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

Optimal. Leaf size=57 \[ \frac {1}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3194, 51, 63, 208} \[ \frac {1}{a f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3194

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

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Mathematica [C]  time = 0.05, size = 46, normalized size = 0.81 \[ \frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \sin ^2(e+f x)}{a}+1\right )}{a f \sqrt {a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.57, size = 225, normalized size = 3.95 \[ \left [\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a}{2 \, {\left (a^{2} b f \cos \left (f x + e\right )^{2} - {\left (a^{3} + a^{2} b\right )} f\right )}}, \frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a}{a^{2} b f \cos \left (f x + e\right )^{2} - {\left (a^{3} + a^{2} b\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 57, normalized size = 1.00 \[ \frac {\arctan \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a f} + \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.58, size = 64, normalized size = 1.12 \[ \frac {1}{a f \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f \,a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 46, normalized size = 0.81 \[ -\frac {\frac {\operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - \frac {1}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {cot}\left (e+f\,x\right )}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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