3.114 \(\int \frac {\cot ^4(e+f x)}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=191 \[ \frac {45 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{8 a^{5/2} f}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{a^{5/2} f}-\frac {19 \cot (e+f x)}{8 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a \sin (e+f x)+a}} \]

[Out]

45/8*arctanh(cos(f*x+e)*a^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/f-4*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+
a*sin(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)-19/8*cot(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)+13/12*cot(f*x+e)*csc(f*x+e
)/a^2/f/(a+a*sin(f*x+e))^(1/2)-1/3*cot(f*x+e)*csc(f*x+e)^2/a^2/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.96, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2717, 2779, 2984, 2985, 2649, 206, 2773, 3044} \[ -\frac {19 \cot (e+f x)}{8 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {45 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{8 a^{5/2} f}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{a^{5/2} f}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(45*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/(8*a^(5/2)*f) - (4*Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[
e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(a^(5/2)*f) - (19*Cot[e + f*x])/(8*a^2*f*Sqrt[a + a*Sin[e + f*x
]]) + (13*Cot[e + f*x]*Csc[e + f*x])/(12*a^2*f*Sqrt[a + a*Sin[e + f*x]]) - (Cot[e + f*x]*Csc[e + f*x]^2)/(3*a^
2*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 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 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 2779

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

Rule 2984

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

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))^{5/2}} \, dx &=\frac {\int \frac {\csc ^4(e+f x) \left (1+\sin ^2(e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}-\frac {2 \int \frac {\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^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {\int \frac {\csc ^3(e+f x) \left (-\frac {a}{2}+\frac {11}{2} a \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{3 a^3}+\frac {\int \frac {\csc ^2(e+f x) (a-3 a \sin (e+f x))}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a^3}\\ &=-\frac {\cot (e+f x)}{2 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {\int \frac {\csc ^2(e+f x) \left (\frac {45 a^2}{4}-\frac {3}{4} a^2 \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{6 a^4}+\frac {\int \frac {\csc (e+f x) \left (-\frac {7 a^2}{2}+\frac {1}{2} a^2 \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a^4}\\ &=-\frac {19 \cot (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {\int \frac {\csc (e+f x) \left (-\frac {51 a^3}{8}+\frac {45}{8} a^3 \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{6 a^5}-\frac {7 \int \csc (e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{4 a^3}+\frac {2 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {19 \cot (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {17 \int \csc (e+f x) \sqrt {a+a \sin (e+f x)} \, dx}{16 a^3}+\frac {2 \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}+\frac {7 \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^2 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^2 f}\\ &=\frac {7 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{2 a^{5/2} 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^{5/2} f}-\frac {19 \cot (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {17 \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^2 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^2 f}\\ &=\frac {45 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{8 a^{5/2} f}-\frac {4 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} f}-\frac {19 \cot (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {13 \cot (e+f x) \csc (e+f x)}{12 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cot (e+f x) \csc ^2(e+f x)}{3 a^2 f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 2.41, size = 332, normalized size = 1.74 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (-\frac {8 \csc ^9\left (\frac {1}{2} (e+f x)\right ) \left (-396 \sin \left (\frac {1}{2} (e+f x)\right )-218 \sin \left (\frac {3}{2} (e+f x)\right )+114 \sin \left (\frac {5}{2} (e+f x)\right )+396 \cos \left (\frac {1}{2} (e+f x)\right )-218 \cos \left (\frac {3}{2} (e+f x)\right )-114 \cos \left (\frac {5}{2} (e+f x)\right )-405 \sin (e+f x) \log \left (-\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )+1\right )+405 \sin (e+f x) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )+135 \sin (3 (e+f x)) \log \left (-\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )+1\right )-135 \sin (3 (e+f x)) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )\right )}{\left (\csc ^2\left (\frac {1}{4} (e+f x)\right )-\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )^3}+(1536+1536 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )\right )}{192 f (a (\sin (e+f x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[e + f*x]^4/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*((1536 + 1536*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[
(e + f*x)/4])] - (8*Csc[(e + f*x)/2]^9*(396*Cos[(e + f*x)/2] - 218*Cos[(3*(e + f*x))/2] - 114*Cos[(5*(e + f*x)
)/2] - 396*Sin[(e + f*x)/2] - 405*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[e + f*x] + 405*Log[1 - Cos[
(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] - 218*Sin[(3*(e + f*x))/2] + 114*Sin[(5*(e + f*x))/2] + 135*Log[
1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[3*(e + f*x)] - 135*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*S
in[3*(e + f*x)]))/(Csc[(e + f*x)/4]^2 - Sec[(e + f*x)/4]^2)^3))/(192*f*(a*(1 + Sin[e + f*x]))^(5/2))

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fricas [B]  time = 0.46, size = 564, normalized size = 2.95 \[ \frac {135 \, {\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 ) + \frac {192 \, \sqrt {2} {\left (a \cos \left (f x + e\right )^{4} - 2 \, a \cos \left (f x + e\right )^{2} - {\left (a \cos \left (f x + e\right )^{3} + a \cos \left (f x + e\right )^{2} - 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 \, {\left (57 \, \cos \left (f x + e\right )^{3} + 83 \, \cos \left (f x + e\right )^{2} - {\left (57 \, \cos \left (f x + e\right )^{2} - 26 \, \cos \left (f x + e\right ) - 91\right )} \sin \left (f x + e\right ) - 65 \, \cos \left (f x + e\right ) - 91\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{96 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f - {\left (a^{3} f \cos \left (f x + e\right )^{3} + a^{3} f \cos \left (f x + e\right )^{2} - a^{3} f \cos \left (f x + e\right ) - a^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/96*(135*(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)) + 192*sqrt(2)*(a*cos(f*x + e)^4 - 2*a*cos(f*x + e)^2 - (a*cos(f*x + e)^3 + 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*(57*cos(f*x + e)^3 + 83*cos(f*x + e)^2 - (57*cos(f*x + e)^2
- 26*cos(f*x + e) - 91)*sin(f*x + e) - 65*cos(f*x + e) - 91)*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^4 -
2*a^3*f*cos(f*x + e)^2 + a^3*f - (a^3*f*cos(f*x + e)^3 + a^3*f*cos(f*x + e)^2 - a^3*f*cos(f*x + e) - a^3*f)*s
in(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)^4/(a+a*sin(f*x+e))^(5/2),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)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^3/sign(tan((f*x+ex
p(1))/2)+1)-5/64/a^3/sign(tan((f*x+exp(1))/2)+1))+37/96/a^3/sign(tan((f*x+exp(1))/2)+1))+2*(1/96*(15*(-sqrt(a)
*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+78*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f
*x+exp(1))/2)^2+a))^4-15*a^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-144*sqrt(a)*a*(-sq
rt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+74*sqrt(a)*a^2)/a^2/((-sqrt(a)*tan((f*x+exp(1))/2
)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-a)^3/sign(tan((f*x+exp(1))/2)+1)-45/32*atan((-sqrt(a)*tan((f*x+exp(1))/2)
+sqrt(a*tan((f*x+exp(1))/2)^2+a))/sqrt(-a))/a^2/sqrt(-a)/sign(tan((f*x+exp(1))/2)+1)+4*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))/sqrt(2)/a^2/sqrt(-a)/sign(tan((f*x+exp(
1))/2)+1)+45/64/sqrt(a)/a^2*ln(abs(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a)))/sign(tan((f*
x+exp(1))/2)+1))+(-945*sqrt(-a)*sqrt(2)*ln(sqrt(2)*sqrt(a)+sqrt(a))-1302*sqrt(-a)*sqrt(2)-1350*sqrt(-a)*ln(sqr
t(2)*sqrt(a)+sqrt(a))-1808*sqrt(-a)+1890*sqrt(2)*sqrt(a)*atan((sqrt(2)*sqrt(a)+sqrt(a))/sqrt(-a))-3840*sqrt(2)
*sqrt(a)*atan(sqrt(a)/sqrt(-a))+2700*sqrt(a)*atan((sqrt(2)*sqrt(a)+sqrt(a))/sqrt(-a))-5376*sqrt(a)*atan(sqrt(a
)/sqrt(-a)))/(672*a^2*sqrt(-a)*sqrt(2)*sqrt(a)+960*a^2*sqrt(-a)*sqrt(a))*sign(tan((f*x+exp(1))/2)+1))

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maple [A]  time = 1.01, size = 182, normalized size = 0.95 \[ \frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-96 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{5} \left (\sin ^{3}\left (f x +e \right )\right )-57 \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {5}{2}}+88 \left (-a \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {7}{2}}-39 \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, a^{\frac {9}{2}}+135 a^{5} \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{\sqrt {a}}\right ) \left (\sin ^{3}\left (f x +e \right )\right )\right )}{24 a^{\frac {15}{2}} \sin \left (f x +e \right )^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+a*sin(f*x+e))^(5/2),x)

[Out]

1/24*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(-96*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/
2))*a^5*sin(f*x+e)^3-57*(-a*(sin(f*x+e)-1))^(5/2)*a^(5/2)+88*(-a*(sin(f*x+e)-1))^(3/2)*a^(7/2)-39*(-a*(sin(f*x
+e)-1))^(1/2)*a^(9/2)+135*a^5*arctanh((-a*(sin(f*x+e)-1))^(1/2)/a^(1/2))*sin(f*x+e)^3)/a^(15/2)/sin(f*x+e)^3/c
os(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^4}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int(cot(e + f*x)^4/(a + a*sin(e + f*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Integral(cot(e + f*x)**4/(a*(sin(e + f*x) + 1))**(5/2), x)

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