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

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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 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]

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*}

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Mathematica [C]  time = 0.19, size = 68, normalized size = 0.59 \[ -\frac {2 a e (e \cot (c+d x))^{3/2} \left (5 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right )+3 \cot (c+d x) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\tan ^2(c+d x)\right )\right )}{15 d} \]

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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fricas [A]  time = 0.90, size = 377, normalized size = 3.25 \[ \left [\frac {15 \, \sqrt {2} {\left (a e^{2} \cos \left (2 \, d x + 2 \, c\right ) - a e^{2}\right )} \sqrt {e} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) + 4 \, {\left (18 \, a e^{2} \cos \left (2 \, d x + 2 \, c\right ) + 5 \, a e^{2} \sin \left (2 \, d x + 2 \, c\right ) - 12 \, a e^{2}\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{30 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}}, \frac {15 \, \sqrt {2} {\left (a e^{2} \cos \left (2 \, d x + 2 \, c\right ) - a e^{2}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {2} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) + 2 \, {\left (18 \, a e^{2} \cos \left (2 \, d x + 2 \, c\right ) + 5 \, a e^{2} \sin \left (2 \, d x + 2 \, c\right ) - 12 \, a e^{2}\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{15 \, {\left (d \cos \left (2 \, d x + 2 \, c\right ) - d\right )}}\right ] \]

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

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)

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maple [B]  time = 0.50, size = 388, normalized size = 3.34 \[ -\frac {2 a \left (e \cot \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d}-\frac {2 a e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{3 d}+\frac {2 a \,e^{2} \sqrt {e \cot \left (d x +c \right )}}{d}-\frac {a \,e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{4 d}-\frac {a \,e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d}+\frac {a \,e^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d}+\frac {a \,e^{3} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{4 d \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a \,e^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a \,e^{3} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \left (e^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cot(d*x+c))^(5/2)*(a+cot(d*x+c)*a),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*a/d*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*a/d*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*a/d*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*a/d*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*a/d*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*a/d*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 [A]  time = 0.76, size = 149, normalized size = 1.28 \[ -\frac {{\left (15 \, a e^{2} {\left (\frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{\sqrt {e}}\right )} - \frac {4 \, {\left (15 \, a e^{2} \sqrt {\frac {e}{\tan \left (d x + c\right )}} - 5 \, a e \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}} - 3 \, a \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {5}{2}}\right )}}{e}\right )} e}{30 \, d} \]

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]

-1/30*(15*a*e^2*(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)) - 4*(15*a*e^2*sqrt(e/tan(d*x + c)) - 5*a*
e*(e/tan(d*x + c))^(3/2) - 3*a*(e/tan(d*x + c))^(5/2))/e)*e/d

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mupad [B]  time = 1.98, size = 144, normalized size = 1.24 \[ \frac {2\,a\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{d}-\frac {2\,a\,e\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{3\,d}-\frac {2\,a\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}}{5\,d}+\frac {{\left (-1\right )}^{1/4}\,a\,e^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {e}}\right )}{d}-\frac {{\left (-1\right )}^{1/4}\,a\,e^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,e^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,\left (1+1{}\mathrm {i}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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