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

Optimal. Leaf size=232 \[ \frac {c^{7/2} \log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}-\frac {c^{7/2} \log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}+\frac {c^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {c^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b}+\frac {2 c^3 \sqrt {c \cot (a+b x)}}{b}-\frac {2 c (c \cot (a+b x))^{5/2}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 232, 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 = {3473, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {2 c^3 \sqrt {c \cot (a+b x)}}{b}+\frac {c^{7/2} \log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}-\frac {c^{7/2} \log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}+\frac {c^{7/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {c^{7/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b}-\frac {2 c (c \cot (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cot[a + b*x])^(7/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.52, size = 175, normalized size = 0.75 \[ \frac {c^3 \sqrt {c \cot (a+b x)} \left (-8 \cot ^{\frac {5}{2}}(a+b x)+40 \sqrt {\cot (a+b x)}+5 \sqrt {2} \log \left (\cot (a+b x)-\sqrt {2} \sqrt {\cot (a+b x)}+1\right )-5 \sqrt {2} \log \left (\cot (a+b x)+\sqrt {2} \sqrt {\cot (a+b x)}+1\right )+10 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (a+b x)}\right )-10 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (a+b x)}+1\right )\right )}{20 b \sqrt {\cot (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cot[a + b*x])^(7/2),x]

[Out]

(c^3*Sqrt[c*Cot[a + b*x]]*(10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[a + b*x]]] - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*S
qrt[Cot[a + b*x]]] + 40*Sqrt[Cot[a + b*x]] - 8*Cot[a + b*x]^(5/2) + 5*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Cot[a + b*x
]] + Cot[a + b*x]] - 5*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[a + b*x]] + Cot[a + b*x]]))/(20*b*Sqrt[Cot[a + b*x]])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.43, size = 200, normalized size = 0.86 \[ -\frac {2 c \left (c \cot \left (b x +a \right )\right )^{\frac {5}{2}}}{5 b}+\frac {2 c^{3} \sqrt {c \cot \left (b x +a \right )}}{b}-\frac {c^{3} \left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )}{4 b}-\frac {c^{3} \left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )}{2 b}+\frac {c^{3} \left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 197, normalized size = 0.85 \[ -\frac {{\left (10 \, \sqrt {2} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} + 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right ) + 10 \, \sqrt {2} c^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} - 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right ) + 5 \, \sqrt {2} c^{\frac {5}{2}} \log \left (\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right ) - 5 \, \sqrt {2} c^{\frac {5}{2}} \log \left (-\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right ) - 40 \, c^{2} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + 8 \, \left (\frac {c}{\tan \left (b x + a\right )}\right )^{\frac {5}{2}}\right )} c}{20 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(10*sqrt(2)*c^(5/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(c) + 2*sqrt(c/tan(b*x + a)))/sqrt(c)) + 10*sqrt(2)*
c^(5/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(c) - 2*sqrt(c/tan(b*x + a)))/sqrt(c)) + 5*sqrt(2)*c^(5/2)*log(sqrt(2
)*sqrt(c)*sqrt(c/tan(b*x + a)) + c + c/tan(b*x + a)) - 5*sqrt(2)*c^(5/2)*log(-sqrt(2)*sqrt(c)*sqrt(c/tan(b*x +
a)) + c + c/tan(b*x + a)) - 40*c^2*sqrt(c/tan(b*x + a)) + 8*(c/tan(b*x + a))^(5/2))*c/b

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mupad [B]  time = 0.84, size = 91, normalized size = 0.39 \[ \frac {2\,c^3\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{b}-\frac {2\,c\,{\left (c\,\mathrm {cot}\left (a+b\,x\right )\right )}^{5/2}}{5\,b}+\frac {{\left (-1\right )}^{1/4}\,c^{7/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{b}+\frac {{\left (-1\right )}^{1/4}\,c^{7/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}\,1{}\mathrm {i}}{\sqrt {c}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*cot(a + b*x))^(7/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*cot(a + b*x))**(7/2), x)

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