3.1.38 \(\int \frac {1}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^3} \, dx\) [38]

Optimal. Leaf size=165 \[ -\frac {11 \text {ArcTan}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d \sqrt {e}}-\frac {\text {ArcTan}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d \sqrt {e}}-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (1+\cot (c+d x))}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2} \]

[Out]

-11/8*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a^3/d/e^(1/2)-1/4*arctan(1/2*(e^(1/2)-cot(d*x+c)*e^(1/2))*2^(1/2)/(
e*cot(d*x+c))^(1/2))/a^3/d*2^(1/2)/e^(1/2)-7/8*(e*cot(d*x+c))^(1/2)/a^3/d/e/(1+cot(d*x+c))-1/4*(e*cot(d*x+c))^
(1/2)/a/d/e/(a+a*cot(d*x+c))^2

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Rubi [A]
time = 0.43, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3650, 3730, 3734, 3613, 211, 3715, 65} \begin {gather*} -\frac {11 \text {ArcTan}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d \sqrt {e}}-\frac {\text {ArcTan}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d \sqrt {e}}-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (\cot (c+d x)+1)}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a \cot (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Cot[c + d*x]]*(a + a*Cot[c + d*x])^3),x]

[Out]

(-11*ArcTan[Sqrt[e*Cot[c + d*x]]/Sqrt[e]])/(8*a^3*d*Sqrt[e]) - ArcTan[(Sqrt[e] - Sqrt[e]*Cot[c + d*x])/(Sqrt[2
]*Sqrt[e*Cot[c + d*x]])]/(2*Sqrt[2]*a^3*d*Sqrt[e]) - (7*Sqrt[e*Cot[c + d*x]])/(8*a^3*d*e*(1 + Cot[c + d*x])) -
Sqrt[e*Cot[c + d*x]]/(4*a*d*e*(a + a*Cot[c + d*x])^2)

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 3650

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 + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*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] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !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 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, 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] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

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 {1}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^3} \, dx &=-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2}-\frac {\int \frac {-\frac {7 a^2 e}{2}+2 a^2 e \cot (c+d x)-\frac {3}{2} a^2 e \cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))^2} \, dx}{4 a^3 e}\\ &=-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (1+\cot (c+d x))}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2}+\frac {\int \frac {\frac {7 a^4 e^2}{2}-4 a^4 e^2 \cot (c+d x)+\frac {7}{2} a^4 e^2 \cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{8 a^6 e^2}\\ &=-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (1+\cot (c+d x))}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2}+\frac {11 \int \frac {1+\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (a+a \cot (c+d x))} \, dx}{16 a^2}+\frac {\int \frac {-4 a^5 e^2-4 a^5 e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}} \, dx}{16 a^8 e^2}\\ &=-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (1+\cot (c+d x))}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2}+\frac {11 \text {Subst}\left (\int \frac {1}{\sqrt {-e x} (a-a x)} \, dx,x,-\cot (c+d x)\right )}{16 a^2 d}-\frac {\left (2 a^2 e^2\right ) \text {Subst}\left (\int \frac {1}{-32 a^{10} e^4-e x^2} \, dx,x,\frac {-4 a^5 e^2+4 a^5 e^2 \cot (c+d x)}{\sqrt {e \cot (c+d x)}}\right )}{d}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d \sqrt {e}}-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (1+\cot (c+d x))}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2}-\frac {11 \text {Subst}\left (\int \frac {1}{a+\frac {a x^2}{e}} \, dx,x,\sqrt {e \cot (c+d x)}\right )}{8 a^2 d e}\\ &=-\frac {11 \tan ^{-1}\left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{8 a^3 d \sqrt {e}}-\frac {\tan ^{-1}\left (\frac {\sqrt {e}-\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{2 \sqrt {2} a^3 d \sqrt {e}}-\frac {7 \sqrt {e \cot (c+d x)}}{8 a^3 d e (1+\cot (c+d x))}-\frac {\sqrt {e \cot (c+d x)}}{4 a d e (a+a \cot (c+d x))^2}\\ \end {align*}

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Mathematica [A]
time = 1.32, size = 217, normalized size = 1.32 \begin {gather*} \frac {\sqrt {\cot (c+d x)} \left (-22 \text {ArcTan}\left (\sqrt {\cot (c+d x)}\right )-9 \sqrt {\cot (c+d x)}+9 \cos (2 (c+d x)) \sqrt {\cot (c+d x)}-4 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right ) (\cos (c+d x)+\sin (c+d x))^2+4 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right ) (\cos (c+d x)+\sin (c+d x))^2-22 \text {ArcTan}\left (\sqrt {\cot (c+d x)}\right ) \sin (2 (c+d x))-7 \sqrt {\cot (c+d x)} \sin (2 (c+d x))\right )}{16 a^3 d \sqrt {e \cot (c+d x)} (\cos (c+d x)+\sin (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Cot[c + d*x]]*(a + a*Cot[c + d*x])^3),x]

[Out]

(Sqrt[Cot[c + d*x]]*(-22*ArcTan[Sqrt[Cot[c + d*x]]] - 9*Sqrt[Cot[c + d*x]] + 9*Cos[2*(c + d*x)]*Sqrt[Cot[c + d
*x]] - 4*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]*(Cos[c + d*x] + Sin[c + d*x])^2 + 4*Sqrt[2]*ArcTan[1 +
Sqrt[2]*Sqrt[Cot[c + d*x]]]*(Cos[c + d*x] + Sin[c + d*x])^2 - 22*ArcTan[Sqrt[Cot[c + d*x]]]*Sin[2*(c + d*x)]
- 7*Sqrt[Cot[c + d*x]]*Sin[2*(c + d*x)]))/(16*a^3*d*Sqrt[e*Cot[c + d*x]]*(Cos[c + d*x] + Sin[c + d*x])^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(136)=272\).
time = 0.60, size = 349, normalized size = 2.12

method	result	size
derivativedivides	\(-\frac {2 e^{4} \left (\frac {-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}}{4 e^{4}}+\frac {\frac {\frac {7 \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{4}+\frac {9 e \sqrt {e \cot \left (d x +c \right )}}{4}}{\left (e \cot \left (d x +c \right )+e \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{4 \sqrt {e}}}{4 e^{4}}\right )}{d \,a^{3}}\)	\(349\)
default	\(-\frac {2 e^{4} \left (\frac {-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}-\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}}{4 e^{4}}+\frac {\frac {\frac {7 \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}{4}+\frac {9 e \sqrt {e \cot \left (d x +c \right )}}{4}}{\left (e \cot \left (d x +c \right )+e \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{4 \sqrt {e}}}{4 e^{4}}\right )}{d \,a^{3}}\)	\(349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-2/d/a^3*e^4*(1/4/e^4*(-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)))+1/4/e^4*((7/4*(e*cot(d*x+c))^(3/2)+9/4*e*(e*cot(d*x+c))^(1/2))/(e*cot(d*x+c)+e)^2+11/4/
e^(1/2)*arctan((e*cot(d*x+c))^(1/2)/e^(1/2))))

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Maxima [A]
time = 0.54, size = 129, normalized size = 0.78 \begin {gather*} -\frac {{\left (\frac {\frac {9}{\sqrt {\tan \left (d x + c\right )}} + \frac {7}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{a^{3} + \frac {2 \, a^{3}}{\tan \left (d x + c\right )} + \frac {a^{3}}{\tan \left (d x + c\right )^{2}}} - \frac {2 \, {\left (\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 )\right )}}{a^{3}} + \frac {11 \, \arctan \left (\frac {1}{\sqrt {\tan \left (d x + c\right )}}\right )}{a^{3}}\right )} e^{\left (-\frac {1}{2}\right )}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/8*((9/sqrt(tan(d*x + c)) + 7/tan(d*x + c)^(3/2))/(a^3 + 2*a^3/tan(d*x + c) + a^3/tan(d*x + c)^2) - 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^3 + 11*arctan(1/sqrt(tan(d*x + c)))/a^3)*e^(-1/2)/d

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Fricas [A]
time = 3.50, size = 233, normalized size = 1.41 \begin {gather*} -\frac {4 \, {\left (\sqrt {2} \sin \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} \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 ) - 22 \, {\left (\sin \left (2 \, d x + 2 \, c\right ) + 1\right )} \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 ) - \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (9 \, \cos \left (2 \, d x + 2 \, c\right ) - 7 \, \sin \left (2 \, d x + 2 \, c\right ) - 9\right )}}{16 \, {\left (a^{3} d e^{\frac {1}{2}} \sin \left (2 \, d x + 2 \, c\right ) + a^{3} d e^{\frac {1}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/16*(4*(sqrt(2)*sin(2*d*x + 2*c) + 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)) - 22*(sin(2*d*x + 2*c) + 1)*
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)) - sqrt((cos(2*d*
x + 2*c) + 1)/sin(2*d*x + 2*c))*(9*cos(2*d*x + 2*c) - 7*sin(2*d*x + 2*c) - 9))/(a^3*d*e^(1/2)*sin(2*d*x + 2*c)
+ a^3*d*e^(1/2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{\sqrt {e \cot {\left (c + d x \right )}} \cot ^{3}{\left (c + d x \right )} + 3 \sqrt {e \cot {\left (c + d x \right )}} \cot ^{2}{\left (c + d x \right )} + 3 \sqrt {e \cot {\left (c + d x \right )}} \cot {\left (c + d x \right )} + \sqrt {e \cot {\left (c + d x \right )}}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))**(1/2)/(a+a*cot(d*x+c))**3,x)

[Out]

Integral(1/(sqrt(e*cot(c + d*x))*cot(c + d*x)**3 + 3*sqrt(e*cot(c + d*x))*cot(c + d*x)**2 + 3*sqrt(e*cot(c + d
*x))*cot(c + d*x) + sqrt(e*cot(c + d*x))), x)/a**3

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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(1/(e*cot(d*x+c))^(1/2)/(a+a*cot(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((a*cot(d*x + c) + a)^3*sqrt(e*cot(d*x + c))), x)

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Mupad [B]
time = 0.94, size = 173, normalized size = 1.05 \begin {gather*} \frac {\sqrt {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 )}{8\,a^3\,d\,\sqrt {e}}-\frac {11\,\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{8\,a^3\,d\,\sqrt {e}}-\frac {\frac {9\,e\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{8}+\frac {7\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}}{8}}{d\,a^3\,e^2\,{\mathrm {cot}\left (c+d\,x\right )}^2+2\,d\,a^3\,e^2\,\mathrm {cot}\left (c+d\,x\right )+d\,a^3\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cot(c + d*x))^(1/2)*(a + a*cot(c + d*x))^3),x)

[Out]

(2^(1/2)*(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)))))/(8*a^3*d*e^(1/2)) - (11*atan((e*cot(c + d*x))^(1/2)/e
^(1/2)))/(8*a^3*d*e^(1/2)) - ((9*e*(e*cot(c + d*x))^(1/2))/8 + (7*(e*cot(c + d*x))^(3/2))/8)/(a^3*d*e^2 + a^3*
d*e^2*cot(c + d*x)^2 + 2*a^3*d*e^2*cot(c + d*x))

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