\(\int \frac {\tan (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [465]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2} f}-\frac {1}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {1}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f-1/3/(a+b)/f/(a 
+b*sin(f*x+e)^2)^(3/2)-1/(a+b)^2/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 = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1-\frac {b \cos ^2(e+f x)}{a+b}\right )}{3 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3673, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-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*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 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. \(962\) vs. \(2(79)=158\).

Time = 0.57 (sec) , antiderivative size = 963, normalized size of antiderivative = 10.58

method	result	size
default	\(\frac {3 \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 ) a^{2} b^{5} \cos \left (f x +e \right )^{4}+3 \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 ) a^{2} b^{5} \cos \left (f x +e \right )^{4}+6 \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {b^{2} a +b^{3}}{b^{2}}}\, \sqrt {a +b}\, a^{2} b^{4} \cos \left (f x +e \right )^{2}-6 \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 ) a^{3} b^{4} \cos \left (f x +e \right )^{2}-6 \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 ) a^{2} b^{5} \cos \left (f x +e \right )^{2}-6 \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 ) a^{3} b^{4} \cos \left (f x +e \right )^{2}-6 \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 ) a^{2} b^{5} \cos \left (f x +e \right )^{2}-8 \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {b^{2} a +b^{3}}{b^{2}}}\, \sqrt {a +b}\, a^{3} b^{3}-8 \sqrt {-b \cos \left (f x +e \right )^{2}+\frac {b^{2} a +b^{3}}{b^{2}}}\, \sqrt {a +b}\, a^{2} b^{4}+3 \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 ) a^{4} b^{3}+6 \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 ) a^{3} b^{4}+3 \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 ) a^{2} b^{5}+3 \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 ) a^{4} b^{3}+6 \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 ) a^{3} b^{4}+3 \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 ) a^{2} b^{5}}{6 b^{3} \sqrt {a +b}\, a^{2} \left (a^{2} b^{2} \cos \left (f x +e \right )^{4}+2 a \,b^{3} \cos \left (f x +e \right )^{4}+b^{4} \cos \left (f x +e \right )^{4}-2 a^{3} b \cos \left (f x +e \right )^{2}-6 a^{2} b^{2} \cos \left (f x +e \right )^{2}-6 a \,b^{3} \cos \left (f x +e \right )^{2}-2 b^{4} \cos \left (f x +e \right )^{2}+a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) f}\)	\(963\)

Input:

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

Output:

1/6/b^3/(a+b)^(1/2)/a^2/(a^2*b^2*cos(f*x+e)^4+2*a*b^3*cos(f*x+e)^4+b^4*cos 
(f*x+e)^4-2*a^3*b*cos(f*x+e)^2-6*a^2*b^2*cos(f*x+e)^2-6*a*b^3*cos(f*x+e)^2 
-2*b^4*cos(f*x+e)^2+a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(3*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))*a^2*b^5*cos(f 
*x+e)^4+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*si 
n(f*x+e)+a))*a^2*b^5*cos(f*x+e)^4+6*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2 
)*(a+b)^(1/2)*a^2*b^4*cos(f*x+e)^2-6*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))*a^3*b^4*cos(f*x+e)^2-6*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))*a^2*b^5* 
cos(f*x+e)^2-6*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^3*b^4*cos(f*x+e)^2-6*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^2*b^5*cos(f*x+e)^2-8*(-b*cos 
(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*a^3*b^3-8*(-b*cos(f*x+e)^2+(a 
*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)*a^2*b^4+3*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))*a^4*b^3+6*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))*a^3*b^4+3*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))*a^ 
2*b^5+3*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^4*b^3+6*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^3*b^4+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (79) = 158\).

Time = 0.17 (sec) , antiderivative size = 534, normalized size of antiderivative = 5.87 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 8 \, a b - 4 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left ({\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f\right )}}, \frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{b \cos \left (f x + e\right )^{2} - a - b}\right ) + {\left (3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 8 \, a b - 4 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left ({\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \] Input:

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

Output:

[1/6*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + 
 b^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*(a*b + b^2)*cos(f*x + e) 
^2 - 4*a^2 - 8*a*b - 4*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b^2 + 3 
*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*x + e)^4 - 2*(a^4*b + 4*a^3*b^2 + 6*a^2* 
b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a 
^2*b^3 + 5*a*b^4 + b^5)*f), 1/3*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos 
(f*x + e)^2 + a^2 + 2*a*b + b^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*(a*b + b^2)*cos(f 
*x + e)^2 - 4*a^2 - 8*a*b - 4*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3* 
b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*x + e)^4 - 2*(a^4*b + 4*a^3*b^2 + 
 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 
 + 10*a^2*b^3 + 5*a*b^4 + b^5)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (79) = 158\).

Time = 0.15 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.23 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a b + \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}} + \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}} - \frac {3 \, \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}}}{6 \, f} \] Input:

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

Output:

-1/6*(2/((b*sin(f*x + e)^2 + a)^(3/2)*a + (b*sin(f*x + e)^2 + a)^(3/2)*b) 
+ 6/(sqrt(b*sin(f*x + e)^2 + a)*a^2 + 2*sqrt(b*sin(f*x + e)^2 + a)*a*b + s 
qrt(b*sin(f*x + e)^2 + a)*b^2) + 3*arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(sin( 
f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1)))/(a + b)^(5/2) - 3*arcsi 
nh(-b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x 
+ e) - 1)))/(a + b)^(5/2))/f
                                                                                    
                                                                                    
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 832 vs. \(2 (79) = 158\).

Time = 0.64 (sec) , antiderivative size = 832, normalized size of antiderivative = 9.14 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/3*(((((4*a^9*b^2 + 33*a^8*b^3 + 120*a^7*b^4 + 252*a^6*b^5 + 336*a^5*b^6 
 + 294*a^4*b^7 + 168*a^3*b^8 + 60*a^2*b^9 + 12*a*b^10 + b^11)*tan(1/2*f*x 
+ 1/2*e)^2/(a^10*b^2 + 10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b^6 
 + 252*a^5*b^7 + 210*a^4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^1 
2) + 3*(4*a^9*b^2 + 37*a^8*b^3 + 152*a^7*b^4 + 364*a^6*b^5 + 560*a^5*b^6 + 
 574*a^4*b^7 + 392*a^3*b^8 + 172*a^2*b^9 + 44*a*b^10 + 5*b^11)/(a^10*b^2 + 
 10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b^6 + 252*a^5*b^7 + 210*a 
^4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^12))*tan(1/2*f*x + 1/2* 
e)^2 + 3*(4*a^9*b^2 + 37*a^8*b^3 + 152*a^7*b^4 + 364*a^6*b^5 + 560*a^5*b^6 
 + 574*a^4*b^7 + 392*a^3*b^8 + 172*a^2*b^9 + 44*a*b^10 + 5*b^11)/(a^10*b^2 
 + 10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b^6 + 252*a^5*b^7 + 210 
*a^4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^12))*tan(1/2*f*x + 1/ 
2*e)^2 + (4*a^9*b^2 + 33*a^8*b^3 + 120*a^7*b^4 + 252*a^6*b^5 + 336*a^5*b^6 
 + 294*a^4*b^7 + 168*a^3*b^8 + 60*a^2*b^9 + 12*a*b^10 + b^11)/(a^10*b^2 + 
10*a^9*b^3 + 45*a^8*b^4 + 120*a^7*b^5 + 210*a^6*b^6 + 252*a^5*b^7 + 210*a^ 
4*b^8 + 120*a^3*b^9 + 45*a^2*b^10 + 10*a*b^11 + b^12))/(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)^(3/2 
) + 6*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) - s 
qrt(a))/sqrt(-a - b))/((a^2 + 2*a*b + b^2)*sqrt(-a - b)))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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