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

Optimal. Leaf size=83 \[ -\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {\cos ^2(a+b x)}{d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3313, 12, 3303, 3299, 3302} \[ -\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {\cos ^2(a+b x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(c + d*x)^2,x]

[Out]

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

Rule 12

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.62, size = 75, normalized size = 0.90 \[ -\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )+b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d \cos ^2(a+b x)}{c+d x}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 127, normalized size = 1.53 \[ -\frac {2 \, d \cos \left (b x + a\right )^{2} + 2 \, {\left (b d x + b c\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.84, size = 534, normalized size = 6.43 \[ -\frac {{\left (2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{3} c \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, a b^{2} d \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) - 2 \, b^{3} c \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) + 2 \, a b^{2} d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) + b^{2} d \cos \left (-\frac {2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )}}{d}\right ) + b^{2} d\right )} d^{2}}{2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} d^{4} + b c d^{4} - a d^{5}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos_integral(2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(
d*x + c)) + b*c - a*d)/d)*sin(-2*(b*c - a*d)/d) + 2*b^3*c*cos_integral(2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(
d*x + c)) + b*c - a*d)/d)*sin(-2*(b*c - a*d)/d) - 2*a*b^2*d*cos_integral(2*((d*x + c)*(b - b*c/(d*x + c) + a*d
/(d*x + c)) + b*c - a*d)/d)*sin(-2*(b*c - a*d)/d) - 2*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos(-2
*(b*c - a*d)/d)*sin_integral(-2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) - 2*b^3*c*cos(-
2*(b*c - a*d)/d)*sin_integral(-2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + 2*a*b^2*d*co
s(-2*(b*c - a*d)/d)*sin_integral(-2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + b^2*d*cos
(-2*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))/d) + b^2*d)*d^2/(((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x +
c))*d^4 + b*c*d^4 - a*d^5)*b)

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maple [A]  time = 0.02, size = 156, normalized size = 1.88 \[ \frac {\frac {b^{2} \left (-\frac {2 \cos \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}\right )}{4}-\frac {b^{2}}{2 \left (\left (b x +a \right ) d -d a +c b \right ) d}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.93, size = 171, normalized size = 2.06 \[ -\frac {16 \, b^{2} {\left (E_{2}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{2}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{2} {\left (16 i \, E_{2}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 16 i \, E_{2}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 32 \, b^{2}}{64 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(16*b^2*(exp_integral_e(2, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(2, -(2*I*b*c + 2*I*
(b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) - b^2*(16*I*exp_integral_e(2, (2*I*b*c + 2*I*(b*x + a)*d - 2*
I*a*d)/d) - 16*I*exp_integral_e(2, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d))*sin(-2*(b*c - a*d)/d) + 32*b^2)/
((b*c*d + (b*x + a)*d^2 - a*d^2)*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(c + d*x)^2,x)

[Out]

int(cos(a + b*x)^2/(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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