3.98 \(\int \frac{a+a \sin (e+f x)}{c+d x} \, dx\)

Optimal. Leaf size=64 \[ \frac{a \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d}+\frac{a \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d}+\frac{a \log (c+d x)}{d} \]

[Out]

(a*Log[c + d*x])/d + (a*CosIntegral[(c*f)/d + f*x]*Sin[e - (c*f)/d])/d + (a*Cos[e - (c*f)/d]*SinIntegral[(c*f)
/d + f*x])/d

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Rubi [A]  time = 0.150089, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3317, 3303, 3299, 3302} \[ \frac{a \text{CosIntegral}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d}+\frac{a \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (x f+\frac{c f}{d}\right )}{d}+\frac{a \log (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])/(c + d*x),x]

[Out]

(a*Log[c + d*x])/d + (a*CosIntegral[(c*f)/d + f*x]*Sin[e - (c*f)/d])/d + (a*Cos[e - (c*f)/d]*SinIntegral[(c*f)
/d + f*x])/d

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{a+a \sin (e+f x)}{c+d x} \, dx &=\int \left (\frac{a}{c+d x}+\frac{a \sin (e+f x)}{c+d x}\right ) \, dx\\ &=\frac{a \log (c+d x)}{d}+a \int \frac{\sin (e+f x)}{c+d x} \, dx\\ &=\frac{a \log (c+d x)}{d}+\left (a \cos \left (e-\frac{c f}{d}\right )\right ) \int \frac{\sin \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx+\left (a \sin \left (e-\frac{c f}{d}\right )\right ) \int \frac{\cos \left (\frac{c f}{d}+f x\right )}{c+d x} \, dx\\ &=\frac{a \log (c+d x)}{d}+\frac{a \text{Ci}\left (\frac{c f}{d}+f x\right ) \sin \left (e-\frac{c f}{d}\right )}{d}+\frac{a \cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (\frac{c f}{d}+f x\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.294371, size = 54, normalized size = 0.84 \[ \frac{a \left (\text{CosIntegral}\left (f \left (\frac{c}{d}+x\right )\right ) \sin \left (e-\frac{c f}{d}\right )+\cos \left (e-\frac{c f}{d}\right ) \text{Si}\left (f \left (\frac{c}{d}+x\right )\right )+\log (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])/(c + d*x),x]

[Out]

(a*(Log[c + d*x] + CosIntegral[f*(c/d + x)]*Sin[e - (c*f)/d] + Cos[e - (c*f)/d]*SinIntegral[f*(c/d + x)]))/d

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Maple [A]  time = 0.014, size = 96, normalized size = 1.5 \begin{align*}{\frac{a}{d}{\it Si} \left ( fx+e+{\frac{cf-de}{d}} \right ) \cos \left ({\frac{cf-de}{d}} \right ) }-{\frac{a}{d}{\it Ci} \left ( fx+e+{\frac{cf-de}{d}} \right ) \sin \left ({\frac{cf-de}{d}} \right ) }+{\frac{a\ln \left ( \left ( fx+e \right ) d+cf-de \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))/(d*x+c),x)

[Out]

a*Si(f*x+e+(c*f-d*e)/d)*cos((c*f-d*e)/d)/d-a*Ci(f*x+e+(c*f-d*e)/d)*sin((c*f-d*e)/d)/d+a*ln((f*x+e)*d+c*f-d*e)/
d

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Maxima [C]  time = 1.20674, size = 231, normalized size = 3.61 \begin{align*} \frac{\frac{2 \, a f \log \left (c + \frac{{\left (f x + e\right )} d}{f} - \frac{d e}{f}\right )}{d} + \frac{{\left (f{\left (-i \, E_{1}\left (\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{1}\left (-\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac{d e - c f}{d}\right ) + f{\left (E_{1}\left (\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{1}\left (-\frac{i \,{\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac{d e - c f}{d}\right )\right )} a}{d}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(d*x+c),x, algorithm="maxima")

[Out]

1/2*(2*a*f*log(c + (f*x + e)*d/f - d*e/f)/d + (f*(-I*exp_integral_e(1, (I*(f*x + e)*d - I*d*e + I*c*f)/d) + I*
exp_integral_e(1, -(I*(f*x + e)*d - I*d*e + I*c*f)/d))*cos(-(d*e - c*f)/d) + f*(exp_integral_e(1, (I*(f*x + e)
*d - I*d*e + I*c*f)/d) + exp_integral_e(1, -(I*(f*x + e)*d - I*d*e + I*c*f)/d))*sin(-(d*e - c*f)/d))*a/d)/f

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Fricas [A]  time = 1.66127, size = 234, normalized size = 3.66 \begin{align*} \frac{2 \, a \cos \left (-\frac{d e - c f}{d}\right ) \operatorname{Si}\left (\frac{d f x + c f}{d}\right ) + 2 \, a \log \left (d x + c\right ) -{\left (a \operatorname{Ci}\left (\frac{d f x + c f}{d}\right ) + a \operatorname{Ci}\left (-\frac{d f x + c f}{d}\right )\right )} \sin \left (-\frac{d e - c f}{d}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(d*x+c),x, algorithm="fricas")

[Out]

1/2*(2*a*cos(-(d*e - c*f)/d)*sin_integral((d*f*x + c*f)/d) + 2*a*log(d*x + c) - (a*cos_integral((d*f*x + c*f)/
d) + a*cos_integral(-(d*f*x + c*f)/d))*sin(-(d*e - c*f)/d))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{\sin{\left (e + f x \right )}}{c + d x}\, dx + \int \frac{1}{c + d x}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(d*x+c),x)

[Out]

a*(Integral(sin(e + f*x)/(c + d*x), x) + Integral(1/(c + d*x), x))

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Giac [C]  time = 1.25651, size = 961, normalized size = 15.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))/(d*x+c),x, algorithm="giac")

[Out]

1/2*(a*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - a*imag_part(cos_integral(-f*x - c*
f/d))*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*a*log(abs(d*x + c))*tan(1/2*c*f/d)^2*tan(1/2*e)^2 + 2*a*sin_integral((
d*f*x + c*f)/d)*tan(1/2*c*f/d)^2*tan(1/2*e)^2 - 2*a*real_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d)^2*tan(
1/2*e) - 2*a*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*c*f/d)^2*tan(1/2*e) + 2*a*real_part(cos_integral(f*
x + c*f/d))*tan(1/2*c*f/d)*tan(1/2*e)^2 + 2*a*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*c*f/d)*tan(1/2*e)^
2 - a*imag_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d)^2 + a*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*
c*f/d)^2 + 2*a*log(abs(d*x + c))*tan(1/2*c*f/d)^2 - 2*a*sin_integral((d*f*x + c*f)/d)*tan(1/2*c*f/d)^2 + 4*a*i
mag_part(cos_integral(f*x + c*f/d))*tan(1/2*c*f/d)*tan(1/2*e) - 4*a*imag_part(cos_integral(-f*x - c*f/d))*tan(
1/2*c*f/d)*tan(1/2*e) + 8*a*sin_integral((d*f*x + c*f)/d)*tan(1/2*c*f/d)*tan(1/2*e) - a*imag_part(cos_integral
(f*x + c*f/d))*tan(1/2*e)^2 + a*imag_part(cos_integral(-f*x - c*f/d))*tan(1/2*e)^2 + 2*a*log(abs(d*x + c))*tan
(1/2*e)^2 - 2*a*sin_integral((d*f*x + c*f)/d)*tan(1/2*e)^2 - 2*a*real_part(cos_integral(f*x + c*f/d))*tan(1/2*
c*f/d) - 2*a*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*c*f/d) + 2*a*real_part(cos_integral(f*x + c*f/d))*t
an(1/2*e) + 2*a*real_part(cos_integral(-f*x - c*f/d))*tan(1/2*e) + a*imag_part(cos_integral(f*x + c*f/d)) - a*
imag_part(cos_integral(-f*x - c*f/d)) + 2*a*log(abs(d*x + c)) + 2*a*sin_integral((d*f*x + c*f)/d))/(d*tan(1/2*
c*f/d)^2*tan(1/2*e)^2 + d*tan(1/2*c*f/d)^2 + d*tan(1/2*e)^2 + d)