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

Optimal. Leaf size=129 \[ \frac{b e^{a+b x} \sin (2 c+2 d x)}{4 \left (b^2+4 d^2\right )}-\frac{b e^{a+b x} \sin (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}-\frac{d e^{a+b x} \cos (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}+\frac{d e^{a+b x} \cos (4 c+4 d x)}{2 \left (b^2+16 d^2\right )} \]

[Out]

-(d*E^(a + b*x)*Cos[2*c + 2*d*x])/(2*(b^2 + 4*d^2)) + (d*E^(a + b*x)*Cos[4*c + 4*d*x])/(2*(b^2 + 16*d^2)) + (b
*E^(a + b*x)*Sin[2*c + 2*d*x])/(4*(b^2 + 4*d^2)) - (b*E^(a + b*x)*Sin[4*c + 4*d*x])/(8*(b^2 + 16*d^2))

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Rubi [A]  time = 0.0877769, antiderivative size = 129, 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} \[ \frac{b e^{a+b x} \sin (2 c+2 d x)}{4 \left (b^2+4 d^2\right )}-\frac{b e^{a+b x} \sin (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}-\frac{d e^{a+b x} \cos (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}+\frac{d e^{a+b x} \cos (4 c+4 d x)}{2 \left (b^2+16 d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*E^(a + b*x)*Cos[2*c + 2*d*x])/(2*(b^2 + 4*d^2)) + (d*E^(a + b*x)*Cos[4*c + 4*d*x])/(2*(b^2 + 16*d^2)) + (b
*E^(a + b*x)*Sin[2*c + 2*d*x])/(4*(b^2 + 4*d^2)) - (b*E^(a + b*x)*Sin[4*c + 4*d*x])/(8*(b^2 + 16*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 (c+d x) \sin ^3(c+d x) \, dx &=\int \left (\frac{1}{4} e^{a+b x} \sin (2 c+2 d x)-\frac{1}{8} e^{a+b x} \sin (4 c+4 d x)\right ) \, dx\\ &=-\left (\frac{1}{8} \int e^{a+b x} \sin (4 c+4 d x) \, dx\right )+\frac{1}{4} \int e^{a+b x} \sin (2 c+2 d x) \, dx\\ &=-\frac{d e^{a+b x} \cos (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}+\frac{d e^{a+b x} \cos (4 c+4 d x)}{2 \left (b^2+16 d^2\right )}+\frac{b e^{a+b x} \sin (2 c+2 d x)}{4 \left (b^2+4 d^2\right )}-\frac{b e^{a+b x} \sin (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.931011, size = 82, normalized size = 0.64 \[ \frac{1}{8} e^{a+b x} \left (\frac{2 (b \sin (2 (c+d x))-2 d \cos (2 (c+d x)))}{b^2+4 d^2}+\frac{4 d \cos (4 (c+d x))-b \sin (4 (c+d x))}{b^2+16 d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(a + b*x)*((2*(-2*d*Cos[2*(c + d*x)] + b*Sin[2*(c + d*x)]))/(b^2 + 4*d^2) + (4*d*Cos[4*(c + d*x)] - b*Sin[4
*(c + d*x)])/(b^2 + 16*d^2)))/8

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Maple [A]  time = 0.017, size = 118, normalized size = 0.9 \begin{align*} -{\frac{d{{\rm e}^{bx+a}}\cos \left ( 2\,dx+2\,c \right ) }{2\,{b}^{2}+8\,{d}^{2}}}+{\frac{d{{\rm e}^{bx+a}}\cos \left ( 4\,dx+4\,c \right ) }{2\,{b}^{2}+32\,{d}^{2}}}+{\frac{b{{\rm e}^{bx+a}}\sin \left ( 2\,dx+2\,c \right ) }{4\,{b}^{2}+16\,{d}^{2}}}-{\frac{b{{\rm e}^{bx+a}}\sin \left ( 4\,dx+4\,c \right ) }{8\,{b}^{2}+128\,{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.14362, 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)*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

1/16*((4*b^2*d*cos(4*c)*e^a + 16*d^3*cos(4*c)*e^a - b^3*e^a*sin(4*c) - 4*b*d^2*e^a*sin(4*c))*cos(4*d*x)*e^(b*x
) + (4*b^2*d*cos(4*c)*e^a + 16*d^3*cos(4*c)*e^a + b^3*e^a*sin(4*c) + 4*b*d^2*e^a*sin(4*c))*cos(4*d*x + 8*c)*e^
(b*x) - 2*(2*b^2*d*cos(4*c)*e^a + 32*d^3*cos(4*c)*e^a + b^3*e^a*sin(4*c) + 16*b*d^2*e^a*sin(4*c))*cos(2*d*x +
6*c)*e^(b*x) - 2*(2*b^2*d*cos(4*c)*e^a + 32*d^3*cos(4*c)*e^a - b^3*e^a*sin(4*c) - 16*b*d^2*e^a*sin(4*c))*cos(2
*d*x - 2*c)*e^(b*x) - (b^3*cos(4*c)*e^a + 4*b*d^2*cos(4*c)*e^a + 4*b^2*d*e^a*sin(4*c) + 16*d^3*e^a*sin(4*c))*e
^(b*x)*sin(4*d*x) - (b^3*cos(4*c)*e^a + 4*b*d^2*cos(4*c)*e^a - 4*b^2*d*e^a*sin(4*c) - 16*d^3*e^a*sin(4*c))*e^(
b*x)*sin(4*d*x + 8*c) + 2*(b^3*cos(4*c)*e^a + 16*b*d^2*cos(4*c)*e^a - 2*b^2*d*e^a*sin(4*c) - 32*d^3*e^a*sin(4*
c))*e^(b*x)*sin(2*d*x + 6*c) + 2*(b^3*cos(4*c)*e^a + 16*b*d^2*cos(4*c)*e^a + 2*b^2*d*e^a*sin(4*c) + 32*d^3*e^a
*sin(4*c))*e^(b*x)*sin(2*d*x - 2*c))/(b^4*cos(4*c)^2 + b^4*sin(4*c)^2 + 64*(cos(4*c)^2 + sin(4*c)^2)*d^4 + 20*
(b^2*cos(4*c)^2 + b^2*sin(4*c)^2)*d^2)

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Fricas [A]  time = 0.487336, size = 302, normalized size = 2.34 \begin{align*} -\frac{{\left ({\left (b^{3} + 4 \, b d^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (b^{3} + 10 \, b d^{2}\right )} \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) -{\left (4 \,{\left (b^{2} d + 4 \, d^{3}\right )} \cos \left (d x + c\right )^{4} + b^{2} d + 10 \, d^{3} -{\left (5 \, b^{2} d + 32 \, d^{3}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )}}{b^{4} + 20 \, b^{2} d^{2} + 64 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((b^3 + 4*b*d^2)*cos(d*x + c)^3 - (b^3 + 10*b*d^2)*cos(d*x + c))*e^(b*x + a)*sin(d*x + c) - (4*(b^2*d + 4*d^
3)*cos(d*x + c)^4 + b^2*d + 10*d^3 - (5*b^2*d + 32*d^3)*cos(d*x + c)^2)*e^(b*x + a))/(b^4 + 20*b^2*d^2 + 64*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)*sin(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20013, size = 150, normalized size = 1.16 \begin{align*} \frac{1}{8} \,{\left (\frac{4 \, d \cos \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}} - \frac{b \sin \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac{1}{4} \,{\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)*sin(d*x+c)^3,x, algorithm="giac")

[Out]

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