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

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

[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))

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Rubi [A]  time = 0.188038, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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 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]

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 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]

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*}

Mathematica [C]  time = 6.11614, size = 417, normalized size = 3.56 \[ -\frac{2 \cos ^3(c+d x) \cot (c+d x) (a \cot (c+d x)+a)^3 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\cot ^2(c+d x)\right )}{3 d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3}+\frac{6 \sin (c+d x) \cos ^2(c+d x) (a \cot (c+d x)+a)^3 \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},-\cot ^2(c+d x)\right )}{d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3}+\frac{2 \sin ^2(c+d x) \cos (c+d x) (a \cot (c+d x)+a)^3 \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\cot ^2(c+d x)\right )}{3 d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3}+\frac{3 \sin ^3(c+d x) \cot ^{\frac{5}{2}}(c+d x) (a \cot (c+d x)+a)^3 \left (\sqrt{2} \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\sqrt{2} \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+2 \left (\sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-\sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )\right )}{4 d (e \cot (c+d x))^{5/2} (\sin (c+d x)+\cos (c+d x))^3} \]

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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Maple [B]  time = 0.03, size = 388, normalized size = 3.3 \begin{align*} -{\frac{{a}^{3}\sqrt{2}}{2\,d{e}^{3}}\sqrt [4]{{e}^{2}}\ln \left ({ \left ( e\cot \left ( dx+c \right ) +\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) \left ( e\cot \left ( dx+c \right ) -\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{{a}^{3}\sqrt{2}}{d{e}^{3}}\sqrt [4]{{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }+{\frac{{a}^{3}\sqrt{2}}{d{e}^{3}}\sqrt [4]{{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }+{\frac{{a}^{3}\sqrt{2}}{2\,d{e}^{2}}\ln \left ({ \left ( e\cot \left ( dx+c \right ) -\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) \left ( e\cot \left ( dx+c \right ) +\sqrt [4]{{e}^{2}}\sqrt{e\cot \left ( dx+c \right ) }\sqrt{2}+\sqrt{{e}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{{a}^{3}\sqrt{2}}{d{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}-{\frac{{a}^{3}\sqrt{2}}{d{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{2\,{a}^{3}}{3\,de} \left ( e\cot \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}+6\,{\frac{{a}^{3}}{d{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cot(d*x+c))^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/(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/(e^2)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-1/d*a^3/e^2/(e^2)^(1/4)*2^(1/2)*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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.59294, size = 917, normalized size = 7.84 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)