3.1.21 \(\int \frac {(a+a \cot (c+d x))^3}{(e \cot (c+d x))^{7/2}} \, dx\) [21]

Optimal. Leaf size=141 \[ -\frac {2 \sqrt {2} a^3 \text {ArcTan}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{7/2}}+\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{5 d e (e \cot (c+d x))^{5/2}} \]

[Out]

8/5*a^3/d/e^2/(e*cot(d*x+c))^(3/2)+2/5*(a^3+a^3*cot(d*x+c))/d/e/(e*cot(d*x+c))^(5/2)-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)/d/e^(7/2)+4*a^3/d/e^3/(e*cot(d*x+c))^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 141, 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 = {3646, 3709, 3610, 3613, 211} \begin {gather*} -\frac {2 \sqrt {2} a^3 \text {ArcTan}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{7/2}}+\frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{5 d e (e \cot (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Cot[c + d*x])^3/(e*Cot[c + d*x])^(7/2),x]

[Out]

(-2*Sqrt[2]*a^3*ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/(d*e^(7/2)) + (8*a^3)
/(5*d*e^2*(e*Cot[c + d*x])^(3/2)) + (4*a^3)/(d*e^3*Sqrt[e*Cot[c + d*x]]) + (2*(a^3 + a^3*Cot[c + d*x]))/(5*d*e
*(e*Cot[c + d*x])^(5/2))

Rule 211

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

Rule 3610

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

Rule 3613

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 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3709

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

Rubi steps

\begin {align*} \int \frac {(a+a \cot (c+d x))^3}{(e \cot (c+d x))^{7/2}} \, dx &=\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{5 d e (e \cot (c+d x))^{5/2}}-\frac {2 \int \frac {-6 a^3 e^2-5 a^3 e^2 \cot (c+d x)-a^3 e^2 \cot ^2(c+d x)}{(e \cot (c+d x))^{5/2}} \, dx}{5 e^3}\\ &=\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{5 d e (e \cot (c+d x))^{5/2}}-\frac {2 \int \frac {-5 a^3 e^3+5 a^3 e^3 \cot (c+d x)}{(e \cot (c+d x))^{3/2}} \, dx}{5 e^5}\\ &=\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{5 d e (e \cot (c+d x))^{5/2}}-\frac {2 \int \frac {5 a^3 e^4+5 a^3 e^4 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{5 e^7}\\ &=\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{5 d e (e \cot (c+d x))^{5/2}}+\frac {\left (20 a^6 e\right ) \text {Subst}\left (\int \frac {1}{-50 a^6 e^8-e x^2} \, dx,x,\frac {5 a^3 e^4-5 a^3 e^4 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=-\frac {2 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{7/2}}+\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{5 d e (e \cot (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 3.44, size = 269, normalized size = 1.91 \begin {gather*} \frac {a^3 \left (120 \cos ^3(c+d x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )+\sin (c+d x) \left (40 \cos ^2(c+d x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )+\sin (c+d x) \left (8 \cos (c+d x) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(c+d x)\right )+5 \sqrt {2} \cot ^{\frac {7}{2}}(c+d x) \left (2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right ) \sin (c+d x)\right )\right )\right ) (1+\tan (c+d x))^3}{20 d e^3 \sqrt {e \cot (c+d x)} (\cos (c+d x)+\sin (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cot[c + d*x])^3/(e*Cot[c + d*x])^(7/2),x]

[Out]

(a^3*(120*Cos[c + d*x]^3*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + Sin[c + d*x]*(40*Cos[c + d*x]^2*Hy
pergeometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2] + Sin[c + d*x]*(8*Cos[c + d*x]*Hypergeometric2F1[-5/4, 1, -1/4,
-Cot[c + d*x]^2] + 5*Sqrt[2]*Cot[c + d*x]^(7/2)*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*ArcTan[1 + Sqrt
[2]*Sqrt[Cot[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[c + d*
x]] + Cot[c + d*x]])*Sin[c + d*x])))*(1 + Tan[c + d*x])^3)/(20*d*e^3*Sqrt[e*Cot[c + d*x]]*(Cos[c + d*x] + Sin[
c + d*x])^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs. \(2(120)=240\).
time = 0.44, size = 323, normalized size = 2.29

method	result	size
derivativedivides	\(-\frac {2 a^{3} \left (-\frac {e}{5 \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {1}{\left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {2}{e \sqrt {e \cot \left (d x +c \right )}}+\frac {-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 e}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (e^{2}\right )^{\frac {1}{4}}}}{e}\right )}{d \,e^{2}}\)	\(323\)
default	\(-\frac {2 a^{3} \left (-\frac {e}{5 \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {1}{\left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {2}{e \sqrt {e \cot \left (d x +c \right )}}+\frac {-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 e}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (e^{2}\right )^{\frac {1}{4}}}}{e}\right )}{d \,e^{2}}\)	\(323\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 101, normalized size = 0.72 \begin {gather*} \frac {2 \, {\left (5 \, {\left (\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right )\right )} a^{3} + {\left (a^{3} + \frac {5 \, a^{3}}{\tan \left (d x + c\right )} + \frac {10 \, a^{3}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}}\right )} e^{\left (-\frac {7}{2}\right )}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/5*(5*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) -
2/sqrt(tan(d*x + c)))))*a^3 + (a^3 + 5*a^3/tan(d*x + c) + 10*a^3/tan(d*x + c)^2)*tan(d*x + c)^(5/2))*e^(-7/2)/
d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (109) = 218\).
time = 3.80, size = 235, normalized size = 1.67 \begin {gather*} -\frac {2 \, {\left (5 \, {\left (\sqrt {2} a^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \sqrt {2} a^{3} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} a^{3}\right )} \arctan \left (-\frac {{\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) - \sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )}}\right ) + {\left (5 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} - 5 \, a^{3} - {\left (9 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + 11 \, a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}}\right )}}{5 \, {\left (d \cos \left (2 \, d x + 2 \, c\right )^{2} e^{\frac {7}{2}} + 2 \, d \cos \left (2 \, d x + 2 \, c\right ) e^{\frac {7}{2}} + d e^{\frac {7}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-2/5*(5*(sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + 2*sqrt(2)*a^3*cos(2*d*x + 2*c) + sqrt(2)*a^3)*arctan(-1/2*(sqrt(2)*c
os(2*d*x + 2*c) - sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))/(cos(2*d*x
+ 2*c) + 1)) + (5*a^3*cos(2*d*x + 2*c)^2 - 5*a^3 - (9*a^3*cos(2*d*x + 2*c) + 11*a^3)*sin(2*d*x + 2*c))*sqrt((
cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c)))/(d*cos(2*d*x + 2*c)^2*e^(7/2) + 2*d*cos(2*d*x + 2*c)*e^(7/2) + d*e^(7
/2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {3 \cot {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {3 \cot ^{2}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))**3/(e*cot(d*x+c))**(7/2),x)

[Out]

a**3*(Integral((e*cot(c + d*x))**(-7/2), x) + Integral(3*cot(c + d*x)/(e*cot(c + d*x))**(7/2), x) + Integral(3
*cot(c + d*x)**2/(e*cot(c + d*x))**(7/2), x) + Integral(cot(c + d*x)**3/(e*cot(c + d*x))**(7/2), x))

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 1.26, size = 126, normalized size = 0.89 \begin {gather*} \frac {4\,e\,a^3\,{\mathrm {cot}\left (c+d\,x\right )}^2+2\,e\,a^3\,\mathrm {cot}\left (c+d\,x\right )+\frac {2\,e\,a^3}{5}}{d\,e^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}}+\frac {\sqrt {2}\,a^3\,\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\,e^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cot(c + d*x))^3/(e*cot(c + d*x))^(7/2),x)

[Out]

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

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