\(\int \frac {\cot ^5(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [457]

Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (8 a^2+24 a b+15 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{7/2} f}+\frac {8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

-1/8*(8*a^2+24*a*b+15*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(7/ 
2)/f+1/8*(8*a^2+24*a*b+15*b^2)/a^3/f/(a+b*sin(f*x+e)^2)^(1/2)+1/8*(8*a+5*b 
)*csc(f*x+e)^2/a^2/f/(a+b*sin(f*x+e)^2)^(1/2)-1/4*csc(f*x+e)^4/a/f/(a+b*si 
n(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 = 0.36 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.56 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {a \csc ^2(e+f x) \left (8 a+5 b-2 a \csc ^2(e+f x)\right )+\left (8 a^2+24 a b+15 b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}} \] Input:

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

Output:

(a*Csc[e + f*x]^2*(8*a + 5*b - 2*a*Csc[e + f*x]^2) + (8*a^2 + 24*a*b + 15* 
b^2)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sin[e + f*x]^2)/a])/(8*a^3*f*S 
qrt[a + b*Sin[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.95, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

\(\displaystyle \frac {\int \frac {\csc ^6(e+f x) \left (1-\sin ^2(e+f x)\right )^2}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 f}\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {\frac {\int -\frac {\csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a+5 b\right )}{2 \left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 a}-\frac {\csc ^4(e+f x)}{2 a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\csc ^4(e+f x) \left (-4 a \sin ^2(e+f x)+8 a+5 b\right )}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{4 a}-\frac {\csc ^4(e+f x)}{2 a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+3 b (8 a+5 b)\right ) \int \frac {\csc ^2(e+f x)}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin ^2(e+f x)}{2 a}-\frac {(8 a+5 b) \csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{4 a}-\frac {\csc ^4(e+f x)}{2 a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+3 b (8 a+5 b)\right ) \left (\frac {\int \frac {\csc ^2(e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}d\sin ^2(e+f x)}{a}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 a}-\frac {(8 a+5 b) \csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{4 a}-\frac {\csc ^4(e+f x)}{2 a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+3 b (8 a+5 b)\right ) \left (\frac {2 \int \frac {1}{\frac {\sin ^4(e+f x)}{b}-\frac {a}{b}}d\sqrt {b \sin ^2(e+f x)+a}}{a b}+\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}\right )}{2 a}-\frac {(8 a+5 b) \csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{4 a}-\frac {\csc ^4(e+f x)}{2 a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {-\frac {\left (8 a^2+3 b (8 a+5 b)\right ) \left (\frac {2}{a \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {(8 a+5 b) \csc ^2(e+f x)}{a \sqrt {a+b \sin ^2(e+f x)}}}{4 a}-\frac {\csc ^4(e+f x)}{2 a \sqrt {a+b \sin ^2(e+f x)}}}{2 f}\)

Input:

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

Output:

(-1/2*Csc[e + f*x]^4/(a*Sqrt[a + b*Sin[e + f*x]^2]) - (-(((8*a + 5*b)*Csc[ 
e + f*x]^2)/(a*Sqrt[a + b*Sin[e + f*x]^2])) - ((8*a^2 + 3*b*(8*a + 5*b))*( 
(-2*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b 
*Sin[e + f*x]^2])))/(2*a))/(4*a))/(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 [A] (verified)

Time = 0.54 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {-\frac {1}{4 a \sin \left (f x +e \right )^{4} \sqrt {a +b \sin \left (f x +e \right )^{2}}}+\frac {5 b}{8 a^{2} \sin \left (f x +e \right )^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}+\frac {15 b^{2}}{8 a^{3} \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {15 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{8 a^{\frac {7}{2}}}+\frac {1}{a \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}+\frac {1}{a \sin \left (f x +e \right )^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}+\frac {3 b}{a^{2} \sqrt {a +b \sin \left (f x +e \right )^{2}}}-\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \sin \left (f x +e \right )^{2}}}{\sin \left (f x +e \right )}\right )}{a^{\frac {5}{2}}}}{f}\)	\(265\)

Input:

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

Output:

(-1/4/a/sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2)+5/8*b/a^2/sin(f*x+e)^2/(a+b* 
sin(f*x+e)^2)^(1/2)+15/8*b^2/a^3/(a+b*sin(f*x+e)^2)^(1/2)-15/8*b^2/a^(7/2) 
*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))+1/a/(a+b*sin(f*x+ 
e)^2)^(1/2)-1/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+ 
e))+1/a/sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)+3*b/a^2/(a+b*sin(f*x+e)^2)^( 
1/2)-3*b/a^(5/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e)))/ 
f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (147) = 294\).

Time = 0.18 (sec) , antiderivative size = 667, normalized size of antiderivative = 3.99 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(((8*a^2*b + 24*a*b^2 + 15*b^3)*cos(f*x + e)^6 - (8*a^3 + 48*a^2*b + 
 87*a*b^2 + 45*b^3)*cos(f*x + e)^4 - 8*a^3 - 32*a^2*b - 39*a*b^2 - 15*b^3 
+ (16*a^3 + 72*a^2*b + 102*a*b^2 + 45*b^3)*cos(f*x + e)^2)*sqrt(a)*log(2*( 
b*cos(f*x + e)^2 + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(c 
os(f*x + e)^2 - 1)) - 2*((8*a^3 + 24*a^2*b + 15*a*b^2)*cos(f*x + e)^4 + 14 
*a^3 + 29*a^2*b + 15*a*b^2 - (24*a^3 + 53*a^2*b + 30*a*b^2)*cos(f*x + e)^2 
)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^4*b*f*cos(f*x + e)^6 - (a^5 + 3*a^4* 
b)*f*cos(f*x + e)^4 + (2*a^5 + 3*a^4*b)*f*cos(f*x + e)^2 - (a^5 + a^4*b)*f 
), -1/8*(((8*a^2*b + 24*a*b^2 + 15*b^3)*cos(f*x + e)^6 - (8*a^3 + 48*a^2*b 
 + 87*a*b^2 + 45*b^3)*cos(f*x + e)^4 - 8*a^3 - 32*a^2*b - 39*a*b^2 - 15*b^ 
3 + (16*a^3 + 72*a^2*b + 102*a*b^2 + 45*b^3)*cos(f*x + e)^2)*sqrt(-a)*arct 
an(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/(b*cos(f*x + e)^2 - a - b)) + 
((8*a^3 + 24*a^2*b + 15*a*b^2)*cos(f*x + e)^4 + 14*a^3 + 29*a^2*b + 15*a*b 
^2 - (24*a^3 + 53*a^2*b + 30*a*b^2)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 
 + a + b))/(a^4*b*f*cos(f*x + e)^6 - (a^5 + 3*a^4*b)*f*cos(f*x + e)^4 + (2 
*a^5 + 3*a^4*b)*f*cos(f*x + e)^2 - (a^5 + a^4*b)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.31 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {8 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {24 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {15 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} - \frac {8}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {24 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {15 \, b^{2}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {8}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{2}} - \frac {5 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} \sin \left (f x + e\right )^{2}} + \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{4}}}{8 \, f} \] Input:

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

Output:

-1/8*(8*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) + 24*b*arcsinh(a/ 
(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2) + 15*b^2*arcsinh(a/(sqrt(a*b)*abs(s 
in(f*x + e))))/a^(7/2) - 8/(sqrt(b*sin(f*x + e)^2 + a)*a) - 24*b/(sqrt(b*s 
in(f*x + e)^2 + a)*a^2) - 15*b^2/(sqrt(b*sin(f*x + e)^2 + a)*a^3) - 8/(sqr 
t(b*sin(f*x + e)^2 + a)*a*sin(f*x + e)^2) - 5*b/(sqrt(b*sin(f*x + e)^2 + a 
)*a^2*sin(f*x + e)^2) + 2/(sqrt(b*sin(f*x + e)^2 + a)*a*sin(f*x + e)^4))/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

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