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

Optimal. Leaf size=113 \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{a^{3/2} f}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{a^{3/2} f}-\frac {\cot (e+f x)}{a f \sqrt {a \sin (e+f x)+a}} \]

[Out]

3*arctanh(cos(f*x+e)*a^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/f-2*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*s
in(f*x+e))^(1/2))/a^(3/2)/f*2^(1/2)-cot(f*x+e)/a/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2715, 2985, 2649, 206, 2773} \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{a^{3/2} f}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{a^{3/2} f}-\frac {\cot (e+f x)}{a f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2715

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e + f
*x])^(m + 1)/(a*f*Tan[e + f*x]), x] + Dist[1/b^2, Int[((a + b*Sin[e + f*x])^(m + 1)*(b*m - a*(m + 1)*Sin[e + f
*x]))/Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && LtQ[m, -1]

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

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

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Mathematica [C]  time = 2.11, size = 206, normalized size = 1.82 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (-\cot \left (\frac {1}{4} (e+f x)\right )+(16+16 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )+2 \left (\sec \left (\frac {1}{2} (e+f x)\right )+\sin ^2\left (\frac {1}{4} (e+f x)\right ) \csc (e+f x)-\sin \left (\frac {3}{4} (e+f x)\right ) \sin \left (\frac {1}{4} (e+f x)\right ) \csc (e+f x)+3 \log \left (-\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )+1\right )-3 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )\right )\right )}{4 f (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*((16 + 16*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e +
f*x)/4])] - Cot[(e + f*x)/4] + 2*(3*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 3*Log[1 - Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2]] + Sec[(e + f*x)/2] + Csc[e + f*x]*Sin[(e + f*x)/4]^2 - Csc[e + f*x]*Sin[(e + f*x)/4]*Sin[
(3*(e + f*x))/4])))/(4*f*(a*(1 + Sin[e + f*x]))^(3/2))

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fricas [B]  time = 0.46, size = 421, normalized size = 3.73 \[ \frac {3 \, {\left (\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} - 9 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + \frac {4 \, \sqrt {2} {\left (a \cos \left (f x + e\right )^{2} - {\left (a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) - a\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {a}} + 4 \, \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{4 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*(cos(f*x + e)^2 - (cos(f*x + e) + 1)*sin(f*x + e) - 1)*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)
^2 + 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*sin(f*x + e) - 2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a
) - 9*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 8*a*cos(f*x + e) - a)*sin(f*x + e) - a)/(cos(f*x + e)^3 + cos(f*x +
e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) - 1)) + 4*sqrt(2)*(a*cos(f*x + e)^2 - (a*cos(f*x + e)
+ a)*sin(f*x + e) - a)*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(f*x + e)
+ a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin
(f*x + e) - cos(f*x + e) - 2))/sqrt(a) + 4*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1))/(a^2*f*
cos(f*x + e)^2 - a^2*f - (a^2*f*cos(f*x + e) + a^2*f)*sin(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2/f*(1/4*sqrt(a*tan((f*x+exp(1))/2)^2+a)/a^2/sign(tan((f*x+exp(1))/2)+1)+2*(-3/4*ata
n((-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))/sqrt(-a))/sqrt(-a)/a/sign(tan((f*x+exp(1))/2)
+1)+3/8*ln(abs(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a)))/sqrt(a)/a/sign(tan((f*x+exp(1))/
2)+1)+sqrt(2)*atan((-sqrt(a)*tan((f*x+exp(1))/2)-sqrt(a)+sqrt(a*tan((f*x+exp(1))/2)^2+a))/sqrt(2)/sqrt(-a))/sq
rt(-a)/a/sign(tan((f*x+exp(1))/2)+1)+1/4/sqrt(a)/((-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a
))^2-a)/sign(tan((f*x+exp(1))/2)+1))+(-3*sqrt(-a)*sqrt(2)*ln(sqrt(2)*sqrt(a)+sqrt(a))-sqrt(-a)*sqrt(2)-3*sqrt(
-a)*ln(sqrt(2)*sqrt(a)+sqrt(a))-3*sqrt(-a)+6*sqrt(2)*sqrt(a)*atan((sqrt(2)*sqrt(a)+sqrt(a))/sqrt(-a))-8*sqrt(2
)*sqrt(a)*atan(sqrt(a)/sqrt(-a))+6*sqrt(a)*atan((sqrt(2)*sqrt(a)+sqrt(a))/sqrt(-a))-16*sqrt(a)*atan(sqrt(a)/sq
rt(-a)))/(4*a*sqrt(-a)*sqrt(2)*sqrt(a)+4*a*sqrt(-a)*sqrt(a))*sign(tan((f*x+exp(1))/2)+1))

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maple [A]  time = 0.80, size = 134, normalized size = 1.19 \[ -\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sin \left (f x +e \right ) a^{2} \left (2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-3 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}}{\sqrt {a}}\right )\right )+\sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}}\right )}{a^{\frac {7}{2}} \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^(7/2)*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(sin(f*x+e)*a^2*(2*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1
/2)*2^(1/2)/a^(1/2))-3*arctanh((a-a*sin(f*x+e))^(1/2)/a^(1/2)))+(a-a*sin(f*x+e))^(1/2)*a^(3/2))/sin(f*x+e)/cos
(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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