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

Optimal. Leaf size=141 \[ -\frac {2 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{7/2}}+\frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{5 d e (e \cot (c+d x))^{5/2}} \]

[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 = 0.23, 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 = {3565, 3628, 3529, 3532, 205} \[ \frac {4 a^3}{d e^3 \sqrt {e \cot (c+d x)}}+\frac {8 a^3}{5 d e^2 (e \cot (c+d x))^{3/2}}-\frac {2 \sqrt {2} a^3 \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{7/2}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{5 d e (e \cot (c+d x))^{5/2}} \]

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 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 3529

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 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 3565

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 3628

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 ) \operatorname {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]  time = 3.34, size = 269, normalized size = 1.91 \[ \frac {a^3 (\tan (c+d x)+1)^3 \left (120 \cos ^3(c+d x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )+\sin (c+d x) \left (40 \cos ^2(c+d x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )+\sin (c+d x) \left (8 \cos (c+d x) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(c+d x)\right )+5 \sqrt {2} \sin (c+d x) \cot ^{\frac {7}{2}}(c+d x) \left (\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )\right )\right )\right )}{20 d e^3 \sqrt {e \cot (c+d x)} (\sin (c+d x)+\cos (c+d x))^3} \]

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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fricas [A]  time = 0.45, size = 485, normalized size = 3.44 \[ \left [\frac {5 \, \sqrt {2} {\left (a^{3} e \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, a^{3} e \cos \left (2 \, d x + 2 \, c\right ) + a^{3} e\right )} \sqrt {-\frac {1}{e}} \log \left (\sqrt {2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sqrt {-\frac {1}{e}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} - 2 \, \sin \left (2 \, d x + 2 \, c\right ) + 1\right ) - 2 \, {\left (5 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} - 5 \, a^{3} - {\left (9 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + 11 \, a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{5 \, {\left (d e^{4} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, d e^{4} \cos \left (2 \, d x + 2 \, c\right ) + d e^{4}\right )}}, -\frac {2 \, {\left (\frac {5 \, \sqrt {2} {\left (a^{3} e \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, a^{3} e \cos \left (2 \, d x + 2 \, c\right ) + a^{3} e\right )} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, \sqrt {e} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )}}\right )}{\sqrt {e}} + {\left (5 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} - 5 \, a^{3} - {\left (9 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + 11 \, a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}\right )}}{5 \, {\left (d e^{4} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, d e^{4} \cos \left (2 \, d x + 2 \, c\right ) + d e^{4}\right )}}\right ] \]

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]

[1/5*(5*sqrt(2)*(a^3*e*cos(2*d*x + 2*c)^2 + 2*a^3*e*cos(2*d*x + 2*c) + a^3*e)*sqrt(-1/e)*log(sqrt(2)*sqrt((e*c
os(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sqrt(-1/e)*(cos(2*d*x + 2*c) + sin(2*d*x + 2*c) - 1) - 2*sin(2*d*x + 2*
c) + 1) - 2*(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((e*co
s(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*e^4*cos(2*d*x + 2*c)^2 + 2*d*e^4*cos(2*d*x + 2*c) + d*e^4), -2/5*(5*
sqrt(2)*(a^3*e*cos(2*d*x + 2*c)^2 + 2*a^3*e*cos(2*d*x + 2*c) + a^3*e)*arctan(-1/2*sqrt(2)*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)/(sqrt(e)*(cos(2*d*x + 2*c) + 1)))/sqrt(e
) + (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((e*cos(2*d*x
+ 2*c) + e)/sin(2*d*x + 2*c)))/(d*e^4*cos(2*d*x + 2*c)^2 + 2*d*e^4*cos(2*d*x + 2*c) + d*e^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \cot \left (d x + c\right ) + a\right )}^{3}}{\left (e \cot \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

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]

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

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maple [B]  time = 0.53, size = 409, normalized size = 2.90 \[ \frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \,e^{4}}+\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{4}}-\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{4}}+\frac {a^{3} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \,e^{3} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{3} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{3} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2 a^{3}}{5 d e \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {2 a^{3}}{d \,e^{2} \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {4 a^{3}}{d \,e^{3} \sqrt {e \cot \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*a^3/e^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*co
t(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^4*(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^4*(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)*arc
tan(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)+2/5/d*a^3/e/(e*cot(d*x+c))^(5/2)+2*a^3/d/e^2/(e*cot(d*x+c))^(3/2)+4*a^3/d/e^3/(e*cot(d*x
+c))^(1/2)

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maxima [A]  time = 0.77, size = 148, normalized size = 1.05 \[ \frac {2 \, e {\left (\frac {5 \, a^{3} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}\right )}}{e^{4}} + \frac {a^{3} e^{2} + \frac {5 \, a^{3} e^{2}}{\tan \left (d x + c\right )} + \frac {10 \, a^{3} e^{2}}{\tan \left (d x + c\right )^{2}}}{e^{4} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}}}\right )}}{5 \, d} \]

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*e*(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)
*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e))/sqrt(e))/e^4 + (a^3*e^2 + 5*a^3*e^2/t
an(d*x + c) + 10*a^3*e^2/tan(d*x + c)^2)/(e^4*(e/tan(d*x + c))^(5/2)))/d

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

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

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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