3.5.91 \(\int \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [491]

Optimal. Leaf size=54 \[ -\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a+b \sin ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 54, 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 = {3273, 52, 65, 214} \begin {gather*} \frac {\sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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*} \int \cot (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac {\sqrt {a+b \sin ^2(e+f x)}}{f}+\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac {\sqrt {a+b \sin ^2(e+f x)}}{f}+\frac {a \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{b f}\\ &=-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a+b \sin ^2(e+f x)}}{f}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 53, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )-\sqrt {a+b \sin ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 8.72, size = 58, normalized size = 1.07

method	result	size
default	\(\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 45, normalized size = 0.83 \begin {gather*} -\frac {\sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - \sqrt {b \sin \left (f x + e\right )^{2} + a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.78, size = 135, normalized size = 2.50 \begin {gather*} \left [\frac {\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}}{2 \, f}, \frac {\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}}{f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \cot {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.51, size = 49, normalized size = 0.91 \begin {gather*} \frac {\frac {a \arctan \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \sqrt {b \sin \left (f x + e\right )^{2} + a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \mathrm {cot}\left (e+f\,x\right )\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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