Integrand size = 24, antiderivative size = 157 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {15 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}+\frac {5 i \cos (c+d x)}{16 a d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{32 a^2 d} \]
15/64*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a ^(3/2)/d*2^(1/2)+5/16*I*cos(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(1/2)-15/32*I*co s(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^2/d+1/4*I*cos(d*x+c)/d/(a+I*a*tan(d*x+ c))^(3/2)
Time = 1.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.76 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sec (c+d x) \left (\frac {30 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-2 (-9+6 \cos (2 (c+d x))+10 i \sin (2 (c+d x)))\right )}{64 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]
(Sec[c + d*x]*((30*E^((2*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x) )]])/Sqrt[1 + E^((2*I)*(c + d*x))] - 2*(-9 + 6*Cos[2*(c + d*x)] + (10*I)*S in[2*(c + d*x)])))/(64*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.63 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3983, 3042, 3983, 3042, 3971, 3042, 3970, 219}
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.
| \(\displaystyle \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x) (a+i a \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {5 \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3983 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3971 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3970 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )}{8 a}+\frac {i \cos (c+d x)}{4 d (a+i a \tan (c+d x))^{3/2}}\) |
((I/4)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (5*(((I/2)*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (3*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Se c[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d))/(4*a)))/(8*a)
3.4.60.3.1 Defintions of rubi rules used
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[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S ymbol] :> Simp[-2*(a/(b*f)) Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 0 ]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (a*f*m)), x] + Simp[a/(2*d^2) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && EqQ[m/2 + n, 0] && GtQ[n, 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n)) Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x ] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* n]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 538 vs. \(2 (126 ) = 252\).
Time = 9.95 (sec) , antiderivative size = 539, normalized size of antiderivative = 3.43
| method | result | size |
| default | \(-\frac {24 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-30 i \cos \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \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 )+24 i \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-40 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-15 i \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \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 )-30 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+30 \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \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 ) \sin \left (d x +c \right )-40 \sin \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+15 i \sec \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \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 )-30 i \sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+15 \tan \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \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 )}{64 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (1+i \tan \left (d x +c \right )\right ) a}\) | \(539\) |
-1/64/d/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(cos(d*x+c)+1)/(a*(1+I*tan(d*x+ c)))^(1/2)/(1+I*tan(d*x+c))/a*(24*I*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2)-30*I*cos(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+ c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+24*I*cos(d*x+c)*(-cos(d*x+c)/(co s(d*x+c)+1))^(1/2)-40*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d* x+c)-15*I*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+ c)/(cos(d*x+c)+1))^(1/2))-30*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+30*arcta n(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2))*sin(d*x+c)-40*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+15* I*sec(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d *x+c)/(cos(d*x+c)+1))^(1/2))-30*I*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^ (1/2)+15*tan(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/ (-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (118) = 236\).
Time = 0.24 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.72 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {15 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{16 \, a d}\right ) + 15 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {15 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{16 \, a d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{64 \, a^{2} d} \]
1/64*(-15*I*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(-15/ 16*(sqrt(2)*sqrt(1/2)*(I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d *x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - I)*e^(-I*d*x - I*c)/(a*d)) + 15*I*sq rt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(-15/16*(sqrt(2)*sq rt(1/2)*(-I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - I)*e^(-I*d*x - I*c)/(a*d)) + sqrt(2)*sqrt(a/(e^(2 *I*d*x + 2*I*c) + 1))*(-8*I*e^(6*I*d*x + 6*I*c) + I*e^(4*I*d*x + 4*I*c) + 11*I*e^(2*I*d*x + 2*I*c) + 2*I))*e^(-4*I*d*x - 4*I*c)/(a^2*d)
\[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1821 vs. \(2 (118) = 236\).
Time = 0.55 (sec) , antiderivative size = 1821, normalized size of antiderivative = 11.60 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
-1/256*(36*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1 /2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin( 4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(3/4)*((-I*sqrt(2)*cos(4*d*x + 4*c) - sqrt(2)*sin(4*d*x + 4*c))*cos(3/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4* c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c) )) + 1)) + (sqrt(2)*cos(4*d*x + 4*c) - I*sqrt(2)*sin(4*d*x + 4*c))*sin(3/2 *arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arc tan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + 4*(cos(1/2*arct an2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4 *c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*((7*I*sqrt(2)*cos(4*d*x + 4*c) + 7*sqrt(2)*sin(4*d*x + 4*c) + 8*I*sqrt(2))*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos (4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1) ) - (7*sqrt(2)*cos(4*d*x + 4*c) - 7*I*sqrt(2)*sin(4*d*x + 4*c) + 8*sqrt(2) )*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), co s(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + 15*(2* sqrt(2)*arctan2((cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2 (sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*sin(1/2*arctan2(sin(1/2*a rctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x ...
\[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]