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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.00, 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, 4433} \[ \frac {d e^{a+b x} \sin (c+d x)}{4 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \sin (3 c+3 d x)}{4 \left (b^2+9 d^2\right )}+\frac {b e^{a+b x} \cos (c+d x)}{4 \left (b^2+d^2\right )}-\frac {b e^{a+b x} \cos (3 c+3 d x)}{4 \left (b^2+9 d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4433

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.66, size = 74, normalized size = 0.62 \[ \frac {1}{4} e^{a+b x} \left (\frac {b \cos (c+d x)+d \sin (c+d x)}{b^2+d^2}-\frac {b \cos (3 (c+d x))+3 d \sin (3 (c+d x))}{b^2+9 d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 109, normalized size = 0.92 \[ \frac {{\left (b^{2} d + 3 \, d^{3} - 3 \, {\left (b^{2} d + d^{3}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) - {\left ({\left (b^{3} + b d^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (b^{3} + 3 \, b d^{2}\right )} \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )}}{b^{4} + 10 \, b^{2} d^{2} + 9 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 98, normalized size = 0.82 \[ -\frac {1}{4} \, {\left (\frac {b \cos \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}} + \frac {3 \, d \sin \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}}\right )} e^{\left (b x + a\right )} + \frac {1}{4} \, {\left (\frac {b \cos \left (d x + c\right )}{b^{2} + d^{2}} + \frac {d \sin \left (d x + c\right )}{b^{2} + d^{2}}\right )} e^{\left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 108, normalized size = 0.91 \[ \frac {b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )}{4 b^{2}+4 d^{2}}-\frac {b \,{\mathrm e}^{b x +a} \cos \left (3 d x +3 c \right )}{4 \left (b^{2}+9 d^{2}\right )}+\frac {d \,{\mathrm e}^{b x +a} \sin \left (d x +c \right )}{4 b^{2}+4 d^{2}}-\frac {3 d \,{\mathrm e}^{b x +a} \sin \left (3 d x +3 c \right )}{4 \left (b^{2}+9 d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 538, normalized size = 4.52 \[ -\frac {{\left (b^{3} \cos \left (3 \, c\right ) e^{a} + b d^{2} \cos \left (3 \, c\right ) e^{a} + 3 \, b^{2} d e^{a} \sin \left (3 \, c\right ) + 3 \, d^{3} e^{a} \sin \left (3 \, c\right )\right )} \cos \left (3 \, d x\right ) e^{\left (b x\right )} + {\left (b^{3} \cos \left (3 \, c\right ) e^{a} + b d^{2} \cos \left (3 \, c\right ) e^{a} - 3 \, b^{2} d e^{a} \sin \left (3 \, c\right ) - 3 \, d^{3} e^{a} \sin \left (3 \, c\right )\right )} \cos \left (3 \, d x + 6 \, c\right ) e^{\left (b x\right )} - {\left (b^{3} \cos \left (3 \, c\right ) e^{a} + 9 \, b d^{2} \cos \left (3 \, c\right ) e^{a} - b^{2} d e^{a} \sin \left (3 \, c\right ) - 9 \, d^{3} e^{a} \sin \left (3 \, c\right )\right )} \cos \left (d x + 4 \, c\right ) e^{\left (b x\right )} - {\left (b^{3} \cos \left (3 \, c\right ) e^{a} + 9 \, b d^{2} \cos \left (3 \, c\right ) e^{a} + b^{2} d e^{a} \sin \left (3 \, c\right ) + 9 \, d^{3} e^{a} \sin \left (3 \, c\right )\right )} \cos \left (d x - 2 \, c\right ) e^{\left (b x\right )} + {\left (3 \, b^{2} d \cos \left (3 \, c\right ) e^{a} + 3 \, d^{3} \cos \left (3 \, c\right ) e^{a} - b^{3} e^{a} \sin \left (3 \, c\right ) - b d^{2} e^{a} \sin \left (3 \, c\right )\right )} e^{\left (b x\right )} \sin \left (3 \, d x\right ) + {\left (3 \, b^{2} d \cos \left (3 \, c\right ) e^{a} + 3 \, d^{3} \cos \left (3 \, c\right ) e^{a} + b^{3} e^{a} \sin \left (3 \, c\right ) + b d^{2} e^{a} \sin \left (3 \, c\right )\right )} e^{\left (b x\right )} \sin \left (3 \, d x + 6 \, c\right ) - {\left (b^{2} d \cos \left (3 \, c\right ) e^{a} + 9 \, d^{3} \cos \left (3 \, c\right ) e^{a} + b^{3} e^{a} \sin \left (3 \, c\right ) + 9 \, b d^{2} e^{a} \sin \left (3 \, c\right )\right )} e^{\left (b x\right )} \sin \left (d x + 4 \, c\right ) - {\left (b^{2} d \cos \left (3 \, c\right ) e^{a} + 9 \, d^{3} \cos \left (3 \, c\right ) e^{a} - b^{3} e^{a} \sin \left (3 \, c\right ) - 9 \, b d^{2} e^{a} \sin \left (3 \, c\right )\right )} e^{\left (b x\right )} \sin \left (d x - 2 \, c\right )}{8 \, {\left (b^{4} \cos \left (3 \, c\right )^{2} + b^{4} \sin \left (3 \, c\right )^{2} + 9 \, {\left (\cos \left (3 \, c\right )^{2} + \sin \left (3 \, c\right )^{2}\right )} d^{4} + 10 \, {\left (b^{2} \cos \left (3 \, c\right )^{2} + b^{2} \sin \left (3 \, c\right )^{2}\right )} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.01, size = 166, normalized size = 1.39 \[ \frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )-\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \relax (c)-\sin \relax (c)\,1{}\mathrm {i}\right )}{8\,\left (b-d\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )+\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )+\sin \left (3\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,\left (-3\,d+b\,1{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )+\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \relax (c)+\sin \relax (c)\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,\left (-d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )-\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )-\sin \left (3\,c\right )\,1{}\mathrm {i}\right )}{8\,\left (b-d\,3{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(a + b*x)*(cos(d*x) - sin(d*x)*1i)*(cos(c) - sin(c)*1i))/(8*(b - d*1i)) - (exp(a + b*x)*(cos(3*d*x) + sin(
3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*1i)/(8*(b*1i - 3*d)) + (exp(a + b*x)*(cos(d*x) + sin(d*x)*1i)*(cos(c) + si
n(c)*1i)*1i)/(8*(b*1i - d)) - (exp(a + b*x)*(cos(3*d*x) - sin(3*d*x)*1i)*(cos(3*c) - sin(3*c)*1i))/(8*(b - d*3
i))

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sympy [A]  time = 44.43, size = 1030, normalized size = 8.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(a)*sin(c)**2*cos(c), Eq(b, 0) & Eq(d, 0)), (I*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**3/8 + 3*x*
exp(a)*exp(-3*I*d*x)*sin(c + d*x)**2*cos(c + d*x)/8 - 3*I*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)*cos(c + d*x)**2/
8 - x*exp(a)*exp(-3*I*d*x)*cos(c + d*x)**3/8 + exp(a)*exp(-3*I*d*x)*sin(c + d*x)**3/(24*d) - exp(a)*exp(-3*I*d
*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) + I*exp(a)*exp(-3*I*d*x)*cos(c + d*x)**3/(8*d), Eq(b, -3*I*d)), (I*x*ex
p(a)*exp(-I*d*x)*sin(c + d*x)**3/8 + x*exp(a)*exp(-I*d*x)*sin(c + d*x)**2*cos(c + d*x)/8 + I*x*exp(a)*exp(-I*d
*x)*sin(c + d*x)*cos(c + d*x)**2/8 + x*exp(a)*exp(-I*d*x)*cos(c + d*x)**3/8 + exp(a)*exp(-I*d*x)*sin(c + d*x)*
*3/(8*d) - exp(a)*exp(-I*d*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) + I*exp(a)*exp(-I*d*x)*cos(c + d*x)**3/(8*d),
Eq(b, -I*d)), (-I*x*exp(a)*exp(I*d*x)*sin(c + d*x)**3/8 + x*exp(a)*exp(I*d*x)*sin(c + d*x)**2*cos(c + d*x)/8
- I*x*exp(a)*exp(I*d*x)*sin(c + d*x)*cos(c + d*x)**2/8 + x*exp(a)*exp(I*d*x)*cos(c + d*x)**3/8 + exp(a)*exp(I*
d*x)*sin(c + d*x)**3/(8*d) - exp(a)*exp(I*d*x)*sin(c + d*x)*cos(c + d*x)**2/(4*d) - I*exp(a)*exp(I*d*x)*cos(c
+ d*x)**3/(8*d), Eq(b, I*d)), (-I*x*exp(a)*exp(3*I*d*x)*sin(c + d*x)**3/8 + 3*x*exp(a)*exp(3*I*d*x)*sin(c + d*
x)**2*cos(c + d*x)/8 + 3*I*x*exp(a)*exp(3*I*d*x)*sin(c + d*x)*cos(c + d*x)**2/8 - x*exp(a)*exp(3*I*d*x)*cos(c
+ d*x)**3/8 + exp(a)*exp(3*I*d*x)*sin(c + d*x)**3/(24*d) - exp(a)*exp(3*I*d*x)*sin(c + d*x)*cos(c + d*x)**2/(4
*d) - I*exp(a)*exp(3*I*d*x)*cos(c + d*x)**3/(8*d), Eq(b, 3*I*d)), (b**3*exp(a)*exp(b*x)*sin(c + d*x)**2*cos(c
+ d*x)/(b**4 + 10*b**2*d**2 + 9*d**4) + b**2*d*exp(a)*exp(b*x)*sin(c + d*x)**3/(b**4 + 10*b**2*d**2 + 9*d**4)
- 2*b**2*d*exp(a)*exp(b*x)*sin(c + d*x)*cos(c + d*x)**2/(b**4 + 10*b**2*d**2 + 9*d**4) + 3*b*d**2*exp(a)*exp(b
*x)*sin(c + d*x)**2*cos(c + d*x)/(b**4 + 10*b**2*d**2 + 9*d**4) + 2*b*d**2*exp(a)*exp(b*x)*cos(c + d*x)**3/(b*
*4 + 10*b**2*d**2 + 9*d**4) + 3*d**3*exp(a)*exp(b*x)*sin(c + d*x)**3/(b**4 + 10*b**2*d**2 + 9*d**4), True))

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