3.1.19 \(\int \frac {(a+a \cot (c+d x))^3}{(e \cot (c+d x))^{3/2}} \, dx\) [19]

Optimal. Leaf size=114 \[ \frac {2 \sqrt {2} a^3 \text {ArcTan}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{3/2}}-\frac {4 a^3 \sqrt {e \cot (c+d x)}}{d e^2}+\frac {2 \left (a^3+a^3 \cot (c+d x)\right )}{d e \sqrt {e \cot (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 114, 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 = {3646, 3711, 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^{3/2}}-\frac {4 a^3 \sqrt {e \cot (c+d x)}}{d e^2}+\frac {2 \left (a^3 \cot (c+d x)+a^3\right )}{d e \sqrt {e \cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Cot[c + d*x])^3/(e*Cot[c + d*x])^(3/2),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^(3/2)) - (4*a^3*S
qrt[e*Cot[c + d*x]])/(d*e^2) + (2*(a^3 + a^3*Cot[c + d*x]))/(d*e*Sqrt[e*Cot[c + d*x]])

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 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 3711

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 3.10, size = 311, normalized size = 2.73 \begin {gather*} \frac {a^3 (1+\cot (c+d x))^3 \left (-4 \cos ^3(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )+\sin (c+d x) \left (-4 \cos ^2(c+d x)+2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right ) \cot ^{\frac {3}{2}}(c+d x) \sin ^2(c+d x)-2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right ) \cot ^{\frac {3}{2}}(c+d x) \sin ^2(c+d x)+\sqrt {2} \cot ^{\frac {3}{2}}(c+d x) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right ) \sin ^2(c+d x)-\sqrt {2} \cot ^{\frac {3}{2}}(c+d x) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right ) \sin ^2(c+d x)+2 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(c+d x)\right ) \sin (2 (c+d x))\right )\right )}{2 d (e \cot (c+d x))^{3/2} (\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])^(3/2),x]

[Out]

(a^3*(1 + Cot[c + d*x])^3*(-4*Cos[c + d*x]^3*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2] + Sin[c + d*x]*(-
4*Cos[c + d*x]^2 + 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]*Cot[c + d*x]^(3/2)*Sin[c + d*x]^2 - 2*Sqrt
[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]*Cot[c + d*x]^(3/2)*Sin[c + d*x]^2 + Sqrt[2]*Cot[c + d*x]^(3/2)*Log[
1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]*Sin[c + d*x]^2 - Sqrt[2]*Cot[c + d*x]^(3/2)*Log[1 + Sqrt[2]*Sqr
t[Cot[c + d*x]] + Cot[c + d*x]]*Sin[c + d*x]^2 + 2*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2]*Sin[2*(c +
d*x)])))/(2*d*(e*Cot[c + d*x])^(3/2)*(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. \(304\) vs. \(2(99)=198\).
time = 0.42, size = 305, normalized size = 2.68

method	result	size
derivativedivides	\(-\frac {2 a^{3} \left (\sqrt {e \cot \left (d x +c \right )}+2 e \left (\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )-\frac {e}{\sqrt {e \cot \left (d x +c \right )}}\right )}{d \,e^{2}}\)	\(305\)
default	\(-\frac {2 a^{3} \left (\sqrt {e \cot \left (d x +c \right )}+2 e \left (\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )-\frac {e}{\sqrt {e \cot \left (d x +c \right )}}\right )}{d \,e^{2}}\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 86, normalized size = 0.75 \begin {gather*} -\frac {2 \, {\left ({\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} - a^{3} \sqrt {\tan \left (d x + c\right )} + \frac {a^{3}}{\sqrt {\tan \left (d x + c\right )}}\right )} e^{\left (-\frac {3}{2}\right )}}{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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.73, size = 178, normalized size = 1.56 \begin {gather*} \frac {2 \, {\left ({\left (\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 (a^{3} \cos \left (2 \, d x + 2 \, c\right ) - a^{3} \sin \left (2 \, d x + 2 \, c\right ) + a^{3}\right )} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}}\right )}}{d \cos \left (2 \, d x + 2 \, c\right ) e^{\frac {3}{2}} + d e^{\frac {3}{2}}} \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))^(3/2),x, algorithm="fricas")

[Out]

2*((sqrt(2)*a^3*cos(2*d*x + 2*c) + sqrt(2)*a^3)*arctan(-1/2*(sqrt(2)*cos(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)) - (a^3*cos(2*d*x + 2*c) -
a^3*sin(2*d*x + 2*c) + a^3)*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c)))/(d*cos(2*d*x + 2*c)*e^(3/2) + d*e^(
3/2))

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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 {3}{2}}}\, dx + \int \frac {3 \cot {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {3 \cot ^{2}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{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))**(3/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.58, size = 119, normalized size = 1.04 \begin {gather*} \frac {2\,a^3}{d\,e\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}-\frac {2\,a^3\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{d\,e^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^{3/2}} \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))^(3/2),x)

[Out]

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

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