3.45 \(\int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx\)

Optimal. Leaf size=183 \[ \frac {d e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {5 d e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}+\frac {b e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {b e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )} \]

[Out]

1/8*b*exp(b*x+a)*cos(d*x+c)/(b^2+d^2)-1/16*b*exp(b*x+a)*cos(3*d*x+3*c)/(b^2+9*d^2)-1/16*b*exp(b*x+a)*cos(5*d*x
+5*c)/(b^2+25*d^2)+1/8*d*exp(b*x+a)*sin(d*x+c)/(b^2+d^2)-3/16*d*exp(b*x+a)*sin(3*d*x+3*c)/(b^2+9*d^2)-5/16*d*e
xp(b*x+a)*sin(5*d*x+5*c)/(b^2+25*d^2)

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Rubi [A]  time = 0.13, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4469, 4433} \[ \frac {d e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {5 d e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}+\frac {b e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {b e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cos[c + d*x]^3*Sin[c + d*x]^2,x]

[Out]

(b*E^(a + b*x)*Cos[c + d*x])/(8*(b^2 + d^2)) - (b*E^(a + b*x)*Cos[3*c + 3*d*x])/(16*(b^2 + 9*d^2)) - (b*E^(a +
b*x)*Cos[5*c + 5*d*x])/(16*(b^2 + 25*d^2)) + (d*E^(a + b*x)*Sin[c + d*x])/(8*(b^2 + d^2)) - (3*d*E^(a + b*x)*
Sin[3*c + 3*d*x])/(16*(b^2 + 9*d^2)) - (5*d*E^(a + b*x)*Sin[5*c + 5*d*x])/(16*(b^2 + 25*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 ^3(c+d x) \sin ^2(c+d x) \, dx &=\int \left (\frac {1}{8} e^{a+b x} \cos (c+d x)-\frac {1}{16} e^{a+b x} \cos (3 c+3 d x)-\frac {1}{16} e^{a+b x} \cos (5 c+5 d x)\right ) \, dx\\ &=-\left (\frac {1}{16} \int e^{a+b x} \cos (3 c+3 d x) \, dx\right )-\frac {1}{16} \int e^{a+b x} \cos (5 c+5 d x) \, dx+\frac {1}{8} \int e^{a+b x} \cos (c+d x) \, dx\\ &=\frac {b e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {b e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}+\frac {d e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {5 d e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 110, normalized size = 0.60 \[ \frac {1}{16} e^{a+b x} \left (\frac {2 (b \cos (c+d x)+d \sin (c+d x))}{b^2+d^2}-\frac {b \cos (3 (c+d x))+3 d \sin (3 (c+d x))}{b^2+9 d^2}-\frac {b \cos (5 (c+d x))+5 d \sin (5 (c+d x))}{b^2+25 d^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Cos[c + d*x]^3*Sin[c + d*x]^2,x]

[Out]

(E^(a + b*x)*((2*(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) - (b*Cos[5*(c + d*x)] + 5*d*Sin[5*(c + d*x)])/(b^2 + 25*d^2)))/16

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fricas [A]  time = 0.66, size = 200, normalized size = 1.09 \[ \frac {{\left (6 \, b^{2} d^{3} + 30 \, d^{5} - 5 \, {\left (b^{4} d + 10 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (b^{4} d + 6 \, b^{2} d^{3} + 5 \, d^{5}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) - {\left ({\left (b^{5} + 10 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cos \left (d x + c\right )^{5} - {\left (b^{5} + 6 \, b^{3} d^{2} + 5 \, b d^{4}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (b^{3} d^{2} + 5 \, b d^{4}\right )} \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )}}{b^{6} + 35 \, b^{4} d^{2} + 259 \, b^{2} d^{4} + 225 \, d^{6}} \]

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)^2,x, algorithm="fricas")

[Out]

((6*b^2*d^3 + 30*d^5 - 5*(b^4*d + 10*b^2*d^3 + 9*d^5)*cos(d*x + c)^4 + 3*(b^4*d + 6*b^2*d^3 + 5*d^5)*cos(d*x +
c)^2)*e^(b*x + a)*sin(d*x + c) - ((b^5 + 10*b^3*d^2 + 9*b*d^4)*cos(d*x + c)^5 - (b^5 + 6*b^3*d^2 + 5*b*d^4)*c
os(d*x + c)^3 - 6*(b^3*d^2 + 5*b*d^4)*cos(d*x + c))*e^(b*x + a))/(b^6 + 35*b^4*d^2 + 259*b^2*d^4 + 225*d^6)

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giac [A]  time = 0.16, size = 152, normalized size = 0.83 \[ -\frac {1}{16} \, {\left (\frac {b \cos \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}} + \frac {5 \, d \sin \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{16} \, {\left (\frac {b \cos \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}} + \frac {3 \, d \sin \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}}\right )} e^{\left (b x + a\right )} + \frac {1}{8} \, {\left (\frac {b \cos \left (d x + c\right )}{b^{2} + d^{2}} + \frac {d \sin \left (d x + c\right )}{b^{2} + d^{2}}\right )} e^{\left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c)^2,x, algorithm="giac")

[Out]

-1/16*(b*cos(5*d*x + 5*c)/(b^2 + 25*d^2) + 5*d*sin(5*d*x + 5*c)/(b^2 + 25*d^2))*e^(b*x + a) - 1/16*(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) + 1/8*(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 = 0.14, size = 166, normalized size = 0.91 \[ \frac {b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )}{8 b^{2}+8 d^{2}}-\frac {b \,{\mathrm e}^{b x +a} \cos \left (3 d x +3 c \right )}{16 \left (b^{2}+9 d^{2}\right )}-\frac {b \,{\mathrm e}^{b x +a} \cos \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}+\frac {d \,{\mathrm e}^{b x +a} \sin \left (d x +c \right )}{8 b^{2}+8 d^{2}}-\frac {3 d \,{\mathrm e}^{b x +a} \sin \left (3 d x +3 c \right )}{16 \left (b^{2}+9 d^{2}\right )}-\frac {5 d \,{\mathrm e}^{b x +a} \sin \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )} \]

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)^2,x)

[Out]

1/8*b*exp(b*x+a)*cos(d*x+c)/(b^2+d^2)-1/16*b*exp(b*x+a)*cos(3*d*x+3*c)/(b^2+9*d^2)-1/16*b*exp(b*x+a)*cos(5*d*x
+5*c)/(b^2+25*d^2)+1/8*d*exp(b*x+a)*sin(d*x+c)/(b^2+d^2)-3/16*d*exp(b*x+a)*sin(3*d*x+3*c)/(b^2+9*d^2)-5/16*d*e
xp(b*x+a)*sin(5*d*x+5*c)/(b^2+25*d^2)

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maxima [B]  time = 0.39, size = 1144, normalized size = 6.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c)^2,x, algorithm="maxima")

[Out]

-1/32*((b^5*cos(5*c)*e^a + 10*b^3*d^2*cos(5*c)*e^a + 9*b*d^4*cos(5*c)*e^a + 5*b^4*d*e^a*sin(5*c) + 50*b^2*d^3*
e^a*sin(5*c) + 45*d^5*e^a*sin(5*c))*cos(5*d*x)*e^(b*x) + (b^5*cos(5*c)*e^a + 10*b^3*d^2*cos(5*c)*e^a + 9*b*d^4
*cos(5*c)*e^a - 5*b^4*d*e^a*sin(5*c) - 50*b^2*d^3*e^a*sin(5*c) - 45*d^5*e^a*sin(5*c))*cos(5*d*x + 10*c)*e^(b*x
) + (b^5*cos(5*c)*e^a + 26*b^3*d^2*cos(5*c)*e^a + 25*b*d^4*cos(5*c)*e^a - 3*b^4*d*e^a*sin(5*c) - 78*b^2*d^3*e^
a*sin(5*c) - 75*d^5*e^a*sin(5*c))*cos(3*d*x + 8*c)*e^(b*x) + (b^5*cos(5*c)*e^a + 26*b^3*d^2*cos(5*c)*e^a + 25*
b*d^4*cos(5*c)*e^a + 3*b^4*d*e^a*sin(5*c) + 78*b^2*d^3*e^a*sin(5*c) + 75*d^5*e^a*sin(5*c))*cos(3*d*x - 2*c)*e^
(b*x) - 2*(b^5*cos(5*c)*e^a + 34*b^3*d^2*cos(5*c)*e^a + 225*b*d^4*cos(5*c)*e^a - b^4*d*e^a*sin(5*c) - 34*b^2*d
^3*e^a*sin(5*c) - 225*d^5*e^a*sin(5*c))*cos(d*x + 6*c)*e^(b*x) - 2*(b^5*cos(5*c)*e^a + 34*b^3*d^2*cos(5*c)*e^a
+ 225*b*d^4*cos(5*c)*e^a + b^4*d*e^a*sin(5*c) + 34*b^2*d^3*e^a*sin(5*c) + 225*d^5*e^a*sin(5*c))*cos(d*x - 4*c
)*e^(b*x) + (5*b^4*d*cos(5*c)*e^a + 50*b^2*d^3*cos(5*c)*e^a + 45*d^5*cos(5*c)*e^a - b^5*e^a*sin(5*c) - 10*b^3*
d^2*e^a*sin(5*c) - 9*b*d^4*e^a*sin(5*c))*e^(b*x)*sin(5*d*x) + (5*b^4*d*cos(5*c)*e^a + 50*b^2*d^3*cos(5*c)*e^a
+ 45*d^5*cos(5*c)*e^a + b^5*e^a*sin(5*c) + 10*b^3*d^2*e^a*sin(5*c) + 9*b*d^4*e^a*sin(5*c))*e^(b*x)*sin(5*d*x +
10*c) + (3*b^4*d*cos(5*c)*e^a + 78*b^2*d^3*cos(5*c)*e^a + 75*d^5*cos(5*c)*e^a + b^5*e^a*sin(5*c) + 26*b^3*d^2
*e^a*sin(5*c) + 25*b*d^4*e^a*sin(5*c))*e^(b*x)*sin(3*d*x + 8*c) + (3*b^4*d*cos(5*c)*e^a + 78*b^2*d^3*cos(5*c)*
e^a + 75*d^5*cos(5*c)*e^a - b^5*e^a*sin(5*c) - 26*b^3*d^2*e^a*sin(5*c) - 25*b*d^4*e^a*sin(5*c))*e^(b*x)*sin(3*
d*x - 2*c) - 2*(b^4*d*cos(5*c)*e^a + 34*b^2*d^3*cos(5*c)*e^a + 225*d^5*cos(5*c)*e^a + b^5*e^a*sin(5*c) + 34*b^
3*d^2*e^a*sin(5*c) + 225*b*d^4*e^a*sin(5*c))*e^(b*x)*sin(d*x + 6*c) - 2*(b^4*d*cos(5*c)*e^a + 34*b^2*d^3*cos(5
*c)*e^a + 225*d^5*cos(5*c)*e^a - b^5*e^a*sin(5*c) - 34*b^3*d^2*e^a*sin(5*c) - 225*b*d^4*e^a*sin(5*c))*e^(b*x)*
sin(d*x - 4*c))/(b^6*cos(5*c)^2 + b^6*sin(5*c)^2 + 225*(cos(5*c)^2 + sin(5*c)^2)*d^6 + 259*(b^2*cos(5*c)^2 + b
^2*sin(5*c)^2)*d^4 + 35*(b^4*cos(5*c)^2 + b^4*sin(5*c)^2)*d^2)

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mupad [B]  time = 3.71, size = 255, normalized size = 1.39 \[ \frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )-\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \relax (c)-\sin \relax (c)\,1{}\mathrm {i}\right )}{16\,\left (b-d\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )+\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )+\sin \left (3\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (-3\,d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )+\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )+\sin \left (5\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (-5\,d+b\,1{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )+\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \relax (c)+\sin \relax (c)\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,\left (-d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )-\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )-\sin \left (3\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (b-d\,3{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )-\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )-\sin \left (5\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (b-d\,5{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*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))/(16*(b - d*1i)) - (exp(a + b*x)*(cos(3*d*x) + sin
(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*1i)/(32*(b*1i - 3*d)) - (exp(a + b*x)*(cos(5*d*x) + sin(5*d*x)*1i)*(cos(5
*c) + sin(5*c)*1i)*1i)/(32*(b*1i - 5*d)) + (exp(a + b*x)*(cos(d*x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*1i)/(16
*(b*1i - d)) - (exp(a + b*x)*(cos(3*d*x) - sin(3*d*x)*1i)*(cos(3*c) - sin(3*c)*1i))/(32*(b - d*3i)) - (exp(a +
b*x)*(cos(5*d*x) - sin(5*d*x)*1i)*(cos(5*c) - sin(5*c)*1i))/(32*(b - d*5i))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**2,x)

[Out]

Timed out

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