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

Optimal. Leaf size=78 \[ \frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\log (c+d x)}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3312, 3303, 3299, 3302} \[ \frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}-\frac {\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d}+\frac {\log (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(c + d*x),x]

[Out]

(Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/(2*d) + Log[c + d*x]/(2*d) - (Sin[2*a - (2*b*c)/d]*SinIn
tegral[(2*b*c)/d + 2*b*x])/(2*d)

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 3312

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.13, size = 65, normalized size = 0.83 \[ \frac {\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )-\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+\log (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2/(c + d*x),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 88, normalized size = 1.13 \[ \frac {{\left (\operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, \log \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((cos_integral(2*(b*d*x + b*c)/d) + cos_integral(-2*(b*d*x + b*c)/d))*cos(-2*(b*c - a*d)/d) - 2*sin(-2*(b*
c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) + 2*log(d*x + c))/d

________________________________________________________________________________________

giac [C]  time = 0.63, size = 610, normalized size = 7.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*log(abs(d*x + c))*tan(a)^2*tan(b*c/d)^2 + real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^
2 + real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 - 2*imag_part(cos_integral(2*b*x + 2*b*c/d
))*tan(a)^2*tan(b*c/d) + 2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d) - 4*sin_integral(2*(b
*d*x + b*c)/d)*tan(a)^2*tan(b*c/d) + 2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d)^2 - 2*imag_p
art(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d)^2 + 4*sin_integral(2*(b*d*x + b*c)/d)*tan(a)*tan(b*c/d)^
2 + 2*log(abs(d*x + c))*tan(a)^2 - real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2 - real_part(cos_integral(
-2*b*x - 2*b*c/d))*tan(a)^2 + 4*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d) + 4*real_part(cos_i
ntegral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d) + 2*log(abs(d*x + c))*tan(b*c/d)^2 - real_part(cos_integral(2*b*x
+ 2*b*c/d))*tan(b*c/d)^2 - real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d)^2 - 2*imag_part(cos_integral(
2*b*x + 2*b*c/d))*tan(a) + 2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a) - 4*sin_integral(2*(b*d*x + b*c)
/d)*tan(a) + 2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d) - 2*imag_part(cos_integral(-2*b*x - 2*b*c/d
))*tan(b*c/d) + 4*sin_integral(2*(b*d*x + b*c)/d)*tan(b*c/d) + 2*log(abs(d*x + c)) + real_part(cos_integral(2*
b*x + 2*b*c/d)) + real_part(cos_integral(-2*b*x - 2*b*c/d)))/(d*tan(a)^2*tan(b*c/d)^2 + d*tan(a)^2 + d*tan(b*c
/d)^2 + d)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 105, normalized size = 1.35 \[ \frac {\Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{2 d}+\frac {\Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{2 d}+\frac {\ln \left (\left (b x +a \right ) d -d a +c b \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [C]  time = 0.99, size = 161, normalized size = 2.06 \[ -\frac {b {\left (E_{1}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{1}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b {\left (i \, E_{1}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - i \, E_{1}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b \log \left (b c + {\left (b x + a\right )} d - a d\right )}{4 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*(exp_integral_e(1, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(1, -(2*I*b*c + 2*I*(b*x +
a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) - b*(I*exp_integral_e(1, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) -
I*exp_integral_e(1, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d))*sin(-2*(b*c - a*d)/d) - 2*b*log(b*c + (b*x + a
)*d - a*d))/(b*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2/(c + d*x),x)

[Out]

int(cos(a + b*x)^2/(c + d*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________