3.40 \(\int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))^3} \, dx\)

Optimal. Leaf size=215 \[ -\frac {59 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d e^{5/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d e^{5/2}}-\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {11}{8 a^3 d e (\cot (c+d x)+1) (e \cot (c+d x))^{3/2}}+\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {1}{4 a d e (a \cot (c+d x)+a)^2 (e \cot (c+d x))^{3/2}} \]

[Out]

-59/8*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^3/d/e^(5/2)+55/24/a^3/d/e/(e*cot(d*x+c))^(3/2)-11/8/a^3/d/e/(e*co
t(d*x+c))^(3/2)/(1+cot(d*x+c))-1/4/a/d/e/(e*cot(d*x+c))^(3/2)/(a+a*cot(d*x+c))^2+1/4*arctan(1/2*(e^(1/2)-cot(d
*x+c)*e^(1/2))*2^(1/2)/(e*cot(d*x+c))^(1/2))/a^3/d/e^(5/2)*2^(1/2)-63/8/a^3/d/e^2/(e*cot(d*x+c))^(1/2)

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Rubi [A]  time = 1.10, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3569, 3649, 3650, 3653, 3532, 205, 3634, 63} \[ -\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {59 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d e^{5/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d e^{5/2}}-\frac {11}{8 a^3 d e (\cot (c+d x)+1) (e \cot (c+d x))^{3/2}}+\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {1}{4 a d e (a \cot (c+d x)+a)^2 (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3),x]

[Out]

(-59*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(8*a^3*d*e^(5/2)) + ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[2
]*Sqrt[e*Cot[c + d*x]])]/(2*Sqrt[2]*a^3*d*e^(5/2)) + 55/(24*a^3*d*e*(e*Cot[c + d*x])^(3/2)) - 63/(8*a^3*d*e^2*
Sqrt[e*Cot[c + d*x]]) - 11/(8*a^3*d*e*(e*Cot[c + d*x])^(3/2)*(1 + Cot[c + d*x])) - 1/(4*a*d*e*(e*Cot[c + d*x])
^(3/2)*(a + a*Cot[c + d*x])^2)

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 205

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

Rule 3532

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

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3650

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

Rule 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rubi steps

\begin {align*} \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))^3} \, dx &=-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}-\frac {\int \frac {-\frac {11 a^2 e}{2}+2 a^2 e \cot (c+d x)-\frac {7}{2} a^2 e \cot ^2(c+d x)}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))^2} \, dx}{4 a^3 e}\\ &=-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}+\frac {\int \frac {\frac {55 a^4 e^2}{2}-4 a^4 e^2 \cot (c+d x)+\frac {55}{2} a^4 e^2 \cot ^2(c+d x)}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx}{8 a^6 e^2}\\ &=\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}+\frac {\int \frac {-\frac {189}{4} a^5 e^4-\frac {165}{4} a^5 e^4 \cot ^2(c+d x)}{(e \cot (c+d x))^{3/2} (a+a \cot (c+d x))} \, dx}{12 a^7 e^5}\\ &=\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}+\frac {\int \frac {\frac {189 a^6 e^6}{8}+3 a^6 e^6 \cot (c+d x)+\frac {189}{8} a^6 e^6 \cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{6 a^8 e^8}\\ &=\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}+\frac {\int \frac {3 a^7 e^6+3 a^7 e^6 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{12 a^{10} e^8}+\frac {59 \int \frac {1+\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{16 a^2 e^2}\\ &=\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}+\frac {59 \operatorname {Subst}\left (\int \frac {1}{\sqrt {-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{16 a^2 d e^2}-\frac {\left (3 a^4 e^4\right ) \operatorname {Subst}\left (\int \frac {1}{-18 a^{14} e^{12}-e x^2} \, dx,x,\frac {3 a^7 e^6-3 a^7 e^6 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{2 d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d e^{5/2}}+\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}-\frac {59 \operatorname {Subst}\left (\int \frac {1}{a+\frac {a x^2}{e}} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{8 a^2 d e^3}\\ &=-\frac {59 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d e^{5/2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d e^{5/2}}+\frac {55}{24 a^3 d e (e \cot (c+d x))^{3/2}}-\frac {63}{8 a^3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {11}{8 a^3 d e (e \cot (c+d x))^{3/2} (1+\cot (c+d x))}-\frac {1}{4 a d e (e \cot (c+d x))^{3/2} (a+a \cot (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 3.19, size = 167, normalized size = 0.78 \[ \frac {\cot ^{\frac {5}{2}}(c+d x) \left (4 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-4 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-118 \tan ^{-1}\left (\sqrt {\cot (c+d x)}\right )-\frac {\sqrt {\cot (c+d x)} \sec ^2(c+d x) (678 \cos (2 (c+d x))+679 \cot (c+d x)+77 \cos (3 (c+d x)) \csc (c+d x)+614)}{6 (\cot (c+d x)+1)^2}\right )}{16 a^3 d (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3),x]

[Out]

(Cot[c + d*x]^(5/2)*(4*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 4*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[
c + d*x]]] - 118*ArcTan[Sqrt[Cot[c + d*x]]] - (Sqrt[Cot[c + d*x]]*(614 + 678*Cos[2*(c + d*x)] + 679*Cot[c + d*
x] + 77*Cos[3*(c + d*x)]*Csc[c + d*x])*Sec[c + d*x]^2)/(6*(1 + Cot[c + d*x])^2)))/(16*a^3*d*(e*Cot[c + d*x])^(
5/2))

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fricas [A]  time = 0.48, size = 718, normalized size = 3.34 \[ \left [-\frac {6 \, \sqrt {2} {\left ({\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {-e} \log \left (\sqrt {2} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} - 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) + 177 \, {\left ({\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {-e} \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) + 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right ) - {\left (339 \, \cos \left (2 \, d x + 2 \, c\right )^{2} - 7 \, {\left (11 \, \cos \left (2 \, d x + 2 \, c\right ) + 43\right )} \sin \left (2 \, d x + 2 \, c\right ) - 32 \, \cos \left (2 \, d x + 2 \, c\right ) - 307\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{48 \, {\left (a^{3} d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} d e^{3} + {\left (a^{3} d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} d e^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )}}, \frac {12 \, \sqrt {2} {\left ({\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {e} \arctan \left (-\frac {\sqrt {2} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) - 354 \, {\left ({\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sqrt {e} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right ) + {\left (339 \, \cos \left (2 \, d x + 2 \, c\right )^{2} - 7 \, {\left (11 \, \cos \left (2 \, d x + 2 \, c\right ) + 43\right )} \sin \left (2 \, d x + 2 \, c\right ) - 32 \, \cos \left (2 \, d x + 2 \, c\right ) - 307\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{48 \, {\left (a^{3} d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} d e^{3} + {\left (a^{3} d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} d e^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/48*(6*sqrt(2)*((cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c) + cos(2*d*x + 2*c) + 1)*sqrt(-e)*log(sqrt(2)*sqrt(-
e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*d*x + 2*c) + sin(2*d*x + 2*c) - 1) - 2*e*sin(2*d*x +
2*c) + e) + 177*((cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c) + cos(2*d*x + 2*c) + 1)*sqrt(-e)*log((e*cos(2*d*x +
2*c) - e*sin(2*d*x + 2*c) + 2*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + e)/(
cos(2*d*x + 2*c) + sin(2*d*x + 2*c) + 1)) - (339*cos(2*d*x + 2*c)^2 - 7*(11*cos(2*d*x + 2*c) + 43)*sin(2*d*x +
2*c) - 32*cos(2*d*x + 2*c) - 307)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a^3*d*e^3*cos(2*d*x + 2*c
) + a^3*d*e^3 + (a^3*d*e^3*cos(2*d*x + 2*c) + a^3*d*e^3)*sin(2*d*x + 2*c)), 1/48*(12*sqrt(2)*((cos(2*d*x + 2*c
) + 1)*sin(2*d*x + 2*c) + cos(2*d*x + 2*c) + 1)*sqrt(e)*arctan(-1/2*sqrt(2)*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) +
e)/sin(2*d*x + 2*c))*(cos(2*d*x + 2*c) - sin(2*d*x + 2*c) + 1)/(e*cos(2*d*x + 2*c) + e)) - 354*((cos(2*d*x +
2*c) + 1)*sin(2*d*x + 2*c) + cos(2*d*x + 2*c) + 1)*sqrt(e)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*
c))/sqrt(e)) + (339*cos(2*d*x + 2*c)^2 - 7*(11*cos(2*d*x + 2*c) + 43)*sin(2*d*x + 2*c) - 32*cos(2*d*x + 2*c) -
307)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a^3*d*e^3*cos(2*d*x + 2*c) + a^3*d*e^3 + (a^3*d*e^3*co
s(2*d*x + 2*c) + a^3*d*e^3)*sin(2*d*x + 2*c))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cot \left (d x + c\right ) + a\right )}^{3} \left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((a*cot(d*x + c) + a)^3*(e*cot(d*x + c))^(5/2)), x)

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maple [B]  time = 0.86, size = 482, normalized size = 2.24 \[ -\frac {15 \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{8 d \,a^{3} e^{2} \left (e \cot \left (d x +c \right )+e \right )^{2}}-\frac {17 \sqrt {e \cot \left (d x +c \right )}}{8 d \,a^{3} e \left (e \cot \left (d x +c \right )+e \right )^{2}}-\frac {59 \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{8 a^{3} d \,e^{\frac {5}{2}}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{16 d \,a^{3} e^{3}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} e^{3}}+\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} e^{3}}-\frac {\sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{16 d \,a^{3} e^{2} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{8 d \,a^{3} e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2}{3 a^{3} d e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {6}{a^{3} d \,e^{2} \sqrt {e \cot \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cot(d*x+c))^(5/2)/(a+cot(d*x+c)*a)^3,x)

[Out]

-15/8/d/a^3/e^2/(e*cot(d*x+c)+e)^2*(e*cot(d*x+c))^(3/2)-17/8/d/a^3/e/(e*cot(d*x+c)+e)^2*(e*cot(d*x+c))^(1/2)-5
9/8*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^3/d/e^(5/2)-1/16/d/a^3/e^3*(e^2)^(1/4)*2^(1/2)*ln((e*cot(d*x+c)+(e^
2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2
)^(1/2)))-1/8/d/a^3/e^3*(e^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/8/d/a^3/e^3*(
e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/16/d/a^3/e^2*2^(1/2)/(e^2)^(1/4)*ln((
e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1
/2)*2^(1/2)+(e^2)^(1/2)))-1/8/d/a^3/e^2*2^(1/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)
+1/8/d/a^3/e^2*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+2/3/a^3/d/e/(e*cot(d*x+
c))^(3/2)-6/a^3/d/e^2/(e*cot(d*x+c))^(1/2)

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maxima [A]  time = 0.78, size = 224, normalized size = 1.04 \[ \frac {e {\left (\frac {16 \, e^{3} - \frac {112 \, e^{3}}{\tan \left (d x + c\right )} - \frac {323 \, e^{3}}{\tan \left (d x + c\right )^{2}} - \frac {189 \, e^{3}}{\tan \left (d x + c\right )^{3}}}{a^{3} e^{5} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}} + 2 \, a^{3} e^{4} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}} + a^{3} e^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {7}{2}}} - \frac {6 \, {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}\right )}}{a^{3} e^{3}} - \frac {177 \, \arctan \left (\frac {\sqrt {\frac {e}{\tan \left (d x + c\right )}}}{\sqrt {e}}\right )}{a^{3} e^{\frac {7}{2}}}\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c))^3,x, algorithm="maxima")

[Out]

1/24*e*((16*e^3 - 112*e^3/tan(d*x + c) - 323*e^3/tan(d*x + c)^2 - 189*e^3/tan(d*x + c)^3)/(a^3*e^5*(e/tan(d*x
+ c))^(3/2) + 2*a^3*e^4*(e/tan(d*x + c))^(5/2) + a^3*e^3*(e/tan(d*x + c))^(7/2)) - 6*(sqrt(2)*arctan(1/2*sqrt(
2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e)
- 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e))/(a^3*e^3) - 177*arctan(sqrt(e/tan(d*x + c))/sqrt(e))/(a^3*e^(7/2))
)/d

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mupad [B]  time = 1.38, size = 193, normalized size = 0.90 \[ -\frac {\frac {63\,e\,{\mathrm {cot}\left (c+d\,x\right )}^3}{8}+\frac {323\,e\,{\mathrm {cot}\left (c+d\,x\right )}^2}{24}+\frac {14\,e\,\mathrm {cot}\left (c+d\,x\right )}{3}-\frac {2\,e}{3}}{a^3\,d\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{7/2}+2\,a^3\,d\,e\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}+a^3\,d\,e^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}-\frac {59\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{8\,a^3\,d\,e^{5/2}}-\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}+\frac {\sqrt {2}\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{2\,e^{3/2}}\right )\right )}{8\,a^3\,d\,e^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cot(c + d*x))^(5/2)*(a + a*cot(c + d*x))^3),x)

[Out]

- ((14*e*cot(c + d*x))/3 - (2*e)/3 + (323*e*cot(c + d*x)^2)/24 + (63*e*cot(c + d*x)^3)/8)/(a^3*d*(e*cot(c + d*
x))^(7/2) + 2*a^3*d*e*(e*cot(c + d*x))^(5/2) + a^3*d*e^2*(e*cot(c + d*x))^(3/2)) - (59*atan((e*cot(c + d*x))^(
1/2)/e^(1/2)))/(8*a^3*d*e^(5/2)) - (2^(1/2)*(2*atan((2^(1/2)*(e*cot(c + d*x))^(1/2))/(2*e^(1/2))) + 2*atan((2^
(1/2)*(e*cot(c + d*x))^(1/2))/(2*e^(1/2)) + (2^(1/2)*(e*cot(c + d*x))^(3/2))/(2*e^(3/2)))))/(8*a^3*d*e^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot ^{3}{\left (c + d x \right )} + 3 \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot ^{2}{\left (c + d x \right )} + 3 \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot {\left (c + d x \right )} + \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))**(5/2)/(a+a*cot(d*x+c))**3,x)

[Out]

Integral(1/((e*cot(c + d*x))**(5/2)*cot(c + d*x)**3 + 3*(e*cot(c + d*x))**(5/2)*cot(c + d*x)**2 + 3*(e*cot(c +
d*x))**(5/2)*cot(c + d*x) + (e*cot(c + d*x))**(5/2)), x)/a**3

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