\(\int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [514]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3273, 79, 65, 214} \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a f} \]

[In]

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

[Out]

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

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 \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = -\frac {\csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a f}-\frac {(2 a+b) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f} \\ & = -\frac {\csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a f}-\frac {(2 a+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 b f} \\ & = \frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a}}{2 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.45

method	result	size
default	\(\frac {\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{\sqrt {a}}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2 a \sin \left (f x +e \right )^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2 a^{\frac {3}{2}}}}{f}\)	\(109\)

[In]

int(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(1/a^(1/2)*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)
+1/2*b/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e)))/f

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.93 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}, -\frac {{\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a}{2 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f\right )}}\right ] \]

[In]

integrate(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(((2*a + b)*cos(f*x + e)^2 - 2*a - b)*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*sqrt(-b*cos(f*x + e)^2 + a + b)*a)/(a^2*f*cos(f*x + e)^2 - a^2*f
), -1/2*(((2*a + b)*cos(f*x + e)^2 - 2*a - b)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - sq
rt(-b*cos(f*x + e)^2 + a + b)*a)/(a^2*f*cos(f*x + e)^2 - a^2*f)]

Sympy [F]

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

[In]

integrate(cot(f*x+e)**3/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\frac {2 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} + \frac {b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}}}{2 \, f} \]

[In]

integrate(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

int(cot(e + f*x)^3/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(cot(e + f*x)^3/(a + b*sin(e + f*x)^2)^(1/2), x)