3.37 \(\int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-2+b c x \log (F)) \sin (d+e x))}{x^3} \, dx\)

Optimal. Leaf size=20 \[ \frac {\sin (d+e x) F^{a c+b c x}}{x^2} \]

[Out]

F^(b*c*x+a*c)*sin(e*x+d)/x^2

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Rubi [A]  time = 1.95, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6741, 6742, 4468, 4467} \[ \frac {\sin (d+e x) F^{a c+b c x}}{x^2} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-2 + b*c*x*Log[F])*Sin[d + e*x]))/x^3,x]

[Out]

(F^(a*c + b*c*x)*Sin[d + e*x])/x^2

Rule 4467

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_)*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[((f*x)^(m +
1)*F^(c*(a + b*x))*Sin[d + e*x])/(f*(m + 1)), x] + (-Dist[e/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Co
s[d + e*x], x], x] - Dist[(b*c*Log[F])/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Sin[d + e*x], x], x]) /;
FreeQ[{F, a, b, c, d, e, f, m}, x] && (LtQ[m, -1] || SumSimplerQ[m, 1])

Rule 4468

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_), x_Symbol] :> Simp[((f*x)^(m +
1)*F^(c*(a + b*x))*Cos[d + e*x])/(f*(m + 1)), x] + (Dist[e/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Sin
[d + e*x], x], x] - Dist[(b*c*Log[F])/(f*(m + 1)), Int[(f*x)^(m + 1)*F^(c*(a + b*x))*Cos[d + e*x], x], x]) /;
FreeQ[{F, a, b, c, d, e, f, m}, x] && (LtQ[m, -1] || SumSimplerQ[m, 1])

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.65, size = 19, normalized size = 0.95 \[ \frac {\sin (d+e x) F^{c (a+b x)}}{x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-2 + b*c*x*Log[F])*Sin[d + e*x]))/x^3,x]

[Out]

(F^(c*(a + b*x))*Sin[d + e*x])/x^2

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fricas [A]  time = 0.58, size = 20, normalized size = 1.00 \[ \frac {F^{b c x + a c} \sin \left (e x + d\right )}{x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*log(F))*sin(e*x+d))/x^3,x, algorithm="fricas")

[Out]

F^(b*c*x + a*c)*sin(e*x + d)/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x \cos \left (e x + d\right ) + {\left (b c x \log \relax (F) - 2\right )} \sin \left (e x + d\right )\right )} F^{{\left (b x + a\right )} c}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*log(F))*sin(e*x+d))/x^3,x, algorithm="giac")

[Out]

integrate((e*x*cos(e*x + d) + (b*c*x*log(F) - 2)*sin(e*x + d))*F^((b*x + a)*c)/x^3, x)

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maple [A]  time = 0.16, size = 40, normalized size = 2.00 \[ \frac {2 \,{\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right ) x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*ln(F))*sin(e*x+d))/x^3,x)

[Out]

2*exp(c*(b*x+a)*ln(F))*tan(1/2*d+1/2*e*x)/(1+tan(1/2*d+1/2*e*x)^2)/x^2

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maxima [C]  time = 1.44, size = 1072, normalized size = 53.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*log(F))*sin(e*x+d))/x^3,x, algorithm="maxima")

[Out]

-1/2*F^(a*c)*b^2*c^2*(-I*conjugate(gamma(-2, -(b*c*log(F) + I*e)*x)) + I*conjugate(gamma(-2, -(b*c*log(F) - I*
e)*x)) + I*gamma(-2, -(b*c*log(F) + I*e)*x) - I*gamma(-2, -(b*c*log(F) - I*e)*x))*cos(d)*log(F)^2 + 1/2*F^(a*c
)*b^2*c^2*(conjugate(gamma(-2, -(b*c*log(F) + I*e)*x)) + conjugate(gamma(-2, -(b*c*log(F) - I*e)*x)) + gamma(-
2, -(b*c*log(F) + I*e)*x) + gamma(-2, -(b*c*log(F) - I*e)*x))*log(F)^2*sin(d) + 1/4*(F^(a*c)*b*c*(I*conjugate(
gamma(-1, -(b*c*log(F) + I*e)*x)) - I*conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) - I*gamma(-1, -(b*c*log(F) +
I*e)*x) + I*gamma(-1, -(b*c*log(F) - I*e)*x))*cos(d)*log(F) + F^(a*c)*b*c*(conjugate(gamma(-1, -(b*c*log(F) +
I*e)*x)) + conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) + gamma(-1, -(b*c*log(F) + I*e)*x) + gamma(-1, -(b*c*l
og(F) - I*e)*x))*log(F)*sin(d) + (F^(a*c)*(conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + conjugate(gamma(-1, -
(b*c*log(F) - I*e)*x)) + gamma(-1, -(b*c*log(F) + I*e)*x) + gamma(-1, -(b*c*log(F) - I*e)*x))*cos(d) + F^(a*c)
*(-I*conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + I*conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) + I*gamma(-1,
-(b*c*log(F) + I*e)*x) - I*gamma(-1, -(b*c*log(F) - I*e)*x))*sin(d))*e)*b*c*log(F) - 1/2*(F^(a*c)*(I*conjugat
e(gamma(-2, -(b*c*log(F) + I*e)*x)) - I*conjugate(gamma(-2, -(b*c*log(F) - I*e)*x)) - I*gamma(-2, -(b*c*log(F)
+ I*e)*x) + I*gamma(-2, -(b*c*log(F) - I*e)*x))*cos(d) + F^(a*c)*(conjugate(gamma(-2, -(b*c*log(F) + I*e)*x))
+ conjugate(gamma(-2, -(b*c*log(F) - I*e)*x)) + gamma(-2, -(b*c*log(F) + I*e)*x) + gamma(-2, -(b*c*log(F) - I
*e)*x))*sin(d))*e^2 + 1/4*(F^(a*c)*b*c*(conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + conjugate(gamma(-1, -(b*
c*log(F) - I*e)*x)) + gamma(-1, -(b*c*log(F) + I*e)*x) + gamma(-1, -(b*c*log(F) - I*e)*x))*cos(d)*log(F) - F^(
a*c)*b*c*(I*conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) - I*conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) - I*ga
mma(-1, -(b*c*log(F) + I*e)*x) + I*gamma(-1, -(b*c*log(F) - I*e)*x))*log(F)*sin(d) - (F^(a*c)*(I*conjugate(gam
ma(-1, -(b*c*log(F) + I*e)*x)) - I*conjugate(gamma(-1, -(b*c*log(F) - I*e)*x)) - I*gamma(-1, -(b*c*log(F) + I*
e)*x) + I*gamma(-1, -(b*c*log(F) - I*e)*x))*cos(d) + F^(a*c)*(conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + co
njugate(gamma(-1, -(b*c*log(F) - I*e)*x)) + gamma(-1, -(b*c*log(F) + I*e)*x) + gamma(-1, -(b*c*log(F) - I*e)*x
))*sin(d))*e)*e + 1/2*(2*F^(a*c)*b*c*(conjugate(gamma(-2, -(b*c*log(F) + I*e)*x)) + conjugate(gamma(-2, -(b*c*
log(F) - I*e)*x)) + gamma(-2, -(b*c*log(F) + I*e)*x) + gamma(-2, -(b*c*log(F) - I*e)*x))*cos(d)*log(F) - F^(a*
c)*b*c*(2*I*conjugate(gamma(-2, -(b*c*log(F) + I*e)*x)) - 2*I*conjugate(gamma(-2, -(b*c*log(F) - I*e)*x)) - 2*
I*gamma(-2, -(b*c*log(F) + I*e)*x) + 2*I*gamma(-2, -(b*c*log(F) - I*e)*x))*log(F)*sin(d))*e

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mupad [B]  time = 2.75, size = 19, normalized size = 0.95 \[ \frac {F^{c\,\left (a+b\,x\right )}\,\sin \left (d+e\,x\right )}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((F^(c*(a + b*x))*(sin(d + e*x)*(b*c*x*log(F) - 2) + e*x*cos(d + e*x)))/x^3,x)

[Out]

(F^(c*(a + b*x))*sin(d + e*x))/x^2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{c \left (a + b x\right )} \left (b c x \log {\relax (F )} \sin {\left (d + e x \right )} + e x \cos {\left (d + e x \right )} - 2 \sin {\left (d + e x \right )}\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*(e*x*cos(e*x+d)+(-2+b*c*x*ln(F))*sin(e*x+d))/x**3,x)

[Out]

Integral(F**(c*(a + b*x))*(b*c*x*log(F)*sin(d + e*x) + e*x*cos(d + e*x) - 2*sin(d + e*x))/x**3, x)

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