3.22 \(\int \frac {(a+a \cot (c+d x))^3}{(e \cot (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{9/2}}-\frac {4 a^3}{d e^4 \sqrt {e \cot (c+d x)}}+\frac {4 a^3}{3 d e^3 (e \cot (c+d x))^{3/2}}+\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{7 d e (e \cot (c+d x))^{7/2}} \]

[Out]

32/35*a^3/d/e^2/(e*cot(d*x+c))^(5/2)+4/3*a^3/d/e^3/(e*cot(d*x+c))^(3/2)+2/7*(a^3+a^3*cot(d*x+c))/d/e/(e*cot(d*
x+c))^(7/2)+2*a^3*arctanh(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^(9/2)-4*a
^3/d/e^4/(e*cot(d*x+c))^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3565, 3628, 3529, 3532, 208} \[ -\frac {4 a^3}{d e^4 \sqrt {e \cot (c+d x)}}+\frac {4 a^3}{3 d e^3 (e \cot (c+d x))^{3/2}}+\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{9/2}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{7 d e (e \cot (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cot[c + d*x])^3/(e*Cot[c + d*x])^(9/2),x]

[Out]

(2*Sqrt[2]*a^3*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/(d*e^(9/2)) + (32*a^3
)/(35*d*e^2*(e*Cot[c + d*x])^(5/2)) + (4*a^3)/(3*d*e^3*(e*Cot[c + d*x])^(3/2)) - (4*a^3)/(d*e^4*Sqrt[e*Cot[c +
d*x]]) + (2*(a^3 + a^3*Cot[c + d*x]))/(7*d*e*(e*Cot[c + d*x])^(7/2))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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))^{9/2}} \, dx &=\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{7 d e (e \cot (c+d x))^{7/2}}-\frac {2 \int \frac {-8 a^3 e^2-7 a^3 e^2 \cot (c+d x)-a^3 e^2 \cot ^2(c+d x)}{(e \cot (c+d x))^{7/2}} \, dx}{7 e^3}\\ &=\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{7 d e (e \cot (c+d x))^{7/2}}-\frac {2 \int \frac {-7 a^3 e^3+7 a^3 e^3 \cot (c+d x)}{(e \cot (c+d x))^{5/2}} \, dx}{7 e^5}\\ &=\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {4 a^3}{3 d e^3 (e \cot (c+d x))^{3/2}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{7 d e (e \cot (c+d x))^{7/2}}-\frac {2 \int \frac {7 a^3 e^4+7 a^3 e^4 \cot (c+d x)}{(e \cot (c+d x))^{3/2}} \, dx}{7 e^7}\\ &=\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {4 a^3}{3 d e^3 (e \cot (c+d x))^{3/2}}-\frac {4 a^3}{d e^4 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{7 d e (e \cot (c+d x))^{7/2}}-\frac {2 \int \frac {7 a^3 e^5-7 a^3 e^5 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{7 e^9}\\ &=\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {4 a^3}{3 d e^3 (e \cot (c+d x))^{3/2}}-\frac {4 a^3}{d e^4 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{7 d e (e \cot (c+d x))^{7/2}}+\frac {\left (28 a^6 e\right ) \operatorname {Subst}\left (\int \frac {1}{98 a^6 e^{10}-e x^2} \, dx,x,\frac {7 a^3 e^5+7 a^3 e^5 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {e}+\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{9/2}}+\frac {32 a^3}{35 d e^2 (e \cot (c+d x))^{5/2}}+\frac {4 a^3}{3 d e^3 (e \cot (c+d x))^{3/2}}-\frac {4 a^3}{d e^4 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{7 d e (e \cot (c+d x))^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 2.00, size = 174, normalized size = 1.05 \[ \frac {2 a^3 \cos (c+d x) (\cot (c+d x)+1)^3 \left (35 \cos ^2(c+d x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )+35 \cos ^2(c+d x) \cot (c+d x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )+5 \sin ^2(c+d x) \, _2F_1\left (-\frac {7}{4},1;-\frac {3}{4};-\cot ^2(c+d x)\right )+\frac {21}{2} \sin (2 (c+d x)) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(c+d x)\right )\right )}{35 d (e \cot (c+d x))^{9/2} (\sin (c+d x)+\cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cot[c + d*x])^3/(e*Cot[c + d*x])^(9/2),x]

[Out]

(2*a^3*Cos[c + d*x]*(1 + Cot[c + d*x])^3*(35*Cos[c + d*x]^2*Hypergeometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2] +
35*Cos[c + d*x]^2*Cot[c + d*x]*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + 5*Hypergeometric2F1[-7/4, 1
, -3/4, -Cot[c + d*x]^2]*Sin[c + d*x]^2 + (21*Hypergeometric2F1[-5/4, 1, -1/4, -Cot[c + d*x]^2]*Sin[2*(c + d*x
)])/2))/(35*d*(e*Cot[c + d*x])^(9/2)*(Cos[c + d*x] + Sin[c + d*x])^3)

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fricas [A]  time = 0.87, size = 514, normalized size = 3.12 \[ \left [\frac {\frac {105 \, \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 )} \log \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 )}}{\sqrt {e}} + 2 \, \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}{\sqrt {e}} - 2 \, {\left (55 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + 30 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) - 85 \, a^{3} + 21 \, {\left (13 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + 7 \, 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 )}}}{105 \, {\left (d e^{5} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, d e^{5} \cos \left (2 \, d x + 2 \, c\right ) + d e^{5}\right )}}, -\frac {2 \, {\left (105 \, \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}} \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 )}} \sqrt {-\frac {1}{e}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )}}\right ) + {\left (55 \, a^{3} \cos \left (2 \, d x + 2 \, c\right )^{2} + 30 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) - 85 \, a^{3} + 21 \, {\left (13 \, a^{3} \cos \left (2 \, d x + 2 \, c\right ) + 7 \, 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 )}}{105 \, {\left (d e^{5} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, d e^{5} \cos \left (2 \, d x + 2 \, c\right ) + d e^{5}\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))^(9/2),x, algorithm="fricas")

[Out]

[1/105*(105*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)*log(-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) + 2*sin(2*d*x + 2*c) + 1)/s
qrt(e) - 2*(55*a^3*cos(2*d*x + 2*c)^2 + 30*a^3*cos(2*d*x + 2*c) - 85*a^3 + 21*(13*a^3*cos(2*d*x + 2*c) + 7*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^5*cos(2*d*x + 2*c)^2 + 2*d*e^5*cos(2
*d*x + 2*c) + d*e^5), -2/105*(105*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)*arctan(1/2*sqrt(2)*sqrt((e*cos(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)/(cos(2*d*x + 2*c) + 1)) + (55*a^3*cos(2*d*x + 2*c)^2 + 30*a^3*cos(2*d*x + 2*c) - 85*a^3 + 21*(1
3*a^3*cos(2*d*x + 2*c) + 7*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^5*cos(
2*d*x + 2*c)^2 + 2*d*e^5*cos(2*d*x + 2*c) + d*e^5)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \cot \left (d x + c\right ) + a\right )}^{3}}{\left (e \cot \left (d x + c\right )\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((a*cot(d*x + c) + a)^3/(e*cot(d*x + c))^(9/2), x)

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maple [B]  time = 0.54, size = 430, normalized size = 2.61 \[ \frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \,e^{5}}+\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{5}}-\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{5}}-\frac {a^{3} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \,e^{4} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{4} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{4} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2 a^{3}}{7 d e \left (e \cot \left (d x +c \right )\right )^{\frac {7}{2}}}+\frac {6 a^{3}}{5 d \,e^{2} \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {4 a^{3}}{d \,e^{4} \sqrt {e \cot \left (d x +c \right )}}+\frac {4 a^{3}}{3 d \,e^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cot(d*x+c)*a)^3/(e*cot(d*x+c))^(9/2),x)

[Out]

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

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maxima [A]  time = 0.85, size = 170, normalized size = 1.03 \[ \frac {e {\left (\frac {105 \, a^{3} {\left (\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )}}{e^{5}} + \frac {2 \, {\left (15 \, a^{3} e^{3} + \frac {63 \, a^{3} e^{3}}{\tan \left (d x + c\right )} + \frac {70 \, a^{3} e^{3}}{\tan \left (d x + c\right )^{2}} - \frac {210 \, a^{3} e^{3}}{\tan \left (d x + c\right )^{3}}\right )}}{e^{5} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {7}{2}}}\right )}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

1/105*e*(105*a^3*(sqrt(2)*log(sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e) - sqrt(2)*log
(-sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/e^5 + 2*(15*a^3*e^3 + 63*a^3*e^3/tan(d*x
+ c) + 70*a^3*e^3/tan(d*x + c)^2 - 210*a^3*e^3/tan(d*x + c)^3)/(e^5*(e/tan(d*x + c))^(7/2)))/d

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mupad [B]  time = 1.92, size = 129, normalized size = 0.78 \[ \frac {-4\,e\,a^3\,{\mathrm {cot}\left (c+d\,x\right )}^3+\frac {4\,e\,a^3\,{\mathrm {cot}\left (c+d\,x\right )}^2}{3}+\frac {6\,e\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{5}+\frac {2\,e\,a^3}{7}}{d\,e^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{7/2}}+\frac {2\,\sqrt {2}\,a^3\,\mathrm {atanh}\left (\frac {32\,\sqrt {2}\,a^6\,d\,e^{9/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{32\,a^6\,d\,e^5+32\,a^6\,d\,e^5\,\mathrm {cot}\left (c+d\,x\right )}\right )}{d\,e^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cot(c + d*x))^3/(e*cot(c + d*x))^(9/2),x)

[Out]

((2*a^3*e)/7 + (4*a^3*e*cot(c + d*x)^2)/3 - 4*a^3*e*cot(c + d*x)^3 + (6*a^3*e*cot(c + d*x))/5)/(d*e^2*(e*cot(c
+ d*x))^(7/2)) + (2*2^(1/2)*a^3*atanh((32*2^(1/2)*a^6*d*e^(9/2)*(e*cot(c + d*x))^(1/2))/(32*a^6*d*e^5 + 32*a^
6*d*e^5*cot(c + d*x))))/(d*e^(9/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {9}{2}}}\, dx + \int \frac {3 \cot {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {9}{2}}}\, dx + \int \frac {3 \cot ^{2}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {9}{2}}}\, dx + \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {9}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))**3/(e*cot(d*x+c))**(9/2),x)

[Out]

a**3*(Integral((e*cot(c + d*x))**(-9/2), x) + Integral(3*cot(c + d*x)/(e*cot(c + d*x))**(9/2), x) + Integral(3
*cot(c + d*x)**2/(e*cot(c + d*x))**(9/2), x) + Integral(cot(c + d*x)**3/(e*cot(c + d*x))**(9/2), x))

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