3.3.92 \(\int \frac {a+b \sin (c+\frac {d}{x})}{(e+f x)^2} \, dx\) [292]

Optimal. Leaf size=94 \[ \frac {a}{e \left (f+\frac {e}{x}\right )}-\frac {b d \cos \left (c-\frac {d f}{e}\right ) \text {Ci}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right )}{e^2}+\frac {b \sin \left (c+\frac {d}{x}\right )}{e \left (f+\frac {e}{x}\right )}+\frac {b d \sin \left (c-\frac {d f}{e}\right ) \text {Si}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right )}{e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3512, 3398, 3378, 3384, 3380, 3383} \begin {gather*} \frac {a}{e \left (\frac {e}{x}+f\right )}-\frac {b d \cos \left (c-\frac {d f}{e}\right ) \text {CosIntegral}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right )}{e^2}+\frac {b d \sin \left (c-\frac {d f}{e}\right ) \text {Si}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right )}{e^2}+\frac {b \sin \left (c+\frac {d}{x}\right )}{e \left (\frac {e}{x}+f\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a/(e*(f + e/x)) - (b*d*Cos[c - (d*f)/e]*CosIntegral[d*(f/e + x^(-1))])/e^2 + (b*Sin[c + d/x])/(e*(f + e/x)) +
(b*d*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))])/e^2

Rule 3378

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 3380

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 3383

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 3384

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 3398

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 3512

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - e*(h/f) + h*(x^(1/n)/f))^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rubi steps

\begin {align*} \int \frac {a+b \sin \left (c+\frac {d}{x}\right )}{(e+f x)^2} \, dx &=-\text {Subst}\left (\int \frac {a+b \sin (c+d x)}{(f+e x)^2} \, dx,x,\frac {1}{x}\right )\\ &=-\text {Subst}\left (\int \left (\frac {a}{(f+e x)^2}+\frac {b \sin (c+d x)}{(f+e x)^2}\right ) \, dx,x,\frac {1}{x}\right )\\ &=\frac {a}{e \left (f+\frac {e}{x}\right )}-b \text {Subst}\left (\int \frac {\sin (c+d x)}{(f+e x)^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {a}{e \left (f+\frac {e}{x}\right )}+\frac {b \sin \left (c+\frac {d}{x}\right )}{e \left (f+\frac {e}{x}\right )}-\frac {(b d) \text {Subst}\left (\int \frac {\cos (c+d x)}{f+e x} \, dx,x,\frac {1}{x}\right )}{e}\\ &=\frac {a}{e \left (f+\frac {e}{x}\right )}+\frac {b \sin \left (c+\frac {d}{x}\right )}{e \left (f+\frac {e}{x}\right )}-\frac {\left (b d \cos \left (c-\frac {d f}{e}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {d f}{e}+d x\right )}{f+e x} \, dx,x,\frac {1}{x}\right )}{e}+\frac {\left (b d \sin \left (c-\frac {d f}{e}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {d f}{e}+d x\right )}{f+e x} \, dx,x,\frac {1}{x}\right )}{e}\\ &=\frac {a}{e \left (f+\frac {e}{x}\right )}-\frac {b d \cos \left (c-\frac {d f}{e}\right ) \text {Ci}\left (\frac {d \left (f+\frac {e}{x}\right )}{e}\right )}{e^2}+\frac {b \sin \left (c+\frac {d}{x}\right )}{e \left (f+\frac {e}{x}\right )}+\frac {b d \sin \left (c-\frac {d f}{e}\right ) \text {Si}\left (\frac {d \left (f+\frac {e}{x}\right )}{e}\right )}{e^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.48, size = 85, normalized size = 0.90 \begin {gather*} \frac {-b d \cos \left (c-\frac {d f}{e}\right ) \text {Ci}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right )+\frac {e \left (-a e+b f x \sin \left (c+\frac {d}{x}\right )\right )}{f (e+f x)}+b d \sin \left (c-\frac {d f}{e}\right ) \text {Si}\left (d \left (\frac {f}{e}+\frac {1}{x}\right )\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*d*Cos[c - (d*f)/e]*CosIntegral[d*(f/e + x^(-1))]) + (e*(-(a*e) + b*f*x*Sin[c + d/x]))/(f*(e + f*x)) + b*d
*Sin[c - (d*f)/e]*SinIntegral[d*(f/e + x^(-1))])/e^2

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 149, normalized size = 1.59

method	result	size
derivativedivides	\(-d \left (-\frac {a}{\left (-c e +d f +e \left (c +\frac {d}{x}\right )\right ) e}+b \left (-\frac {\sin \left (c +\frac {d}{x}\right )}{\left (-c e +d f +e \left (c +\frac {d}{x}\right )\right ) e}+\frac {-\frac {\sinIntegral \left (-\frac {d}{x}-c -\frac {-c e +d f}{e}\right ) \sin \left (\frac {-c e +d f}{e}\right )}{e}+\frac {\cosineIntegral \left (\frac {d}{x}+c +\frac {-c e +d f}{e}\right ) \cos \left (\frac {-c e +d f}{e}\right )}{e}}{e}\right )\right )\)	\(149\)
default	\(-d \left (-\frac {a}{\left (-c e +d f +e \left (c +\frac {d}{x}\right )\right ) e}+b \left (-\frac {\sin \left (c +\frac {d}{x}\right )}{\left (-c e +d f +e \left (c +\frac {d}{x}\right )\right ) e}+\frac {-\frac {\sinIntegral \left (-\frac {d}{x}-c -\frac {-c e +d f}{e}\right ) \sin \left (\frac {-c e +d f}{e}\right )}{e}+\frac {\cosineIntegral \left (\frac {d}{x}+c +\frac {-c e +d f}{e}\right ) \cos \left (\frac {-c e +d f}{e}\right )}{e}}{e}\right )\right )\)	\(149\)
risch	\(-\frac {a}{f \left (f x +e \right )}+\frac {b d \,{\mathrm e}^{-\frac {i \left (c e -d f \right )}{e}} \expIntegral \left (1, \frac {i d}{x}+i c -\frac {i \left (c e -d f \right )}{e}\right )}{2 e^{2}}+\frac {b d \,{\mathrm e}^{\frac {i \left (c e -d f \right )}{e}} \expIntegral \left (1, -\frac {i d}{x}-i c -\frac {-i c e +i d f}{e}\right )}{2 e^{2}}+\frac {i b d \sin \left (\frac {c x +d}{x}\right )}{e \left (-i c e +e \left (i c +\frac {i d}{x}\right )+i d f \right )}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 165, normalized size = 1.76 \begin {gather*} \frac {2 \, b f x e \sin \left (\frac {c x + d}{x}\right ) + 2 \, {\left (b d f^{2} x + b d f e\right )} \sin \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (\frac {{\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right ) - {\left ({\left (b d f^{2} x + b d f e\right )} \operatorname {Ci}\left (\frac {{\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right ) + {\left (b d f^{2} x + b d f e\right )} \operatorname {Ci}\left (-\frac {{\left (d f x + d e\right )} e^{\left (-1\right )}}{x}\right )\right )} \cos \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) - 2 \, a e^{2}}{2 \, {\left (f^{2} x e^{2} + f e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (88) = 176\).
time = 5.63, size = 347, normalized size = 3.69 \begin {gather*} -\frac {b d^{3} f \cos \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Ci}\left ({\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) - b c d^{2} \cos \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Ci}\left ({\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) e + b d^{3} f \sin \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (-{\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) - b c d^{2} e \sin \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (-{\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) + \frac {{\left (c x + d\right )} b d^{2} \cos \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Ci}\left ({\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right ) e}{x} + \frac {{\left (c x + d\right )} b d^{2} e \sin \left (-{\left (d f - c e\right )} e^{\left (-1\right )}\right ) \operatorname {Si}\left (-{\left (d f - c e + \frac {{\left (c x + d\right )} e}{x}\right )} e^{\left (-1\right )}\right )}{x} - b d^{2} e \sin \left (\frac {c x + d}{x}\right ) - a d^{2} e}{{\left (d f e^{2} - c e^{3} + \frac {{\left (c x + d\right )} e^{3}}{x}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________