\(\int \frac {\tan ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [464]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{7/2} f}+\frac {2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 a-3 b}{2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

-1/2*(2*a-3*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(7/2)/f 
+1/6*(2*a-3*b)/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(3/2)+1/2*sec(f*x+e)^2/(a+b)/f 
/(a+b*sin(f*x+e)^2)^(3/2)+1/2*(2*a-3*b)/(a+b)^3/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 = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.50 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a-3 b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin ^2(e+f x)}{a+b}\right )+3 (a+b) \sec ^2(e+f x)}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

((2*a - 3*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[e + f*x]^2)/(a + 
b)] + 3*(a + b)*Sec[e + f*x]^2)/(6*(a + b)^2*f*(a + b*Sin[e + f*x]^2)^(3/2 
))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3673, 87, 61, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

\(\displaystyle \frac {\int \frac {\sin ^2(e+f x)}{\left (1-\sin ^2(e+f x)\right )^2 \left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {1}{(a+b) \left (1-\sin ^2(e+f x)\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(2 a-3 b) \int \frac {1}{\left (1-\sin ^2(e+f x)\right ) \left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin ^2(e+f x)}{2 (a+b)}}{2 f}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(833\) vs. \(2(133)=266\).

Time = 0.62 (sec) , antiderivative size = 834, normalized size of antiderivative = 5.45

method	result	size
default	\(\frac {\frac {b^{2} \left (-\frac {\sqrt {a +b -b \cos \left (f x +e \right )^{2}}}{\left (a +b \right ) \left (\sin \left (f x +e \right )-1\right )}+\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )}{4 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2}}-\frac {b^{2} \left (-\frac {\sqrt {a +b -b \cos \left (f x +e \right )^{2}}}{\left (a +b \right ) \left (1+\sin \left (f x +e \right )\right )}-\frac {b \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{\left (a +b \right )^{\frac {3}{2}}}\right )}{4 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2}}-\frac {b^{3} \sqrt {-a b}\, \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {b^{2} a +b^{3}}{b^{2}}}\, \left (\sqrt {-a b}\, \sin \left (f x +e \right )+2 a \right )}{12 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2} a \left (-\sqrt {-a b}\, b^{2} \cos \left (f x +e \right )^{2}+2 a \,b^{2} \sin \left (f x +e \right )+\left (-a b \right )^{\frac {3}{2}}+b^{2} \sqrt {-a b}\right )}+\frac {b^{3} \sqrt {-a b}\, \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {b^{2} a +b^{3}}{b^{2}}}\, \left (\sqrt {-a b}\, \sin \left (f x +e \right )-2 a \right )}{12 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2} a \left (-\sqrt {-a b}\, b^{2} \cos \left (f x +e \right )^{2}-2 a \,b^{2} \sin \left (f x +e \right )+\left (-a b \right )^{\frac {3}{2}}+b^{2} \sqrt {-a b}\right )}+\frac {b^{3} \left (a -b \right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {a +b}}+\frac {b^{3} \left (a -b \right ) \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \cos \left (f x +e \right )^{2}}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {a +b}}-\frac {b^{3} \left (a -b \right ) \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {a b +b^{2}}{b}}}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}+\frac {b^{3} \left (a -b \right ) \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {a b +b^{2}}{b}}}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}}{f}\)	\(834\)

Input:

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

Output:

(1/4*b^2/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2*(-1/(a+b)/(sin(f*x+e)-1)*( 
a+b-b*cos(f*x+e)^2)^(1/2)+b/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e 
)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1)))-1/4*b^2/(b+(-a*b)^(1/2))^2 
/(-b+(-a*b)^(1/2))^2*(-1/(a+b)/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)-b 
/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2 
*a)/(1+sin(f*x+e))))-1/12*b^3*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^( 
1/2))^2/a/(-(-a*b)^(1/2)*b^2*cos(f*x+e)^2+2*a*b^2*sin(f*x+e)+(-a*b)^(3/2)+ 
b^2*(-a*b)^(1/2))*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*((-a*b)^(1/2)*si 
n(f*x+e)+2*a)+1/12*b^3*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2 
/a/(-(-a*b)^(1/2)*b^2*cos(f*x+e)^2-2*a*b^2*sin(f*x+e)+(-a*b)^(3/2)+b^2*(-a 
*b)^(1/2))*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*((-a*b)^(1/2)*sin(f*x+e 
)-2*a)+1/2*b^3*(a-b)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln 
((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e) 
-1))+1/2*b^3*(a-b)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln(( 
2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e) 
))-1/2*b^3*(a-b)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(sin( 
f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+1/2*b^3*(a-b)/( 
b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2 
)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2))/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (133) = 266\).

Time = 0.23 (sec) , antiderivative size = 782, normalized size of antiderivative = 5.11 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/12*(3*((2*a*b^2 - 3*b^3)*cos(f*x + e)^6 - 2*(2*a^2*b - a*b^2 - 3*b^3)* 
cos(f*x + e)^4 + (2*a^3 + a^2*b - 4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*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*(3*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e 
)^4 - 3*a^3 - 9*a^2*b - 9*a*b^2 - 3*b^3 - 4*(2*a^3 + a^2*b - 4*a*b^2 - 3*b 
^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b^2 + 4*a^3*b^3 
 + 6*a^2*b^4 + 4*a*b^5 + b^6)*f*cos(f*x + e)^6 - 2*(a^5*b + 5*a^4*b^2 + 10 
*a^3*b^3 + 10*a^2*b^4 + 5*a*b^5 + b^6)*f*cos(f*x + e)^4 + (a^6 + 6*a^5*b + 
 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)*f*cos(f*x + e)^2), 
-1/6*(3*((2*a*b^2 - 3*b^3)*cos(f*x + e)^6 - 2*(2*a^2*b - a*b^2 - 3*b^3)*co 
s(f*x + e)^4 + (2*a^3 + a^2*b - 4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(-a - 
 b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(b*cos(f*x + e)^2 
- a - b)) + (3*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 - 3*a^3 - 9*a^2*b 
- 9*a*b^2 - 3*b^3 - 4*(2*a^3 + a^2*b - 4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*sq 
rt(-b*cos(f*x + e)^2 + a + b))/((a^4*b^2 + 4*a^3*b^3 + 6*a^2*b^4 + 4*a*b^5 
 + b^6)*f*cos(f*x + e)^6 - 2*(a^5*b + 5*a^4*b^2 + 10*a^3*b^3 + 10*a^2*b^4 
+ 5*a*b^5 + b^6)*f*cos(f*x + e)^4 + (a^6 + 6*a^5*b + 15*a^4*b^2 + 20*a^3*b 
^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)*f*cos(f*x + e)^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.71 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {3 \, {\left (2 \, a b^{2} - 3 \, b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a + b}} - \frac {2 \, {\left (2 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 2 \, a b^{4} - 3 \, {\left (2 \, a b^{2} - 3 \, b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} + 2 \, {\left (2 \, a^{2} b^{2} - a b^{3} - 3 \, b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} - {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}}{12 \, b^{2} f} \] Input:

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

Output:

1/12*(3*(2*a*b^2 - 3*b^3)*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*(2*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 - 3*(2*a*b^2 - 3*b^3)* 
(b*sin(f*x + e)^2 + a)^2 + 2*(2*a^2*b^2 - a*b^3 - 3*b^4)*(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) - (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)))/(b 
^2*f)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1746 vs. \(2 (133) = 266\).

Time = 0.93 (sec) , antiderivative size = 1746, normalized size of antiderivative = 11.41 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/3*(3*(2*a - 3*b)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta 
n(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*((((2*a^13*b^2 + 21*a^12*b^3 + 99*a^11*b^4 + 275*a^10*b^5 + 49 
5*a^9*b^6 + 594*a^8*b^7 + 462*a^7*b^8 + 198*a^6*b^9 - 55*a^4*b^11 - 33*a^3 
*b^12 - 9*a^2*b^13 - a*b^14)*tan(1/2*f*x + 1/2*e)^2/(a^14*b^2 + 14*a^13*b^ 
3 + 91*a^12*b^4 + 364*a^11*b^5 + 1001*a^10*b^6 + 2002*a^9*b^7 + 3003*a^8*b 
^8 + 3432*a^7*b^9 + 3003*a^6*b^10 + 2002*a^5*b^11 + 1001*a^4*b^12 + 364*a^ 
3*b^13 + 91*a^2*b^14 + 14*a*b^15 + b^16) + 3*(2*a^13*b^2 + 23*a^12*b^3 + 1 
19*a^11*b^4 + 363*a^10*b^5 + 715*a^9*b^6 + 924*a^8*b^7 + 726*a^7*b^8 + 198 
*a^6*b^9 - 264*a^5*b^10 - 385*a^4*b^11 - 253*a^3*b^12 - 97*a^2*b^13 - 21*a 
*b^14 - 2*b^15)/(a^14*b^2 + 14*a^13*b^3 + 91*a^12*b^4 + 364*a^11*b^5 + 100 
1*a^10*b^6 + 2002*a^9*b^7 + 3003*a^8*b^8 + 3432*a^7*b^9 + 3003*a^6*b^10 + 
2002*a^5*b^11 + 1001*a^4*b^12 + 364*a^3*b^13 + 91*a^2*b^14 + 14*a*b^15 + b 
^16))*tan(1/2*f*x + 1/2*e)^2 + 3*(2*a^13*b^2 + 23*a^12*b^3 + 119*a^11*b^4 
+ 363*a^10*b^5 + 715*a^9*b^6 + 924*a^8*b^7 + 726*a^7*b^8 + 198*a^6*b^9 - 2 
64*a^5*b^10 - 385*a^4*b^11 - 253*a^3*b^12 - 97*a^2*b^13 - 21*a*b^14 - 2*b^ 
15)/(a^14*b^2 + 14*a^13*b^3 + 91*a^12*b^4 + 364*a^11*b^5 + 1001*a^10*b^6 + 
 2002*a^9*b^7 + 3003*a^8*b^8 + 3432*a^7*b^9 + 3003*a^6*b^10 + 2002*a^5*b^1 
1 + 1001*a^4*b^12 + 364*a^3*b^13 + 91*a^2*b^14 + 14*a*b^15 + b^16))*tan...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right )^{2} b +a}\, \tan \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{6} b^{3}+3 \sin \left (f x +e \right )^{4} a \,b^{2}+3 \sin \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*tan(e + f*x)**3)/(sin(e + f*x)**6*b**3 + 
3*sin(e + f*x)**4*a*b**2 + 3*sin(e + f*x)**2*a**2*b + a**3),x)