3.261 \(\int \frac {\cos ^2(c+d x)}{(e+f x) (a+a \sin (c+d x))} \, dx\)

Optimal. Leaf size=72 \[ -\frac {\sin \left (c-\frac {d e}{f}\right ) \text {Ci}\left (\frac {d e}{f}+d x\right )}{a f}-\frac {\cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f}+\frac {\log (e+f x)}{a f} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4523, 31, 3303, 3299, 3302} \[ -\frac {\sin \left (c-\frac {d e}{f}\right ) \text {CosIntegral}\left (\frac {d e}{f}+d x\right )}{a f}-\frac {\cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f}+\frac {\log (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, 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 4523

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

Rubi steps

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

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Mathematica [A]  time = 0.31, size = 58, normalized size = 0.81 \[ \frac {-\sin \left (c-\frac {d e}{f}\right ) \text {Ci}\left (d \left (\frac {e}{f}+x\right )\right )-\cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (d \left (\frac {e}{f}+x\right )\right )+\log (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 89, normalized size = 1.24 \[ -\frac {{\left (\operatorname {Ci}\left (\frac {d f x + d e}{f}\right ) + \operatorname {Ci}\left (-\frac {d f x + d e}{f}\right )\right )} \sin \left (-\frac {d e - c f}{f}\right ) + 2 \, \cos \left (-\frac {d e - c f}{f}\right ) \operatorname {Si}\left (\frac {d f x + d e}{f}\right ) - 2 \, \log \left (f x + e\right )}{2 \, a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 1.31, size = 716, normalized size = 9.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(imag_part(cos_integral(d*x + d*e/f))*tan(1/2*c)^2*tan(1/2*d*e/f)^2 - imag_part(cos_integral(-d*x - d*e/f
))*tan(1/2*c)^2*tan(1/2*d*e/f)^2 - 2*log(abs(f*x + e))*tan(1/2*c)^2*tan(1/2*d*e/f)^2 + 2*sin_integral((d*f*x +
d*e)/f)*tan(1/2*c)^2*tan(1/2*d*e/f)^2 + 2*real_part(cos_integral(d*x + d*e/f))*tan(1/2*c)^2*tan(1/2*d*e/f) +
2*real_part(cos_integral(-d*x - d*e/f))*tan(1/2*c)^2*tan(1/2*d*e/f) - 2*real_part(cos_integral(d*x + d*e/f))*t
an(1/2*c)*tan(1/2*d*e/f)^2 - 2*real_part(cos_integral(-d*x - d*e/f))*tan(1/2*c)*tan(1/2*d*e/f)^2 - imag_part(c
os_integral(d*x + d*e/f))*tan(1/2*c)^2 + imag_part(cos_integral(-d*x - d*e/f))*tan(1/2*c)^2 - 2*log(abs(f*x +
e))*tan(1/2*c)^2 - 2*sin_integral((d*f*x + d*e)/f)*tan(1/2*c)^2 + 4*imag_part(cos_integral(d*x + d*e/f))*tan(1
/2*c)*tan(1/2*d*e/f) - 4*imag_part(cos_integral(-d*x - d*e/f))*tan(1/2*c)*tan(1/2*d*e/f) + 8*sin_integral((d*f
*x + d*e)/f)*tan(1/2*c)*tan(1/2*d*e/f) - imag_part(cos_integral(d*x + d*e/f))*tan(1/2*d*e/f)^2 + imag_part(cos
_integral(-d*x - d*e/f))*tan(1/2*d*e/f)^2 - 2*log(abs(f*x + e))*tan(1/2*d*e/f)^2 - 2*sin_integral((d*f*x + d*e
)/f)*tan(1/2*d*e/f)^2 + 2*real_part(cos_integral(d*x + d*e/f))*tan(1/2*c) + 2*real_part(cos_integral(-d*x - d*
e/f))*tan(1/2*c) - 2*real_part(cos_integral(d*x + d*e/f))*tan(1/2*d*e/f) - 2*real_part(cos_integral(-d*x - d*e
/f))*tan(1/2*d*e/f) + imag_part(cos_integral(d*x + d*e/f)) - imag_part(cos_integral(-d*x - d*e/f)) - 2*log(abs
(f*x + e)) + 2*sin_integral((d*f*x + d*e)/f))/(a*f*tan(1/2*c)^2*tan(1/2*d*e/f)^2 + a*f*tan(1/2*c)^2 + a*f*tan(
1/2*d*e/f)^2 + a*f)

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maple [A]  time = 0.11, size = 102, normalized size = 1.42 \[ -\frac {\Si \left (d x +c +\frac {-c f +d e}{f}\right ) \cos \left (\frac {-c f +d e}{f}\right )}{a f}+\frac {\Ci \left (d x +c +\frac {-c f +d e}{f}\right ) \sin \left (\frac {-c f +d e}{f}\right )}{a f}+\frac {\ln \left (\left (d x +c \right ) f -c f +d e \right )}{a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.41, size = 163, normalized size = 2.26 \[ \frac {d {\left (i \, E_{1}\left (\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right ) - i \, E_{1}\left (-\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right )\right )} \cos \left (-\frac {d e - c f}{f}\right ) + d {\left (E_{1}\left (\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right ) + E_{1}\left (-\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right )\right )} \sin \left (-\frac {d e - c f}{f}\right ) + 2 \, d \log \left (d e + {\left (d x + c\right )} f - c f\right )}{2 \, a d f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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