\(\int \frac {\tan ^5(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [452]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\left (8 a^2-8 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{7/2} f}-\frac {8 a^2-8 a b-b^2}{8 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(8 a+3 b) \sec ^2(e+f x)}{8 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^4(e+f x)}{4 (a+b) f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

1/8*(8*a^2-8*a*b-b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^ 
(7/2)/f-1/8*(8*a^2-8*a*b-b^2)/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(1/2)-1/8*(8*a+ 
3*b)*sec(f*x+e)^2/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+1/4*sec(f*x+e)^4/(a+b 
)/f/(a+b*sin(f*x+e)^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\left (-8 a^2+8 a b+b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \sin ^2(e+f x)}{a+b}\right )-\frac {1}{2} (a+b) (4 a-b+(8 a+3 b) \cos (2 (e+f x))) \sec ^4(e+f x)}{8 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}} \] Input:

Integrate[Tan[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

((-8*a^2 + 8*a*b + b^2)*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Sin[e + f*x 
]^2)/(a + b)] - ((a + b)*(4*a - b + (8*a + 3*b)*Cos[2*(e + f*x)])*Sec[e + 
f*x]^4)/2)/(8*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3673, 100, 27, 87, 61, 73, 221}

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.

\(\displaystyle \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^5}{\left (a+b \sin (e+f x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3673

\(\displaystyle \frac {\int \frac {\sin ^4(e+f x)}{\left (1-\sin ^2(e+f x)\right )^3 \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {1}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {8 a+3 b}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (8 a^2-8 a b-b^2\right ) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\frac {1}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {8 a+3 b}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (8 a^2-8 a b-b^2\right ) \left (\frac {\int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a+b}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {8 a+3 b}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (8 a^2-8 a b-b^2\right ) \left (\frac {2 \int \frac {1}{\frac {a+b}{b}-\frac {\sin ^4(e+f x)}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{b (a+b)}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{2 (a+b) \left (1-\sin ^2(e+f x)\right )^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {\frac {8 a+3 b}{(a+b) \left (1-\sin ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (8 a^2-8 a b-b^2\right ) \left (\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {2}{(a+b) \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 (a+b)}}{4 (a+b)}}{2 f}\)

Input:

Int[Tan[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]
 

Output:

(1/(2*(a + b)*(1 - Sin[e + f*x]^2)^2*Sqrt[a + b*Sin[e + f*x]^2]) - ((8*a + 
 3*b)/((a + b)*(1 - Sin[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2]) - ((8*a^2 
- 8*a*b - b^2)*((2*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b 
)^(3/2) - 2/((a + b)*Sqrt[a + b*Sin[e + f*x]^2])))/(2*(a + b)))/(4*(a + b) 
))/(2*f)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3673

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ 
(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m 
 + 1)/2)/(2*f)   Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 
)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ 
erQ[(m - 1)/2]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3766\) vs. \(2(157)=314\).

Time = 0.82 (sec) , antiderivative size = 3767, normalized size of antiderivative = 21.28

method	result	size
default	\(\text {Expression too large to display}\)	\(3767\)

Input:

int(tan(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/(a^4*b^2*cos(f*x+e)^4+4*a^3*b^3*cos(f*x+e)^4+6*a^2*b^4*cos(f*x+e)^4+4 
*a*b^5*cos(f*x+e)^4+b^6*cos(f*x+e)^4-2*a^5*b*cos(f*x+e)^2-10*a^4*b^2*cos(f 
*x+e)^2-20*a^3*b^3*cos(f*x+e)^2-20*a^2*b^4*cos(f*x+e)^2-10*a*b^5*cos(f*x+e 
)^2-2*b^6*cos(f*x+e)^2+a^6+6*a^5*b+15*a^4*b^2+20*a^3*b^3+15*b^4*a^2+6*b^5* 
a+b^6)/cos(f*x+e)^4/(a+b)^(3/2)*(-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b* 
cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^6*cos(f*x+e)^4+8*ln(2/(sin(f*x+e)-1 
)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^6*cos(f*x+e)^ 
4+32*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2*cos(f*x+e)^4+8*(a+b)^(3/ 
2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*a*b^2*cos(f*x+e)^4-16*(a+b)^(3/ 
2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a^3*b*cos(f*x+e)^4+18*(a+b)^(3/2)*(a+b-b*cos 
(f*x+e)^2)^(1/2)*a^2*b^2*cos(f*x+e)^4+20*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^ 
(1/2)*a*b^3*cos(f*x+e)^4-24*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^ 
(1/2)*a^3*b*cos(f*x+e)^4-72*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^ 
(1/2)*a^2*b^2*cos(f*x+e)^4-40*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2 
)^(1/2)*a*b^3*cos(f*x+e)^4-38*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^2*b 
*cos(f*x+e)^2-28*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2*cos(f*x+e)^2 
+8*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^4*cos(f*x+e)^4-16 
*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^3*cos(f*x+e)^2-6*(a+b)^(3/2)*(a+ 
b-b*cos(f*x+e)^2)^(3/2)*b^3*cos(f*x+e)^2+12*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^ 
2)^(3/2)*a^2*b+12*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2-32*ln(2/...
 

Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 606, normalized size of antiderivative = 3.42 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (8 \, a^{2} b - 8 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left ({\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} + {\left (8 \, a^{3} + 19 \, a^{2} b + 14 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left ({\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4}\right )}}, \frac {{\left ({\left (8 \, a^{2} b - 8 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + {\left ({\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} + {\left (8 \, a^{3} + 19 \, a^{2} b + 14 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left ({\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4}\right )}}\right ] \] Input:

integrate(tan(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/16*(((8*a^2*b - 8*a*b^2 - b^3)*cos(f*x + e)^6 - (8*a^3 - 9*a*b^2 - b^3 
)*cos(f*x + e)^4)*sqrt(a + b)*log((b*cos(f*x + e)^2 + 2*sqrt(-b*cos(f*x + 
e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*((8*a^3 - 9*a*b 
^2 - b^3)*cos(f*x + e)^4 - 2*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + (8*a^3 + 19 
*a^2*b + 14*a*b^2 + 3*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b) 
)/((a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^6 - (a^5 
 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*f*cos(f*x + e)^4), 1 
/8*(((8*a^2*b - 8*a*b^2 - b^3)*cos(f*x + e)^6 - (8*a^3 - 9*a*b^2 - b^3)*co 
s(f*x + e)^4)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a 
- b)/(b*cos(f*x + e)^2 - a - b)) + ((8*a^3 - 9*a*b^2 - b^3)*cos(f*x + e)^4 
 - 2*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + (8*a^3 + 19*a^2*b + 14*a*b^2 + 3*b^ 
3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + 4*a^3*b^2 + 
6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^6 - (a^5 + 5*a^4*b + 10*a^3*b^2 
+ 10*a^2*b^3 + 5*a*b^4 + b^5)*f*cos(f*x + e)^4)]
 

Sympy [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(tan(f*x+e)**5/(a+b*sin(f*x+e)**2)**(3/2),x)
 

Output:

Integral(tan(e + f*x)**5/(a + b*sin(e + f*x)**2)**(3/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (157) = 314\).

Time = 0.12 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.89 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {{\left (8 \, a^{2} b^{3} - 8 \, a b^{4} - b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a + b}} + \frac {2 \, {\left (8 \, a^{4} b^{3} + 16 \, a^{3} b^{4} + 8 \, a^{2} b^{5} + {\left (8 \, a^{2} b^{3} - 8 \, a b^{4} - b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} - {\left (16 \, a^{3} b^{3} + 8 \, a^{2} b^{4} - 7 \, a b^{5} + b^{6}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} - 2 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}}{16 \, b^{3} f} \] Input:

integrate(tan(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
 

Output:

-1/16*((8*a^2*b^3 - 8*a*b^4 - b^5)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt( 
a + b))/(sqrt(b*sin(f*x + e)^2 + a) + sqrt(a + b)))/((a^3 + 3*a^2*b + 3*a* 
b^2 + b^3)*sqrt(a + b)) + 2*(8*a^4*b^3 + 16*a^3*b^4 + 8*a^2*b^5 + (8*a^2*b 
^3 - 8*a*b^4 - b^5)*(b*sin(f*x + e)^2 + a)^2 - (16*a^3*b^3 + 8*a^2*b^4 - 7 
*a*b^5 + b^6)*(b*sin(f*x + e)^2 + a))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b* 
sin(f*x + e)^2 + a)^(5/2) - 2*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)* 
(b*sin(f*x + e)^2 + a)^(3/2) + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 
5*a*b^4 + b^5)*sqrt(b*sin(f*x + e)^2 + a)))/(b^3*f)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2644 vs. \(2 (157) = 314\).

Time = 1.11 (sec) , antiderivative size = 2644, normalized size of antiderivative = 14.94 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
 

Output:

-1/4*(4*((a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*tan(1/2*f*x 
 + 1/2*e)^2/(a^7*b + 7*a^6*b^2 + 21*a^5*b^3 + 35*a^4*b^4 + 35*a^3*b^5 + 21 
*a^2*b^6 + 7*a*b^7 + b^8) + (a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a 
^2*b^5)/(a^7*b + 7*a^6*b^2 + 21*a^5*b^3 + 35*a^4*b^4 + 35*a^3*b^5 + 21*a^2 
*b^6 + 7*a*b^7 + b^8))/sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1 
/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) + (8*a^2 - 8*a*b - b^2)*arctan(- 
1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a* 
tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(- 
a - b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(-a - b)) - 2*(8*(sqrt(a)*tan 
(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/ 
2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^7*a^2 - (sqrt(a)*tan(1/2*f*x + 1 
/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b 
*tan(1/2*f*x + 1/2*e)^2 + a))^7*b^2 - 56*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - 
 sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2* 
f*x + 1/2*e)^2 + a))^6*a^(5/2) - 32*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt 
(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 
 1/2*e)^2 + a))^6*a^(3/2)*b - 25*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a* 
tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/ 
2*e)^2 + a))^6*sqrt(a)*b^2 - 120*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a* 
tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x +...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \] Input:

int(tan(e + f*x)^5/(a + b*sin(e + f*x)^2)^(3/2),x)
 

Output:

int(tan(e + f*x)^5/(a + b*sin(e + f*x)^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{5}}{\sin \left (f x +e \right )^{4} b^{2}+2 \sin \left (f x +e \right )^{2} a b +a^{2}}d x \] Input:

int(tan(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x)
 

Output:

int((sqrt(sin(e + f*x)**2*b + a)*tan(e + f*x)**5)/(sin(e + f*x)**4*b**2 + 
2*sin(e + f*x)**2*a*b + a**2),x)