\(\int \frac {\cot (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [535]

Optimal result

Integrand size = 23, antiderivative size = 83 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {1}{a^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f+1/3/a/f/(a+b*sin(f*x+e)^2)^(3/2)+1/a^2/f/(a+b*sin(f*x+e)^
2)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3273, 53, 65, 214} \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {1}{a^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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 3273

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[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] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 a f} \\ & = \frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {1}{a^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 a^2 f} \\ & = \frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {1}{a^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{a^2 b f} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}+\frac {1}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {1}{a^2 f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.59 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(259\) vs. \(2(71)=142\).

Time = 1.64 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.13

method	result	size
default	\(\frac {-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {5}{2}}}-\frac {7 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 a^{2} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}+\frac {7 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 a^{2} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}-\frac {\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 a^{2} b \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{12 a^{2} b \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )^{2}}}{f}\)	\(260\)

[In]

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

[Out]

(-1/a^(5/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-7/12/a^2/(-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)+7/12/a^2/(-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)-1/12/a^2/b/(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/a^2/b/(s
in(f*x+e)+(-a*b)^(1/2)/b)^2*(-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. 178 vs. \(2 (71) = 142\).

Time = 0.35 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.60 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (3 \, a b \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}, \frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (3 \, a b \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left (a^{3} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}}\right ] \]

[In]

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

[Out]

[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)*log(2*(b*cos(f*x + e)^
2 + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1)) - 2*(3*a*b*cos(f*x + e)^2 - 4*a
^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b^2*f*cos(f*x + e)^4 - 2*(a^4*b + a^3*b^2)*f*cos(f*x + e)^2
+ (a^5 + 2*a^4*b + a^3*b^2)*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)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (3*a*b*cos(f*x + e)^2 - 4*a^2 - 3*a*b)*sqrt(-b*
cos(f*x + e)^2 + a + b))/(a^3*b^2*f*cos(f*x + e)^4 - 2*(a^4*b + a^3*b^2)*f*cos(f*x + e)^2 + (a^5 + 2*a^4*b + a
^3*b^2)*f)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {\cot (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {3}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {1}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a}}{3 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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