Integrand size = 23, antiderivative size = 147 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {43 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {11 a^2 \sqrt {a+a \sec (c+d x)}}{16 d (1-\sec (c+d x))} \]
2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-43/32*a^(5/2)*arctanh( 1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-1/4*a^2*(a+a*sec(d*x +c))^(1/2)/d/(1-sec(d*x+c))^2-11/16*a^2*(a+a*sec(d*x+c))^(1/2)/d/(1-sec(d* x+c))
Time = 0.69 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.74 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {(a (1+\sec (c+d x)))^{5/2} \left (64 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-43 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\frac {2 \sqrt {1+\sec (c+d x)} (-15+11 \sec (c+d x))}{(-1+\sec (c+d x))^2}\right )}{32 d (1+\sec (c+d x))^{5/2}} \]
((a*(1 + Sec[c + d*x]))^(5/2)*(64*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 43*Sqr t[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]] + (2*Sqrt[1 + Sec[c + d*x]]*( -15 + 11*Sec[c + d*x]))/(-1 + Sec[c + d*x])^2))/(32*d*(1 + Sec[c + d*x])^( 5/2))
Time = 0.34 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 25, 4368, 25, 27, 114, 27, 168, 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.
\(\displaystyle \int \cot ^5(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{5/2}}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}dx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {a^6 \int -\frac {\cos (c+d x)}{a^3 (1-\sec (c+d x))^3 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a^6 \int \frac {\cos (c+d x)}{a^3 (1-\sec (c+d x))^3 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 \int \frac {\cos (c+d x)}{(1-\sec (c+d x))^3 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 114 |
\(\displaystyle -\frac {a^3 \left (\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}-\frac {\int -\frac {a \cos (c+d x) (3 \sec (c+d x)+8)}{2 (1-\sec (c+d x))^2 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{4 a}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \int \frac {\cos (c+d x) (3 \sec (c+d x)+8)}{(1-\sec (c+d x))^2 \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 168 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}-\frac {\int -\frac {a \cos (c+d x) (11 \sec (c+d x)+32)}{2 (1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{2 a}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \int \frac {\cos (c+d x) (11 \sec (c+d x)+32)}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (43 \int \frac {1}{(1-\sec (c+d x)) \sqrt {\sec (c+d x) a+a}}d\sec (c+d x)+32 \int \frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {86 \int \frac {1}{2-\frac {\sec (c+d x) a+a}{a}}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {64 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {64 \int \frac {1}{\frac {\sec (c+d x) a+a}{a}-1}d\sqrt {\sec (c+d x) a+a}}{a}+\frac {43 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {a^3 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {43 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {64 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {11 \sqrt {a \sec (c+d x)+a}}{2 a (1-\sec (c+d x))}\right )+\frac {\sqrt {a \sec (c+d x)+a}}{4 a (1-\sec (c+d x))^2}\right )}{d}\) |
-((a^3*(Sqrt[a + a*Sec[c + d*x]]/(4*a*(1 - Sec[c + d*x])^2) + (((-64*ArcTa nh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + (43*Sqrt[2]*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a])/4 + (11*Sqrt[a + a*Sec[c + d*x]])/(2*a*(1 - Sec[c + d*x])))/8))/d)
3.2.64.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -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]
Leaf count of result is larger than twice the leaf count of optimal. \(271\) vs. \(2(122)=244\).
Time = 88.52 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.85
method | result | size |
default | \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-43 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-64 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+43 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+64 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+30 \cos \left (d x +c \right ) \cot \left (d x +c \right )^{2}+8 \cot \left (d x +c \right )^{2}-22 \cot \left (d x +c \right ) \csc \left (d x +c \right )\right )}{32 d \left (\cos \left (d x +c \right )-1\right )}\) | \(272\) |
1/32/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)-1)*(-43*arctan(1/2*2^(1/2) /(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*2^(1/2)*(-cos(d*x+c)/(cos( d*x+c)+1))^(1/2)-64*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)* (-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+43*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))+64*arctan((- cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+30*co s(d*x+c)*cot(d*x+c)^2+8*cot(d*x+c)^2-22*cot(d*x+c)*csc(d*x+c))
Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (118) = 236\).
Time = 0.33 (sec) , antiderivative size = 503, normalized size of antiderivative = 3.42 \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {64 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \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 ) + 43 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + \sqrt {2} a^{2}\right )} \sqrt {a} \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 ) - 4 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{64 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}}, \frac {43 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + \sqrt {2} a^{2}\right )} \sqrt {-a} \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 ) - 64 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \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 ) - 2 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{32 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/64*(64*(a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(a)*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) + 43*(sqrt(2)*a^2*cos(d*x + c)^2 - 2*sqrt(2)*a^2*cos(d*x + c) + sqrt(2)*a^2)*sqrt(a)*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)) - 4*(1 5*a^2*cos(d*x + c)^2 - 11*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos( d*x + c)))/(d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d), 1/32*(43*(sqrt(2)*a^ 2*cos(d*x + c)^2 - 2*sqrt(2)*a^2*cos(d*x + c) + sqrt(2)*a^2)*sqrt(-a)*arct an(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)) - 64*(a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)* sqrt(-a)*arctan(sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 2*(15*a^2*cos(d*x + c)^2 - 11*a^2*cos(d*x + c) )*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d)]
Timed out. \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]