Integrand size = 34, antiderivative size = 325 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\frac {3 e^{\frac {5}{3} (a-i d)+\frac {2}{3} i (d+b x)}}{b (f+i g)}-\frac {\sqrt {3} e^{\frac {5}{3} (a-i d)} \sqrt [3]{f-i g} \arctan \left (\frac {1+\frac {2 e^{\frac {2}{3} i (d+b x)} \sqrt [3]{f+i g}}{\sqrt [3]{f-i g}}}{\sqrt {3}}\right )}{b (f+i g)^{4/3}}+\frac {e^{\frac {5}{3} (a-i d)} \sqrt [3]{f-i g} \log \left (\sqrt [3]{f-i g}-e^{\frac {2}{3} i (d+b x)} \sqrt [3]{f+i g}\right )}{b (f+i g)^{4/3}}-\frac {e^{\frac {5}{3} (a-i d)} \sqrt [3]{f-i g} \log \left ((f-i g)^{2/3}+e^{\frac {2}{3} i (d+b x)} \sqrt [3]{f-i g} \sqrt [3]{f+i g}+e^{\frac {4}{3} i (d+b x)} (f+i g)^{2/3}\right )}{2 b (f+i g)^{4/3}} \] Output:
3*exp(5/3*a-5/3*I*d+2/3*I*(b*x+d))/b/(f+I*g)-3^(1/2)*exp(5/3*a-5/3*I*d)*(f -I*g)^(1/3)*arctan(1/3*(1+2*exp(2/3*I*(b*x+d))*(f+I*g)^(1/3)/(f-I*g)^(1/3) )*3^(1/2))/b/(f+I*g)^(4/3)+exp(5/3*a-5/3*I*d)*(f-I*g)^(1/3)*ln((f-I*g)^(1/ 3)-exp(2/3*I*(b*x+d))*(f+I*g)^(1/3))/b/(f+I*g)^(4/3)-1/2*exp(5/3*a-5/3*I*d )*(f-I*g)^(1/3)*ln((f-I*g)^(2/3)+exp(2/3*I*(b*x+d))*(f-I*g)^(1/3)*(f+I*g)^ (1/3)+exp(4/3*I*(b*x+d))*(f+I*g)^(2/3))/b/(f+I*g)^(4/3)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.51 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.38 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\frac {e^{\frac {5}{3} (a-i d)} \left (18 e^{\frac {2}{3} i (d+b x)} (f+i g)+(-i f-g) \text {RootSum}\left [i f+g-i f \text {$\#$1}^6+g \text {$\#$1}^6\&,\frac {2 d+2 b x+3 i \log \left (\left (e^{\frac {1}{3} i (d+b x)}-\text {$\#$1}\right )^2\right )}{\text {$\#$1}^4}\&\right ]\right )}{6 b (f+i g)^2} \] Input:
Integrate[E^((5*(a + I*b*x))/3)/(g*Cos[d + b*x] + f*Sin[d + b*x]),x]
Output:
(E^((5*(a - I*d))/3)*(18*E^(((2*I)/3)*(d + b*x))*(f + I*g) + ((-I)*f - g)* RootSum[I*f + g - I*f*#1^6 + g*#1^6 & , (2*d + 2*b*x + (3*I)*Log[(E^((I/3) *(d + b*x)) - #1)^2])/#1^4 & ]))/(6*b*(f + I*g)^2)
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.
\(\displaystyle \int \frac {e^{\frac {5}{3} (a+i b x)}}{f \sin (b x+d)+g \cos (b x+d)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{f \sin (b x+d)+g \cos (b x+d)}dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \frac {e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{f \sin (b x+d)+g \cos (b x+d)}dx\) |
Input:
Int[E^((5*(a + I*b*x))/3)/(g*Cos[d + b*x] + f*Sin[d + b*x]),x]
Output:
$Aborted
\[\int \frac {{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{g \cos \left (b x +d \right )+f \sin \left (b x +d \right )}d x\]
Input:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x)
Output:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (218) = 436\).
Time = 0.09 (sec) , antiderivative size = 585, normalized size of antiderivative = 1.80 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx =\text {Too large to display} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x, algorithm="f ricas")
Output:
((I*b*f - b*g - sqrt(3)*(b*f + I*b*g))*((f - I*g)/(b^3*f^4 + 4*I*b^3*f^3*g - 6*b^3*f^2*g^2 - 4*I*b^3*f*g^3 + b^3*g^4))^(1/3)*e^(5/3*a - 5/3*I*d)*log (1/2*((b*f + I*b*g + sqrt(3)*(I*b*f - b*g))*((f - I*g)/(b^3*f^4 + 4*I*b^3* f^3*g - 6*b^3*f^2*g^2 - 4*I*b^3*f*g^3 + b^3*g^4))^(1/3)*e^(5/3*a - 5/3*I*d ) + 2*e^(2/3*I*b*x + 5/3*a - I*d))*e^(-5/3*a + 5/3*I*d)) + (I*b*f - b*g + sqrt(3)*(b*f + I*b*g))*((f - I*g)/(b^3*f^4 + 4*I*b^3*f^3*g - 6*b^3*f^2*g^2 - 4*I*b^3*f*g^3 + b^3*g^4))^(1/3)*e^(5/3*a - 5/3*I*d)*log(1/2*((b*f + I*b *g + sqrt(3)*(-I*b*f + b*g))*((f - I*g)/(b^3*f^4 + 4*I*b^3*f^3*g - 6*b^3*f ^2*g^2 - 4*I*b^3*f*g^3 + b^3*g^4))^(1/3)*e^(5/3*a - 5/3*I*d) + 2*e^(2/3*I* b*x + 5/3*a - I*d))*e^(-5/3*a + 5/3*I*d)) - 2*(I*b*f - b*g)*((f - I*g)/(b^ 3*f^4 + 4*I*b^3*f^3*g - 6*b^3*f^2*g^2 - 4*I*b^3*f*g^3 + b^3*g^4))^(1/3)*e^ (5/3*a - 5/3*I*d)*log(-((b*f + I*b*g)*((f - I*g)/(b^3*f^4 + 4*I*b^3*f^3*g - 6*b^3*f^2*g^2 - 4*I*b^3*f*g^3 + b^3*g^4))^(1/3)*e^(5/3*a - 5/3*I*d) - e^ (2/3*I*b*x + 5/3*a - I*d))*e^(-5/3*a + 5/3*I*d)) - 6*I*e^(2/3*I*b*x + 5/3* a - I*d))/(-2*I*b*f + 2*b*g)
Timed out. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\text {Timed out} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x)
Output:
Timed out
Exception generated. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x, algorithm="m axima")
Output:
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found %i
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7419 vs. \(2 (218) = 436\).
Time = 108.16 (sec) , antiderivative size = 7419, normalized size of antiderivative = 22.83 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x, algorithm="g iac")
Output:
-I*((I*f*e^(5/3*a) + g*e^(5/3*a))*((f - I*g)/(f*e^(2*I*d) + I*g*e^(2*I*d)) )^(1/3)*log(-((f - I*g)/(f*e^(2*I*d) + I*g*e^(2*I*d)))^(1/3) + e^(2/3*I*b* x))/(f^2*e^(I*d) + g^2*e^(I*d)) + 2*I*(sqrt(3)*f*cos(-1/6*pi*sgn(f*cos(d)^ 2 - 2*g*cos(d)*sin(d) - f*sin(d)^2)*sgn(-g*cos(d)^2 - 2*f*cos(d)*sin(d) + g*sin(d)^2) + 1/6*pi*sgn(f)*sgn(g) - 2/3*pi*floor(-1/4*sgn(f*cos(d)^2 - 2* g*cos(d)*sin(d) - f*sin(d)^2)*sgn(-g*cos(d)^2 - 2*f*cos(d)*sin(d) + g*sin( d)^2) + 1/4*sgn(f)*sgn(g) + 1/2*arctan(-g*tan(d)^2/(f*tan(d)^2 + 2*g*tan(d ) - f) + 2*f*tan(d)/(f*tan(d)^2 + 2*g*tan(d) - f) + g/(f*tan(d)^2 + 2*g*ta n(d) - f))/pi - 1/2*arctan(g/f)/pi + 1/4*sgn(-g*cos(d)^2 - 2*f*cos(d)*sin( d) + g*sin(d)^2) - 1/4*sgn(g) + 1/2) + 1/6*pi*sgn(-g*cos(d)^2 - 2*f*cos(d) *sin(d) + g*sin(d)^2) - 1/6*pi*sgn(g) + 1/3*arctan(-g*tan(d)^2/(f*tan(d)^2 + 2*g*tan(d) - f) + 2*f*tan(d)/(f*tan(d)^2 + 2*g*tan(d) - f) + g/(f*tan(d )^2 + 2*g*tan(d) - f)) - 1/3*arctan(g/f))*cos(d)*e^(5/3*a) - sqrt(3)*g*cos (-1/6*pi*sgn(f*cos(d)^2 - 2*g*cos(d)*sin(d) - f*sin(d)^2)*sgn(-g*cos(d)^2 - 2*f*cos(d)*sin(d) + g*sin(d)^2) + 1/6*pi*sgn(f)*sgn(g) - 2/3*pi*floor(-1 /4*sgn(f*cos(d)^2 - 2*g*cos(d)*sin(d) - f*sin(d)^2)*sgn(-g*cos(d)^2 - 2*f* cos(d)*sin(d) + g*sin(d)^2) + 1/4*sgn(f)*sgn(g) + 1/2*arctan(-g*tan(d)^2/( f*tan(d)^2 + 2*g*tan(d) - f) + 2*f*tan(d)/(f*tan(d)^2 + 2*g*tan(d) - f) + g/(f*tan(d)^2 + 2*g*tan(d) - f))/pi - 1/2*arctan(g/f)/pi + 1/4*sgn(-g*cos( d)^2 - 2*f*cos(d)*sin(d) + g*sin(d)^2) - 1/4*sgn(g) + 1/2) + 1/6*pi*sgn...
Timed out. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}}{g\,\cos \left (d+b\,x\right )+f\,\sin \left (d+b\,x\right )} \,d x \] Input:
int(exp((5*a)/3 + (b*x*5i)/3)/(g*cos(d + b*x) + f*sin(d + b*x)),x)
Output:
int(exp((5*a)/3 + (b*x*5i)/3)/(g*cos(d + b*x) + f*sin(d + b*x)), x)
\[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{g \cos (d+b x)+f \sin (d+b x)} \, dx=\frac {3 e^{\frac {5 b i x}{3}+\frac {5 a}{3}} i -10 \left (\int \frac {e^{\frac {5 b i x}{3}+\frac {5 a}{3}}}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} g -2 \tan \left (\frac {b x}{2}+\frac {d}{2}\right ) f -g}d x \right ) b g -10 \left (\int \frac {e^{\frac {5 b i x}{3}+\frac {5 a}{3}} \tan \left (\frac {b x}{2}+\frac {d}{2}\right )}{\tan \left (\frac {b x}{2}+\frac {d}{2}\right )^{2} g -2 \tan \left (\frac {b x}{2}+\frac {d}{2}\right ) f -g}d x \right ) b f}{5 b g} \] Input:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d)),x)
Output:
(3*e**((5*a + 5*b*i*x)/3)*i - 10*int(e**((5*a + 5*b*i*x)/3)/(tan((b*x + d) /2)**2*g - 2*tan((b*x + d)/2)*f - g),x)*b*g - 10*int((e**((5*a + 5*b*i*x)/ 3)*tan((b*x + d)/2))/(tan((b*x + d)/2)**2*g - 2*tan((b*x + d)/2)*f - g),x) *b*f)/(5*b*g)