3.51 \(\int \cos (\frac{a+b x}{c+d x}) \, dx\)

Optimal. Leaf size=101 \[ -\frac{\sin \left (\frac{b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}+\frac{\cos \left (\frac{b}{d}\right ) (b c-a d) \text{Si}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \cos \left (\frac{a+b x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.132663, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4564, 3297, 3303, 3299, 3302} \[ -\frac{\sin \left (\frac{b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}+\frac{\cos \left (\frac{b}{d}\right ) (b c-a d) \text{Si}\left (\frac{b c-a d}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \cos \left (\frac{a+b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[(a + b*x)/(c + d*x)],x]

[Out]

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

Rule 4564

Int[Cos[((e_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_))]^(n_.), x_Symbol] :> -Dist[d^(-1), Subst[Int[Cos[(b*
e)/d - (e*(b*c - a*d)*x)/d]^n/x^2, x], x, 1/(c + d*x)], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && NeQ[b*c
- a*d, 0]

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

Mathematica [C]  time = 5.1352, size = 260, normalized size = 2.57 \[ \frac{-4 \sin \left (\frac{b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{a d-b c}{d (c+d x)}\right )+d \exp \left (-\frac{i (a d+2 b c+b d x)}{d (c+d x)}\right ) \left (2 c \left (e^{2 i \left (\frac{a}{c+d x}+\frac{b}{d}\right )}+e^{\frac{2 i b c}{d (c+d x)}}\right )+d x \left (1+e^{\frac{2 i b}{d}}\right ) \left (e^{\frac{2 i a}{c+d x}}+e^{\frac{2 i b c}{d (c+d x)}}\right )-4 d x \sin \left (\frac{b}{d}\right ) e^{\frac{i (a d+2 b c+b d x)}{d (c+d x)}} \sin \left (\frac{a d-b c}{d (c+d x)}\right )\right )-4 \cos \left (\frac{b}{d}\right ) (b c-a d) \text{Si}\left (\frac{a d-b c}{d (c+d x)}\right )}{4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[(a + b*x)/(c + d*x)],x]

[Out]

(-4*(b*c - a*d)*CosIntegral[(-(b*c) + a*d)/(d*(c + d*x))]*Sin[b/d] + (d*(2*c*(E^(((2*I)*b*c)/(d*(c + d*x))) +
E^((2*I)*(b/d + a/(c + d*x)))) + d*(1 + E^(((2*I)*b)/d))*(E^(((2*I)*a)/(c + d*x)) + E^(((2*I)*b*c)/(d*(c + d*x
))))*x - 4*d*E^((I*(2*b*c + a*d + b*d*x))/(d*(c + d*x)))*x*Sin[b/d]*Sin[(-(b*c) + a*d)/(d*(c + d*x))]))/E^((I*
(2*b*c + a*d + b*d*x))/(d*(c + d*x))) - 4*(b*c - a*d)*Cos[b/d]*SinIntegral[(-(b*c) + a*d)/(d*(c + d*x))])/(4*d
^2)

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Maple [A]  time = 0.016, size = 142, normalized size = 1.4 \begin{align*} - \left ( ad-cb \right ) \left ( -{\frac{1}{d}\cos \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \left ( d \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) -b \right ) ^{-1}}-{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ({\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \cos \left ({\frac{b}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ({\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \sin \left ({\frac{b}{d}} \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos((b*x+a)/(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (\frac{b x + a}{d x + c}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos((b*x+a)/(d*x+c)),x, algorithm="maxima")

[Out]

integrate(cos((b*x + a)/(d*x + c)), x)

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Fricas [A]  time = 2.47045, size = 323, normalized size = 3.2 \begin{align*} -\frac{2 \,{\left (b c - a d\right )} \cos \left (\frac{b}{d}\right ) \operatorname{Si}\left (-\frac{b c - a d}{d^{2} x + c d}\right ) - 2 \,{\left (d^{2} x + c d\right )} \cos \left (\frac{b x + a}{d x + c}\right ) +{\left ({\left (b c - a d\right )} \operatorname{Ci}\left (\frac{b c - a d}{d^{2} x + c d}\right ) +{\left (b c - a d\right )} \operatorname{Ci}\left (-\frac{b c - a d}{d^{2} x + c d}\right )\right )} \sin \left (\frac{b}{d}\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos((b*x+a)/(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos((b*x+a)/(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (\frac{b x + a}{d x + c}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos((b*x+a)/(d*x+c)),x, algorithm="giac")

[Out]

integrate(cos((b*x + a)/(d*x + c)), x)