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

Optimal. Leaf size=111 \[ -\frac{2 e^2 \sqrt{e \cot (c+d x)}}{a d}+\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{a d}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e}-\sqrt{e} \cot (c+d x)}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{\sqrt{2} a d} \]

[Out]

(e^(5/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(a*d) - (e^(5/2)*ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[
2]*Sqrt[e*Cot[c + d*x]])])/(Sqrt[2]*a*d) - (2*e^2*Sqrt[e*Cot[c + d*x]])/(a*d)

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Rubi [A]  time = 0.451404, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3566, 3653, 3532, 205, 3634, 63} \[ -\frac{2 e^2 \sqrt{e \cot (c+d x)}}{a d}+\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e \cot (c+d x)}}{\sqrt{e}}\right )}{a d}-\frac{e^{5/2} \tan ^{-1}\left (\frac{\sqrt{e}-\sqrt{e} \cot (c+d x)}{\sqrt{2} \sqrt{e \cot (c+d x)}}\right )}{\sqrt{2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^(5/2)*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(a*d) - (e^(5/2)*ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[
2]*Sqrt[e*Cot[c + d*x]])])/(Sqrt[2]*a*d) - (2*e^2*Sqrt[e*Cot[c + d*x]])/(a*d)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
 x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3653

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

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
 /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rubi steps

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

Mathematica [A]  time = 0.847899, size = 110, normalized size = 0.99 \[ -\frac{(e \cot (c+d x))^{5/2} \left (4 \sqrt{\cot (c+d x)}+\sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-\sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-2 \tan ^{-1}\left (\sqrt{\cot (c+d x)}\right )\right )}{2 a d \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*ArcTan[
Sqrt[Cot[c + d*x]]] + 4*Sqrt[Cot[c + d*x]])*(e*Cot[c + d*x])^(5/2))/(2*a*d*Cot[c + d*x]^(5/2))

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Maple [B]  time = 0.033, size = 394, normalized size = 3.6 \begin{align*} -2\,{\frac{{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}{ad}}+{\frac{{e}^{2}\sqrt{2}}{8\,ad}\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{{e}^{2}\sqrt{2}}{4\,ad}\sqrt [4]{{e}^{2}}\arctan \left ({\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }-{\frac{{e}^{2}\sqrt{2}}{4\,ad}\sqrt [4]{{e}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{e}^{2}}}}}+1 \right ) }+{\frac{{e}^{3}\sqrt{2}}{8\,ad}\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{{e}^{3}\sqrt{2}}{4\,ad}\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{{e}^{3}\sqrt{2}}{4\,ad}\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{1}{ad}{e}^{{\frac{5}{2}}}\arctan \left ({\sqrt{e\cot \left ( dx+c \right ) }{\frac{1}{\sqrt{e}}}} \right ) } \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*e^2*(e*cot(d*x+c))^(1/2)/a/d+1/8/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)))+1/4/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^2*(e^2)^(1/4)*2^(1/2)*arctan(-2^(
1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+1/8/d/a*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*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+1/4/d/a
*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)-1/4/d/a*e^3/(e^2)^(1/4)*2^(1/2)*ar
ctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)+e^(5/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d

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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 = 1.40146, size = 1019, normalized size = 9.18 \begin{align*} \left [\frac{\sqrt{2} \sqrt{-e} e^{2} \log \left ({\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{-e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} - 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) + 2 \, \sqrt{-e} e^{2} \log \left (\frac{e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) + 2 \, \sqrt{-e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right ) - 8 \, e^{2} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{4 \, a d}, -\frac{\sqrt{2} e^{\frac{5}{2}} \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{e} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \,{\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) - 2 \, e^{\frac{5}{2}} \arctan \left (\frac{\sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt{e}}\right ) + 4 \, e^{2} \sqrt{\frac{e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{2 \, a d}\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]

[1/4*(sqrt(2)*sqrt(-e)*e^2*log((sqrt(2)*cos(2*d*x + 2*c) + sqrt(2)*sin(2*d*x + 2*c) - sqrt(2))*sqrt(-e)*sqrt((
e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)) - 2*e*sin(2*d*x + 2*c) + e) + 2*sqrt(-e)*e^2*log((e*cos(2*d*x + 2*c)
 - e*sin(2*d*x + 2*c) + 2*sqrt(-e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + e)/(cos(
2*d*x + 2*c) + sin(2*d*x + 2*c) + 1)) - 8*e^2*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a*d), -1/2*(sq
rt(2)*e^(5/2)*arctan(-1/2*(sqrt(2)*cos(2*d*x + 2*c) - sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*sqrt(e)*sqrt((e*cos(
2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/(e*cos(2*d*x + 2*c) + e)) - 2*e^(5/2)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)
/sin(2*d*x + 2*c))/sqrt(e)) + 4*e^2*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(a*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 \frac{\left (e \cot \left (d x + c\right )\right )^{\frac{5}{2}}}{a \cot \left (d x + c\right ) + a}\,{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((e*cot(d*x + c))^(5/2)/(a*cot(d*x + c) + a), x)