3.16 \(\int \frac {1}{(c \cot (a+b x))^{7/2}} \, dx\)

Optimal. Leaf size=234 \[ -\frac {\log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b c^{7/2}}+\frac {\log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b c^{7/2}}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b c^{7/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b c^{7/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}+\frac {2}{5 b c (c \cot (a+b x))^{5/2}} \]

[Out]

2/5/b/c/(c*cot(b*x+a))^(5/2)+1/2*arctan(1-2^(1/2)*(c*cot(b*x+a))^(1/2)/c^(1/2))/b/c^(7/2)*2^(1/2)-1/2*arctan(1
+2^(1/2)*(c*cot(b*x+a))^(1/2)/c^(1/2))/b/c^(7/2)*2^(1/2)-1/4*ln(c^(1/2)+cot(b*x+a)*c^(1/2)-2^(1/2)*(c*cot(b*x+
a))^(1/2))/b/c^(7/2)*2^(1/2)+1/4*ln(c^(1/2)+cot(b*x+a)*c^(1/2)+2^(1/2)*(c*cot(b*x+a))^(1/2))/b/c^(7/2)*2^(1/2)
-2/b/c^3/(c*cot(b*x+a))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}-\frac {\log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b c^{7/2}}+\frac {\log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b c^{7/2}}+\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b c^{7/2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b c^{7/2}}+\frac {2}{5 b c (c \cot (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cot[a + b*x])^(-7/2),x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[c*Cot[a + b*x]])/Sqrt[c]]/(Sqrt[2]*b*c^(7/2)) - ArcTan[1 + (Sqrt[2]*Sqrt[c*Cot[a + b*
x]])/Sqrt[c]]/(Sqrt[2]*b*c^(7/2)) + 2/(5*b*c*(c*Cot[a + b*x])^(5/2)) - 2/(b*c^3*Sqrt[c*Cot[a + b*x]]) - Log[Sq
rt[c] + Sqrt[c]*Cot[a + b*x] - Sqrt[2]*Sqrt[c*Cot[a + b*x]]]/(2*Sqrt[2]*b*c^(7/2)) + Log[Sqrt[c] + Sqrt[c]*Cot
[a + b*x] + Sqrt[2]*Sqrt[c*Cot[a + b*x]]]/(2*Sqrt[2]*b*c^(7/2))

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 \frac {1}{(c \cot (a+b x))^{7/2}} \, dx &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {\int \frac {1}{(c \cot (a+b x))^{3/2}} \, dx}{c^2}\\ &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}+\frac {\int \sqrt {c \cot (a+b x)} \, dx}{c^4}\\ &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{c^2+x^2} \, dx,x,c \cot (a+b x)\right )}{b c^3}\\ &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b c^3}\\ &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}+\frac {\operatorname {Subst}\left (\int \frac {c-x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b c^3}-\frac {\operatorname {Subst}\left (\int \frac {c+x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b c^3}\\ &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {c}+2 x}{-c-\sqrt {2} \sqrt {c} x-x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b c^{7/2}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {c}-2 x}{-c+\sqrt {2} \sqrt {c} x-x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b c^{7/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{c-\sqrt {2} \sqrt {c} x+x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 b c^3}-\frac {\operatorname {Subst}\left (\int \frac {1}{c+\sqrt {2} \sqrt {c} x+x^2} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{2 b c^3}\\ &=\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}-\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b c^{7/2}}+\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b c^{7/2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b c^{7/2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b c^{7/2}}\\ &=\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b c^{7/2}}-\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b c^{7/2}}+\frac {2}{5 b c (c \cot (a+b x))^{5/2}}-\frac {2}{b c^3 \sqrt {c \cot (a+b x)}}-\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b c^{7/2}}+\frac {\log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b c^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 40, normalized size = 0.17 \[ \frac {2 \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(a+b x)\right )}{5 b c (c \cot (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cot[a + b*x])^(-7/2),x]

[Out]

(2*Hypergeometric2F1[-5/4, 1, -1/4, -Cot[a + b*x]^2])/(5*b*c*(c*Cot[a + b*x])^(5/2))

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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(1/(c*cot(b*x+a))^(7/2),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 \[ \int \frac {1}{\left (c \cot \left (b x + a\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cot(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate((c*cot(b*x + a))^(-7/2), x)

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maple [A]  time = 0.45, size = 202, normalized size = 0.86 \[ -\frac {\sqrt {2}\, \ln \left (\frac {c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )}{4 b \,c^{3} \left (c^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )}{2 b \,c^{3} \left (c^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )}{2 b \,c^{3} \left (c^{2}\right )^{\frac {1}{4}}}+\frac {2}{5 b c \left (c \cot \left (b x +a \right )\right )^{\frac {5}{2}}}-\frac {2}{b \,c^{3} \sqrt {c \cot \left (b x +a \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cot(b*x+a))^(7/2),x)

[Out]

-1/4/b/c^3/(c^2)^(1/4)*2^(1/2)*ln((c*cot(b*x+a)-(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)*2^(1/2)+(c^2)^(1/2))/(c*cot(b
*x+a)+(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)*2^(1/2)+(c^2)^(1/2)))-1/2/b/c^3/(c^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c^2
)^(1/4)*(c*cot(b*x+a))^(1/2)+1)+1/2/b/c^3/(c^2)^(1/4)*2^(1/2)*arctan(-2^(1/2)/(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)
+1)+2/5/b/c/(c*cot(b*x+a))^(5/2)-2/b/c^3/(c*cot(b*x+a))^(1/2)

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maxima [A]  time = 0.42, size = 205, normalized size = 0.88 \[ -\frac {c {\left (\frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} + 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} - 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right )}{\sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right )}{\sqrt {c}} + \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right )}{\sqrt {c}}\right )}}{c^{4}} - \frac {8 \, {\left (c^{2} - \frac {5 \, c^{2}}{\tan \left (b x + a\right )^{2}}\right )}}{c^{4} \left (\frac {c}{\tan \left (b x + a\right )}\right )^{\frac {5}{2}}}\right )}}{20 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cot(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

-1/20*c*(5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(c) + 2*sqrt(c/tan(b*x + a)))/sqrt(c))/sqrt(c) + 2*sqrt(
2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(c) - 2*sqrt(c/tan(b*x + a)))/sqrt(c))/sqrt(c) - sqrt(2)*log(sqrt(2)*sqrt(
c)*sqrt(c/tan(b*x + a)) + c + c/tan(b*x + a))/sqrt(c) + sqrt(2)*log(-sqrt(2)*sqrt(c)*sqrt(c/tan(b*x + a)) + c
+ c/tan(b*x + a))/sqrt(c))/c^4 - 8*(c^2 - 5*c^2/tan(b*x + a)^2)/(c^4*(c/tan(b*x + a))^(5/2)))/b

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mupad [B]  time = 0.68, size = 91, normalized size = 0.39 \[ \frac {\frac {2}{5\,c}-\frac {2\,{\mathrm {cot}\left (a+b\,x\right )}^2}{c}}{b\,{\left (c\,\mathrm {cot}\left (a+b\,x\right )\right )}^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )}{b\,c^{7/2}}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )}{b\,c^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cot(a + b*x))^(7/2),x)

[Out]

(2/(5*c) - (2*cot(a + b*x)^2)/c)/(b*(c*cot(a + b*x))^(5/2)) - ((-1)^(1/4)*atan(((-1)^(1/4)*(c*cot(a + b*x))^(1
/2))/c^(1/2)))/(b*c^(7/2)) + ((-1)^(1/4)*atanh(((-1)^(1/4)*(c*cot(a + b*x))^(1/2))/c^(1/2)))/(b*c^(7/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \cot {\left (a + b x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cot(b*x+a))**(7/2),x)

[Out]

Integral((c*cot(a + b*x))**(-7/2), x)

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