\(\int \frac {\cot ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [525]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 110 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {2 a+3 b}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^2(e+f x)}{2 a f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 53, 65, 214} \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {2 a+3 b}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^2(e+f x)}{2 a f \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

((2*a + 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*a^(5/2)*f) - (2*a + 3*b)/(2*a^2*f*Sqrt[a + b*Sin[
e + f*x]^2]) - Csc[e + f*x]^2/(2*a*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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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}{x^2 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = -\frac {\csc ^2(e+f x)}{2 a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+3 b) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f} \\ & = -\frac {2 a+3 b}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^2(e+f x)}{2 a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+3 b) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 a^2 f} \\ & = -\frac {2 a+3 b}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^2(e+f x)}{2 a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+3 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{2 a^2 b f} \\ & = \frac {(2 a+3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {2 a+3 b}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^2(e+f x)}{2 a 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 = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.64 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-a \csc ^2(e+f x)-(2 a+3 b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sin ^2(e+f x)}{a}\right )}{2 a^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.35

method result size
default \(\frac {-\frac {1}{a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}-\frac {1}{2 a \sin \left (f x +e \right )^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {3 b}{2 a^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2 a^{\frac {5}{2}}}}{f}\) \(148\)

[In]

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

[Out]

(-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))-1/2/a/sin(f*x
+e)^2/(a+b*sin(f*x+e)^2)^(1/2)-3/2/a^2*b/(a+b*sin(f*x+e)^2)^(1/2)+3/2/a^(5/2)*b*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. 190 vs. \(2 (94) = 188\).

Time = 0.38 (sec) , antiderivative size = 406, normalized size of antiderivative = 3.69 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{2} + 7 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, {\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{4 \, {\left (a^{3} b f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + a^{3} b\right )} f\right )}}, -\frac {{\left ({\left (2 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (2 \, a^{2} + 7 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, {\left (a^{3} b f \cos \left (f x + e\right )^{4} - {\left (a^{4} + 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + a^{3} b\right )} f\right )}}\right ] \]

[In]

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

[Out]

[1/4*(((2*a*b + 3*b^2)*cos(f*x + e)^4 - (2*a^2 + 7*a*b + 6*b^2)*cos(f*x + e)^2 + 2*a^2 + 5*a*b + 3*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*((2
*a^2 + 3*a*b)*cos(f*x + e)^2 - 3*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b*f*cos(f*x + e)^4 - (a^4
+ 2*a^3*b)*f*cos(f*x + e)^2 + (a^4 + a^3*b)*f), -1/2*(((2*a*b + 3*b^2)*cos(f*x + e)^4 - (2*a^2 + 7*a*b + 6*b^2
)*cos(f*x + e)^2 + 2*a^2 + 5*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - ((2*a^
2 + 3*a*b)*cos(f*x + e)^2 - 3*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*b*f*cos(f*x + e)^4 - (a^4 + 2
*a^3*b)*f*cos(f*x + e)^2 + (a^4 + a^3*b)*f)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \frac {\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {2 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {3 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{2}}}{2 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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