Integrand size = 24, antiderivative size = 79 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {e^{a+b x}}{8 b}-\frac {b e^{a+b x} \cos (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}-\frac {d e^{a+b x} \sin (4 c+4 d x)}{2 \left (b^2+16 d^2\right )} \] Output:
1/8*exp(b*x+a)/b-b*exp(b*x+a)*cos(4*d*x+4*c)/(8*b^2+128*d^2)-d*exp(b*x+a)* sin(4*d*x+4*c)/(2*b^2+32*d^2)
Time = 0.38 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.72 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {e^{a+b x} \left (b^2+16 d^2-b^2 \cos (4 (c+d x))-4 b d \sin (4 (c+d x))\right )}{8 \left (b^3+16 b d^2\right )} \] Input:
Integrate[E^(a + b*x)*Cos[c + d*x]^2*Sin[c + d*x]^2,x]
Output:
(E^(a + b*x)*(b^2 + 16*d^2 - b^2*Cos[4*(c + d*x)] - 4*b*d*Sin[4*(c + d*x)] ))/(8*(b^3 + 16*b*d^2))
Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4972, 2009}
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.
\(\displaystyle \int e^{a+b x} \sin ^2(c+d x) \cos ^2(c+d x) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \left (\frac {1}{8} e^{a+b x}-\frac {1}{8} e^{a+b x} \cos (4 c+4 d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d e^{a+b x} \sin (4 c+4 d x)}{2 \left (b^2+16 d^2\right )}-\frac {b e^{a+b x} \cos (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}+\frac {e^{a+b x}}{8 b}\) |
Input:
Int[E^(a + b*x)*Cos[c + d*x]^2*Sin[c + d*x]^2,x]
Output:
E^(a + b*x)/(8*b) - (b*E^(a + b*x)*Cos[4*c + 4*d*x])/(8*(b^2 + 16*d^2)) - (d*E^(a + b*x)*Sin[4*c + 4*d*x])/(2*(b^2 + 16*d^2))
Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_ .) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 1.36 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.76
method | result | size |
parallelrisch | \(-\frac {{\mathrm e}^{b x +a} \left (b^{2} \cos \left (4 d x +4 c \right )+4 b d \sin \left (4 d x +4 c \right )-b^{2}-16 d^{2}\right )}{8 \left (b^{2}+16 d^{2}\right ) b}\) | \(60\) |
risch | \(\frac {{\mathrm e}^{b x +a} \left (-2 b^{2}-32 d^{2}+2 b^{2} \cos \left (4 d x +4 c \right )+8 b d \sin \left (4 d x +4 c \right )\right )}{16 b \left (4 i d +b \right ) \left (4 i d -b \right )}\) | \(68\) |
default | \(\frac {{\mathrm e}^{b x +a}}{8 b}-\frac {b \,{\mathrm e}^{b x +a} \cos \left (4 d x +4 c \right )}{8 \left (b^{2}+16 d^{2}\right )}-\frac {d \,{\mathrm e}^{b x +a} \sin \left (4 d x +4 c \right )}{2 \left (b^{2}+16 d^{2}\right )}\) | \(71\) |
orering | \(\frac {\left (3 b^{2}+16 d^{2}\right ) {\mathrm e}^{b x +a} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}}{\left (b^{2}+16 d^{2}\right ) b}-\frac {3 \left (b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}-2 \,{\mathrm e}^{b x +a} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} d +2 \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) d \right )}{b^{2}+16 d^{2}}+\frac {b^{2} {\mathrm e}^{b x +a} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}-4 b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} d +4 b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) d +2 \,{\mathrm e}^{b x +a} d^{2} \sin \left (d x +c \right )^{4}-12 \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2} d^{2}+2 \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )^{4} d^{2}}{\left (b^{2}+16 d^{2}\right ) b}\) | \(289\) |
norman | \(\frac {-\frac {4 d \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}+16 d^{2}}+\frac {28 d \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{2}+16 d^{2}}-\frac {28 d \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{b^{2}+16 d^{2}}+\frac {4 d \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{2}+16 d^{2}}+\frac {2 d^{2} {\mathrm e}^{b x +a}}{\left (b^{2}+16 d^{2}\right ) b}+\frac {2 d^{2} {\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{\left (b^{2}+16 d^{2}\right ) b}+\frac {4 \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (b^{2}+16 d^{2}\right ) b}+\frac {4 \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{\left (b^{2}+16 d^{2}\right ) b}-\frac {4 \left (2 b^{2}-3 d^{2}\right ) {\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{\left (b^{2}+16 d^{2}\right ) b}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(329\) |
Input:
int(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/8*exp(b*x+a)*(b^2*cos(4*d*x+4*c)+4*b*d*sin(4*d*x+4*c)-b^2-16*d^2)/(b^2+ 16*d^2)/b
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {2 \, {\left (2 \, b d \cos \left (d x + c\right )^{3} - b d \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) + {\left (b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, d^{2}\right )} e^{\left (b x + a\right )}}{b^{3} + 16 \, b d^{2}} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x, algorithm="fricas")
Output:
-(2*(2*b*d*cos(d*x + c)^3 - b*d*cos(d*x + c))*e^(b*x + a)*sin(d*x + c) + ( b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*d^2)*e^(b*x + a))/(b^3 + 16*b* d^2)
Result contains complex when optimal does not.
Time = 6.47 (sec) , antiderivative size = 850, normalized size of antiderivative = 10.76 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx =\text {Too large to display} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)**2*sin(d*x+c)**2,x)
Output:
Piecewise((x*exp(a)*sin(c)**2*cos(c)**2, Eq(b, 0) & Eq(d, 0)), ((x*sin(c + d*x)**4/8 + x*sin(c + d*x)**2*cos(c + d*x)**2/4 + x*cos(c + d*x)**4/8 + s in(c + d*x)**3*cos(c + d*x)/(8*d) - sin(c + d*x)*cos(c + d*x)**3/(8*d))*ex p(a), Eq(b, 0)), (-x*exp(a)*exp(-4*I*d*x)*sin(c + d*x)**4/16 + I*x*exp(a)* exp(-4*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/4 + 3*x*exp(a)*exp(-4*I*d*x)*si n(c + d*x)**2*cos(c + d*x)**2/8 - I*x*exp(a)*exp(-4*I*d*x)*sin(c + d*x)*co s(c + d*x)**3/4 - x*exp(a)*exp(-4*I*d*x)*cos(c + d*x)**4/16 + I*exp(a)*exp (-4*I*d*x)*sin(c + d*x)**4/(24*d) + 5*exp(a)*exp(-4*I*d*x)*sin(c + d*x)**3 *cos(c + d*x)/(48*d) - 5*exp(a)*exp(-4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3 /(48*d) + I*exp(a)*exp(-4*I*d*x)*cos(c + d*x)**4/(24*d), Eq(b, -4*I*d)), ( -x*exp(a)*exp(4*I*d*x)*sin(c + d*x)**4/16 - I*x*exp(a)*exp(4*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/4 + 3*x*exp(a)*exp(4*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**2/8 + I*x*exp(a)*exp(4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/4 - x*ex p(a)*exp(4*I*d*x)*cos(c + d*x)**4/16 - I*exp(a)*exp(4*I*d*x)*sin(c + d*x)* *4/(24*d) + 5*exp(a)*exp(4*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/(48*d) - 5* exp(a)*exp(4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/(48*d) - I*exp(a)*exp(4*I *d*x)*cos(c + d*x)**4/(24*d), Eq(b, 4*I*d)), (b**2*exp(a)*exp(b*x)*sin(c + d*x)**2*cos(c + d*x)**2/(b**3 + 16*b*d**2) + 2*b*d*exp(a)*exp(b*x)*sin(c + d*x)**3*cos(c + d*x)/(b**3 + 16*b*d**2) - 2*b*d*exp(a)*exp(b*x)*sin(c + d*x)*cos(c + d*x)**3/(b**3 + 16*b*d**2) + 2*d**2*exp(a)*exp(b*x)*sin(c ...
Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (70) = 140\).
Time = 0.06 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.99 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {{\left (b^{2} \cos \left (4 \, c\right ) e^{a} + 4 \, b d e^{a} \sin \left (4 \, c\right )\right )} \cos \left (4 \, d x\right ) e^{\left (b x\right )} + {\left (b^{2} \cos \left (4 \, c\right ) e^{a} - 4 \, b d e^{a} \sin \left (4 \, c\right )\right )} \cos \left (4 \, d x + 8 \, c\right ) e^{\left (b x\right )} + {\left (4 \, b d \cos \left (4 \, c\right ) e^{a} - b^{2} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (4 \, d x\right ) + {\left (4 \, b d \cos \left (4 \, c\right ) e^{a} + b^{2} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (4 \, d x + 8 \, c\right ) - 2 \, {\left (b^{2} \cos \left (4 \, c\right )^{2} e^{a} + b^{2} e^{a} \sin \left (4 \, c\right )^{2} + 16 \, {\left (\cos \left (4 \, c\right )^{2} e^{a} + e^{a} \sin \left (4 \, c\right )^{2}\right )} d^{2}\right )} e^{\left (b x\right )}}{16 \, {\left (b^{3} \cos \left (4 \, c\right )^{2} + b^{3} \sin \left (4 \, c\right )^{2} + 16 \, {\left (b \cos \left (4 \, c\right )^{2} + b \sin \left (4 \, c\right )^{2}\right )} d^{2}\right )}} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x, algorithm="maxima")
Output:
-1/16*((b^2*cos(4*c)*e^a + 4*b*d*e^a*sin(4*c))*cos(4*d*x)*e^(b*x) + (b^2*c os(4*c)*e^a - 4*b*d*e^a*sin(4*c))*cos(4*d*x + 8*c)*e^(b*x) + (4*b*d*cos(4* c)*e^a - b^2*e^a*sin(4*c))*e^(b*x)*sin(4*d*x) + (4*b*d*cos(4*c)*e^a + b^2* e^a*sin(4*c))*e^(b*x)*sin(4*d*x + 8*c) - 2*(b^2*cos(4*c)^2*e^a + b^2*e^a*s in(4*c)^2 + 16*(cos(4*c)^2*e^a + e^a*sin(4*c)^2)*d^2)*e^(b*x))/(b^3*cos(4* c)^2 + b^3*sin(4*c)^2 + 16*(b*cos(4*c)^2 + b*sin(4*c)^2)*d^2)
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {1}{8} \, {\left (\frac {b \cos \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}} + \frac {4 \, d \sin \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}}\right )} e^{\left (b x + a\right )} + \frac {e^{\left (b x + a\right )}}{8 \, b} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x, algorithm="giac")
Output:
-1/8*(b*cos(4*d*x + 4*c)/(b^2 + 16*d^2) + 4*d*sin(4*d*x + 4*c)/(b^2 + 16*d ^2))*e^(b*x + a) + 1/8*e^(b*x + a)/b
Time = 0.45 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}\,\left (b^2+16\,d^2-b^2\,\cos \left (4\,c+4\,d\,x\right )-4\,b\,d\,\sin \left (4\,c+4\,d\,x\right )\right )}{8\,b\,\left (b^2+16\,d^2\right )} \] Input:
int(cos(c + d*x)^2*exp(a + b*x)*sin(c + d*x)^2,x)
Output:
(exp(a + b*x)*(b^2 + 16*d^2 - b^2*cos(4*c + 4*d*x) - 4*b*d*sin(4*c + 4*d*x )))/(8*b*(b^2 + 16*d^2))
Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {e^{b x +a} \left (4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b d -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b d -\sin \left (d x +c \right )^{4} b^{2}+\sin \left (d x +c \right )^{2} b^{2}+2 d^{2}\right )}{b \left (b^{2}+16 d^{2}\right )} \] Input:
int(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x)
Output:
(e**(a + b*x)*(4*cos(c + d*x)*sin(c + d*x)**3*b*d - 2*cos(c + d*x)*sin(c + d*x)*b*d - sin(c + d*x)**4*b**2 + sin(c + d*x)**2*b**2 + 2*d**2))/(b*(b** 2 + 16*d**2))