Integrand size = 30, antiderivative size = 132 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d} \] Output:
8/15*I*a*(e*cos(d*x+c))^(5/2)*sec(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)-2/5* I*(e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2)/d-16/15*I*(e*cos(d*x+c))^( 5/2)*sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d
Time = 1.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.48 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i e^2 \sqrt {e \cos (c+d x)} (-15+\cos (2 (c+d x))-4 i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{15 d} \] Input:
Integrate[(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
((I/15)*e^2*Sqrt[e*Cos[c + d*x]]*(-15 + Cos[2*(c + d*x)] - (4*I)*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/d
Time = 0.71 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.15, 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, 3978, 3042, 3983, 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.
\(\displaystyle \int \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {4 a \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {4 a \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {4 a \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {4 a \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}dx}{3 a}+\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2} \left (\frac {4 a \left (\frac {2 i}{3 d \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)}}{3 a d \sqrt {e \sec (c+d x)}}\right )}{5 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{5 d (e \sec (c+d x))^{5/2}}\right )\) |
Input:
Int[(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]],x]
Output:
(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)*((((-2*I)/5)*Sqrt[a + I*a*Ta n[c + d*x]])/(d*(e*Sec[c + d*x])^(5/2)) + (4*a*(((2*I)/3)/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sqrt[a + I*a*Tan[c + d*x] ])/(a*d*Sqrt[e*Sec[c + d*x]])))/(5*e^2))
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 = 9.75 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {2 \left (i \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 i\right ) e^{2} \sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{15 d}\) | \(62\) |
risch | \(-\frac {i e^{2} \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (30-2 \cos \left (2 d x +2 c \right )+8 i \sin \left (2 d x +2 c \right )\right )}{30 d}\) | \(74\) |
Input:
int((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/15/d*(I*cos(d*x+c)^2+4*cos(d*x+c)*sin(d*x+c)-8*I)*e^2*(e*cos(d*x+c))^(1/ 2)*(a*(1+I*tan(d*x+c)))^(1/2)
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-3 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{30 \, d} \] Input:
integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fric as")
Output:
1/30*sqrt(2)*sqrt(1/2)*(-3*I*e^2*e^(4*I*d*x + 4*I*c) - 30*I*e^2*e^(2*I*d*x + 2*I*c) + 5*I*e^2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-3/2*I*d*x - 3/2*I*c)/d
Timed out. \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((e*cos(d*x+c))**(5/2)*(a+I*a*tan(d*x+c))**(1/2),x)
Output:
Timed out
Time = 0.25 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (5 i \, e^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, e^{2} \cos \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 30 i \, e^{2} \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 5 \, e^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, e^{2} \sin \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 30 \, e^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a} \sqrt {e}}{30 \, d} \] Input:
integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxi ma")
Output:
1/30*(5*I*e^2*cos(3/2*d*x + 3/2*c) - 3*I*e^2*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 30*I*e^2*cos(1/3*arctan2(sin(3/2*d*x + 3 /2*c), cos(3/2*d*x + 3/2*c))) + 5*e^2*sin(3/2*d*x + 3/2*c) + 3*e^2*sin(5/3 *arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 30*e^2*sin(1/3*arc tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sqrt(a)*sqrt(e)/d
Exception generated. \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 0.86 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {e^2\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+4\,\sin \left (2\,c+2\,d\,x\right )-15{}\mathrm {i}\right )}{15\,d} \] Input:
int((e*cos(c + d*x))^(5/2)*(a + a*tan(c + d*x)*1i)^(1/2),x)
Output:
(e^2*(e*cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(2*c + 2*d*x)*1i + 4*sin(2*c + 2*d*x ) - 15i))/(15*d)
\[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\sqrt {e}\, \sqrt {a}\, \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) e^{2} \] Input:
int((e*cos(d*x+c))^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x)
Output:
sqrt(e)*sqrt(a)*int(sqrt(tan(c + d*x)*i + 1)*sqrt(cos(c + d*x))*cos(c + d* x)**2,x)*e**2