3.268 \(\int \frac {\cos ^3(c+d x)}{(e+f x)^2 (a+a \sin (c+d x))} \, dx\)

Optimal. Leaf size=175 \[ -\frac {d \sin \left (c-\frac {d e}{f}\right ) \text {Ci}\left (\frac {d e}{f}+d x\right )}{a f^2}-\frac {d \cos \left (2 c-\frac {2 d e}{f}\right ) \text {Ci}\left (\frac {2 d e}{f}+2 d x\right )}{a f^2}+\frac {d \sin \left (2 c-\frac {2 d e}{f}\right ) \text {Si}\left (\frac {2 d e}{f}+2 d x\right )}{a f^2}-\frac {d \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f^2}+\frac {\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac {\cos (c+d x)}{a f (e+f x)} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4523, 3297, 3303, 3299, 3302, 4406, 12} \[ -\frac {d \sin \left (c-\frac {d e}{f}\right ) \text {CosIntegral}\left (\frac {d e}{f}+d x\right )}{a f^2}-\frac {d \cos \left (2 c-\frac {2 d e}{f}\right ) \text {CosIntegral}\left (\frac {2 d e}{f}+2 d x\right )}{a f^2}+\frac {d \sin \left (2 c-\frac {2 d e}{f}\right ) \text {Si}\left (\frac {2 d e}{f}+2 d x\right )}{a f^2}-\frac {d \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f^2}+\frac {\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac {\cos (c+d x)}{a f (e+f x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/((e + f*x)^2*(a + a*Sin[c + d*x])),x]

[Out]

-(Cos[c + d*x]/(a*f*(e + f*x))) - (d*Cos[2*c - (2*d*e)/f]*CosIntegral[(2*d*e)/f + 2*d*x])/(a*f^2) - (d*CosInte
gral[(d*e)/f + d*x]*Sin[c - (d*e)/f])/(a*f^2) + Sin[2*c + 2*d*x]/(2*a*f*(e + f*x)) - (d*Cos[c - (d*e)/f]*SinIn
tegral[(d*e)/f + d*x])/(a*f^2) + (d*Sin[2*c - (2*d*e)/f]*SinIntegral[(2*d*e)/f + 2*d*x])/(a*f^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 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 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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 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 ^3(c+d x)}{(e+f x)^2 (a+a \sin (c+d x))} \, dx &=\frac {\int \frac {\cos (c+d x)}{(e+f x)^2} \, dx}{a}-\frac {\int \frac {\cos (c+d x) \sin (c+d x)}{(e+f x)^2} \, dx}{a}\\ &=-\frac {\cos (c+d x)}{a f (e+f x)}-\frac {\int \frac {\sin (2 c+2 d x)}{2 (e+f x)^2} \, dx}{a}-\frac {d \int \frac {\sin (c+d x)}{e+f x} \, dx}{a f}\\ &=-\frac {\cos (c+d x)}{a f (e+f x)}-\frac {\int \frac {\sin (2 c+2 d x)}{(e+f x)^2} \, dx}{2 a}-\frac {\left (d \cos \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 {\left (d \sin \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 {\cos (c+d x)}{a f (e+f x)}-\frac {d \text {Ci}\left (\frac {d e}{f}+d x\right ) \sin \left (c-\frac {d e}{f}\right )}{a f^2}+\frac {\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac {d \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f^2}-\frac {d \int \frac {\cos (2 c+2 d x)}{e+f x} \, dx}{a f}\\ &=-\frac {\cos (c+d x)}{a f (e+f x)}-\frac {d \text {Ci}\left (\frac {d e}{f}+d x\right ) \sin \left (c-\frac {d e}{f}\right )}{a f^2}+\frac {\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac {d \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f^2}-\frac {\left (d \cos \left (2 c-\frac {2 d e}{f}\right )\right ) \int \frac {\cos \left (\frac {2 d e}{f}+2 d x\right )}{e+f x} \, dx}{a f}+\frac {\left (d \sin \left (2 c-\frac {2 d e}{f}\right )\right ) \int \frac {\sin \left (\frac {2 d e}{f}+2 d x\right )}{e+f x} \, dx}{a f}\\ &=-\frac {\cos (c+d x)}{a f (e+f x)}-\frac {d \cos \left (2 c-\frac {2 d e}{f}\right ) \text {Ci}\left (\frac {2 d e}{f}+2 d x\right )}{a f^2}-\frac {d \text {Ci}\left (\frac {d e}{f}+d x\right ) \sin \left (c-\frac {d e}{f}\right )}{a f^2}+\frac {\sin (2 c+2 d x)}{2 a f (e+f x)}-\frac {d \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (\frac {d e}{f}+d x\right )}{a f^2}+\frac {d \sin \left (2 c-\frac {2 d e}{f}\right ) \text {Si}\left (\frac {2 d e}{f}+2 d x\right )}{a f^2}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 203, normalized size = 1.16 \[ \frac {-2 d (e+f x) \sin \left (c-\frac {d e}{f}\right ) \text {Ci}\left (d \left (\frac {e}{f}+x\right )\right )-2 d (e+f x) \cos \left (2 c-\frac {2 d e}{f}\right ) \text {Ci}\left (\frac {2 d (e+f x)}{f}\right )+2 d e \sin \left (2 c-\frac {2 d e}{f}\right ) \text {Si}\left (\frac {2 d (e+f x)}{f}\right )+2 d f x \sin \left (2 c-\frac {2 d e}{f}\right ) \text {Si}\left (\frac {2 d (e+f x)}{f}\right )-2 d e \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (d \left (\frac {e}{f}+x\right )\right )-2 d f x \cos \left (c-\frac {d e}{f}\right ) \text {Si}\left (d \left (\frac {e}{f}+x\right )\right )+f \sin (2 (c+d x))-2 f \cos (c+d x)}{2 a f^2 (e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/((e + f*x)^2*(a + a*Sin[c + d*x])),x]

[Out]

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

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fricas [A]  time = 0.47, size = 242, normalized size = 1.38 \[ \frac {2 \, f \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, {\left (d f x + d e\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{f}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d f x + d e\right )}}{f}\right ) - 2 \, {\left (d f x + d e\right )} \cos \left (-\frac {d e - c f}{f}\right ) \operatorname {Si}\left (\frac {d f x + d e}{f}\right ) - 2 \, f \cos \left (d x + c\right ) - {\left ({\left (d f x + d e\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d f x + d e\right )}}{f}\right ) + {\left (d f x + d e\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d f x + d e\right )}}{f}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{f}\right ) - {\left ({\left (d f x + d e\right )} \operatorname {Ci}\left (\frac {d f x + d e}{f}\right ) + {\left (d f x + d e\right )} \operatorname {Ci}\left (-\frac {d f x + d e}{f}\right )\right )} \sin \left (-\frac {d e - c f}{f}\right )}{2 \, {\left (a f^{3} x + a e f^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(f*x+e)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(f*x+e)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.11, size = 230, normalized size = 1.31 \[ -\frac {d \left (-\frac {\sin \left (2 d x +2 c \right )}{2 \left (\left (d x +c \right ) f -c f +d e \right ) f}+\frac {\frac {2 \Si \left (2 d x +2 c +\frac {-2 c f +2 d e}{f}\right ) \sin \left (\frac {-2 c f +2 d e}{f}\right )}{f}+\frac {2 \Ci \left (2 d x +2 c +\frac {-2 c f +2 d e}{f}\right ) \cos \left (\frac {-2 c f +2 d e}{f}\right )}{f}}{2 f}+\frac {\cos \left (d x +c \right )}{\left (\left (d x +c \right ) f -c f +d e \right ) f}+\frac {\frac {\Si \left (d x +c +\frac {-c f +d e}{f}\right ) \cos \left (\frac {-c f +d e}{f}\right )}{f}-\frac {\Ci \left (d x +c +\frac {-c f +d e}{f}\right ) \sin \left (\frac {-c f +d e}{f}\right )}{f}}{f}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(f*x+e)^2/(a+a*sin(d*x+c)),x)

[Out]

-d/a*(-1/2*sin(2*d*x+2*c)/((d*x+c)*f-c*f+d*e)/f+1/2*(2*Si(2*d*x+2*c+2*(-c*f+d*e)/f)*sin(2*(-c*f+d*e)/f)/f+2*Ci
(2*d*x+2*c+2*(-c*f+d*e)/f)*cos(2*(-c*f+d*e)/f)/f)/f+cos(d*x+c)/((d*x+c)*f-c*f+d*e)/f+(Si(d*x+c+(-c*f+d*e)/f)*c
os((-c*f+d*e)/f)/f-Ci(d*x+c+(-c*f+d*e)/f)*sin((-c*f+d*e)/f)/f)/f)

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maxima [C]  time = 0.51, size = 307, normalized size = 1.75 \[ -\frac {2 \, d^{2} {\left (E_{2}\left (\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right ) + E_{2}\left (-\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right )\right )} \cos \left (-\frac {d e - c f}{f}\right ) - d^{2} {\left (i \, E_{2}\left (\frac {2 i \, d e + 2 i \, {\left (d x + c\right )} f - 2 i \, c f}{f}\right ) - i \, E_{2}\left (-\frac {2 i \, d e + 2 i \, {\left (d x + c\right )} f - 2 i \, c f}{f}\right )\right )} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{f}\right ) - d^{2} {\left (2 i \, E_{2}\left (\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right ) - 2 i \, E_{2}\left (-\frac {i \, d e + i \, {\left (d x + c\right )} f - i \, c f}{f}\right )\right )} \sin \left (-\frac {d e - c f}{f}\right ) - d^{2} {\left (E_{2}\left (\frac {2 i \, d e + 2 i \, {\left (d x + c\right )} f - 2 i \, c f}{f}\right ) + E_{2}\left (-\frac {2 i \, d e + 2 i \, {\left (d x + c\right )} f - 2 i \, c f}{f}\right )\right )} \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{f}\right )}{4 \, {\left (a d e f + {\left (d x + c\right )} a f^{2} - a c f^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(f*x+e)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(2*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))*cos(-(d*e - c*f)/f) - d^2*(I*exp_integral_e(2, (2*I*d*e + 2*I*(d*x + c)*f - 2*I*c*f)/f) - I*exp_i
ntegral_e(2, -(2*I*d*e + 2*I*(d*x + c)*f - 2*I*c*f)/f))*cos(-2*(d*e - c*f)/f) - d^2*(2*I*exp_integral_e(2, (I*
d*e + I*(d*x + c)*f - I*c*f)/f) - 2*I*exp_integral_e(2, -(I*d*e + I*(d*x + c)*f - I*c*f)/f))*sin(-(d*e - c*f)/
f) - d^2*(exp_integral_e(2, (2*I*d*e + 2*I*(d*x + c)*f - 2*I*c*f)/f) + exp_integral_e(2, -(2*I*d*e + 2*I*(d*x
+ c)*f - 2*I*c*f)/f))*sin(-2*(d*e - c*f)/f))/((a*d*e*f + (d*x + c)*a*f^2 - a*c*f^2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (e+f\,x\right )}^2\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/((e + f*x)^2*(a + a*sin(c + d*x))),x)

[Out]

int(cos(c + d*x)^3/((e + f*x)^2*(a + a*sin(c + d*x))), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(f*x+e)**2/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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