Integrand size = 21, antiderivative size = 95 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {4 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 a^2 \sqrt {a+a \sec (c+d x)}}{d} \]
2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-4*a^(5/2)*arctanh(1/2* (a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d+2*a^2*(a+a*sec(d*x+c))^( 1/2)/d
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 (a (1+\sec (c+d x)))^{5/2} \left (\text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\sqrt {1+\sec (c+d x)}\right )}{d (1+\sec (c+d x))^{5/2}} \]
(2*(a*(1 + Sec[c + d*x]))^(5/2)*(ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 2*Sqrt[ 2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]] + Sqrt[1 + Sec[c + d*x]]))/(d*( 1 + Sec[c + d*x])^(5/2))
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 25, 4368, 25, 27, 95, 25, 27, 174, 73, 219, 220}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cot (c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{5/2}}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
| \(\Big \downarrow \) 4368 |
| \(\displaystyle \frac {a^2 \int -\frac {\cos (c+d x) (\sec (c+d x) a+a)^{3/2}}{a (1-\sec (c+d x))}d\sec (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^2 \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{3/2}}{a (1-\sec (c+d x))}d\sec (c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{3/2}}{1-\sec (c+d x)}d\sec (c+d x)}{d}\) |
| \(\Big \downarrow \) 95 |
| \(\displaystyle -\frac {a \left (-\int -\frac {a^2 \cos (c+d x) (3 \sec (c+d x)+1)}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a \left (\int \frac {a^2 \cos (c+d x) (3 \sec (c+d x)+1)}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \left (a^2 \int \frac {\cos (c+d x) (3 \sec (c+d x)+1)}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle -\frac {a \left (a^2 \left (4 \int \frac {1}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+\int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)\right )-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a \left (a^2 \left (\frac {8 \int \frac {1}{2-\frac {\sec (c+d x) a+a}{a}}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}\right )-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {a \left (a^2 \left (\frac {2 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}\right )-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {a \left (a^2 \left (\frac {4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )-2 a \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
-((a*(a^2*((-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (4*Sqr t[2]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a]) - 2*a*S qrt[a + a*Sec[c + d*x]]))/d)
3.2.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[f*((e + f*x)^(p - 1)/(b*d*(p - 1))), x] + Simp[1/(b*d) Int[(b *d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p - 2)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1) Subst[Int[(-a + b*x)^((m - 1)/2 )*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && !IntegerQ[n]
Time = 5.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(-\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-1\right )}{d}\) | \(116\) |
-2/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*(2*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)*arctan(1/2*2^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+arctan((-cos( d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-1)
Time = 0.35 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.16 \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {2 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + a^{\frac {5}{2}} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right ) + 2 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{d}, \frac {2 \, {\left (2 \, \sqrt {2} \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) + a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}\right )}}{d}\right ] \]
[(2*sqrt(2)*a^(5/2)*log(-(2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos( d*x + c))*cos(d*x + c) - 3*a*cos(d*x + c) - a)/(cos(d*x + c) - 1)) + a^(5/ 2)*log(-2*a*cos(d*x + c) - 2*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c ))*cos(d*x + c) - a) + 2*a^2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/d, 2 *(2*sqrt(2)*sqrt(-a)*a^2*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a) /cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - sqrt(-a)*a^2*arctan(sq rt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c ) + a)) + a^2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/d]
Timed out. \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right ) \,d x } \]
\[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cot (c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int \mathrm {cot}\left (c+d\,x\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]