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

Optimal. Leaf size=117 \[ -\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{5/2}}+\frac {16 a^3}{3 d e^2 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{3 d e (e \cot (c+d x))^{3/2}} \]

[Out]

2/3*(a^3+a^3*cot(d*x+c))/d/e/(e*cot(d*x+c))^(3/2)-2*a^3*arctanh(1/2*(e^(1/2)+cot(d*x+c)*e^(1/2))*2^(1/2)/(e*co
t(d*x+c))^(1/2))*2^(1/2)/d/e^(5/2)+16/3*a^3/d/e^2/(e*cot(d*x+c))^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3565, 3628, 3532, 208} \[ \frac {16 a^3}{3 d e^2 \sqrt {e \cot (c+d x)}}-\frac {2 \sqrt {2} a^3 \tanh ^{-1}\left (\frac {\sqrt {e} \cot (c+d x)+\sqrt {e}}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{5/2}}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{3 d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cot[c + d*x])^3/(e*Cot[c + d*x])^(5/2),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^(5/2)) + (16*a^
3)/(3*d*e^2*Sqrt[e*Cot[c + d*x]]) + (2*(a^3 + a^3*Cot[c + d*x]))/(3*d*e*(e*Cot[c + d*x])^(3/2))

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 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))^{5/2}} \, dx &=\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{3 d e (e \cot (c+d x))^{3/2}}-\frac {2 \int \frac {-4 a^3 e^2-3 a^3 e^2 \cot (c+d x)-a^3 e^2 \cot ^2(c+d x)}{(e \cot (c+d x))^{3/2}} \, dx}{3 e^3}\\ &=\frac {16 a^3}{3 d e^2 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{3 d e (e \cot (c+d x))^{3/2}}-\frac {2 \int \frac {-3 a^3 e^3+3 a^3 e^3 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{3 e^5}\\ &=\frac {16 a^3}{3 d e^2 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{3 d e (e \cot (c+d x))^{3/2}}+\frac {\left (12 a^6 e\right ) \operatorname {Subst}\left (\int \frac {1}{18 a^6 e^6-e x^2} \, dx,x,\frac {-3 a^3 e^3-3 a^3 e^3 \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^{5/2}}+\frac {16 a^3}{3 d e^2 \sqrt {e \cot (c+d x)}}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{3 d e (e \cot (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 6.11, size = 417, normalized size = 3.56 \[ -\frac {2 \cos ^3(c+d x) \cot (c+d x) (a \cot (c+d x)+a)^3 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )}{3 d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3}+\frac {6 \sin (c+d x) \cos ^2(c+d x) (a \cot (c+d x)+a)^3 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right )}{d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3}+\frac {2 \sin ^2(c+d x) \cos (c+d x) (a \cot (c+d x)+a)^3 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )}{3 d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3}+\frac {3 \sin ^3(c+d x) \cot ^{\frac {5}{2}}(c+d x) (a \cot (c+d x)+a)^3 \left (\sqrt {2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )\right )}{4 d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^3*Cot[c + d*x]*(a + a*Cot[c + d*x])^3*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2])/(3*d*(
e*Cot[c + d*x])^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3) + (6*Cos[c + d*x]^2*(a + a*Cot[c + d*x])^3*Hypergeometr
ic2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2]*Sin[c + d*x])/(d*(e*Cot[c + d*x])^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3)
+ (2*Cos[c + d*x]*(a + a*Cot[c + d*x])^3*Hypergeometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2]*Sin[c + d*x]^2)/(3*d
*(e*Cot[c + d*x])^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3) + (3*Cot[c + d*x]^(5/2)*(a + a*Cot[c + d*x])^3*(2*(Sq
rt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]) + Sqrt[2]*Log[1
- Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])*Si
n[c + d*x]^3)/(4*d*(e*Cot[c + d*x])^(5/2)*(Cos[c + d*x] + Sin[c + d*x])^3)

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fricas [A]  time = 1.54, size = 378, normalized size = 3.23 \[ \left [\frac {\frac {3 \, \sqrt {2} {\left (a^{3} e \cos \left (2 \, d x + 2 \, c\right ) + a^{3} e\right )} \log \left (\frac {\sqrt {2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1\right )}}{\sqrt {e}} + 2 \, \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}{\sqrt {e}} - 2 \, {\left (a^{3} \cos \left (2 \, d x + 2 \, c\right ) - 9 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - a^{3}\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{3 \, {\left (d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + d e^{3}\right )}}, \frac {2 \, {\left (3 \, \sqrt {2} {\left (a^{3} e \cos \left (2 \, d x + 2 \, c\right ) + a^{3} e\right )} \sqrt {-\frac {1}{e}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sqrt {-\frac {1}{e}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )}}\right ) - {\left (a^{3} \cos \left (2 \, d x + 2 \, c\right ) - 9 \, a^{3} \sin \left (2 \, d x + 2 \, c\right ) - a^{3}\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}\right )}}{3 \, {\left (d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + d e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))^3/(e*cot(d*x+c))^(5/2),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 = 0.50, size = 388, normalized size = 3.32 \[ -\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \,e^{3}}-\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{3}}+\frac {a^{3} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{3}}+\frac {a^{3} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{2 d \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{d \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2 a^{3}}{3 d e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {6 a^{3}}{d \,e^{2} \sqrt {e \cot \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a^3/e^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*c
ot(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*(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^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*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*2^(1/2)/(e^2)^(1/4)*ar
ctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/d*a^3/e^2*2^(1/2)/(e^2)^(1/4)*arctan(-2^(1/2)/(e^2)^(1/4)*(
e*cot(d*x+c))^(1/2)+1)+2/3/d*a^3/e/(e*cot(d*x+c))^(3/2)+6*a^3/d/e^2/(e*cot(d*x+c))^(1/2)

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maxima [A]  time = 0.79, size = 133, normalized size = 1.14 \[ -\frac {e {\left (\frac {3 \, a^{3} {\left (\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )}}{e^{3}} - \frac {2 \, {\left (a^{3} e + \frac {9 \, a^{3} e}{\tan \left (d x + c\right )}\right )}}{e^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*e*(3*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(-s
qrt(2)*sqrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/sqrt(e))/e^3 - 2*(a^3*e + 9*a^3*e/tan(d*x + c))/(e^3
*(e/tan(d*x + c))^(3/2)))/d

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mupad [B]  time = 0.71, size = 101, normalized size = 0.86 \[ \frac {\frac {2\,a^3\,e}{3}+6\,a^3\,e\,\mathrm {cot}\left (c+d\,x\right )}{d\,e^2\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}-\frac {2\,\sqrt {2}\,a^3\,\mathrm {atanh}\left (\frac {32\,\sqrt {2}\,a^6\,d\,e^{5/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{32\,a^6\,d\,e^3+32\,a^6\,d\,e^3\,\mathrm {cot}\left (c+d\,x\right )}\right )}{d\,e^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {3 \cot {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {3 \cot ^{2}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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