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

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

[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]])

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Rubi [A]  time = 0.134629, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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]

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]

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*}

Mathematica [C]  time = 0.430751, size = 203, normalized size = 2.05 \[ \frac{a \left (-8 \tan ^{\frac{3}{2}}(c+d x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\tan ^2(c+d x)\right )+8 \tan ^{\frac{3}{2}}(c+d x)+6 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )-6 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )+24 \sqrt{\tan (c+d x)}+3 \sqrt{2} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-3 \sqrt{2} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )}{12 d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}} \]

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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Maple [B]  time = 0.02, size = 374, normalized size = 3.8 \begin{align*}{\frac{a\sqrt{2}}{4\,d{e}^{3}}\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\sqrt{2}}{2\,d{e}^{3}}\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\sqrt{2}}{2\,d{e}^{3}}\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\sqrt{2}}{4\,d{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{1}{\sqrt [4]{{e}^{2}}}}}+{\frac{a\sqrt{2}}{2\,d{e}^{2}}\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\sqrt{2}}{2\,d{e}^{2}}\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}}}}}+2\,{\frac{a}{d{e}^{2}\sqrt{e\cot \left ( dx+c \right ) }}}+{\frac{2\,a}{3\,de} \left ( e\cot \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/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)))+1/2/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)-1/2/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^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)+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 [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((a+a*cot(d*x+c))/(e*cot(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.04732, size = 898, normalized size = 9.07 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a \cot \left (d x + c\right ) + a}{\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((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)