\(\int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx\) [103]

Optimal result

Integrand size = 23, antiderivative size = 150 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {67 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{64 \sqrt {2} \sqrt {a} f}-\frac {\sec (e+f x) (53+127 \sin (e+f x))}{192 f \sqrt {a+a \sin (e+f x)}}+\frac {a \sin (e+f x) \tan (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}+\frac {\tan ^3(e+f x)}{3 f \sqrt {a+a \sin (e+f x)}} \] Output:

-67/128*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-1/192*sec(f*x+e)*(53+127*sin(f*x+e))/f/(a+a*sin(f*x+e))^(1/ 
2)+1/24*a*sin(f*x+e)*tan(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)+1/3*tan(f*x+e)^3/ 
f/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.34 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {(804+804 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-\sec ^3(e+f x) (90+122 \cos (2 (e+f x))-41 \sin (e+f x)+183 \sin (3 (e+f x)))}{768 f \sqrt {a (1+\sin (e+f x))}} \] Input:

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

Output:

((804 + 804*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]) - Sec[e + f*x]^3*(90 + 122*C 
os[2*(e + f*x)] - 41*Sin[e + f*x] + 183*Sin[3*(e + f*x)]))/(768*f*Sqrt[a*( 
1 + Sin[e + f*x])])
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.61, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3193, 3042, 3128, 219, 4901, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3193

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4901

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f}+\frac {61 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{64 \sqrt {2} \sqrt {a} f}+\frac {61 a \cos (e+f x)}{64 f (a \sin (e+f x)+a)^{3/2}}+\frac {7 \sec ^3(e+f x) \sqrt {a \sin (e+f x)+a}}{12 a f}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {61 \sec (e+f x)}{48 f \sqrt {a \sin (e+f x)+a}}+\frac {7 a \sec (e+f x)}{24 f (a \sin (e+f x)+a)^{3/2}}\)

Input:

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

Output:

(61*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(6 
4*Sqrt[2]*Sqrt[a]*f) - (Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sq 
rt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f) + (61*a*Cos[e + f*x])/(64*f*(a + a*S 
in[e + f*x])^(3/2)) + (7*a*Sec[e + f*x])/(24*f*(a + a*Sin[e + f*x])^(3/2)) 
 - (61*Sec[e + f*x])/(48*f*Sqrt[a + a*Sin[e + f*x]]) - (5*Sec[e + f*x]^3)/ 
(6*f*Sqrt[a + a*Sin[e + f*x]]) + (7*Sec[e + f*x]^3*Sqrt[a + a*Sin[e + f*x] 
])/(12*a*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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 3193

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^4, 
x_Symbol] :> Int[(a + b*Sin[e + f*x])^m, x] - Int[(a + b*Sin[e + f*x])^m*(( 
1 - 2*Sin[e + f*x]^2)/Cos[e + f*x]^4), x] /; FreeQ[{a, b, e, f, m}, x] && E 
qQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]
 

rule 4901

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; 
 !InertTrigFreeQ[u]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.61

method	result	size
default	\(-\frac {-366 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a^{\frac {7}{2}}-122 \cos \left (f x +e \right )^{2} a^{\frac {7}{2}}+201 \cos \left (f x +e \right )^{2} \sqrt {2}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+112 \sin \left (f x +e \right ) a^{\frac {7}{2}}-402 \sin \left (f x +e \right ) \sqrt {2}\, \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+16 a^{\frac {7}{2}}-402 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -\sin \left (f x +e \right ) a \right )^{\frac {3}{2}} a^{2}}{384 a^{\frac {7}{2}} \left (-1+\sin \left (f x +e \right )\right ) \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\)	\(241\)

Input:

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

Output:

-1/384*(-366*cos(f*x+e)^2*sin(f*x+e)*a^(7/2)-122*cos(f*x+e)^2*a^(7/2)+201* 
cos(f*x+e)^2*2^(1/2)*(a-sin(f*x+e)*a)^(3/2)*arctanh(1/2*(a-sin(f*x+e)*a)^( 
1/2)*2^(1/2)/a^(1/2))*a^2+112*sin(f*x+e)*a^(7/2)-402*sin(f*x+e)*2^(1/2)*(a 
-sin(f*x+e)*a)^(3/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/2))*a 
^2+16*a^(7/2)-402*2^(1/2)*arctanh(1/2*(a-sin(f*x+e)*a)^(1/2)*2^(1/2)/a^(1/ 
2))*(a-sin(f*x+e)*a)^(3/2)*a^2)/a^(7/2)/(-1+sin(f*x+e))/(1+sin(f*x+e))/cos 
(f*x+e)/(a+sin(f*x+e)*a)^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.53 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {201 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (61 \, \cos \left (f x + e\right )^{2} + {\left (183 \, \cos \left (f x + e\right )^{2} - 56\right )} \sin \left (f x + e\right ) - 8\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{768 \, {\left (a f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a f \cos \left (f x + e\right )^{3}\right )}} \] Input:

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

Output:

1/768*(201*sqrt(2)*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*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)*si 
n(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*(61*cos(f*x + e)^2 + (183*cos(f*x + e)^2 - 56)*sin(f*x 
+ e) - 8)*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e)^3*sin(f*x + e) + a*f 
*cos(f*x + e)^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(tan(f*x + e)^4/sqrt(a*sin(f*x + e) + a), x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {\frac {201 \, \sqrt {2} \log \left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {201 \, \sqrt {2} \log \left (-\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {6 \, \sqrt {2} {\left (21 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 19 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} \sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {16 \, \sqrt {2} {\left (15 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}}{768 \, f} \] Input:

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

Output:

-1/768*(201*sqrt(2)*log(sin(3/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(a)*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))) - 201*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))) + 6*sqrt(2)*(21*s 
in(3/4*pi + 1/2*f*x + 1/2*e)^3 - 19*sin(3/4*pi + 1/2*f*x + 1/2*e))/((sin(3 
/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2* 
e))) + 16*sqrt(2)*(15*sin(3/4*pi + 1/2*f*x + 1/2*e)^2 - 1)/(sqrt(a)*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi + 1/2*f*x + 1/2*e)^3))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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