Integrand size = 23, antiderivative size = 96 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d}-\frac {2 a \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{3 d} \]
2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-2/3*a*cot(d* x+c)^3*(a+a*sec(d*x+c))^(3/2)/d+2*a^2*cot(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {2 \left (\frac {1}{1+\cos (c+d x)}\right )^{3/2} \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sec (c+d x)))^{5/2}}{3 d \sqrt {\frac {1}{1+\sec (c+d x)}}} \]
(-2*((1 + Cos[c + d*x])^(-1))^(3/2)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, -3/2, -1/2, 2*Sin[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/2))/(3*d*Sqrt [(1 + Sec[c + d*x])^(-1)])
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 264, 264, 216}
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.
\(\displaystyle \int \cot ^4(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4375 |
\(\displaystyle -\frac {2 a \int \frac {\cot ^4(c+d x) (\sec (c+d x) a+a)^2}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {2 a \left (\frac {1}{3} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-a \int \frac {\cot ^2(c+d x) (\sec (c+d x) a+a)}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {2 a \left (\frac {1}{3} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-a \left (\cot (c+d x) \sqrt {a \sec (c+d x)+a}-a \int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )\right )}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 a \left (\frac {1}{3} \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}-a \left (\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )+\cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )\right )}{d}\) |
(-2*a*((Cot[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2))/3 - a*(Sqrt[a]*ArcTan[( Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] + Cot[c + d*x]*Sqrt[a + a* Sec[c + d*x]])))/d
3.2.69.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d) Subst[Int[x^m*((2 + a*x^2 )^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] ]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I ntegerQ[n - 1/2]
Leaf count of result is larger than twice the leaf count of optimal. \(179\) vs. \(2(84)=168\).
Time = 40.39 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.88
method | result | size |
default | \(\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-3 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4 \cot \left (d x +c \right ) \cos \left (d x +c \right )-3 \cot \left (d x +c \right )\right )}{3 d \left (\cos \left (d x +c \right )-1\right )}\) | \(180\) |
2/3/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)-1)*(3*(-cos(d*x+c)/(cos(d*x +c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2))*cos(d*x+c)-3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+ c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4*cot(d*x+c)*cos(d*x +c)-3*cot(d*x+c))
Time = 0.34 (sec) , antiderivative size = 355, normalized size of antiderivative = 3.70 \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}, \frac {3 \, {\left (a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )}\right ] \]
[1/6*(3*(a^2*cos(d*x + c) - a^2)*sqrt(-a)*log(-(8*a*cos(d*x + c)^3 - 4*(2* cos(d*x + c)^2 - cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c) + a)/(cos(d*x + c) + 1))*sin(d*x + c ) + 4*(4*a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a )/cos(d*x + c)))/((d*cos(d*x + c) - d)*sin(d*x + c)), 1/3*(3*(a^2*cos(d*x + c) - a^2)*sqrt(a)*arctan(2*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c ))*cos(d*x + c)*sin(d*x + c)/(2*a*cos(d*x + c)^2 + a*cos(d*x + c) - a))*si n(d*x + c) + 2*(4*a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((d*cos(d*x + c) - d)*sin(d*x + c))]
Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \]
\[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{4} \,d x } \]
\[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \cot ^4(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]