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

Optimal. Leaf size=99 \[ -\frac {\sqrt {2} a \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{5/2}}+\frac {2 a}{d e^2 \sqrt {e \cot (c+d x)}}+\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3529, 3532, 205} \[ \frac {2 a}{d e^2 \sqrt {e \cot (c+d x)}}-\frac {\sqrt {2} a \tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{d e^{5/2}}+\frac {2 a}{3 d e (e \cot (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -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]

Rubi steps

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

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Mathematica [C]  time = 0.42, size = 203, normalized size = 2.05 \[ \frac {a \left (-8 \tan ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(c+d x)\right )+6 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-6 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )+8 \tan ^{\frac {3}{2}}(c+d x)+24 \sqrt {\tan (c+d x)}+3 \sqrt {2} \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-3 \sqrt {2} \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )}{12 d \tan ^{\frac {5}{2}}(c+d x) (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 6*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 3*Sq
rt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - 3*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[
c + d*x]] + 24*Sqrt[Tan[c + d*x]] + 8*Tan[c + d*x]^(3/2) - 8*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2]*T
an[c + d*x]^(3/2)))/(12*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))

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fricas [B]  time = 0.82, size = 358, normalized size = 3.62 \[ \left [\frac {3 \, \sqrt {2} {\left (a e \cos \left (2 \, d x + 2 \, c\right ) + a e\right )} \sqrt {-\frac {1}{e}} \log \left (\sqrt {2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sqrt {-\frac {1}{e}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} - 2 \, \sin \left (2 \, d x + 2 \, c\right ) + 1\right ) - 4 \, {\left (a \cos \left (2 \, d x + 2 \, c\right ) - 3 \, a \sin \left (2 \, d x + 2 \, c\right ) - a\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{6 \, {\left (d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + d e^{3}\right )}}, -\frac {\frac {3 \, \sqrt {2} {\left (a e \cos \left (2 \, d x + 2 \, c\right ) + a e\right )} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, \sqrt {e} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )}}\right )}{\sqrt {e}} + 2 \, {\left (a \cos \left (2 \, d x + 2 \, c\right ) - 3 \, a \sin \left (2 \, d x + 2 \, c\right ) - a\right )} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{3 \, {\left (d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + d e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(2)*(a*e*cos(2*d*x + 2*c) + a*e)*sqrt(-1/e)*log(sqrt(2)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x +
2*c))*sqrt(-1/e)*(cos(2*d*x + 2*c) + sin(2*d*x + 2*c) - 1) - 2*sin(2*d*x + 2*c) + 1) - 4*(a*cos(2*d*x + 2*c) -
3*a*sin(2*d*x + 2*c) - a)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*e^3*cos(2*d*x + 2*c) + d*e^3),
-1/3*(3*sqrt(2)*(a*e*cos(2*d*x + 2*c) + a*e)*arctan(-1/2*sqrt(2)*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)/(sqrt(e)*(cos(2*d*x + 2*c) + 1)))/sqrt(e) + 2*(a*cos(2*d*x + 2*c)
- 3*a*sin(2*d*x + 2*c) - a)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c)))/(d*e^3*cos(2*d*x + 2*c) + d*e^3)
]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cot(d*x+c))/(e*cot(d*x+c))^(5/2),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 = 0.36, size = 374, normalized size = 3.78 \[ \frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{4 d \,e^{3}}+\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \,e^{3}}-\frac {a \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \,e^{3}}+\frac {a \sqrt {2}\, \ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )}{4 d \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {a \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}-\frac {a \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \,e^{2} \left (e^{2}\right )^{\frac {1}{4}}}+\frac {2 a}{d \,e^{2} \sqrt {e \cot \left (d x +c \right )}}+\frac {2 a}{3 d e \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.60, size = 123, normalized size = 1.24 \[ \frac {e {\left (\frac {3 \, a {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{\sqrt {e}}\right )}}{e^{3}} + \frac {2 \, {\left (a e + \frac {3 \, a e}{\tan \left (d x + c\right )}\right )}}{e^{3} \left (\frac {e}{\tan \left (d x + c\right )}\right )^{\frac {3}{2}}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.48, size = 103, normalized size = 1.04 \[ \frac {2\,a}{d\,e^2\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}+\frac {2\,a}{3\,d\,e\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,\left (1-\mathrm {i}\right )}{d\,e^{5/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )\,\left (-1-\mathrm {i}\right )}{d\,e^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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