3.110 \(\int \frac {\cot ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{8 a^{3/2} f}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a \sin (e+f x)+a}}{3 a^2 f}-\frac {\cot (e+f x)}{8 a f \sqrt {a \sin (e+f x)+a}}+\frac {11 \cot (e+f x) \csc (e+f x)}{12 a f \sqrt {a \sin (e+f x)+a}} \]

[Out]

-1/8*arctanh(cos(f*x+e)*a^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/f-1/8*cot(f*x+e)/a/f/(a+a*sin(f*x+e))^(1/2)+11
/12*cot(f*x+e)*csc(f*x+e)/a/f/(a+a*sin(f*x+e))^(1/2)-1/3*cot(f*x+e)*csc(f*x+e)^2*(a+a*sin(f*x+e))^(1/2)/a^2/f

________________________________________________________________________________________

Rubi [A]  time = 0.55, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2717, 2772, 2773, 206, 3044, 2980} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{8 a^{3/2} f}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a \sin (e+f x)+a}}{3 a^2 f}-\frac {\cot (e+f x)}{8 a f \sqrt {a \sin (e+f x)+a}}+\frac {11 \cot (e+f x) \csc (e+f x)}{12 a f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

-ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]]/(8*a^(3/2)*f) - Cot[e + f*x]/(8*a*f*Sqrt[a + a*Sin[e
+ f*x]]) + (11*Cot[e + f*x]*Csc[e + f*x])/(12*a*f*Sqrt[a + a*Sin[e + f*x]]) - (Cot[e + f*x]*Csc[e + f*x]^2*Sq
rt[a + a*Sin[e + f*x]])/(3*a^2*f)

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 2717

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Dist[-2/(a*b), Int[(a
+ b*Sin[e + f*x])^(m + 2)/Sin[e + f*x]^3, x], x] + Dist[1/a^2, Int[((a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e +
f*x]^2))/Sin[e + f*x]^4, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && LtQ[m,
-1]

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

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 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rubi steps

\begin {align*} \int \frac {\cot ^4(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx &=\frac {\int \csc ^4(e+f x) \sqrt {a+a \sin (e+f x)} \left (1+\sin ^2(e+f x)\right ) \, dx}{a^2}-\frac {2 \int \csc ^3(e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{a^2}\\ &=\frac {\cot (e+f x) \csc (e+f x)}{a f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{3 a^2 f}+\frac {\int \csc ^3(e+f x) \sqrt {a+a \sin (e+f x)} \left (\frac {a}{2}+\frac {9}{2} a \sin (e+f x)\right ) \, dx}{3 a^3}-\frac {3 \int \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{2 a^2}\\ &=\frac {3 \cot (e+f x)}{2 a f \sqrt {a+a \sin (e+f x)}}+\frac {11 \cot (e+f x) \csc (e+f x)}{12 a f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{3 a^2 f}-\frac {3 \int \csc (e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{4 a^2}+\frac {13 \int \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{8 a^2}\\ &=-\frac {\cot (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)}}+\frac {11 \cot (e+f x) \csc (e+f x)}{12 a f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{3 a^2 f}+\frac {13 \int \csc (e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{16 a^2}+\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 )}{2 a f}\\ &=\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{2 a^{3/2} f}-\frac {\cot (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)}}+\frac {11 \cot (e+f x) \csc (e+f x)}{12 a f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{3 a^2 f}-\frac {13 \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 )}{8 a f}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{8 a^{3/2} f}-\frac {\cot (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)}}+\frac {11 \cot (e+f x) \csc (e+f x)}{12 a f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x) \sqrt {a+a \sin (e+f x)}}{3 a^2 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.76, size = 294, normalized size = 2.04 \[ \frac {\csc ^9\left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (132 \sin \left (\frac {1}{2} (e+f x)\right )+62 \sin \left (\frac {3}{2} (e+f x)\right )-6 \sin \left (\frac {5}{2} (e+f x)\right )-132 \cos \left (\frac {1}{2} (e+f x)\right )+62 \cos \left (\frac {3}{2} (e+f x)\right )+6 \cos \left (\frac {5}{2} (e+f x)\right )-9 \sin (e+f x) \log \left (-\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )+1\right )+9 \sin (e+f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )+3 \sin (3 (e+f x)) \log \left (-\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )+1\right )-3 \sin (3 (e+f x)) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )\right )}{24 f (a (\sin (e+f x)+1))^{3/2} \left (\csc ^2\left (\frac {1}{4} (e+f x)\right )-\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(3/2),x]

[Out]

(Csc[(e + f*x)/2]^9*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-132*Cos[(e + f*x)/2] + 62*Cos[(3*(e + f*x))/2] +
6*Cos[(5*(e + f*x))/2] + 132*Sin[(e + f*x)/2] - 9*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[e + f*x] +
9*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] + 62*Sin[(3*(e + f*x))/2] - 6*Sin[(5*(e + f*x))/2
] + 3*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[3*(e + f*x)] - 3*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*
x)/2]]*Sin[3*(e + f*x)]))/(24*f*(Csc[(e + f*x)/4]^2 - Sec[(e + f*x)/4]^2)^3*(a*(1 + Sin[e + f*x]))^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 383, normalized size = 2.66 \[ \frac {3 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} - 9 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (f x + e\right )^{3} + 17 \, \cos \left (f x + e\right )^{2} - {\left (3 \, \cos \left (f x + e\right )^{2} - 14 \, \cos \left (f x + e\right ) - 25\right )} \sin \left (f x + e\right ) - 11 \, \cos \left (f x + e\right ) - 25\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{96 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f - {\left (a^{2} f \cos \left (f x + e\right )^{3} + a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/96*(3*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 - (cos(f*x + e)^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*(3*cos(f*x + e)^3 + 17*cos(f*x + e)^2 - (3*cos(f*x + e)^2 - 14*cos(f*x + e) - 25)*sin(f*x + e) - 11
*cos(f*x + e) - 25)*sqrt(a*sin(f*x + e) + a))/(a^2*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f - (a^2*f*
cos(f*x + e)^3 + a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - a^2*f)*sin(f*x + e))

________________________________________________________________________________________

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)^4/(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)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*(2*sqrt(a*tan((f*x+exp(1))/2)^2+a)*(
tan((f*x+exp(1))/2)*(1/96*tan((f*x+exp(1))/2)/a^2/sign(tan((f*x+exp(1))/2)+1)-3/64/a^2/sign(tan((f*x+exp(1))/2
)+1))+7/96/a^2/sign(tan((f*x+exp(1))/2)+1))+2*(1/96*(9*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2
)^2+a))^5+18*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-9*a^2*(-sqrt(a)*tan((f*x
+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-24*sqrt(a)*a*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1)
)/2)^2+a))^2+14*sqrt(a)*a^2)/((-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-a)^3/a/sign(tan
((f*x+exp(1))/2)+1)+1/32*atan((-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)-1/64*ln(abs(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a)))/sqr
t(a)/a/sign(tan((f*x+exp(1))/2)+1))+(15*sqrt(-a)*sqrt(2)*ln(sqrt(2)*sqrt(a)+sqrt(a))-280*sqrt(-a)*sqrt(2)+21*s
qrt(-a)*ln(sqrt(2)*sqrt(a)+sqrt(a))-402*sqrt(-a)-30*sqrt(2)*sqrt(a)*atan((sqrt(2)*sqrt(a)+sqrt(a))/sqrt(-a))-4
2*sqrt(a)*atan((sqrt(2)*sqrt(a)+sqrt(a))/sqrt(-a)))/(480*a*sqrt(-a)*sqrt(2)*sqrt(a)+672*a*sqrt(-a)*sqrt(a))*si
gn(tan((f*x+exp(1))/2)+1))

________________________________________________________________________________________

maple [A]  time = 0.89, size = 144, normalized size = 1.00 \[ -\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (3 \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}+3 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{\sqrt {a}}\right ) a^{4} \left (\sin ^{3}\left (f x +e \right )\right )+8 \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-3 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, a^{\frac {7}{2}}\right )}{24 a^{\frac {11}{2}} \sin \left (f x +e \right )^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+a*sin(f*x+e))^(3/2),x)

[Out]

-1/24/a^(11/2)*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(3*(-a*(sin(f*x+e)-1))^(5/2)*a^(3/2)+3*arctanh((-a*(si
n(f*x+e)-1))^(1/2)/a^(1/2))*a^4*sin(f*x+e)^3+8*(-a*(sin(f*x+e)-1))^(3/2)*a^(5/2)-3*(-a*(sin(f*x+e)-1))^(1/2)*a
^(7/2))/sin(f*x+e)^3/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(3/2),x)

[Out]

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4/(a+a*sin(f*x+e))**(3/2),x)

[Out]

Integral(cot(e + f*x)**4/(a*(sin(e + f*x) + 1))**(3/2), x)

________________________________________________________________________________________