3.46 \(\int e^{a+b x} \cos ^3(c+d x) \sin ^3(c+d x) \, dx\)

Optimal. Leaf size=129 \[ \frac{3 b e^{a+b x} \sin (2 c+2 d x)}{32 \left (b^2+4 d^2\right )}-\frac{b e^{a+b x} \sin (6 c+6 d x)}{32 \left (b^2+36 d^2\right )}-\frac{3 d e^{a+b x} \cos (2 c+2 d x)}{16 \left (b^2+4 d^2\right )}+\frac{3 d e^{a+b x} \cos (6 c+6 d x)}{16 \left (b^2+36 d^2\right )} \]

[Out]

(-3*d*E^(a + b*x)*Cos[2*c + 2*d*x])/(16*(b^2 + 4*d^2)) + (3*d*E^(a + b*x)*Cos[6*c + 6*d*x])/(16*(b^2 + 36*d^2)
) + (3*b*E^(a + b*x)*Sin[2*c + 2*d*x])/(32*(b^2 + 4*d^2)) - (b*E^(a + b*x)*Sin[6*c + 6*d*x])/(32*(b^2 + 36*d^2
))

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Rubi [A]  time = 0.100633, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {4469, 4432} \[ \frac{3 b e^{a+b x} \sin (2 c+2 d x)}{32 \left (b^2+4 d^2\right )}-\frac{b e^{a+b x} \sin (6 c+6 d x)}{32 \left (b^2+36 d^2\right )}-\frac{3 d e^{a+b x} \cos (2 c+2 d x)}{16 \left (b^2+4 d^2\right )}+\frac{3 d e^{a+b x} \cos (6 c+6 d x)}{16 \left (b^2+36 d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cos[c + d*x]^3*Sin[c + d*x]^3,x]

[Out]

(-3*d*E^(a + b*x)*Cos[2*c + 2*d*x])/(16*(b^2 + 4*d^2)) + (3*d*E^(a + b*x)*Cos[6*c + 6*d*x])/(16*(b^2 + 36*d^2)
) + (3*b*E^(a + b*x)*Sin[2*c + 2*d*x])/(32*(b^2 + 4*d^2)) - (b*E^(a + b*x)*Sin[6*c + 6*d*x])/(32*(b^2 + 36*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 ^3(c+d x) \sin ^3(c+d x) \, dx &=\int \left (\frac{3}{32} e^{a+b x} \sin (2 c+2 d x)-\frac{1}{32} e^{a+b x} \sin (6 c+6 d x)\right ) \, dx\\ &=-\left (\frac{1}{32} \int e^{a+b x} \sin (6 c+6 d x) \, dx\right )+\frac{3}{32} \int e^{a+b x} \sin (2 c+2 d x) \, dx\\ &=-\frac{3 d e^{a+b x} \cos (2 c+2 d x)}{16 \left (b^2+4 d^2\right )}+\frac{3 d e^{a+b x} \cos (6 c+6 d x)}{16 \left (b^2+36 d^2\right )}+\frac{3 b e^{a+b x} \sin (2 c+2 d x)}{32 \left (b^2+4 d^2\right )}-\frac{b e^{a+b x} \sin (6 c+6 d x)}{32 \left (b^2+36 d^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.948675, size = 111, normalized size = 0.86 \[ \frac{e^{a+b x} \left (-6 d \left (b^2+36 d^2\right ) \cos (2 (c+d x))+6 d \left (b^2+4 d^2\right ) \cos (6 (c+d x))-2 b \sin (2 (c+d x)) \left (\left (b^2+4 d^2\right ) \cos (4 (c+d x))-b^2-52 d^2\right )\right )}{32 \left (40 b^2 d^2+b^4+144 d^4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Cos[c + d*x]^3*Sin[c + d*x]^3,x]

[Out]

(E^(a + b*x)*(-6*d*(b^2 + 36*d^2)*Cos[2*(c + d*x)] + 6*d*(b^2 + 4*d^2)*Cos[6*(c + d*x)] - 2*b*(-b^2 - 52*d^2 +
(b^2 + 4*d^2)*Cos[4*(c + d*x)])*Sin[2*(c + d*x)]))/(32*(b^4 + 40*b^2*d^2 + 144*d^4))

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Maple [A]  time = 0.022, size = 118, normalized size = 0.9 \begin{align*} -{\frac{3\,d{{\rm e}^{bx+a}}\cos \left ( 2\,dx+2\,c \right ) }{16\,{b}^{2}+64\,{d}^{2}}}+{\frac{3\,d{{\rm e}^{bx+a}}\cos \left ( 6\,dx+6\,c \right ) }{16\,{b}^{2}+576\,{d}^{2}}}+{\frac{3\,b{{\rm e}^{bx+a}}\sin \left ( 2\,dx+2\,c \right ) }{32\,{b}^{2}+128\,{d}^{2}}}-{\frac{b{{\rm e}^{bx+a}}\sin \left ( 6\,dx+6\,c \right ) }{32\,{b}^{2}+1152\,{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c)^3,x)

[Out]

-3/16*d*exp(b*x+a)*cos(2*d*x+2*c)/(b^2+4*d^2)+3/16*d*exp(b*x+a)*cos(6*d*x+6*c)/(b^2+36*d^2)+3/32*b*exp(b*x+a)*
sin(2*d*x+2*c)/(b^2+4*d^2)-1/32*b*exp(b*x+a)*sin(6*d*x+6*c)/(b^2+36*d^2)

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Maxima [B]  time = 1.1529, size = 743, normalized size = 5.76 \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)^3*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

1/64*((6*b^2*d*cos(6*c)*e^a + 24*d^3*cos(6*c)*e^a - b^3*e^a*sin(6*c) - 4*b*d^2*e^a*sin(6*c))*cos(6*d*x)*e^(b*x
) + (6*b^2*d*cos(6*c)*e^a + 24*d^3*cos(6*c)*e^a + b^3*e^a*sin(6*c) + 4*b*d^2*e^a*sin(6*c))*cos(6*d*x + 12*c)*e
^(b*x) - 3*(2*b^2*d*cos(6*c)*e^a + 72*d^3*cos(6*c)*e^a + b^3*e^a*sin(6*c) + 36*b*d^2*e^a*sin(6*c))*cos(2*d*x +
8*c)*e^(b*x) - 3*(2*b^2*d*cos(6*c)*e^a + 72*d^3*cos(6*c)*e^a - b^3*e^a*sin(6*c) - 36*b*d^2*e^a*sin(6*c))*cos(
2*d*x - 4*c)*e^(b*x) - (b^3*cos(6*c)*e^a + 4*b*d^2*cos(6*c)*e^a + 6*b^2*d*e^a*sin(6*c) + 24*d^3*e^a*sin(6*c))*
e^(b*x)*sin(6*d*x) - (b^3*cos(6*c)*e^a + 4*b*d^2*cos(6*c)*e^a - 6*b^2*d*e^a*sin(6*c) - 24*d^3*e^a*sin(6*c))*e^
(b*x)*sin(6*d*x + 12*c) + 3*(b^3*cos(6*c)*e^a + 36*b*d^2*cos(6*c)*e^a - 2*b^2*d*e^a*sin(6*c) - 72*d^3*e^a*sin(
6*c))*e^(b*x)*sin(2*d*x + 8*c) + 3*(b^3*cos(6*c)*e^a + 36*b*d^2*cos(6*c)*e^a + 2*b^2*d*e^a*sin(6*c) + 72*d^3*e
^a*sin(6*c))*e^(b*x)*sin(2*d*x - 4*c))/(b^4*cos(6*c)^2 + b^4*sin(6*c)^2 + 144*(cos(6*c)^2 + sin(6*c)^2)*d^4 +
40*(b^2*cos(6*c)^2 + b^2*sin(6*c)^2)*d^2)

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Fricas [A]  time = 0.493848, size = 356, normalized size = 2.76 \begin{align*} -\frac{{\left ({\left (b^{3} + 4 \, b d^{2}\right )} \cos \left (d x + c\right )^{5} - 6 \, b d^{2} \cos \left (d x + c\right ) -{\left (b^{3} + 4 \, b d^{2}\right )} \cos \left (d x + c\right )^{3}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) - 3 \,{\left (2 \,{\left (b^{2} d + 4 \, d^{3}\right )} \cos \left (d x + c\right )^{6} + b^{2} d \cos \left (d x + c\right )^{2} - 3 \,{\left (b^{2} d + 4 \, d^{3}\right )} \cos \left (d x + c\right )^{4} + 2 \, d^{3}\right )} e^{\left (b x + a\right )}}{b^{4} + 40 \, b^{2} d^{2} + 144 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

-(((b^3 + 4*b*d^2)*cos(d*x + c)^5 - 6*b*d^2*cos(d*x + c) - (b^3 + 4*b*d^2)*cos(d*x + c)^3)*e^(b*x + a)*sin(d*x
+ c) - 3*(2*(b^2*d + 4*d^3)*cos(d*x + c)^6 + b^2*d*cos(d*x + c)^2 - 3*(b^2*d + 4*d^3)*cos(d*x + c)^4 + 2*d^3)
*e^(b*x + a))/(b^4 + 40*b^2*d^2 + 144*d^4)

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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)**3*sin(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.14695, size = 150, normalized size = 1.16 \begin{align*} \frac{1}{32} \,{\left (\frac{6 \, d \cos \left (6 \, d x + 6 \, c\right )}{b^{2} + 36 \, d^{2}} - \frac{b \sin \left (6 \, d x + 6 \, c\right )}{b^{2} + 36 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac{3}{32} \,{\left (\frac{2 \, d \cos \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}} - \frac{b \sin \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, 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)^3*sin(d*x+c)^3,x, algorithm="giac")

[Out]

1/32*(6*d*cos(6*d*x + 6*c)/(b^2 + 36*d^2) - b*sin(6*d*x + 6*c)/(b^2 + 36*d^2))*e^(b*x + a) - 3/32*(2*d*cos(2*d
*x + 2*c)/(b^2 + 4*d^2) - b*sin(2*d*x + 2*c)/(b^2 + 4*d^2))*e^(b*x + a)