Integrand size = 30, antiderivative size = 126 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i (e \cos (c+d x))^{3/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {8 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{15 a d} \]
2/5*I*(e*cos(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)+16/15*I*(e*cos(d*x+c ))^(3/2)*sec(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)-8/15*I*(e*cos(d*x+c))^(3/ 2)*(a+I*a*tan(d*x+c))^(1/2)/a/d
Time = 1.60 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.50 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i e^2 (-15+\cos (2 (c+d x))+4 i \sin (2 (c+d x)))}{15 d \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]
((-1/15*I)*e^2*(-15 + Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(d*Sqrt[ e*Cos[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.69 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3998, 3042, 3983, 3042, 3978, 3042, 3969}
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.
\(\displaystyle \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2} \left (\frac {4 \left (\frac {4 i a \sqrt {e \sec (c+d x)}}{3 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )\) |
(e*Cos[c + d*x])^(3/2)*(e*Sec[c + d*x])^(3/2)*(((2*I)/5)/(d*(e*Sec[c + d*x ])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (4*((((4*I)/3)*a*Sqrt[e*Sec[c + d*x ]])/(d*e^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/3)*Sqrt[a + I*a*Tan[c + d *x]])/(d*(e*Sec[c + d*x])^(3/2))))/(5*a))
3.7.82.3.1 Defintions of rubi rules used
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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ [Simplify[m + n], 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*((m + n)/(m*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] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
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]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
Time = 8.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.47
method | result | size |
default | \(-\frac {2 e \sqrt {e \cos \left (d x +c \right )}\, \left (i \cos \left (d x +c \right )-4 \sin \left (d x +c \right )-8 i \sec \left (d x +c \right )\right )}{15 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(59\) |
-2/15/d*e*(e*cos(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(I*cos(d*x+c)-4* sin(d*x+c)-8*I*sec(d*x+c))
Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-5 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{30 \, a d} \]
1/30*sqrt(2)*sqrt(1/2)*(-5*I*e*e^(4*I*d*x + 4*I*c) + 30*I*e*e^(2*I*d*x + 2 *I*c) + 3*I*e)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-5/2*I*d*x - 5/2*I*c)/(a*d)
\[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Time = 0.69 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (3 i \, e \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, e \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 i \, e \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 3 \, e \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, e \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 \, e \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} \sqrt {e}}{30 \, \sqrt {a} d} \]
1/30*(3*I*e*cos(5/2*d*x + 5/2*c) - 5*I*e*cos(3/5*arctan2(sin(5/2*d*x + 5/2 *c), cos(5/2*d*x + 5/2*c))) + 30*I*e*cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 3*e*sin(5/2*d*x + 5/2*c) + 5*e*sin(3/5*arctan2(s in(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*e*sin(1/5*arctan2(sin(5/2 *d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*sqrt(e)/(sqrt(a)*d)
\[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Time = 1.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {e\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (35\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,25{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}\right )}{30\,a\,d} \]