3.155 \(\int \frac {a+b \sin (e+f x)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=88 \[ -\frac {a}{d (c+d x)}+\frac {b f \text {Ci}\left (x f+\frac {c f}{d}\right ) \cos \left (e-\frac {c f}{d}\right )}{d^2}-\frac {b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}-\frac {b \sin (e+f x)}{d (c+d x)} \]

[Out]

-a/d/(d*x+c)+b*f*Ci(c*f/d+f*x)*cos(-e+c*f/d)/d^2+b*f*Si(c*f/d+f*x)*sin(-e+c*f/d)/d^2-b*sin(f*x+e)/d/(d*x+c)

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Rubi [A]  time = 0.16, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3317, 3297, 3303, 3299, 3302} \[ -\frac {a}{d (c+d x)}+\frac {b f \text {CosIntegral}\left (\frac {c f}{d}+f x\right ) \cos \left (e-\frac {c f}{d}\right )}{d^2}-\frac {b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (x f+\frac {c f}{d}\right )}{d^2}-\frac {b \sin (e+f x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])/(c + d*x)^2,x]

[Out]

-(a/(d*(c + d*x))) + (b*f*Cos[e - (c*f)/d]*CosIntegral[(c*f)/d + f*x])/d^2 - (b*Sin[e + f*x])/(d*(c + d*x)) -
(b*f*Sin[e - (c*f)/d]*SinIntegral[(c*f)/d + f*x])/d^2

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

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]

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 72, normalized size = 0.82 \[ \frac {-\frac {d (a+b \sin (e+f x))}{c+d x}+b f \text {Ci}\left (f \left (\frac {c}{d}+x\right )\right ) \cos \left (e-\frac {c f}{d}\right )-b f \sin \left (e-\frac {c f}{d}\right ) \text {Si}\left (f \left (\frac {c}{d}+x\right )\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])/(c + d*x)^2,x]

[Out]

(b*f*Cos[e - (c*f)/d]*CosIntegral[f*(c/d + x)] - (d*(a + b*Sin[e + f*x]))/(c + d*x) - b*f*Sin[e - (c*f)/d]*Sin
Integral[f*(c/d + x)])/d^2

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fricas [A]  time = 0.75, size = 135, normalized size = 1.53 \[ -\frac {2 \, b d \sin \left (f x + e\right ) - 2 \, {\left (b d f x + b c f\right )} \sin \left (-\frac {d e - c f}{d}\right ) \operatorname {Si}\left (\frac {d f x + c f}{d}\right ) + 2 \, a d - {\left ({\left (b d f x + b c f\right )} \operatorname {Ci}\left (\frac {d f x + c f}{d}\right ) + {\left (b d f x + b c f\right )} \operatorname {Ci}\left (-\frac {d f x + c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/2*(2*b*d*sin(f*x + e) - 2*(b*d*f*x + b*c*f)*sin(-(d*e - c*f)/d)*sin_integral((d*f*x + c*f)/d) + 2*a*d - ((b
*d*f*x + b*c*f)*cos_integral((d*f*x + c*f)/d) + (b*d*f*x + b*c*f)*cos_integral(-(d*f*x + c*f)/d))*cos(-(d*e -
c*f)/d))/(d^3*x + c*d^2)

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giac [B]  time = 2.53, size = 578, normalized size = 6.57 \[ \frac {{\left ({\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} f^{2} \cos \left (\frac {c f - d e}{d}\right ) \operatorname {Ci}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) - c f^{3} \cos \left (\frac {c f - d e}{d}\right ) \operatorname {Ci}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) + d f^{2} \cos \left (\frac {c f - d e}{d}\right ) \operatorname {Ci}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) e + {\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} f^{2} \sin \left (\frac {c f - d e}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) - c f^{3} \sin \left (\frac {c f - d e}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) + d f^{2} e \sin \left (\frac {c f - d e}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c f + d e}{d}\right ) - d f^{2} \sin \left (\frac {{\left (d x + c\right )} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )}}{d}\right )\right )} b d^{2}}{{\left ({\left (d x + c\right )} d^{4} {\left (\frac {c f}{d x + c} - f - \frac {d e}{d x + c}\right )} - c d^{4} f + d^{5} e\right )} f} - \frac {a}{{\left (d x + c\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(d*x+c)^2,x, algorithm="giac")

[Out]

((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c))*f^2*cos((c*f - d*e)/d)*cos_integral(-((d*x + c)*(c*f/(d*x + c)
- f - d*e/(d*x + c)) - c*f + d*e)/d) - c*f^3*cos((c*f - d*e)/d)*cos_integral(-((d*x + c)*(c*f/(d*x + c) - f -
d*e/(d*x + c)) - c*f + d*e)/d) + d*f^2*cos((c*f - d*e)/d)*cos_integral(-((d*x + c)*(c*f/(d*x + c) - f - d*e/(d
*x + c)) - c*f + d*e)/d)*e + (d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c))*f^2*sin((c*f - d*e)/d)*sin_integral
(-((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) - c*f^3*sin((c*f - d*e)/d)*sin_integral(-((d*
x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) + d*f^2*e*sin((c*f - d*e)/d)*sin_integral(-((d*x +
c)*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*f + d*e)/d) - d*f^2*sin((d*x + c)*(c*f/(d*x + c) - f - d*e/(d*x + c
))/d))*b*d^2/(((d*x + c)*d^4*(c*f/(d*x + c) - f - d*e/(d*x + c)) - c*d^4*f + d^5*e)*f) - a/((d*x + c)*d)

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maple [A]  time = 0.02, size = 141, normalized size = 1.60 \[ \frac {-\frac {a \,f^{2}}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+f^{2} b \left (-\frac {\sin \left (f x +e \right )}{\left (\left (f x +e \right ) d +c f -d e \right ) d}+\frac {\frac {\Si \left (f x +e +\frac {c f -d e}{d}\right ) \sin \left (\frac {c f -d e}{d}\right )}{d}+\frac {\Ci \left (f x +e +\frac {c f -d e}{d}\right ) \cos \left (\frac {c f -d e}{d}\right )}{d}}{d}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))/(d*x+c)^2,x)

[Out]

1/f*(-a*f^2/((f*x+e)*d+c*f-d*e)/d+f^2*b*(-sin(f*x+e)/((f*x+e)*d+c*f-d*e)/d+(Si(f*x+e+(c*f-d*e)/d)*sin((c*f-d*e
)/d)/d+Ci(f*x+e+(c*f-d*e)/d)*cos((c*f-d*e)/d)/d)/d))

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maxima [C]  time = 0.50, size = 196, normalized size = 2.23 \[ -\frac {\frac {2 \, a f^{2}}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f} - \frac {{\left (f^{2} {\left (-i \, E_{2}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + i \, E_{2}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \cos \left (-\frac {d e - c f}{d}\right ) + f^{2} {\left (E_{2}\left (\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right ) + E_{2}\left (-\frac {i \, {\left (f x + e\right )} d - i \, d e + i \, c f}{d}\right )\right )} \sin \left (-\frac {d e - c f}{d}\right )\right )} b}{{\left (f x + e\right )} d^{2} - d^{2} e + c d f}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\sin \left (e+f\,x\right )}{{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))/(c + d*x)^2,x)

[Out]

int((a + b*sin(e + f*x))/(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))/(d*x+c)**2,x)

[Out]

Integral((a + b*sin(e + f*x))/(c + d*x)**2, x)

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