3.1.37 \(\int \frac {F^{c (a+b x)} (e x \cos (d+e x)+(-2+b c x \log (F)) \sin (d+e x))}{x^3} \, dx\) [37]

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

[Out]

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

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Rubi [A]
time = 1.39, 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 = {6873, 6874, 4556, 4555} \begin {gather*} \frac {\sin (d+e x) F^{a c+b c x}}{x^2} \end {gather*}

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 4555

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_)*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[((f*x)^(m +
1)/(f*(m + 1)))*F^(c*(a + b*x))*Sin[d + e*x], 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 4556

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_), x_Symbol] :> Simp[((f*x)^(m +
1)/(f*(m + 1)))*F^(c*(a + b*x))*Cos[d + e*x], 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 6873

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

Rule 6874

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 = 0.43, size = 19, normalized size = 0.95 \begin {gather*} \frac {F^{c (a+b x)} \sin (d+e x)}{x^2} \end {gather*}

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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Maple [A]
time = 0.17, size = 20, normalized size = 1.00

method	result	size
risch	\(\frac {\sin \left (e x +d \right ) F^{c \left (b x +a \right )}}{x^{2}}\)	\(20\)
norman	\(\frac {2 \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d}{2}+\frac {e x}{2}\right )\right ) x^{2}}\)	\(40\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.97, size = 1189, normalized size = 59.45 \begin {gather*} \text {Too large to display} \end {gather*}

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]

1/4*((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))*log(F) + (conjugate(gamma(-1, -(
b*c*log(F) + I*e)*x))*e + conjugate(gamma(-1, -(b*c*log(F) - I*e)*x))*e + e*gamma(-1, -(b*c*log(F) + I*e)*x) +
e*gamma(-1, -(b*c*log(F) - I*e)*x))*F^(a*c))*cos(d) + (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))*log(F) + (-I*conjugate(gamma(-1, -(b*c*log(F) + I*e)*x))*e + I*conjugate(gamma(-1, -(b*c*log(F) - I*
e)*x))*e + I*e*gamma(-1, -(b*c*log(F) + I*e)*x) - I*e*gamma(-1, -(b*c*log(F) - I*e)*x))*F^(a*c))*sin(d))*b*c*l
og(F) + 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))*log(F)^2 + 2*(conjuga
te(gamma(-2, -(b*c*log(F) + I*e)*x))*e + conjugate(gamma(-2, -(b*c*log(F) - I*e)*x))*e + e*gamma(-2, -(b*c*log
(F) + I*e)*x) + e*gamma(-2, -(b*c*log(F) - I*e)*x))*F^(a*c)*b*c*log(F) + (-I*conjugate(gamma(-2, -(b*c*log(F)
+ I*e)*x))*e^2 + I*conjugate(gamma(-2, -(b*c*log(F) - I*e)*x))*e^2 + I*e^2*gamma(-2, -(b*c*log(F) + I*e)*x) -
I*e^2*gamma(-2, -(b*c*log(F) - I*e)*x))*F^(a*c))*cos(d) + 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))*log(F) + (-I*conjugate(gamma(-1, -(b*c*log(F) + I*e)*x))*e + I*conjugate(gamma(-1, -(b*c*log
(F) - I*e)*x))*e + I*e*gamma(-1, -(b*c*log(F) + I*e)*x) - I*e*gamma(-1, -(b*c*log(F) - I*e)*x))*F^(a*c))*cos(d
) + (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))*log(F) - (conjugate(gamma(-1, -(
b*c*log(F) + I*e)*x))*e + conjugate(gamma(-1, -(b*c*log(F) - I*e)*x))*e + e*gamma(-1, -(b*c*log(F) + I*e)*x) +
e*gamma(-1, -(b*c*log(F) - I*e)*x))*F^(a*c))*sin(d))*e + 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 - 2*(I*conjugate(gamma(-2, -(b*c*log(F) + I*e)*x))*e - I*conjugate(gamma(-2, -(b
*c*log(F) - I*e)*x))*e - I*e*gamma(-2, -(b*c*log(F) + I*e)*x) + I*e*gamma(-2, -(b*c*log(F) - I*e)*x))*F^(a*c)*
b*c*log(F) - (conjugate(gamma(-2, -(b*c*log(F) + I*e)*x))*e^2 + conjugate(gamma(-2, -(b*c*log(F) - I*e)*x))*e^
2 + e^2*gamma(-2, -(b*c*log(F) + I*e)*x) + e^2*gamma(-2, -(b*c*log(F) - I*e)*x))*F^(a*c))*sin(d)

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Fricas [A]
time = 2.62, size = 21, normalized size = 1.05 \begin {gather*} \frac {F^{b c x + a c} \sin \left (x e + d\right )}{x^{2}} \end {gather*}

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(x*e + d)/x^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{c \left (a + b x\right )} \left (b c x \log {\left (F \right )} \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 \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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