Integrand size = 23, antiderivative size = 66 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}-\frac {4 a^2 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{d} \] Output:
-2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-4*a^2*cot(d *x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.51 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.88 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 \cos (c+d x)-\frac {\text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) (-1+\cos (c+d x))}{\sqrt {1-\sec (c+d x)}}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{3/2} (a (1+\sec (c+d x)))^{5/2}}{d \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \] Input:
Integrate[Cot[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2),x]
Output:
-((Sqrt[2]*Cot[c + d*x]*Sec[(c + d*x)/2]^2*(2*Cos[c + d*x] - (ArcTanh[Sqrt [1 - Sec[c + d*x]]]*(-1 + Cos[c + d*x]))/Sqrt[1 - Sec[c + d*x]])*((1 + Sec [c + d*x])^(-1))^(3/2)*(a*(1 + Sec[c + d*x]))^(5/2))/(d*Sqrt[1 - Tan[(c + d*x)/2]^2]))
Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4375, 359, 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.
\(\displaystyle \int \cot ^2(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\cot \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 4375 |
\(\displaystyle -\frac {2 a^2 \int \frac {\cot ^2(c+d x) (\sec (c+d x) a+a) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {2 a^2 \left (2 \cot (c+d x) \sqrt {a \sec (c+d x)+a}-a \int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 a^2 \left (\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )+2 \cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )}{d}\) |
Input:
Int[Cot[c + d*x]^2*(a + a*Sec[c + d*x])^(5/2),x]
Output:
(-2*a^2*(Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] + 2*Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]]))/d
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]
Time = 5.75 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.64
method | result | size |
default | \(\frac {\left (-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )-4 \cot \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d}\) | \(108\) |
Input:
int(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/d*(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*arctanh(2^(1/2)/(cot(d* x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c )))-4*cot(d*x+c))*a^2*(a*(1+sec(d*x+c)))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (58) = 116\).
Time = 0.12 (sec) , antiderivative size = 270, normalized size of antiderivative = 4.09 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\left [\frac {\sqrt {-a} a^{2} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) - 8 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )}, -\frac {a^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 4 \, a^{2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{d \sin \left (d x + c\right )}\right ] \] Input:
integrate(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[1/2*(sqrt(-a)*a^2*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) - 8*a^2*sqrt((a*cos(d *x + c) + a)/cos(d*x + c))*cos(d*x + c))/(d*sin(d*x + c)), -(a^(5/2)*arcta n(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))*sin(d*x + c) + 4*a^2*sqrt(( a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c))/(d*sin(d*x + c))]
Timed out. \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**2*(a+a*sec(d*x+c))**(5/2),x)
Output:
Timed out
\[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{2} \,d x } \] Input:
integrate(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^(5/2)*cot(d*x + c)^2, x)
Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (58) = 116\).
Time = 0.89 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.91 \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\frac {\sqrt {2} \sqrt {-a} a^{4} {\left (\frac {\sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{a {\left | a \right |}} + \frac {8}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} a}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{2 \, d} \] Input:
integrate(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
1/2*sqrt(2)*sqrt(-a)*a^4*(sqrt(2)*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2* (sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6*a))/(a*abs(a)) + 8/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)*a))*sgn(cos(d*x + c))/d
Timed out. \[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:
int(cot(c + d*x)^2*(a + a/cos(c + d*x))^(5/2),x)
Output:
int(cot(c + d*x)^2*(a + a/cos(c + d*x))^(5/2), x)
\[ \int \cot ^2(c+d x) (a+a \sec (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right )+\int \sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2}d x \right ) \] Input:
int(cot(d*x+c)^2*(a+a*sec(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**2*sec(c + d*x)**2,x ) + 2*int(sqrt(sec(c + d*x) + 1)*cot(c + d*x)**2*sec(c + d*x),x) + int(sqr t(sec(c + d*x) + 1)*cot(c + d*x)**2,x))