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

Optimal. Leaf size=116 \[ \frac{2 a e^2 \sqrt{e \cot (c+d x)}}{d}-\frac{\sqrt{2} a e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} \cot (c+d x)+\sqrt{e}}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{d}-\frac{2 a e (e \cot (c+d x))^{3/2}}{3 d}-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d} \]

[Out]

-((Sqrt[2]*a*e^(5/2)*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/d) + (2*a*e^2*S
qrt[e*Cot[c + d*x]])/d - (2*a*e*(e*Cot[c + d*x])^(3/2))/(3*d) - (2*a*(e*Cot[c + d*x])^(5/2))/(5*d)

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Rubi [A]  time = 0.157382, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3528, 3532, 208} \[ \frac{2 a e^2 \sqrt{e \cot (c+d x)}}{d}-\frac{\sqrt{2} a e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} \cot (c+d x)+\sqrt{e}}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{d}-\frac{2 a e (e \cot (c+d x))^{3/2}}{3 d}-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*a*e^(5/2)*ArcTanh[(Sqrt[e] + Sqrt[e]*Cot[c + d*x])/(Sqrt[2]*Sqrt[e*Cot[c + d*x]])])/d) + (2*a*e^2*S
qrt[e*Cot[c + d*x]])/d - (2*a*e*(e*Cot[c + d*x])^(3/2))/(3*d) - (2*a*(e*Cot[c + d*x])^(5/2))/(5*d)

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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 (e \cot (c+d x))^{5/2} (a+a \cot (c+d x)) \, dx &=-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d}+\int (e \cot (c+d x))^{3/2} (-a e+a e \cot (c+d x)) \, dx\\ &=-\frac{2 a e (e \cot (c+d x))^{3/2}}{3 d}-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d}+\int \sqrt{e \cot (c+d x)} \left (-a e^2-a e^2 \cot (c+d x)\right ) \, dx\\ &=\frac{2 a e^2 \sqrt{e \cot (c+d x)}}{d}-\frac{2 a e (e \cot (c+d x))^{3/2}}{3 d}-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d}+\int \frac{a e^3-a e^3 \cot (c+d x)}{\sqrt{e \cot (c+d x)}} \, dx\\ &=\frac{2 a e^2 \sqrt{e \cot (c+d x)}}{d}-\frac{2 a e (e \cot (c+d x))^{3/2}}{3 d}-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d}-\frac{\left (2 a^2 e^6\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2 e^6-e x^2} \, dx,x,\frac{a e^3+a e^3 \cot (c+d x)}{\sqrt{e \cot (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} a e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e}+\sqrt{e} \cot (c+d x)}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{d}+\frac{2 a e^2 \sqrt{e \cot (c+d x)}}{d}-\frac{2 a e (e \cot (c+d x))^{3/2}}{3 d}-\frac{2 a (e \cot (c+d x))^{5/2}}{5 d}\\ \end{align*}

Mathematica [C]  time = 0.181304, size = 68, normalized size = 0.59 \[ -\frac{2 a e (e \cot (c+d x))^{3/2} \left (5 \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )+3 \cot (c+d x) \text{Hypergeometric2F1}\left (-\frac{5}{4},1,-\frac{1}{4},-\tan ^2(c+d x)\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*e*(e*Cot[c + d*x])^(3/2)*(3*Cot[c + d*x]*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[c + d*x]^2] + 5*Hypergeom
etric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2]))/(15*d)

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Maple [B]  time = 0.041, size = 388, normalized size = 3.3 \begin{align*} -{\frac{2\,a}{5\,d} \left ( e\cot \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{2\,ae}{3\,d} \left ( e\cot \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+2\,{\frac{a{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}{d}}-{\frac{a{e}^{2}\sqrt{2}}{4\,d}\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{e}^{2}\sqrt{2}}{2\,d}\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{e}^{2}\sqrt{2}}{2\,d}\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{e}^{3}\sqrt{2}}{4\,d}\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{e}^{3}\sqrt{2}}{2\,d}\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{e}^{3}\sqrt{2}}{2\,d}\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}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95518, size = 936, normalized size = 8.07 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left (a e^{2} \cos \left (2 \, d x + 2 \, c\right ) - a e^{2}\right )} \sqrt{e} \log \left (\sqrt{2} \sqrt{e} \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 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) + 4 \,{\left (18 \, a e^{2} \cos \left (2 \, d x + 2 \, c\right ) + 5 \, a e^{2} \sin \left (2 \, d x + 2 \, c\right ) - 12 \, a e^{2}\right )} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{30 \,{\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}}, \frac{15 \, \sqrt{2}{\left (a e^{2} \cos \left (2 \, d x + 2 \, c\right ) - a e^{2}\right )} \sqrt{-e} \arctan \left (\frac{\sqrt{2} \sqrt{-e} \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 \,{\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) + 2 \,{\left (18 \, a e^{2} \cos \left (2 \, d x + 2 \, c\right ) + 5 \, a e^{2} \sin \left (2 \, d x + 2 \, c\right ) - 12 \, a e^{2}\right )} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{15 \,{\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*sqrt(2)*(a*e^2*cos(2*d*x + 2*c) - a*e^2)*sqrt(e)*log(sqrt(2)*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) + e)/s
in(2*d*x + 2*c))*(cos(2*d*x + 2*c) - sin(2*d*x + 2*c) - 1) + 2*e*sin(2*d*x + 2*c) + e) + 4*(18*a*e^2*cos(2*d*x
+ 2*c) + 5*a*e^2*sin(2*d*x + 2*c) - 12*a*e^2)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*cos(2*d*x +
2*c) - d), 1/15*(15*sqrt(2)*(a*e^2*cos(2*d*x + 2*c) - a*e^2)*sqrt(-e)*arctan(1/2*sqrt(2)*sqrt(-e)*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)/(e*cos(2*d*x + 2*c) + e)) + 2*(
18*a*e^2*cos(2*d*x + 2*c) + 5*a*e^2*sin(2*d*x + 2*c) - 12*a*e^2)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c
)))/(d*cos(2*d*x + 2*c) - d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cot \left (d x + c\right ) + a\right )} \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((e*cot(d*x+c))^(5/2)*(a+a*cot(d*x+c)),x, algorithm="giac")

[Out]

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