

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x, algorithm="fricas")
 

Output:

-1/16*(15*cos(f*x + e)^4*log(-(sin(f*x + e) - 1)/(sin(f*x + e) + 1)) + 2*( 
8*cos(f*x + e)^4 + 9*cos(f*x + e)^2 - 2)*sin(f*x + e))*sqrt(a*cos(f*x + e) 
^2)/(f*cos(f*x + e)^5)
 

Sympy [F]

\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{6}{\left (e + f x \right )}\, dx \] Input:

integrate((a-a*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**6,x)
 

Output:

Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**6, x 
)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1955 vs. \(2 (116) = 232\).

Time = 0.88 (sec) , antiderivative size = 1955, normalized size of antiderivative = 15.04 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x, algorithm="maxima")
 

Output:

1/16*(8*(sin(9*f*x + 9*e) + 4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*si 
n(3*f*x + 3*e) + sin(f*x + e))*cos(10*f*x + 10*e) - 20*(3*sin(8*f*x + 8*e) 
 + sin(6*f*x + 6*e) - sin(4*f*x + 4*e) - 3*sin(2*f*x + 2*e))*cos(9*f*x + 9 
*e) + 60*(4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + s 
in(f*x + e))*cos(8*f*x + 8*e) - 80*(sin(6*f*x + 6*e) - sin(4*f*x + 4*e) - 
3*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) + 20*(6*sin(5*f*x + 5*e) + 4*sin(3*f* 
x + 3*e) + sin(f*x + e))*cos(6*f*x + 6*e) + 120*(sin(4*f*x + 4*e) + 3*sin( 
2*f*x + 2*e))*cos(5*f*x + 5*e) - 20*(4*sin(3*f*x + 3*e) + sin(f*x + e))*co 
s(4*f*x + 4*e) + 15*(2*(4*cos(7*f*x + 7*e) + 6*cos(5*f*x + 5*e) + 4*cos(3* 
f*x + 3*e) + cos(f*x + e))*cos(9*f*x + 9*e) + cos(9*f*x + 9*e)^2 + 8*(6*co 
s(5*f*x + 5*e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(7*f*x + 7*e) + 16* 
cos(7*f*x + 7*e)^2 + 12*(4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(5*f*x + 5* 
e) + 36*cos(5*f*x + 5*e)^2 + 16*cos(3*f*x + 3*e)^2 + 8*cos(3*f*x + 3*e)*co 
s(f*x + e) + cos(f*x + e)^2 + 2*(4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 
 4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(9*f*x + 9*e) + sin(9*f*x + 9*e)^2 
+ 8*(6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(7*f*x + 7 
*e) + 16*sin(7*f*x + 7*e)^2 + 12*(4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5 
*f*x + 5*e) + 36*sin(5*f*x + 5*e)^2 + 16*sin(3*f*x + 3*e)^2 + 8*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) - 15*(2*(4*cos(7*f*x + 7*e) + 6*cos(5*f*x + 5*e) ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (116) = 232\).

Time = 0.52 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.80 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=-\frac {{\left (15 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 15 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \frac {32 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - \frac {4 \, {\left (7 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 36 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )\right )}}{{\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} - 4\right )}^{2}}\right )} \sqrt {a}}{16 \, f} \] Input:

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x, algorithm="giac")
 

Output:

-1/16*(15*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) - 15*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) - 32*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)) - 4*(7*(1/ta 
n(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))^3*sgn(tan(1/2*f*x + 1/2*e)^4 - 
1) - 36*(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))*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))^2 - 4)^2)* 
sqrt(a)/f
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^6\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \] Input:

int(tan(e + f*x)^6*(a - a*sin(e + f*x)^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(tan(e + f*x)^6*(a - a*sin(e + f*x)^2)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx=\sqrt {a}\, \left (\int \sqrt {-\sin \left (f x +e \right )^{2}+1}\, \tan \left (f x +e \right )^{6}d x \right ) \] Input:

int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x)
 

Output:

sqrt(a)*int(sqrt( - sin(e + f*x)**2 + 1)*tan(e + f*x)**6,x)