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

Optimal. Leaf size=95 \[ -\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{CosIntegral}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{d \sin \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{\sin (c+d x)}{a f (e+f x)}-\frac{1}{a f (e+f x)} \]

[Out]

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

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Rubi [A]  time = 0.200108, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4523, 32, 3297, 3303, 3299, 3302} \[ -\frac{d \cos \left (c-\frac{d e}{f}\right ) \text{CosIntegral}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{d \sin \left (c-\frac{d e}{f}\right ) \text{Si}\left (\frac{d e}{f}+d x\right )}{a f^2}+\frac{\sin (c+d x)}{a f (e+f x)}-\frac{1}{a f (e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

Mathematica [A]  time = 0.415589, size = 80, normalized size = 0.84 \[ \frac{-d (e+f x) \cos \left (c-\frac{d e}{f}\right ) \text{CosIntegral}\left (d \left (\frac{e}{f}+x\right )\right )+d (e+f x) \sin \left (c-\frac{d e}{f}\right ) \text{Si}\left (d \left (\frac{e}{f}+x\right )\right )+f (\sin (c+d x)-1)}{a f^2 (e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(d*(e + f*x)*Cos[c - (d*e)/f]*CosIntegral[d*(e/f + x)]) + f*(-1 + Sin[c + d*x]) + d*(e + f*x)*Sin[c - (d*e)/
f]*SinIntegral[d*(e/f + x)])/(a*f^2*(e + f*x))

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Maple [A]  time = 0.053, size = 132, normalized size = 1.4 \begin{align*}{\frac{d}{a} \left ({\frac{\sin \left ( dx+c \right ) }{ \left ( \left ( dx+c \right ) f-cf+de \right ) f}}-{\frac{1}{f} \left ({\frac{1}{f}{\it Si} \left ( dx+c+{\frac{-cf+de}{f}} \right ) \sin \left ({\frac{-cf+de}{f}} \right ) }+{\frac{1}{f}{\it Ci} \left ( dx+c+{\frac{-cf+de}{f}} \right ) \cos \left ({\frac{-cf+de}{f}} \right ) } \right ) }-{\frac{1}{ \left ( \left ( dx+c \right ) f-cf+de \right ) f}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.41665, size = 232, normalized size = 2.44 \begin{align*} \frac{d^{2}{\left (i \, E_{2}\left (\frac{i \, d e + i \,{\left (d x + c\right )} f - i \, c f}{f}\right ) - i \, E_{2}\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^{2}{\left (E_{2}\left (\frac{i \, d e + i \,{\left (d x + c\right )} f - i \, c f}{f}\right ) + E_{2}\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^{2}}{2 \,{\left (a d e f +{\left (d x + c\right )} a f^{2} - a c f^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.61146, size = 315, normalized size = 3.32 \begin{align*} \frac{2 \,{\left (d f x + d e\right )} \sin \left (-\frac{d e - c f}{f}\right ) \operatorname{Si}\left (\frac{d f x + d e}{f}\right ) -{\left ({\left (d f x + d e\right )} \operatorname{Ci}\left (\frac{d f x + d e}{f}\right ) +{\left (d f x + d e\right )} \operatorname{Ci}\left (-\frac{d f x + d e}{f}\right )\right )} \cos \left (-\frac{d e - c f}{f}\right ) + 2 \, f \sin \left (d x + c\right ) - 2 \, f}{2 \,{\left (a f^{3} x + a e f^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2}}{{\left (f x + e\right )}^{2}{\left (a \sin \left (d x + c\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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