Integrand size = 24, antiderivative size = 65 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {2 (-1)^{3/4} a \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 i a \sqrt {\cot (c+d x)}}{d}-\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d} \] Output:
-2*(-1)^(3/4)*a*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-2*I*a*cot(d*x+c)^(1 /2)/d-2/3*a*cot(d*x+c)^(3/2)/d
Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {2 a \left (3 (-1)^{3/4} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )+\sqrt {\cot (c+d x)} (3 i+\cot (c+d x))\right )}{3 d} \] Input:
Integrate[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x]),x]
Output:
(-2*a*(3*(-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]] + Sqrt[Cot[c + d*x]]*(3*I + Cot[c + d*x])))/(3*d)
Time = 0.49 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4156, 3042, 4011, 3042, 4011, 3042, 4016, 221}
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.
\(\displaystyle \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cot (c+d x)^{5/2} (a+i a \tan (c+d x))dx\) |
\(\Big \downarrow \) 4156 |
\(\displaystyle \int \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-\tan \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-a \tan \left (c+d x+\frac {\pi }{2}\right )+i a\right )dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle -\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\int \sqrt {\cot (c+d x)} (i a \cot (c+d x)-a)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d}+\int \sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (-i \tan \left (c+d x+\frac {\pi }{2}\right ) a-a\right )dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {-\cot (c+d x) a-i a}{\sqrt {\cot (c+d x)}}dx-\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 i a \sqrt {\cot (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \tan \left (c+d x+\frac {\pi }{2}\right )-i a}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 i a \sqrt {\cot (c+d x)}}{d}\) |
\(\Big \downarrow \) 4016 |
\(\displaystyle -\frac {2 a^2 \int \frac {1}{i a-a \cot (c+d x)}d\sqrt {\cot (c+d x)}}{d}-\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 i a \sqrt {\cot (c+d x)}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 (-1)^{3/4} a \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a \cot ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 i a \sqrt {\cot (c+d x)}}{d}\) |
Input:
Int[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x]),x]
Output:
(-2*(-1)^(3/4)*a*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - ((2*I)*a*Sqrt [Cot[c + d*x]])/d - (2*a*Cot[c + d*x]^(3/2))/(3*d)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2*(c^2/f) Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b *Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]
Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p) Int[(d*Cot[e + f*x])^(m - n*p )*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && !IntegerQ[m] && IntegersQ[n, p]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (52 ) = 104\).
Time = 0.44 (sec) , antiderivative size = 269, normalized size of antiderivative = 4.14
method | result | size |
derivativedivides | \(-\frac {a \left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \left (6 i \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+6 i \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+3 i \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+3 \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \ln \left (-\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right )+6 \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+6 \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+24 i \tan \left (d x +c \right )+8\right )}{12 d}\) | \(269\) |
default | \(-\frac {a \left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \left (6 i \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+6 i \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+3 i \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+3 \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \ln \left (-\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right )+6 \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+6 \sqrt {2}\, \tan \left (d x +c \right )^{\frac {3}{2}} \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+24 i \tan \left (d x +c \right )+8\right )}{12 d}\) | \(269\) |
Input:
int(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/12*a/d*(1/tan(d*x+c))^(5/2)*tan(d*x+c)*(6*I*2^(1/2)*tan(d*x+c)^(3/2)*ar ctan(1+2^(1/2)*tan(d*x+c)^(1/2))+6*I*2^(1/2)*tan(d*x+c)^(3/2)*arctan(-1+2^ (1/2)*tan(d*x+c)^(1/2))+3*I*2^(1/2)*tan(d*x+c)^(3/2)*ln(-(2^(1/2)*tan(d*x+ c)^(1/2)-tan(d*x+c)-1)/(tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1))+3*2^(1/2)* tan(d*x+c)^(3/2)*ln(-(tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1)/(2^(1/2)*tan( d*x+c)^(1/2)-tan(d*x+c)-1))+6*2^(1/2)*tan(d*x+c)^(3/2)*arctan(1+2^(1/2)*ta n(d*x+c)^(1/2))+6*2^(1/2)*tan(d*x+c)^(3/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1 /2))+24*I*tan(d*x+c)+8)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (51) = 102\).
Time = 0.11 (sec) , antiderivative size = 287, normalized size of antiderivative = 4.42 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{2}}} \log \left (\frac {{\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{2}}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{2}}} \log \left (\frac {{\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{2}}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + 16 \, {\left (2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:
integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-1/12*(3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-4*I*a^2/d^2)*log(((I*d*e^(2*I*d *x + 2*I*c) - I*d)*sqrt(-4*I*a^2/d^2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^ (2*I*d*x + 2*I*c) - 1)) + 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/ a) - 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-4*I*a^2/d^2)*log(((-I*d*e^(2*I*d* x + 2*I*c) + I*d)*sqrt(-4*I*a^2/d^2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^( 2*I*d*x + 2*I*c) - 1)) + 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a ) + 16*(2*I*a*e^(2*I*d*x + 2*I*c) - I*a)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/ (e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(2*I*d*x + 2*I*c) - d)
\[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=i a \left (\int \left (- i \cot ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int \tan {\left (c + d x \right )} \cot ^{\frac {5}{2}}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cot(d*x+c)**(5/2)*(a+I*a*tan(d*x+c)),x)
Output:
I*a*(Integral(-I*cot(c + d*x)**(5/2), x) + Integral(tan(c + d*x)*cot(c + d *x)**(5/2), x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (51) = 102\).
Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.14 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {3 \, {\left (-\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \left (i - 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \left (i - 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a + \frac {24 i \, a}{\sqrt {\tan \left (d x + c\right )}} + \frac {8 \, a}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \] Input:
integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
-1/12*(3*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) - (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - (I - 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + (I - 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a + 24*I*a/sqrt(tan(d*x + c)) + 8*a/tan(d*x + c)^(3/2))/d
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {i \, {\left (-\left (3 i - 3\right ) \, \sqrt {2} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) + \frac {2 \, {\left (3 \, \tan \left (d x + c\right ) - i\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}\right )} a}{3 \, d} \] Input:
integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
-1/3*I*(-(3*I - 3)*sqrt(2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c) )) + 2*(3*tan(d*x + c) - I)/tan(d*x + c)^(3/2))*a/d
Timed out. \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \] Input:
int(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i),x)
Output:
int(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i), x)
\[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \, dx=a \left (\left (\int \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{2} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{2}d x \right ) \] Input:
int(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c)),x)
Output:
a*(int(sqrt(cot(c + d*x))*cot(c + d*x)**2*tan(c + d*x),x)*i + int(sqrt(cot (c + d*x))*cot(c + d*x)**2,x))