\(\int (c \cot (a+b x))^{7/2} \, dx\) [9]

Optimal result

Integrand size = 12, antiderivative size = 232 \[ \int (c \cot (a+b x))^{7/2} \, dx=\frac {c^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {c^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}+\frac {2 c^3 \sqrt {c \cot (a+b x)}}{b}-\frac {2 c (c \cot (a+b x))^{5/2}}{5 b}+\frac {c^{7/2} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b}-\frac {c^{7/2} \log \left (\sqrt {c}+\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}\right )}{2 \sqrt {2} b} \]

[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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3554, 3557, 335, 217, 1179, 642, 1176, 631, 210} \[ \int (c \cot (a+b x))^{7/2} \, dx=\frac {c^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {c^{7/2} \arctan \left (\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}+1\right )}{\sqrt {2} b}+\frac {c^{7/2} \log \left (\sqrt {c} \cot (a+b x)-\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}-\frac {c^{7/2} \log \left (\sqrt {c} \cot (a+b x)+\sqrt {2} \sqrt {c \cot (a+b x)}+\sqrt {c}\right )}{2 \sqrt {2} b}+\frac {2 c^3 \sqrt {c \cot (a+b x)}}{b}-\frac {2 c (c \cot (a+b x))^{5/2}}{5 b} \]

[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 210

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

Rule 217

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 335

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 631

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 642

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 1176

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 1179

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 3554

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 3557

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*} \text {integral}& = -\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 \text {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 ) \text {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 \text {Subst}\left (\int \frac {c-x^2}{c^2+x^4} \, dx,x,\sqrt {c \cot (a+b x)}\right )}{b}-\frac {c^4 \text {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} \text {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} \text {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 \text {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 \text {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} \text {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} \text {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} \arctan \left (1-\frac {\sqrt {2} \sqrt {c \cot (a+b x)}}{\sqrt {c}}\right )}{\sqrt {2} b}-\frac {c^{7/2} \arctan \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*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.75 \[ \int (c \cot (a+b x))^{7/2} \, dx=-\frac {(c \cot (a+b x))^{7/2} \left (-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (a+b x)}\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} \sqrt {\cot (a+b x)}\right )}{\sqrt {2}}-2 \sqrt {\cot (a+b x)}+\frac {2}{5} \cot ^{\frac {5}{2}}(a+b x)-\frac {\log \left (1-\sqrt {2} \sqrt {\cot (a+b x)}+\cot (a+b x)\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {\cot (a+b x)}+\cot (a+b x)\right )}{2 \sqrt {2}}\right )}{b \cot ^{\frac {7}{2}}(a+b x)} \]

[In]

Integrate[(c*Cot[a + b*x])^(7/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(-\frac {2 c \left (\frac {\left (c \cot \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-c^{2} \sqrt {c \cot \left (b x +a \right )}+\frac {c^{2} \left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{b}\)	\(169\)
default	\(-\frac {2 c \left (\frac {\left (c \cot \left (b x +a \right )\right )^{\frac {5}{2}}}{5}-c^{2} \sqrt {c \cot \left (b x +a \right )}+\frac {c^{2} \left (c^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {c \cot \left (b x +a \right )+\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}{c \cot \left (b x +a \right )-\left (c^{2}\right )^{\frac {1}{4}} \sqrt {c \cot \left (b x +a \right )}\, \sqrt {2}+\sqrt {c^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {c \cot \left (b x +a \right )}}{\left (c^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{b}\)	\(169\)

[In]

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

[Out]

-2/b*c*(1/5*(c*cot(b*x+a))^(5/2)-c^2*(c*cot(b*x+a))^(1/2)+1/8*c^2*(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)))+2*arctan(2^(1/2)/(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)+1)-2*arctan(-2^(1/2)/(c^2)^(1/4)*(c*cot(b*x+a))^(1/2)
+1)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.56 \[ \int (c \cot (a+b x))^{7/2} \, dx=-\frac {5 \, \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} {\left (b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \log \left (c^{3} \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}} + \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} b\right ) + 5 \, \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} {\left (i \, b \cos \left (2 \, b x + 2 \, a\right ) - i \, b\right )} \log \left (c^{3} \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}} + i \, \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} b\right ) + 5 \, \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} {\left (-i \, b \cos \left (2 \, b x + 2 \, a\right ) + i \, b\right )} \log \left (c^{3} \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}} - i \, \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} b\right ) - 5 \, \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} {\left (b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \log \left (c^{3} \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}} - \left (-\frac {c^{14}}{b^{4}}\right )^{\frac {1}{4}} b\right ) - 8 \, {\left (3 \, c^{3} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, c^{3}\right )} \sqrt {\frac {c \cos \left (2 \, b x + 2 \, a\right ) + c}{\sin \left (2 \, b x + 2 \, a\right )}}}{10 \, {\left (b \cos \left (2 \, b x + 2 \, a\right ) - b\right )}} \]

[In]

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

[Out]

-1/10*(5*(-c^14/b^4)^(1/4)*(b*cos(2*b*x + 2*a) - b)*log(c^3*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a)) +
(-c^14/b^4)^(1/4)*b) + 5*(-c^14/b^4)^(1/4)*(I*b*cos(2*b*x + 2*a) - I*b)*log(c^3*sqrt((c*cos(2*b*x + 2*a) + c)/
sin(2*b*x + 2*a)) + I*(-c^14/b^4)^(1/4)*b) + 5*(-c^14/b^4)^(1/4)*(-I*b*cos(2*b*x + 2*a) + I*b)*log(c^3*sqrt((c
*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a)) - I*(-c^14/b^4)^(1/4)*b) - 5*(-c^14/b^4)^(1/4)*(b*cos(2*b*x + 2*a) -
b)*log(c^3*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a)) - (-c^14/b^4)^(1/4)*b) - 8*(3*c^3*cos(2*b*x + 2*a)
- 2*c^3)*sqrt((c*cos(2*b*x + 2*a) + c)/sin(2*b*x + 2*a)))/(b*cos(2*b*x + 2*a) - b)

Sympy [F]

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

[In]

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

[Out]

Integral((c*cot(a + b*x))**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.85 \[ \int (c \cot (a+b x))^{7/2} \, dx=-\frac {{\left (10 \, \sqrt {2} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} + 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right ) + 10 \, \sqrt {2} c^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {c} - 2 \, \sqrt {\frac {c}{\tan \left (b x + a\right )}}\right )}}{2 \, \sqrt {c}}\right ) + 5 \, \sqrt {2} c^{\frac {5}{2}} \log \left (\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right ) - 5 \, \sqrt {2} c^{\frac {5}{2}} \log \left (-\sqrt {2} \sqrt {c} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + c + \frac {c}{\tan \left (b x + a\right )}\right ) - 40 \, c^{2} \sqrt {\frac {c}{\tan \left (b x + a\right )}} + 8 \, \left (\frac {c}{\tan \left (b x + a\right )}\right )^{\frac {5}{2}}\right )} c}{20 \, b} \]

[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

Giac [F]

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

[In]

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

[Out]

integrate((c*cot(b*x + a))^(7/2), x)

Mupad [B] (verification not implemented)

Time = 12.90 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.39 \[ \int (c \cot (a+b x))^{7/2} \, dx=\frac {2\,c^3\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{b}-\frac {2\,c\,{\left (c\,\mathrm {cot}\left (a+b\,x\right )\right )}^{5/2}}{5\,b}+\frac {{\left (-1\right )}^{1/4}\,c^{7/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{b}+\frac {{\left (-1\right )}^{1/4}\,c^{7/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {c\,\mathrm {cot}\left (a+b\,x\right )}\,1{}\mathrm {i}}{\sqrt {c}}\right )}{b} \]

[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