\(\int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx\) [42]

Optimal result

Integrand size = 24, antiderivative size = 79 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {e^{a+b x}}{8 b}-\frac {b e^{a+b x} \cos (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}-\frac {d e^{a+b x} \sin (4 c+4 d x)}{2 \left (b^2+16 d^2\right )} \]

[Out]

1/8*exp(b*x+a)/b-1/8*b*exp(b*x+a)*cos(4*d*x+4*c)/(b^2+16*d^2)-1/2*d*exp(b*x+a)*sin(4*d*x+4*c)/(b^2+16*d^2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4557, 2225, 4518} \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {d e^{a+b x} \sin (4 c+4 d x)}{2 \left (b^2+16 d^2\right )}-\frac {b e^{a+b x} \cos (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}+\frac {e^{a+b x}}{8 b} \]

[In]

Int[E^(a + b*x)*Cos[c + d*x]^2*Sin[c + d*x]^2,x]

[Out]

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))

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 4518

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(C
os[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] + Simp[e*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 4557

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8} e^{a+b x}-\frac {1}{8} e^{a+b x} \cos (4 c+4 d x)\right ) \, dx \\ & = \frac {1}{8} \int e^{a+b x} \, dx-\frac {1}{8} \int e^{a+b x} \cos (4 c+4 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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.72 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {e^{a+b x} \left (b^2+16 d^2-b^2 \cos (4 (c+d x))-4 b d \sin (4 (c+d x))\right )}{8 \left (b^3+16 b d^2\right )} \]

[In]

Integrate[E^(a + b*x)*Cos[c + d*x]^2*Sin[c + d*x]^2,x]

[Out]

(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))

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{x b +a} \left (4 b d \sin \left (4 d x +4 c \right )+b^{2} \cos \left (4 d x +4 c \right )-b^{2}-16 d^{2}\right )}{8 \left (b^{2}+16 d^{2}\right ) b}\)	\(60\)
risch	\(\frac {{\mathrm e}^{x b +a} \left (-2 b^{2}-32 d^{2}+2 b^{2} \cos \left (4 d x +4 c \right )+8 b d \sin \left (4 d x +4 c \right )\right )}{16 b \left (4 i d +b \right ) \left (4 i d -b \right )}\)	\(68\)
default	\(\frac {{\mathrm e}^{x b +a}}{8 b}-\frac {b \,{\mathrm e}^{x b +a} \cos \left (4 d x +4 c \right )}{8 \left (b^{2}+16 d^{2}\right )}-\frac {d \,{\mathrm e}^{x b +a} \sin \left (4 d x +4 c \right )}{2 \left (b^{2}+16 d^{2}\right )}\)	\(71\)
norman	\(\frac {-\frac {4 d \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}+16 d^{2}}+\frac {28 d \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{2}+16 d^{2}}-\frac {28 d \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{b^{2}+16 d^{2}}+\frac {4 d \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{2}+16 d^{2}}+\frac {2 d^{2} {\mathrm e}^{x b +a}}{\left (b^{2}+16 d^{2}\right ) b}+\frac {2 d^{2} {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{\left (b^{2}+16 d^{2}\right ) b}+\frac {4 \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (b^{2}+16 d^{2}\right ) b}+\frac {4 \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{\left (b^{2}+16 d^{2}\right ) b}-\frac {4 \left (2 b^{2}-3 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{\left (b^{2}+16 d^{2}\right ) b}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(329\)

[In]

int(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

-1/8*exp(b*x+a)*(4*b*d*sin(4*d*x+4*c)+b^2*cos(4*d*x+4*c)-b^2-16*d^2)/(b^2+16*d^2)/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {2 \, {\left (2 \, b d \cos \left (d x + c\right )^{3} - b d \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) + {\left (b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, d^{2}\right )} e^{\left (b x + a\right )}}{b^{3} + 16 \, b d^{2}} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x, algorithm="fricas")

[Out]

-(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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.58 (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=\begin {cases} x e^{a} \sin ^{2}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\left (\frac {x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {\sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {\sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d}\right ) e^{a} & \text {for}\: b = 0 \\- \frac {x e^{a} e^{- 4 i d x} \sin ^{4}{\left (c + d x \right )}}{16} + \frac {i x e^{a} e^{- 4 i d x} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4} + \frac {3 x e^{a} e^{- 4 i d x} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} - \frac {i x e^{a} e^{- 4 i d x} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4} - \frac {x e^{a} e^{- 4 i d x} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {i e^{a} e^{- 4 i d x} \sin ^{4}{\left (c + d x \right )}}{24 d} + \frac {5 e^{a} e^{- 4 i d x} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{48 d} - \frac {5 e^{a} e^{- 4 i d x} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{48 d} + \frac {i e^{a} e^{- 4 i d x} \cos ^{4}{\left (c + d x \right )}}{24 d} & \text {for}\: b = - 4 i d \\- \frac {x e^{a} e^{4 i d x} \sin ^{4}{\left (c + d x \right )}}{16} - \frac {i x e^{a} e^{4 i d x} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4} + \frac {3 x e^{a} e^{4 i d x} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {i x e^{a} e^{4 i d x} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4} - \frac {x e^{a} e^{4 i d x} \cos ^{4}{\left (c + d x \right )}}{16} - \frac {i e^{a} e^{4 i d x} \sin ^{4}{\left (c + d x \right )}}{24 d} + \frac {5 e^{a} e^{4 i d x} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{48 d} - \frac {5 e^{a} e^{4 i d x} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{48 d} - \frac {i e^{a} e^{4 i d x} \cos ^{4}{\left (c + d x \right )}}{24 d} & \text {for}\: b = 4 i d \\\frac {b^{2} e^{a} e^{b x} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} + 16 b d^{2}} + \frac {2 b d e^{a} e^{b x} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{b^{3} + 16 b d^{2}} - \frac {2 b d e^{a} e^{b x} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{b^{3} + 16 b d^{2}} + \frac {2 d^{2} e^{a} e^{b x} \sin ^{4}{\left (c + d x \right )}}{b^{3} + 16 b d^{2}} + \frac {4 d^{2} e^{a} e^{b x} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{b^{3} + 16 b d^{2}} + \frac {2 d^{2} e^{a} e^{b x} \cos ^{4}{\left (c + d x \right )}}{b^{3} + 16 b d^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)**2*sin(d*x+c)**2,x)

[Out]

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 + sin(c + d*x)**3*cos(c + d*x)/(8*d) - sin(c + d*x)*cos(c + d*x)**3/(8*d))*
exp(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)*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*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*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)), (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 + d*x)**4/(b**3 + 16*b*d**2) + 4*d**2*exp(a)*exp(b*x)*sin(c + d
*x)**2*cos(c + d*x)**2/(b**3 + 16*b*d**2) + 2*d**2*exp(a)*exp(b*x)*cos(c + d*x)**4/(b**3 + 16*b*d**2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (70) = 140\).

Time = 0.21 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.99 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {{\left (b^{2} \cos \left (4 \, c\right ) e^{a} + 4 \, b d e^{a} \sin \left (4 \, c\right )\right )} \cos \left (4 \, d x\right ) e^{\left (b x\right )} + {\left (b^{2} \cos \left (4 \, c\right ) e^{a} - 4 \, b d e^{a} \sin \left (4 \, c\right )\right )} \cos \left (4 \, d x + 8 \, c\right ) e^{\left (b x\right )} + {\left (4 \, b d \cos \left (4 \, c\right ) e^{a} - b^{2} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (4 \, d x\right ) + {\left (4 \, b d \cos \left (4 \, c\right ) e^{a} + b^{2} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (4 \, d x + 8 \, c\right ) - 2 \, {\left (b^{2} \cos \left (4 \, c\right )^{2} e^{a} + b^{2} e^{a} \sin \left (4 \, c\right )^{2} + 16 \, {\left (\cos \left (4 \, c\right )^{2} e^{a} + e^{a} \sin \left (4 \, c\right )^{2}\right )} d^{2}\right )} e^{\left (b x\right )}}{16 \, {\left (b^{3} \cos \left (4 \, c\right )^{2} + b^{3} \sin \left (4 \, c\right )^{2} + 16 \, {\left (b \cos \left (4 \, c\right )^{2} + b \sin \left (4 \, c\right )^{2}\right )} d^{2}\right )}} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x, algorithm="maxima")

[Out]

-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*cos(4*c)*e^a - 4*b*d*e^a*sin(4*c))*co
s(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*sin(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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=-\frac {1}{8} \, {\left (\frac {b \cos \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}} + \frac {4 \, d \sin \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}}\right )} e^{\left (b x + a\right )} + \frac {e^{\left (b x + a\right )}}{8 \, b} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^2,x, algorithm="giac")

[Out]

-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

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^2(c+d x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}\,\left (b^2+16\,d^2-b^2\,\cos \left (4\,c+4\,d\,x\right )-4\,b\,d\,\sin \left (4\,c+4\,d\,x\right )\right )}{8\,b\,\left (b^2+16\,d^2\right )} \]

[In]

int(cos(c + d*x)^2*exp(a + b*x)*sin(c + d*x)^2,x)

[Out]

(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))