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

Optimal. Leaf size=186 \[ \frac {2 \sqrt {2} a^3 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d}+\frac {4 a^3 e^2 \sqrt {e \cot (c+d x)}}{d}-\frac {2 \left (a^3 \cot (c+d x)+a^3\right ) (e \cot (c+d x))^{7/2}}{9 d e}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}+\frac {4 a^3 e (e \cot (c+d x))^{3/2}}{3 d} \]

[Out]

4/3*a^3*e*(e*cot(d*x+c))^(3/2)/d-4/5*a^3*(e*cot(d*x+c))^(5/2)/d-40/63*a^3*(e*cot(d*x+c))^(7/2)/d/e-2/9*(e*cot(
d*x+c))^(7/2)*(a^3+a^3*cot(d*x+c))/d/e+2*a^3*e^(5/2)*arctan(1/2*(e^(1/2)-cot(d*x+c)*e^(1/2))*2^(1/2)/(e*cot(d*
x+c))^(1/2))*2^(1/2)/d+4*a^3*e^2*(e*cot(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.30, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3566, 3630, 3528, 3532, 205} \[ \frac {4 a^3 e^2 \sqrt {e \cot (c+d x)}}{d}+\frac {2 \sqrt {2} a^3 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d}-\frac {2 \left (a^3 \cot (c+d x)+a^3\right ) (e \cot (c+d x))^{7/2}}{9 d e}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}+\frac {4 a^3 e (e \cot (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*a^3*e^(5/2)*ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/d + (4*a^3*e^2
*Sqrt[e*Cot[c + d*x]])/d + (4*a^3*e*(e*Cot[c + d*x])^(3/2))/(3*d) - (4*a^3*(e*Cot[c + d*x])^(5/2))/(5*d) - (40
*a^3*(e*Cot[c + d*x])^(7/2))/(63*d*e) - (2*(e*Cot[c + d*x])^(7/2)*(a^3 + a^3*Cot[c + d*x]))/(9*d*e)

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 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

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 3566

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 - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

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

Rubi steps

\begin {align*} \int (e \cot (c+d x))^{5/2} (a+a \cot (c+d x))^3 \, dx &=-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}-\frac {2 \int (e \cot (c+d x))^{5/2} \left (-a^3 e-9 a^3 e \cot (c+d x)-10 a^3 e \cot ^2(c+d x)\right ) \, dx}{9 e}\\ &=-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}-\frac {2 \int (e \cot (c+d x))^{5/2} \left (9 a^3 e-9 a^3 e \cot (c+d x)\right ) \, dx}{9 e}\\ &=-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}-\frac {2 \int (e \cot (c+d x))^{3/2} \left (9 a^3 e^2+9 a^3 e^2 \cot (c+d x)\right ) \, dx}{9 e}\\ &=\frac {4 a^3 e (e \cot (c+d x))^{3/2}}{3 d}-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}-\frac {2 \int \sqrt {e \cot (c+d x)} \left (-9 a^3 e^3+9 a^3 e^3 \cot (c+d x)\right ) \, dx}{9 e}\\ &=\frac {4 a^3 e^2 \sqrt {e \cot (c+d x)}}{d}+\frac {4 a^3 e (e \cot (c+d x))^{3/2}}{3 d}-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}-\frac {2 \int \frac {-9 a^3 e^4-9 a^3 e^4 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{9 e}\\ &=\frac {4 a^3 e^2 \sqrt {e \cot (c+d x)}}{d}+\frac {4 a^3 e (e \cot (c+d x))^{3/2}}{3 d}-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}+\frac {\left (36 a^6 e^7\right ) \operatorname {Subst}\left (\int \frac {1}{-162 a^6 e^8-e x^2} \, dx,x,\frac {-9 a^3 e^4+9 a^3 e^4 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=\frac {2 \sqrt {2} a^3 e^{5/2} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d}+\frac {4 a^3 e^2 \sqrt {e \cot (c+d x)}}{d}+\frac {4 a^3 e (e \cot (c+d x))^{3/2}}{3 d}-\frac {4 a^3 (e \cot (c+d x))^{5/2}}{5 d}-\frac {40 a^3 (e \cot (c+d x))^{7/2}}{63 d e}-\frac {2 (e \cot (c+d x))^{7/2} \left (a^3+a^3 \cot (c+d x)\right )}{9 d e}\\ \end {align*}

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Mathematica [C]  time = 6.10, size = 729, normalized size = 3.92 \[ -\frac {4 \sin ^3(c+d x) \tan (c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )}{3 d (\sin (c+d x)+\cos (c+d x))^3}-\frac {2 \sin (c+d x) \cos ^2(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2}}{9 d (\sin (c+d x)+\cos (c+d x))^3}-\frac {4 \sin ^3(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2}}{5 d (\sin (c+d x)+\cos (c+d x))^3}-\frac {6 \sin ^2(c+d x) \cos (c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2}}{7 d (\sin (c+d x)+\cos (c+d x))^3}+\frac {\sin ^3(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \cot ^{\frac {5}{2}}(c+d x) (\sin (c+d x)+\cos (c+d x))^3}-\frac {\sin ^3(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \cot ^{\frac {5}{2}}(c+d x) (\sin (c+d x)+\cos (c+d x))^3}+\frac {\sqrt {2} \sin ^3(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{d \cot ^{\frac {5}{2}}(c+d x) (\sin (c+d x)+\cos (c+d x))^3}-\frac {\sqrt {2} \sin ^3(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{d \cot ^{\frac {5}{2}}(c+d x) (\sin (c+d x)+\cos (c+d x))^3}+\frac {4 \sin ^3(c+d x) \tan ^2(c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2}}{d (\sin (c+d x)+\cos (c+d x))^3}+\frac {4 \sin ^3(c+d x) \tan (c+d x) (a \cot (c+d x)+a)^3 (e \cot (c+d x))^{5/2}}{3 d (\sin (c+d x)+\cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^2*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[c + d*x])/(9*d*(Cos[c + d*x] + Sin[c + d*
x])^3) - (6*Cos[c + d*x]*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[c + d*x]^2)/(7*d*(Cos[c + d*x] + Si
n[c + d*x])^3) - (4*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[c + d*x]^3)/(5*d*(Cos[c + d*x] + Sin[c +
d*x])^3) + (Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[
c + d*x]^3)/(d*Cot[c + d*x]^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3) - (Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c +
d*x]]]*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[c + d*x]^3)/(d*Cot[c + d*x]^(5/2)*(Cos[c + d*x] + Sin
[c + d*x])^3) + ((e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*
x]]*Sin[c + d*x]^3)/(Sqrt[2]*d*Cot[c + d*x]^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3) - ((e*Cot[c + d*x])^(5/2)*(
a + a*Cot[c + d*x])^3*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]*Sin[c + d*x]^3)/(Sqrt[2]*d*Cot[c + d*
x]^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3) + (4*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[c + d*x]^3*Ta
n[c + d*x])/(3*d*(Cos[c + d*x] + Sin[c + d*x])^3) - (4*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Hypergeom
etric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2]*Sin[c + d*x]^3*Tan[c + d*x])/(3*d*(Cos[c + d*x] + Sin[c + d*x])^3) + (4
*(e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])^3*Sin[c + d*x]^3*Tan[c + d*x]^2)/(d*(Cos[c + d*x] + Sin[c + d*x])
^3)

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fricas [A]  time = 0.95, size = 535, normalized size = 2.88 \[ \left [\frac {315 \, \sqrt {2} {\left (a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} e^{2}\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 ) + 2 \, {\left (721 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} - 1330 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right ) + 469 \, a^{3} e^{2} - 15 \, {\left (23 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, a^{3} e^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{315 \, {\left (d \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, d \cos \left (2 \, d x + 2 \, c\right ) + d\right )}}, \frac {2 \, {\left (315 \, \sqrt {2} {\left (a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right ) + a^{3} e^{2}\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 ) + {\left (721 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} - 1330 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right ) + 469 \, a^{3} e^{2} - 15 \, {\left (23 \, a^{3} e^{2} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, a^{3} e^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}\right )}}{315 \, {\left (d \cos \left (2 \, d x + 2 \, c\right )^{2} - 2 \, d \cos \left (2 \, d x + 2 \, c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/315*(315*sqrt(2)*(a^3*e^2*cos(2*d*x + 2*c)^2 - 2*a^3*e^2*cos(2*d*x + 2*c) + a^3*e^2)*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) + 2*(721*a^3*e^2*cos(2*d*x + 2*c)^2 - 1330*a^3*e^2*cos(2*d*x + 2*c) + 469*a^3*e^2 - 15*(23*a^
3*e^2*cos(2*d*x + 2*c) - 5*a^3*e^2)*sin(2*d*x + 2*c))*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*cos(
2*d*x + 2*c)^2 - 2*d*cos(2*d*x + 2*c) + d), 2/315*(315*sqrt(2)*(a^3*e^2*cos(2*d*x + 2*c)^2 - 2*a^3*e^2*cos(2*d
*x + 2*c) + a^3*e^2)*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)) + (721*a^3*e^2*cos(2*d*x + 2*c)^2 - 1330*a^3*e^
2*cos(2*d*x + 2*c) + 469*a^3*e^2 - 15*(23*a^3*e^2*cos(2*d*x + 2*c) - 5*a^3*e^2)*sin(2*d*x + 2*c))*sqrt((e*cos(
2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*cos(2*d*x + 2*c)^2 - 2*d*cos(2*d*x + 2*c) + d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.79, size = 446, normalized size = 2.40 \[ -\frac {2 a^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {9}{2}}}{9 d \,e^{2}}-\frac {6 a^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d e}-\frac {4 a^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}+\frac {4 a^{3} e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}+\frac {4 a^{3} e^{2} \sqrt {e \cot \left (d x +c \right )}}{d}-\frac {a^{3} e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d}-\frac {a^{3} e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d}+\frac {a^{3} e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d}-\frac {a^{3} e^{3} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} e^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} e^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9/d*a^3/e^2*(e*cot(d*x+c))^(9/2)-6/7*a^3*(e*cot(d*x+c))^(7/2)/d/e-4/5*a^3*(e*cot(d*x+c))^(5/2)/d+4/3*a^3*e*
(e*cot(d*x+c))^(3/2)/d+4*a^3*e^2*(e*cot(d*x+c))^(1/2)/d-1/2/d*a^3*e^2*(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/d*a^3*e^2*(e^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/d*a^3*e^2*(e^2)
^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/2/d*a^3*e^3*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/d*a^3*e^3*2^(1/2)/(e^2)^(1/4)*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/d*a^
3*e^3*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)

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maxima [A]  time = 0.50, size = 193, normalized size = 1.04 \[ -\frac {2 \, {\left (315 \, a^{3} e^{2} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}\right )} - \frac {630 \, a^{3} e^{4} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + 210 \, a^{3} e^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}} - 126 \, a^{3} e^{2} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}} - 135 \, a^{3} e \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {7}{2}} - 35 \, a^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {9}{2}}}{e^{3}}\right )} e}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(315*a^3*e^2*(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)) - (630*a^3*e^4*sqrt(e
/tan(d*x + c)) + 210*a^3*e^3*(e/tan(d*x + c))^(3/2) - 126*a^3*e^2*(e/tan(d*x + c))^(5/2) - 135*a^3*e*(e/tan(d*
x + c))^(7/2) - 35*a^3*(e/tan(d*x + c))^(9/2))/e^3)*e/d

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mupad [B]  time = 2.44, size = 177, normalized size = 0.95 \[ \frac {4\,a^3\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{d}-\frac {4\,a^3\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}}{5\,d}-\frac {6\,a^3\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{7/2}}{7\,d\,e}-\frac {2\,a^3\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{9/2}}{9\,d\,e^2}+\frac {4\,a^3\,e\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{3\,d}-\frac {\sqrt {2}\,a^3\,e^{5/2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}+\frac {\sqrt {2}\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{2\,e^{3/2}}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^3*e^2*(e*cot(c + d*x))^(1/2))/d - (4*a^3*(e*cot(c + d*x))^(5/2))/(5*d) - (6*a^3*(e*cot(c + d*x))^(7/2))/(
7*d*e) - (2*a^3*(e*cot(c + d*x))^(9/2))/(9*d*e^2) + (4*a^3*e*(e*cot(c + d*x))^(3/2))/(3*d) - (2^(1/2)*a^3*e^(5
/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)))))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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