\(\int \frac {\cot ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [467]

Optimal result

Integrand size = 25, antiderivative size = 143 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a+5 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {2 a+5 b}{2 a^3 f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

1/2*(2*a+5*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(7/2)/f-1/6*(2*a 
+5*b)/a^2/f/(a+b*sin(f*x+e)^2)^(3/2)-1/2*csc(f*x+e)^2/a/f/(a+b*sin(f*x+e)^ 
2)^(3/2)-1/2*(2*a+5*b)/a^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 = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.48 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {3 a \csc ^2(e+f x)+(2 a+5 b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \] Input:

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

Output:

-1/6*(3*a*Csc[e + f*x]^2 + (2*a + 5*b)*Hypergeometric2F1[-3/2, 1, -1/2, 1 
+ (b*Sin[e + f*x]^2)/a])/(a^2*f*(a + b*Sin[e + f*x]^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3673

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-(Csc[e + f*x]^2/(a*(a + b*Sin[e + f*x]^2)^(3/2))) - ((2*a + 5*b)*(2/(3*a 
*(a + b*Sin[e + f*x]^2)^(3/2)) + ((-2*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/S 
qrt[a]])/a^(3/2) + 2/(a*Sqrt[a + b*Sin[e + f*x]^2]))/a))/(2*a))/(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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1037\) vs. \(2(123)=246\).

Time = 0.60 (sec) , antiderivative size = 1038, normalized size of antiderivative = 7.26

method	result	size
default	\(\text {Expression too large to display}\)	\(1038\)

Input:

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

Output:

1/6/a^(13/2)/b^2/(cos(f*x+e)^6*b^2-2*cos(f*x+e)^4*a*b-3*cos(f*x+e)^4*b^2+c 
os(f*x+e)^2*a^2+4*cos(f*x+e)^2*a*b+3*cos(f*x+e)^2*b^2-a^2-2*a*b-b^2)*(3*(a 
+b-b*cos(f*x+e)^2)^(1/2)*a^(11/2)*b^2-6*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*co 
s(f*x+e)^2)^(1/2)+a))*a^6*b^2+3*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(7/2)*b^4+8*( 
-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(11/2)*b^2+20*(-b*cos(f*x+e)^2+(a 
*b^2+b^3)/b^2)^(1/2)*a^(9/2)*b^3+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*b^3+ 
12*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(7/2)*b^4-27*ln(2/sin(f*x+e)* 
(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^5*b^3-36*ln(2/sin(f*x+e)*(a^(1/2 
)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^4*b^4-15*ln(2/sin(f*x+e)*(a^(1/2)*(a+b- 
b*cos(f*x+e)^2)^(1/2)+a))*a^3*b^5+3*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f* 
x+e)^2)^(1/2)+a))*a^3*b^4*(5*b+2*a)*cos(f*x+e)^6+3*cos(f*x+e)^4*b^3*(2*(-b 
*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(9/2)+4*(-b*cos(f*x+e)^2+(a*b^2+b^3 
)/b^2)^(1/2)*a^(7/2)*b+(a+b-b*cos(f*x+e)^2)^(1/2)*a^(7/2)*b-4*ln(2/sin(f*x 
+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^5-16*ln(2/sin(f*x+e)*(a^(1/2 
)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^4*b-15*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b* 
cos(f*x+e)^2)^(1/2)+a))*a^3*b^2)-cos(f*x+e)^2*b^2*(8*(-b*cos(f*x+e)^2+(a*b 
^2+b^3)/b^2)^(1/2)*a^(11/2)+26*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^( 
9/2)*b+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*b+24*(-b*cos(f*x+e)^2+(a*b^2+b 
^3)/b^2)^(1/2)*a^(7/2)*b^2+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(7/2)*b^2-6*ln(2 
/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^6-39*ln(2/sin(f*x...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (123) = 246\).

Time = 0.17 (sec) , antiderivative size = 681, normalized size of antiderivative = 4.76 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/12*(3*((2*a*b^2 + 5*b^3)*cos(f*x + e)^6 - (4*a^2*b + 16*a*b^2 + 15*b^3) 
*cos(f*x + e)^4 - 2*a^3 - 9*a^2*b - 12*a*b^2 - 5*b^3 + (2*a^3 + 13*a^2*b + 
 26*a*b^2 + 15*b^3)*cos(f*x + e)^2)*sqrt(a)*log(2*(b*cos(f*x + e)^2 - 2*sq 
rt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1)) + 2 
*(3*(2*a^2*b + 5*a*b^2)*cos(f*x + e)^4 + 11*a^3 + 26*a^2*b + 15*a*b^2 - 2* 
(4*a^3 + 16*a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + 
 b))/(a^4*b^2*f*cos(f*x + e)^6 - (2*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^4 + 
(a^6 + 4*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^2 - (a^6 + 2*a^5*b + a^4*b^2)*f 
), 1/6*(3*((2*a*b^2 + 5*b^3)*cos(f*x + e)^6 - (4*a^2*b + 16*a*b^2 + 15*b^3 
)*cos(f*x + e)^4 - 2*a^3 - 9*a^2*b - 12*a*b^2 - 5*b^3 + (2*a^3 + 13*a^2*b 
+ 26*a*b^2 + 15*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^ 
2 + a + b)*sqrt(-a)/(b*cos(f*x + e)^2 - a - b)) + (3*(2*a^2*b + 5*a*b^2)*c 
os(f*x + e)^4 + 11*a^3 + 26*a^2*b + 15*a*b^2 - 2*(4*a^3 + 16*a^2*b + 15*a* 
b^2)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^4*b^2*f*cos(f*x + 
 e)^6 - (2*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^4 + (a^6 + 4*a^5*b + 3*a^4*b^ 
2)*f*cos(f*x + e)^2 - (a^6 + 2*a^5*b + a^4*b^2)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {6 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {15 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} - \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} - \frac {15 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {5 \, b}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {3}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )^{2}}}{6 \, f} \] Input:

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

Output:

1/6*(6*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2) + 15*b*arcsinh(a/( 
sqrt(a*b)*abs(sin(f*x + e))))/a^(7/2) - 6/(sqrt(b*sin(f*x + e)^2 + a)*a^2) 
 - 2/((b*sin(f*x + e)^2 + a)^(3/2)*a) - 15*b/(sqrt(b*sin(f*x + e)^2 + a)*a 
^3) - 5*b/((b*sin(f*x + e)^2 + a)^(3/2)*a^2) - 3/((b*sin(f*x + e)^2 + a)^( 
3/2)*a*sin(f*x + e)^2))/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(e + f*x)**2*b + a)*cot(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)