3.17 \(\int \sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^3 \, dx\)

Optimal. Leaf size=138 \[ -\frac {8 a^3 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {4 a^3 \sqrt {e \cot (c+d x)}}{d}-\frac {2 \left (a^3 \cot (c+d x)+a^3\right ) (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 \sqrt {2} a^3 \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d} \]

[Out]

-8/5*a^3*(e*cot(d*x+c))^(3/2)/d/e-2/5*(e*cot(d*x+c))^(3/2)*(a^3+a^3*cot(d*x+c))/d/e-2*a^3*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)*e^(1/2)/d-4*a^3*(e*cot(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.20, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {8 a^3 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {4 a^3 \sqrt {e \cot (c+d x)}}{d}-\frac {2 \left (a^3 \cot (c+d x)+a^3\right ) (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 \sqrt {2} a^3 \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a^3*Sqrt[e]*ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/d - (4*a^3*Sq
rt[e*Cot[c + d*x]])/d - (8*a^3*(e*Cot[c + d*x])^(3/2))/(5*d*e) - (2*(e*Cot[c + d*x])^(3/2)*(a^3 + a^3*Cot[c +
d*x]))/(5*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 \sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^3 \, dx &=-\frac {2 (e \cot (c+d x))^{3/2} \left (a^3+a^3 \cot (c+d x)\right )}{5 d e}-\frac {2 \int \sqrt {e \cot (c+d x)} \left (-a^3 e-5 a^3 e \cot (c+d x)-6 a^3 e \cot ^2(c+d x)\right ) \, dx}{5 e}\\ &=-\frac {8 a^3 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 (e \cot (c+d x))^{3/2} \left (a^3+a^3 \cot (c+d x)\right )}{5 d e}-\frac {2 \int \sqrt {e \cot (c+d x)} \left (5 a^3 e-5 a^3 e \cot (c+d x)\right ) \, dx}{5 e}\\ &=-\frac {4 a^3 \sqrt {e \cot (c+d x)}}{d}-\frac {8 a^3 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 (e \cot (c+d x))^{3/2} \left (a^3+a^3 \cot (c+d x)\right )}{5 d e}-\frac {2 \int \frac {5 a^3 e^2+5 a^3 e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{5 e}\\ &=-\frac {4 a^3 \sqrt {e \cot (c+d x)}}{d}-\frac {8 a^3 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 (e \cot (c+d x))^{3/2} \left (a^3+a^3 \cot (c+d x)\right )}{5 d e}+\frac {\left (20 a^6 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{-50 a^6 e^4-e x^2} \, dx,x,\frac {5 a^3 e^2-5 a^3 e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=-\frac {2 \sqrt {2} a^3 \sqrt {e} \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 \sqrt {e \cot (c+d x)}}{d}-\frac {8 a^3 (e \cot (c+d x))^{3/2}}{5 d e}-\frac {2 (e \cot (c+d x))^{3/2} \left (a^3+a^3 \cot (c+d x)\right )}{5 d e}\\ \end {align*}

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Mathematica [C]  time = 1.55, size = 315, normalized size = 2.28 \[ -\frac {a^3 \sin (c+d x) (\cot (c+d x)+1)^3 \sqrt {e \cot (c+d x)} \left (3 \left (4 \cos ^2(c+d x) \sqrt {\cot (c+d x)}+40 \sin ^2(c+d x) \sqrt {\cot (c+d x)}+10 \sin (2 (c+d x)) \sqrt {\cot (c+d x)}+5 \sqrt {2} \sin ^2(c+d x) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-5 \sqrt {2} \sin ^2(c+d x) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+10 \sqrt {2} \sin ^2(c+d x) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-10 \sqrt {2} \sin ^2(c+d x) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )-20 \sin (2 (c+d x)) \sqrt {\cot (c+d x)} \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )\right )}{30 d \sqrt {\cot (c+d x)} (\sin (c+d x)+\cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(a^3*Sqrt[e*Cot[c + d*x]]*(1 + Cot[c + d*x])^3*Sin[c + d*x]*(-20*Sqrt[Cot[c + d*x]]*Hypergeometric2F1[3/
4, 1, 7/4, -Cot[c + d*x]^2]*Sin[2*(c + d*x)] + 3*(4*Cos[c + d*x]^2*Sqrt[Cot[c + d*x]] + 10*Sqrt[2]*ArcTan[1 -
Sqrt[2]*Sqrt[Cot[c + d*x]]]*Sin[c + d*x]^2 - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]*Sin[c + d*x]^2
+ 40*Sqrt[Cot[c + d*x]]*Sin[c + d*x]^2 + 5*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]*Sin[c +
d*x]^2 - 5*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]*Sin[c + d*x]^2 + 10*Sqrt[Cot[c + d*x]]*S
in[2*(c + d*x)])))/(d*Sqrt[Cot[c + d*x]]*(Cos[c + d*x] + Sin[c + d*x])^3)

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fricas [A]  time = 0.72, size = 366, normalized size = 2.65 \[ \left [\frac {5 \, \sqrt {2} {\left (a^{3} \cos \left (2 \, d x + 2 \, c\right ) - a^{3}\right )} \sqrt {-e} \log \left (\sqrt {2} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} - 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) - 2 \, {\left (9 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - 11 \, a^{3}\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{5 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}}, -\frac {2 \, {\left (5 \, \sqrt {2} {\left (a^{3} \cos \left (2 \, d x + 2 \, c\right ) - a^{3}\right )} \sqrt {e} \arctan \left (-\frac {\sqrt {2} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) + {\left (9 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) - 5 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - 11 \, a^{3}\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}\right )}}{5 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5*(5*sqrt(2)*(a^3*cos(2*d*x + 2*c) - a^3)*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*(9*a^3*cos(2*d*x + 2*c)
- 5*a^3*sin(2*d*x + 2*c) - 11*a^3)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*cos(2*d*x + 2*c) - d),
-2/5*(5*sqrt(2)*(a^3*cos(2*d*x + 2*c) - a^3)*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)) + (9*a^3*cos(2*d*x + 2
*c) - 5*a^3*sin(2*d*x + 2*c) - 11*a^3)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*cos(2*d*x + 2*c) -
d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*cot(d*x + c) + a)^3*sqrt(e*cot(d*x + c)), x)

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maple [B]  time = 0.96, size = 391, normalized size = 2.83 \[ -\frac {2 a^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d \,e^{2}}-\frac {2 a^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{d e}-\frac {4 a^{3} \sqrt {e \cot \left (d x +c \right )}}{d}+\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d}+\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d}-\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d}+\frac {a^{3} e \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} e \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} e \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/d*a^3/e^2*(e*cot(d*x+c))^(5/2)-2*a^3*(e*cot(d*x+c))^(3/2)/d/e-4*a^3*(e*cot(d*x+c))^(1/2)/d+1/2/d*a^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/d*a^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/d*a^3*(e^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/2/d*a^3*e*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*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*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 = 0.76, size = 149, normalized size = 1.08 \[ \frac {2 \, {\left (5 \, a^{3} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}\right )} - \frac {10 \, a^{3} e^{2} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + 5 \, a^{3} e \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}} + a^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}}}{e^{3}}\right )} e}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(5*a^3*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e) + sqrt(2)*a
rctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e)) - (10*a^3*e^2*sqrt(e/tan(d*x +
c)) + 5*a^3*e*(e/tan(d*x + c))^(3/2) + a^3*(e/tan(d*x + c))^(5/2))/e^3)*e/d

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mupad [B]  time = 0.99, size = 136, normalized size = 0.99 \[ \frac {\sqrt {2}\,a^3\,\sqrt {e}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{2\,\sqrt {e}}+\frac {\sqrt {2}\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{2\,e^{3/2}}\right )\right )}{d}-\frac {2\,a^3\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{d\,e}-\frac {2\,a^3\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}}{5\,d\,e^2}-\frac {4\,a^3\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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