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

Optimal. Leaf size=119 \[ \frac{b e^{a+b x} \sin (c+d x)}{4 \left (b^2+d^2\right )}+\frac{b e^{a+b x} \sin (3 c+3 d x)}{4 \left (b^2+9 d^2\right )}-\frac{d e^{a+b x} \cos (c+d x)}{4 \left (b^2+d^2\right )}-\frac{3 d e^{a+b x} \cos (3 c+3 d x)}{4 \left (b^2+9 d^2\right )} \]

[Out]

-(d*E^(a + b*x)*Cos[c + d*x])/(4*(b^2 + d^2)) - (3*d*E^(a + b*x)*Cos[3*c + 3*d*x])/(4*(b^2 + 9*d^2)) + (b*E^(a
+ b*x)*Sin[c + d*x])/(4*(b^2 + d^2)) + (b*E^(a + b*x)*Sin[3*c + 3*d*x])/(4*(b^2 + 9*d^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0825251, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4469, 4432} \[ \frac{b e^{a+b x} \sin (c+d x)}{4 \left (b^2+d^2\right )}+\frac{b e^{a+b x} \sin (3 c+3 d x)}{4 \left (b^2+9 d^2\right )}-\frac{d e^{a+b x} \cos (c+d x)}{4 \left (b^2+d^2\right )}-\frac{3 d e^{a+b x} \cos (3 c+3 d x)}{4 \left (b^2+9 d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*E^(a + b*x)*Cos[c + d*x])/(4*(b^2 + d^2)) - (3*d*E^(a + b*x)*Cos[3*c + 3*d*x])/(4*(b^2 + 9*d^2)) + (b*E^(a
+ b*x)*Sin[c + d*x])/(4*(b^2 + d^2)) + (b*E^(a + b*x)*Sin[3*c + 3*d*x])/(4*(b^2 + 9*d^2))

Rule 4469

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]

Rule 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[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]

Rubi steps

\begin{align*} \int e^{a+b x} \cos ^2(c+d x) \sin (c+d x) \, dx &=\int \left (\frac{1}{4} e^{a+b x} \sin (c+d x)+\frac{1}{4} e^{a+b x} \sin (3 c+3 d x)\right ) \, dx\\ &=\frac{1}{4} \int e^{a+b x} \sin (c+d x) \, dx+\frac{1}{4} \int e^{a+b x} \sin (3 c+3 d x) \, dx\\ &=-\frac{d e^{a+b x} \cos (c+d x)}{4 \left (b^2+d^2\right )}-\frac{3 d e^{a+b x} \cos (3 c+3 d x)}{4 \left (b^2+9 d^2\right )}+\frac{b e^{a+b x} \sin (c+d x)}{4 \left (b^2+d^2\right )}+\frac{b e^{a+b x} \sin (3 c+3 d x)}{4 \left (b^2+9 d^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.664618, size = 74, normalized size = 0.62 \[ \frac{1}{4} e^{a+b x} \left (\frac{b \sin (c+d x)-d \cos (c+d x)}{b^2+d^2}+\frac{b \sin (3 (c+d x))-3 d \cos (3 (c+d x))}{b^2+9 d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(a + b*x)*((-(d*Cos[c + d*x]) + b*Sin[c + d*x])/(b^2 + d^2) + (-3*d*Cos[3*(c + d*x)] + b*Sin[3*(c + d*x)])/
(b^2 + 9*d^2)))/4

________________________________________________________________________________________

Maple [A]  time = 0.016, size = 108, normalized size = 0.9 \begin{align*} -{\frac{d{{\rm e}^{bx+a}}\cos \left ( dx+c \right ) }{4\,{b}^{2}+4\,{d}^{2}}}-{\frac{3\,d{{\rm e}^{bx+a}}\cos \left ( 3\,dx+3\,c \right ) }{4\,{b}^{2}+36\,{d}^{2}}}+{\frac{b{{\rm e}^{bx+a}}\sin \left ( dx+c \right ) }{4\,{b}^{2}+4\,{d}^{2}}}+{\frac{b{{\rm e}^{bx+a}}\sin \left ( 3\,dx+3\,c \right ) }{4\,{b}^{2}+36\,{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d*exp(b*x+a)*cos(d*x+c)/(b^2+d^2)-3/4*d*exp(b*x+a)*cos(3*d*x+3*c)/(b^2+9*d^2)+1/4*b*exp(b*x+a)*sin(d*x+c)
/(b^2+d^2)+1/4*b*exp(b*x+a)*sin(3*d*x+3*c)/(b^2+9*d^2)

________________________________________________________________________________________

Maxima [B]  time = 1.19727, size = 726, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((3*b^2*d*cos(3*c)*e^a + 3*d^3*cos(3*c)*e^a - b^3*e^a*sin(3*c) - b*d^2*e^a*sin(3*c))*cos(3*d*x)*e^(b*x) +
(3*b^2*d*cos(3*c)*e^a + 3*d^3*cos(3*c)*e^a + b^3*e^a*sin(3*c) + b*d^2*e^a*sin(3*c))*cos(3*d*x + 6*c)*e^(b*x)
+ (b^2*d*cos(3*c)*e^a + 9*d^3*cos(3*c)*e^a + b^3*e^a*sin(3*c) + 9*b*d^2*e^a*sin(3*c))*cos(d*x + 4*c)*e^(b*x) +
(b^2*d*cos(3*c)*e^a + 9*d^3*cos(3*c)*e^a - b^3*e^a*sin(3*c) - 9*b*d^2*e^a*sin(3*c))*cos(d*x - 2*c)*e^(b*x) -
(b^3*cos(3*c)*e^a + b*d^2*cos(3*c)*e^a + 3*b^2*d*e^a*sin(3*c) + 3*d^3*e^a*sin(3*c))*e^(b*x)*sin(3*d*x) - (b^3*
cos(3*c)*e^a + b*d^2*cos(3*c)*e^a - 3*b^2*d*e^a*sin(3*c) - 3*d^3*e^a*sin(3*c))*e^(b*x)*sin(3*d*x + 6*c) - (b^3
*cos(3*c)*e^a + 9*b*d^2*cos(3*c)*e^a - b^2*d*e^a*sin(3*c) - 9*d^3*e^a*sin(3*c))*e^(b*x)*sin(d*x + 4*c) - (b^3*
cos(3*c)*e^a + 9*b*d^2*cos(3*c)*e^a + b^2*d*e^a*sin(3*c) + 9*d^3*e^a*sin(3*c))*e^(b*x)*sin(d*x - 2*c))/(b^4*co
s(3*c)^2 + b^4*sin(3*c)^2 + 9*(cos(3*c)^2 + sin(3*c)^2)*d^4 + 10*(b^2*cos(3*c)^2 + b^2*sin(3*c)^2)*d^2)

________________________________________________________________________________________

Fricas [A]  time = 0.486486, size = 224, normalized size = 1.88 \begin{align*} \frac{{\left (2 \, b d^{2} +{\left (b^{3} + b d^{2}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) +{\left (2 \, b^{2} d \cos \left (d x + c\right ) - 3 \,{\left (b^{2} d + d^{3}\right )} \cos \left (d x + c\right )^{3}\right )} e^{\left (b x + a\right )}}{b^{4} + 10 \, b^{2} d^{2} + 9 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*b*d^2 + (b^3 + b*d^2)*cos(d*x + c)^2)*e^(b*x + a)*sin(d*x + c) + (2*b^2*d*cos(d*x + c) - 3*(b^2*d + d^3)*c
os(d*x + c)^3)*e^(b*x + a))/(b^4 + 10*b^2*d^2 + 9*d^4)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.15325, size = 135, normalized size = 1.13 \begin{align*} -\frac{1}{4} \,{\left (\frac{3 \, d \cos \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}} - \frac{b \sin \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac{1}{4} \,{\left (\frac{d \cos \left (d x + c\right )}{b^{2} + d^{2}} - \frac{b \sin \left (d x + c\right )}{b^{2} + d^{2}}\right )} e^{\left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*d*cos(3*d*x + 3*c)/(b^2 + 9*d^2) - b*sin(3*d*x + 3*c)/(b^2 + 9*d^2))*e^(b*x + a) - 1/4*(d*cos(d*x + c)
/(b^2 + d^2) - b*sin(d*x + c)/(b^2 + d^2))*e^(b*x + a)