3.52 \(\int \cos ^2(\frac{a+b x}{c+d x}) \, dx\)

Optimal. Leaf size=107 \[ -\frac{\sin \left (\frac{2 b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{2 (b c-a d)}{d (c+d x)}\right )}{d^2}+\frac{\cos \left (\frac{2 b}{d}\right ) (b c-a d) \text{Si}\left (\frac{2 (b c-a d)}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \cos ^2\left (\frac{a+b x}{c+d x}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.160648, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4564, 3313, 12, 3303, 3299, 3302} \[ -\frac{\sin \left (\frac{2 b}{d}\right ) (b c-a d) \text{CosIntegral}\left (\frac{2 (b c-a d)}{d (c+d x)}\right )}{d^2}+\frac{\cos \left (\frac{2 b}{d}\right ) (b c-a d) \text{Si}\left (\frac{2 (b c-a d)}{d (c+d x)}\right )}{d^2}+\frac{(c+d x) \cos ^2\left (\frac{a+b x}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[(a + b*x)/(c + d*x)]^2,x]

[Out]

((c + d*x)*Cos[(a + b*x)/(c + d*x)]^2)/d - ((b*c - a*d)*CosIntegral[(2*(b*c - a*d))/(d*(c + d*x))]*Sin[(2*b)/d
])/d^2 + ((b*c - a*d)*Cos[(2*b)/d]*SinIntegral[(2*(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 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [C]  time = 6.11297, size = 400, normalized size = 3.74 \[ \frac{\left (a c d-b c^2\right ) \left (\frac{\left (-1+e^{\frac{4 i b}{d}}\right ) \left (e^{\frac{4 i b c}{d (c+d x)}}-e^{\frac{4 i a}{c+d x}}\right ) \exp \left (-\frac{2 i (a d+2 b c+b d x)}{d (c+d x)}\right )}{8 (b c-a d)}-\frac{\left (1+e^{\frac{4 i b}{d}}\right ) \left (e^{\frac{4 i a}{c+d x}}+e^{\frac{4 i b c}{d (c+d x)}}\right ) \exp \left (-\frac{2 i (a d+2 b c+b d x)}{d (c+d x)}\right )}{8 (b c-a d)}\right )}{d}+\frac{2 a d \sin \left (\frac{2 b}{d}\right ) \text{CosIntegral}\left (\frac{2 (a d-b c)}{d (c+d x)}\right )-2 b c \sin \left (\frac{2 b}{d}\right ) \text{CosIntegral}\left (\frac{2 (a d-b c)}{d (c+d x)}\right )+2 a d \cos \left (\frac{2 b}{d}\right ) \text{Si}\left (\frac{2 (a d-b c)}{d (c+d x)}\right )-2 b c \cos \left (\frac{2 b}{d}\right ) \text{Si}\left (\frac{2 (a d-b c)}{d (c+d x)}\right )+d^2 x}{2 d^2}-\frac{1}{2} x \sin \left (\frac{2 b}{d}\right ) \sin \left (\frac{2 (a d-b c)}{d (c+d x)}\right )+\frac{1}{2} x \cos \left (\frac{2 b}{d}\right ) \cos \left (\frac{2 (a d-b c)}{d (c+d x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-(b*c^2) + a*c*d)*(((-1 + E^(((4*I)*b)/d))*(-E^(((4*I)*a)/(c + d*x)) + E^(((4*I)*b*c)/(d*(c + d*x)))))/(8*(b
*c - a*d)*E^(((2*I)*(2*b*c + a*d + b*d*x))/(d*(c + d*x)))) - ((1 + E^(((4*I)*b)/d))*(E^(((4*I)*a)/(c + d*x)) +
E^(((4*I)*b*c)/(d*(c + d*x)))))/(8*(b*c - a*d)*E^(((2*I)*(2*b*c + a*d + b*d*x))/(d*(c + d*x))))))/d + (x*Cos[
(2*b)/d]*Cos[(2*(-(b*c) + a*d))/(d*(c + d*x))])/2 - (x*Sin[(2*b)/d]*Sin[(2*(-(b*c) + a*d))/(d*(c + d*x))])/2 +
(d^2*x - 2*b*c*CosIntegral[(2*(-(b*c) + a*d))/(d*(c + d*x))]*Sin[(2*b)/d] + 2*a*d*CosIntegral[(2*(-(b*c) + a*
d))/(d*(c + d*x))]*Sin[(2*b)/d] - 2*b*c*Cos[(2*b)/d]*SinIntegral[(2*(-(b*c) + a*d))/(d*(c + d*x))] + 2*a*d*Cos
[(2*b)/d]*SinIntegral[(2*(-(b*c) + a*d))/(d*(c + d*x))])/(2*d^2)

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Maple [A]  time = 0.02, size = 195, normalized size = 1.8 \begin{align*} -{\frac{ad-cb}{{d}^{2}} \left ({\frac{{d}^{2}}{4} \left ( -2\,{\frac{1}{d}\cos \left ( 2\,{\frac{ad-cb}{d \left ( dx+c \right ) }}+2\,{\frac{b}{d}} \right ) \left ( d \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) -b \right ) ^{-1}}-2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \cos \left ( 2\,{\frac{b}{d}} \right ) }+2\,{\frac{1}{d}{\it Ci} \left ( 2\,{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) \sin \left ( 2\,{\frac{b}{d}} \right ) } \right ) } \right ) }-{\frac{d}{2} \left ( d \left ({\frac{b}{d}}+{\frac{ad-cb}{d \left ( dx+c \right ) }} \right ) -b \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos((b*x+a)/(d*x+c))^2,x)

[Out]

-1/d^2*(a*d-b*c)*(1/4*d^2*(-2*cos(2*(a*d-b*c)/d/(d*x+c)+2*b/d)/(d*(b/d+(a*d-b*c)/d/(d*x+c))-b)/d-2*(2*Si(2*(a*
d-b*c)/d/(d*x+c))*cos(2*b/d)/d+2*Ci(2*(a*d-b*c)/d/(d*x+c))*sin(2*b/d)/d)/d)-1/2*d/(d*(b/d+(a*d-b*c)/d/(d*x+c))
-b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x + \frac{1}{2} \, \int \cos \left (\frac{2 \,{\left (b x + a\right )}}{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))^2,x, algorithm="maxima")

[Out]

1/2*x + 1/2*integrate(cos(2*(b*x + a)/(d*x + c)), x)

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Fricas [A]  time = 2.583, size = 338, normalized size = 3.16 \begin{align*} \frac{2 \,{\left (d^{2} x + c d\right )} \cos \left (\frac{b x + a}{d x + c}\right )^{2} - 2 \,{\left (b c - a d\right )} \cos \left (\frac{2 \, b}{d}\right ) \operatorname{Si}\left (-\frac{2 \,{\left (b c - a d\right )}}{d^{2} x + c d}\right ) -{\left ({\left (b c - a d\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b c - a d\right )}}{d^{2} x + c d}\right ) +{\left (b c - a d\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (b c - a d\right )}}{d^{2} x + c d}\right )\right )} \sin \left (\frac{2 \, 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))^2,x, algorithm="fricas")

[Out]

1/2*(2*(d^2*x + c*d)*cos((b*x + a)/(d*x + c))^2 - 2*(b*c - a*d)*cos(2*b/d)*sin_integral(-2*(b*c - a*d)/(d^2*x
+ c*d)) - ((b*c - a*d)*cos_integral(2*(b*c - a*d)/(d^2*x + c*d)) + (b*c - a*d)*cos_integral(-2*(b*c - a*d)/(d^
2*x + c*d)))*sin(2*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))**2,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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos((b*x + a)/(d*x + c))^2, x)