3.206 \(\int \cot ^2(e+f x) (d \cot (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=232 \[ \frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}+\frac {2 d \sqrt {d \cot (e+f x)}}{f} \]

[Out]

-2/5*(d*cot(f*x+e))^(5/2)/d/f+1/2*d^(3/2)*arctan(1-2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/2*d^(3/2)
*arctan(1+2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)+1/4*d^(3/2)*ln(d^(1/2)+cot(f*x+e)*d^(1/2)-2^(1/2)*(d
*cot(f*x+e))^(1/2))/f*2^(1/2)-1/4*d^(3/2)*ln(d^(1/2)+cot(f*x+e)*d^(1/2)+2^(1/2)*(d*cot(f*x+e))^(1/2))/f*2^(1/2
)+2*d*(d*cot(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.20, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {16, 3473, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}+\frac {2 d \sqrt {d \cot (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2*(d*Cot[e + f*x])^(3/2),x]

[Out]

(d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (2*d*Sqrt[d*Cot[e + f*x]])/f - (2*(d*Cot[e + f*x])^(5/2))/(5*d*f) + (d
^(3/2)*Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x] - Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f) - (d^(3/2)*Log[Sqrt[d
] + Sqrt[d]*Cot[e + f*x] + Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \cot ^2(e+f x) (d \cot (e+f x))^{3/2} \, dx &=\frac {\int (d \cot (e+f x))^{7/2} \, dx}{d^2}\\ &=-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}-\int (d \cot (e+f x))^{3/2} \, dx\\ &=\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}+d^2 \int \frac {1}{\sqrt {d \cot (e+f x)}} \, dx\\ &=\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}-\frac {d^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \cot (e+f x)\right )}{f}\\ &=\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}-\frac {\left (2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}+\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 f}\\ &=\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}\\ &=\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {2 d \sqrt {d \cot (e+f x)}}{f}-\frac {2 (d \cot (e+f x))^{5/2}}{5 d f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 172, normalized size = 0.74 \[ \frac {(d \cot (e+f x))^{3/2} \left (-8 \cot ^{\frac {5}{2}}(e+f x)+40 \sqrt {\cot (e+f x)}+5 \sqrt {2} \log \left (\cot (e+f x)-\sqrt {2} \sqrt {\cot (e+f x)}+1\right )-5 \sqrt {2} \log \left (\cot (e+f x)+\sqrt {2} \sqrt {\cot (e+f x)}+1\right )+10 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (e+f x)}\right )-10 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (e+f x)}+1\right )\right )}{20 f \cot ^{\frac {3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2*(d*Cot[e + f*x])^(3/2),x]

[Out]

((d*Cot[e + f*x])^(3/2)*(10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[e + f*x]]] - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqr
t[Cot[e + f*x]]] + 40*Sqrt[Cot[e + f*x]] - 8*Cot[e + f*x]^(5/2) + 5*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[e + f*x]]
+ Cot[e + f*x]] - 5*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[e + f*x]] + Cot[e + f*x]]))/(20*f*Cot[e + f*x]^(3/2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cot \left (f x + e\right )\right )^{\frac {3}{2}} \cot \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(d*cot(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*cot(f*x + e))^(3/2)*cot(f*x + e)^2, x)

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maple [A]  time = 0.16, size = 194, normalized size = 0.84 \[ -\frac {2 \left (d \cot \left (f x +e \right )\right )^{\frac {5}{2}}}{5 d f}+\frac {2 d \sqrt {d \cot \left (f x +e \right )}}{f}-\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {d \cot \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}+\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \cot \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 f}-\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {d \cot \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \cot \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \cot \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \cot \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )}{4 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2*(d*cot(f*x+e))^(3/2),x)

[Out]

-2/5*(d*cot(f*x+e))^(5/2)/d/f+2*d*(d*cot(f*x+e))^(1/2)/f-1/2/f*d*(d^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(d^2)^(1/4
)*(d*cot(f*x+e))^(1/2)+1)+1/2/f*d*(d^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(d^2)^(1/4)*(d*cot(f*x+e))^(1/2)+1)-1/4/
f*d*(d^2)^(1/4)*2^(1/2)*ln((d*cot(f*x+e)+(d^2)^(1/4)*(d*cot(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*cot(f*x+e)-(
d^2)^(1/4)*(d*cot(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))

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maxima [A]  time = 0.51, size = 199, normalized size = 0.86 \[ -\frac {10 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right ) + 10 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right ) + 5 \, \sqrt {2} d^{\frac {5}{2}} \log \left (\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right ) - 5 \, \sqrt {2} d^{\frac {5}{2}} \log \left (-\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right ) - 40 \, d^{2} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + 8 \, \left (\frac {d}{\tan \left (f x + e\right )}\right )^{\frac {5}{2}}}{20 \, d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(10*sqrt(2)*d^(5/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d/tan(f*x + e)))/sqrt(d)) + 10*sqrt(2)*
d^(5/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d/tan(f*x + e)))/sqrt(d)) + 5*sqrt(2)*d^(5/2)*log(sqrt(2
)*sqrt(d)*sqrt(d/tan(f*x + e)) + d + d/tan(f*x + e)) - 5*sqrt(2)*d^(5/2)*log(-sqrt(2)*sqrt(d)*sqrt(d/tan(f*x +
e)) + d + d/tan(f*x + e)) - 40*d^2*sqrt(d/tan(f*x + e)) + 8*(d/tan(f*x + e))^(5/2))/(d*f)

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mupad [B]  time = 2.94, size = 91, normalized size = 0.39 \[ \frac {2\,d\,\sqrt {d\,\mathrm {cot}\left (e+f\,x\right )}}{f}-\frac {2\,{\left (d\,\mathrm {cot}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {cot}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {cot}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^2*(d*cot(e + f*x))^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cot {\left (e + f x \right )}\right )^{\frac {3}{2}} \cot ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2*(d*cot(f*x+e))**(3/2),x)

[Out]

Integral((d*cot(e + f*x))**(3/2)*cot(e + f*x)**2, x)

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