3.36 \(\int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-1+b c x \log (F)) \sin (d+e x))}{x^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 1.73, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6741, 6742, 4467} \[ \frac {\sin (d+e x) F^{a c+b c x}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-1 + b*c*x*Log[F])*Sin[d + e*x]))/x^2,x]

[Out]

(F^(a*c + b*c*x)*Sin[d + e*x])/x

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

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Mathematica [A]  time = 0.61, size = 19, normalized size = 0.95 \[ \frac {\sin (d+e x) F^{c (a+b x)}}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c*(a + b*x))*(e*x*Cos[d + e*x] + (-1 + b*c*x*Log[F])*Sin[d + e*x]))/x^2,x]

[Out]

(F^(c*(a + b*x))*Sin[d + e*x])/x

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fricas [A]  time = 0.94, size = 20, normalized size = 1.00 \[ \frac {F^{b c x + a c} \sin \left (e x + d\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*log(F))*sin(e*x+d))/x^2,x, algorithm="fricas")

[Out]

F^(b*c*x + a*c)*sin(e*x + d)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x \cos \left (e x + d\right ) + {\left (b c x \log \relax (F) - 1\right )} \sin \left (e x + d\right )\right )} F^{{\left (b x + a\right )} c}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*log(F))*sin(e*x+d))/x^2,x, algorithm="giac")

[Out]

integrate((e*x*cos(e*x + d) + (b*c*x*log(F) - 1)*sin(e*x + d))*F^((b*x + a)*c)/x^2, x)

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maple [A]  time = 0.13, size = 40, normalized size = 2.00 \[ \frac {2 \,{\mathrm e}^{c \left (b x +a \right ) \ln \relax (F )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right ) x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*ln(F))*sin(e*x+d))/x^2,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

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maxima [C]  time = 1.03, size = 564, normalized size = 28.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*log(F))*sin(e*x+d))/x^2,x, algorithm="maxima")

[Out]

-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) - 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*lo
g(F) + I*e)*x) + gamma(-1, -(b*c*log(F) - I*e)*x))*log(F)*sin(d) - 1/4*(F^(a*c)*(I*Ei((b*c*log(F) + I*e)*x) -
I*Ei((b*c*log(F) - I*e)*x) - I*conjugate(Ei((b*c*log(F) + I*e)*x)) + I*conjugate(Ei((b*c*log(F) - I*e)*x)))*co
s(d) - F^(a*c)*(Ei((b*c*log(F) + I*e)*x) + Ei((b*c*log(F) - I*e)*x) + conjugate(Ei((b*c*log(F) + I*e)*x)) + co
njugate(Ei((b*c*log(F) - I*e)*x)))*sin(d))*b*c*log(F) + 1/4*(F^(a*c)*(Ei((b*c*log(F) + I*e)*x) + Ei((b*c*log(F
) - I*e)*x) + conjugate(Ei((b*c*log(F) + I*e)*x)) + conjugate(Ei((b*c*log(F) - I*e)*x)))*cos(d) - F^(a*c)*(-I*
Ei((b*c*log(F) + I*e)*x) + I*Ei((b*c*log(F) - I*e)*x) + I*conjugate(Ei((b*c*log(F) + I*e)*x)) - I*conjugate(Ei
((b*c*log(F) - I*e)*x)))*sin(d))*e - 1/4*(F^(a*c)*(conjugate(gamma(-1, -(b*c*log(F) + I*e)*x)) + conjugate(gam
ma(-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*g
amma(-1, -(b*c*log(F) + I*e)*x) - I*gamma(-1, -(b*c*log(F) - I*e)*x))*sin(d))*e

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mupad [B]  time = 2.74, size = 19, normalized size = 0.95 \[ \frac {F^{c\,\left (a+b\,x\right )}\,\sin \left (d+e\,x\right )}{x} \]

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) - 1) + e*x*cos(d + e*x)))/x^2,x)

[Out]

(F^(c*(a + b*x))*sin(d + e*x))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*(e*x*cos(e*x+d)+(-1+b*c*x*ln(F))*sin(e*x+d))/x**2,x)

[Out]

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

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