Integrand size = 34, antiderivative size = 429 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=-\frac {8 e^{\frac {5}{3} (a-i d)+\frac {2}{3} i (d+b x)}}{3 b \left (f-e^{2 i (d+b x)} (f+i g)-i g\right ) (f+i g)^2}+\frac {2 e^{\frac {5}{3} (a-i d)+\frac {8}{3} i (d+b x)}}{b \left (f-e^{2 i (d+b x)} (f+i g)-i g\right )^2 (f+i g)}+\frac {8 e^{\frac {5}{3} (a-i d)} \arctan \left (\frac {1+\frac {2 e^{\frac {2}{3} i (d+b x)} \sqrt [3]{f+i g}}{\sqrt [3]{f-i g}}}{\sqrt {3}}\right )}{3 \sqrt {3} b (f-i g)^{2/3} (f+i g)^{7/3}}-\frac {8 e^{\frac {5}{3} (a-i d)} \log \left (\sqrt [3]{f-i g}-e^{\frac {2}{3} i (d+b x)} \sqrt [3]{f+i g}\right )}{9 b (f-i g)^{2/3} (f+i g)^{7/3}}+\frac {4 e^{\frac {5}{3} (a-i d)} \log \left ((f-i g)^{2/3}+e^{\frac {2}{3} i (d+b x)} \sqrt [3]{f-i g} \sqrt [3]{f+i g}+e^{\frac {4}{3} i (d+b x)} (f+i g)^{2/3}\right )}{9 b (f-i g)^{2/3} (f+i g)^{7/3}} \] Output:
-8/3*exp(5/3*a-5/3*I*d+2/3*I*(b*x+d))/b/(f-exp(2*I*(b*x+d))*(f+I*g)-I*g)/( f+I*g)^2+2*exp(5/3*a-5/3*I*d+8/3*I*(b*x+d))/b/(f-exp(2*I*(b*x+d))*(f+I*g)- I*g)^2/(f+I*g)+8/9*exp(5/3*a-5/3*I*d)*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))*3^(1/2)/b/(f-I*g)^(2/3)/(f+I*g)^(7/3) -8/9*exp(5/3*a-5/3*I*d)*ln((f-I*g)^(1/3)-exp(2/3*I*(b*x+d))*(f+I*g)^(1/3)) /b/(f-I*g)^(2/3)/(f+I*g)^(7/3)+4/9*exp(5/3*a-5/3*I*d)*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)^(2/3)/(f+I*g)^(7/3)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.02 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.45 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\frac {2 e^{\frac {5}{3} (a-i d)} \left (\frac {9 e^{\frac {2}{3} i (d+b x)} \left (\left (-4+7 e^{2 i (d+b x)}\right ) f+i \left (4+7 e^{2 i (d+b x)}\right ) g\right )}{(f+i g)^2 \left (\left (-1+e^{2 i (d+b x)}\right ) f+i \left (1+e^{2 i (d+b x)}\right ) g\right )^2}+\frac {2 i \text {RootSum}\left [i f+g-i f \text {$\#$1}^6+g \text {$\#$1}^6\&,\frac {2 d+2 b x+3 i \log \left (\left (e^{\frac {1}{3} i (d+b x)}-\text {$\#$1}\right )^2\right )}{\text {$\#$1}^4}\&\right ]}{(f+i g)^3}\right )}{27 b} \] Input:
Integrate[E^((5*(a + I*b*x))/3)/(g*Cos[d + b*x] + f*Sin[d + b*x])^3,x]
Output:
(2*E^((5*(a - I*d))/3)*((9*E^(((2*I)/3)*(d + b*x))*((-4 + 7*E^((2*I)*(d + b*x)))*f + I*(4 + 7*E^((2*I)*(d + b*x)))*g))/((f + I*g)^2*((-1 + E^((2*I)* (d + b*x)))*f + I*(1 + E^((2*I)*(d + b*x)))*g)^2) + ((2*I)*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 & ])/(f + I*g)^3))/(27*b)
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.
\(\displaystyle \int \frac {e^{\frac {5}{3} (a+i b x)}}{(f \sin (b x+d)+g \cos (b x+d))^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{(f \sin (b x+d)+g \cos (b x+d))^3}dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \frac {e^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{(f \sin (b x+d)+g \cos (b x+d))^3}dx\) |
Input:
Int[E^((5*(a + I*b*x))/3)/(g*Cos[d + b*x] + f*Sin[d + b*x])^3,x]
Output:
$Aborted
\[\int \frac {{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{3}}d x\]
Input:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x)
Output:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1426 vs. \(2 (291) = 582\).
Time = 0.10 (sec) , antiderivative size = 1426, normalized size of antiderivative = 3.32 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x, algorithm= "fricas")
Output:
-1/6*(3*(b*f^4 + 2*b*f^2*g^2 + b*g^4 + (b*f^4 + 4*I*b*f^3*g - 6*b*f^2*g^2 - 4*I*b*f*g^3 + b*g^4 + sqrt(3)*(I*b*f^4 - 4*b*f^3*g - 6*I*b*f^2*g^2 + 4*b *f*g^3 + I*b*g^4))*e^(4*I*b*x + 4*I*d) - 2*(b*f^4 + 2*I*b*f^3*g + 2*I*b*f* g^3 - b*g^4 - sqrt(3)*(-I*b*f^4 + 2*b*f^3*g + 2*b*f*g^3 + I*b*g^4))*e^(2*I *b*x + 2*I*d) + sqrt(3)*(I*b*f^4 + 2*I*b*f^2*g^2 + I*b*g^4))*(512*I*e^(5*a - 5*I*d)/(-729*I*b^3*f^9 + 3645*b^3*f^8*g + 5832*I*b^3*f^7*g^2 + 10206*I* b^3*f^5*g^4 - 10206*b^3*f^4*g^5 - 5832*b^3*f^2*g^7 - 3645*I*b^3*f*g^8 + 72 9*b^3*g^9))^(1/3)*log(-1/16*(9*(b*f^3 + I*b*f^2*g + b*f*g^2 + I*b*g^3 + sq rt(3)*(I*b*f^3 - b*f^2*g + I*b*f*g^2 - b*g^3))*(512*I*e^(5*a - 5*I*d)/(-72 9*I*b^3*f^9 + 3645*b^3*f^8*g + 5832*I*b^3*f^7*g^2 + 10206*I*b^3*f^5*g^4 - 10206*b^3*f^4*g^5 - 5832*b^3*f^2*g^7 - 3645*I*b^3*f*g^8 + 729*b^3*g^9))^(1 /3) - 16*e^(2/3*I*b*x + 5/3*a - I*d))*e^(-5/3*a + 5/3*I*d)) + 3*(b*f^4 + 2 *b*f^2*g^2 + b*g^4 + (b*f^4 + 4*I*b*f^3*g - 6*b*f^2*g^2 - 4*I*b*f*g^3 + b* g^4 + sqrt(3)*(-I*b*f^4 + 4*b*f^3*g + 6*I*b*f^2*g^2 - 4*b*f*g^3 - I*b*g^4) )*e^(4*I*b*x + 4*I*d) - 2*(b*f^4 + 2*I*b*f^3*g + 2*I*b*f*g^3 - b*g^4 - sqr t(3)*(I*b*f^4 - 2*b*f^3*g - 2*b*f*g^3 - I*b*g^4))*e^(2*I*b*x + 2*I*d) + sq rt(3)*(-I*b*f^4 - 2*I*b*f^2*g^2 - I*b*g^4))*(512*I*e^(5*a - 5*I*d)/(-729*I *b^3*f^9 + 3645*b^3*f^8*g + 5832*I*b^3*f^7*g^2 + 10206*I*b^3*f^5*g^4 - 102 06*b^3*f^4*g^5 - 5832*b^3*f^2*g^7 - 3645*I*b^3*f*g^8 + 729*b^3*g^9))^(1/3) *log(-1/16*(9*(b*f^3 + I*b*f^2*g + b*f*g^2 + I*b*g^3 + sqrt(3)*(-I*b*f^...
Time = 11.31 (sec) , antiderivative size = 512, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\frac {\left (- 8 f e^{\frac {5 a}{3}} + 8 i g e^{\frac {5 a}{3}}\right ) e^{\frac {2 i b x}{3}} + \left (14 f e^{\frac {5 a}{3}} e^{2 i d} + 14 i g e^{\frac {5 a}{3}} e^{2 i d}\right ) e^{\frac {8 i b x}{3}}}{3 b f^{4} e^{i d} + 6 b f^{2} g^{2} e^{i d} + 3 b g^{4} e^{i d} + \left (- 6 b f^{4} e^{3 i d} - 12 i b f^{3} g e^{3 i d} - 12 i b f g^{3} e^{3 i d} + 6 b g^{4} e^{3 i d}\right ) e^{2 i b x} + \left (3 b f^{4} e^{5 i d} + 12 i b f^{3} g e^{5 i d} - 18 b f^{2} g^{2} e^{5 i d} - 12 i b f g^{3} e^{5 i d} + 3 b g^{4} e^{5 i d}\right ) e^{4 i b x}} + \operatorname {RootSum} {\left (z^{3} \cdot \left (729 b^{3} f^{9} e^{5 i d} + 3645 i b^{3} f^{8} g e^{5 i d} - 5832 b^{3} f^{7} g^{2} e^{5 i d} - 10206 b^{3} f^{5} g^{4} e^{5 i d} - 10206 i b^{3} f^{4} g^{5} e^{5 i d} - 5832 i b^{3} f^{2} g^{7} e^{5 i d} + 3645 b^{3} f g^{8} e^{5 i d} + 729 i b^{3} g^{9} e^{5 i d}\right ) + 512 e^{5 a}, \left ( i \mapsto i \log {\left (\frac {\left (9 i b f^{3} e^{i d} + 9 i i b f^{2} g e^{i d} + 9 i b f g^{2} e^{i d} + 9 i i b g^{3} e^{i d}\right ) e^{- \frac {5 a}{3}}}{8} + e^{\frac {2 i b x}{3}} \right )} \right )\right )} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))**3,x)
Output:
((-8*f*exp(5*a/3) + 8*I*g*exp(5*a/3))*exp(2*I*b*x/3) + (14*f*exp(5*a/3)*ex p(2*I*d) + 14*I*g*exp(5*a/3)*exp(2*I*d))*exp(8*I*b*x/3))/(3*b*f**4*exp(I*d ) + 6*b*f**2*g**2*exp(I*d) + 3*b*g**4*exp(I*d) + (-6*b*f**4*exp(3*I*d) - 1 2*I*b*f**3*g*exp(3*I*d) - 12*I*b*f*g**3*exp(3*I*d) + 6*b*g**4*exp(3*I*d))* exp(2*I*b*x) + (3*b*f**4*exp(5*I*d) + 12*I*b*f**3*g*exp(5*I*d) - 18*b*f**2 *g**2*exp(5*I*d) - 12*I*b*f*g**3*exp(5*I*d) + 3*b*g**4*exp(5*I*d))*exp(4*I *b*x)) + RootSum(_z**3*(729*b**3*f**9*exp(5*I*d) + 3645*I*b**3*f**8*g*exp( 5*I*d) - 5832*b**3*f**7*g**2*exp(5*I*d) - 10206*b**3*f**5*g**4*exp(5*I*d) - 10206*I*b**3*f**4*g**5*exp(5*I*d) - 5832*I*b**3*f**2*g**7*exp(5*I*d) + 3 645*b**3*f*g**8*exp(5*I*d) + 729*I*b**3*g**9*exp(5*I*d)) + 512*exp(5*a), L ambda(_i, _i*log((9*_i*b*f**3*exp(I*d) + 9*_i*I*b*f**2*g*exp(I*d) + 9*_i*b *f*g**2*exp(I*d) + 9*_i*I*b*g**3*exp(I*d))*exp(-5*a/3)/8 + exp(2*I*b*x/3)) ))
Exception generated. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x, algorithm= "maxima")
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. 8533 vs. \(2 (291) = 582\).
Time = 101.40 (sec) , antiderivative size = 8533, normalized size of antiderivative = 19.89 \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\text {Too large to display} \] Input:
integrate(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x, algorithm= "giac")
Output:
-8/9*(2*((sqrt(3)*cos(d)^4 - 6*sqrt(3)*cos(d)^2*sin(d)^2 + sqrt(3)*sin(d)^ 4)*f*cos(-1/6*pi*sgn(f*cos(d)^2 - 2*g*cos(d)*sin(d) - f*sin(d)^2)*sgn(-g*c os(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*ta n(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*p i*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)) - 4*(sqrt(3)*cos(d)^3*sin(d) - sqrt(3)*cos(d)*sin(d)^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)*si n(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(-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/(...
Timed out. \[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\int \frac {{\mathrm {e}}^{\frac {5\,a}{3}+\frac {b\,x\,5{}\mathrm {i}}{3}}}{{\left (g\,\cos \left (d+b\,x\right )+f\,\sin \left (d+b\,x\right )\right )}^3} \,d x \] Input:
int(exp((5*a)/3 + (b*x*5i)/3)/(g*cos(d + b*x) + f*sin(d + b*x))^3,x)
Output:
int(exp((5*a)/3 + (b*x*5i)/3)/(g*cos(d + b*x) + f*sin(d + b*x))^3, x)
\[ \int \frac {e^{\frac {5}{3} (a+i b x)}}{(g \cos (d+b x)+f \sin (d+b x))^3} \, dx=\int \frac {{\mathrm e}^{\frac {5 a}{3}+\frac {5 i b x}{3}}}{\left (g \cos \left (b x +d \right )+f \sin \left (b x +d \right )\right )^{3}}d x \] Input:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x)
Output:
int(exp(5/3*a+5/3*I*b*x)/(g*cos(b*x+d)+f*sin(b*x+d))^3,x)