3.1.23 \(\int \frac {(e \cot (c+d x))^{5/2}}{a+a \cot (c+d x)} \, dx\) [23]

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

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3647, 3734, 3613, 211, 3715, 65} \begin {gather*} \frac {e^{5/2} \text {ArcTan}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d}-\frac {e^{5/2} \text {ArcTan}\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 {gather*}

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 65

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]

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 3647

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 3715

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 3734

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]

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 \text {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 ) \text {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 \text {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*}

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Mathematica [A]
time = 0.93, size = 110, normalized size = 0.99 \begin {gather*} -\frac {\left (\sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-\sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \text {ArcTan}\left (\sqrt {\cot (c+d x)}\right )+4 \sqrt {\cot (c+d x)}\right ) (e \cot (c+d x))^{5/2}}{2 a d \cot ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(311\) vs. \(2(92)=184\).
time = 0.52, size = 312, normalized size = 2.81

method	result	size
derivativedivides	\(-\frac {2 e^{2} \left (\sqrt {e \cot \left (d x +c \right )}-\frac {e \left (\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{2}-\frac {\sqrt {e}\, \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2}\right )}{d a}\)	\(312\)
default	\(-\frac {2 e^{2} \left (\sqrt {e \cot \left (d x +c \right )}-\frac {e \left (\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}\right )}{2}-\frac {\sqrt {e}\, \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2}\right )}{d a}\)	\(312\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 88, normalized size = 0.79 \begin {gather*} \frac {{\left (\frac {\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 )}{a} + \frac {2 \, \arctan \left (\frac {1}{\sqrt {\tan \left (d x + c\right )}}\right )}{a} - \frac {4}{a \sqrt {\tan \left (d x + c\right )}}\right )} e^{\frac {5}{2}}}{2 \, d} \end {gather*}

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]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (80) = 160\).
time = 4.07, size = 169, normalized size = 1.52 \begin {gather*} -\frac {\sqrt {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 {\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 ) e^{\frac {5}{2}} + 2 \, \arctan \left (\frac {\sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right )}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) e^{\frac {5}{2}} + 4 \, \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} e^{\frac {5}{2}}}{2 \, a d} \end {gather*}

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]

-1/2*(sqrt(2)*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))*e^(5/2) + 2*arctan(sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2
*c))*sin(2*d*x + 2*c)/(cos(2*d*x + 2*c) + 1))*e^(5/2) + 4*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*e^(5/2
))/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\cot {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

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]

Integral((e*cot(c + d*x))**(5/2)/(cot(c + d*x) + 1), x)/a

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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((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)

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Mupad [B]
time = 0.68, size = 123, normalized size = 1.11 \begin {gather*} \frac {e^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d}-\frac {2\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{a\,d}+\frac {\sqrt {2}\,e^{5/2}\,\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 )}{4\,a\,d} \end {gather*}

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]

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

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