3.22 \(\int \frac {\cos ^3(a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=184 \[ -\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {3 b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {3 b \sin (a+b x) \cos ^2(a+b x)}{2 d^2 (c+d x)}-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2} \]

[Out]

-9/8*b^2*Ci(3*b*c/d+3*b*x)*cos(3*a-3*b*c/d)/d^3-3/8*b^2*Ci(b*c/d+b*x)*cos(a-b*c/d)/d^3-1/2*cos(b*x+a)^3/d/(d*x
+c)^2+9/8*b^2*Si(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^3+3/8*b^2*Si(b*c/d+b*x)*sin(a-b*c/d)/d^3+3/2*b*cos(b*x+a)^2
*sin(b*x+a)/d^2/(d*x+c)

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Rubi [A]  time = 0.34, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3314, 3303, 3299, 3302, 3312} \[ -\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {3 b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {3 b \sin (a+b x) \cos ^2(a+b x)}{2 d^2 (c+d x)}-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/(c + d*x)^3,x]

[Out]

-Cos[a + b*x]^3/(2*d*(c + d*x)^2) - (3*b^2*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(8*d^3) - (9*b^2*Cos[3
*a - (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*b*x])/(8*d^3) + (3*b*Cos[a + b*x]^2*Sin[a + b*x])/(2*d^2*(c + d*x))
+ (3*b^2*Sin[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(8*d^3) + (9*b^2*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*c
)/d + 3*b*x])/(8*d^3)

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])
)

Rule 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

Rubi steps

\begin {align*} \int \frac {\cos ^3(a+b x)}{(c+d x)^3} \, dx &=-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2}+\frac {3 b \cos ^2(a+b x) \sin (a+b x)}{2 d^2 (c+d x)}+\frac {\left (3 b^2\right ) \int \frac {\cos (a+b x)}{c+d x} \, dx}{d^2}-\frac {\left (9 b^2\right ) \int \frac {\cos ^3(a+b x)}{c+d x} \, dx}{2 d^2}\\ &=-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2}+\frac {3 b \cos ^2(a+b x) \sin (a+b x)}{2 d^2 (c+d x)}-\frac {\left (9 b^2\right ) \int \left (\frac {3 \cos (a+b x)}{4 (c+d x)}+\frac {\cos (3 a+3 b x)}{4 (c+d x)}\right ) \, dx}{2 d^2}+\frac {\left (3 b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{d^2}-\frac {\left (3 b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{d^2}\\ &=-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2}+\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{d^3}+\frac {3 b \cos ^2(a+b x) \sin (a+b x)}{2 d^2 (c+d x)}-\frac {3 b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^3}-\frac {\left (9 b^2\right ) \int \frac {\cos (3 a+3 b x)}{c+d x} \, dx}{8 d^2}-\frac {\left (27 b^2\right ) \int \frac {\cos (a+b x)}{c+d x} \, dx}{8 d^2}\\ &=-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2}+\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{d^3}+\frac {3 b \cos ^2(a+b x) \sin (a+b x)}{2 d^2 (c+d x)}-\frac {3 b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^3}-\frac {\left (9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{8 d^2}-\frac {\left (27 b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{8 d^2}+\frac {\left (9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{8 d^2}+\frac {\left (27 b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{8 d^2}\\ &=-\frac {\cos ^3(a+b x)}{2 d (c+d x)^2}-\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {3 b \cos ^2(a+b x) \sin (a+b x)}{2 d^2 (c+d x)}+\frac {3 b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.83, size = 221, normalized size = 1.20 \[ \frac {-6 b^2 (c+d x)^2 \left (\cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )-\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )-3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )\right )+6 d \cos (b x) (b \sin (a) (c+d x)-d \cos (a))+2 d \cos (3 b x) (3 b \sin (3 a) (c+d x)-d \cos (3 a))+6 d \sin (b x) (b \cos (a) (c+d x)+d \sin (a))+2 d \sin (3 b x) (3 b \cos (3 a) (c+d x)+d \sin (3 a))}{16 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3/(c + d*x)^3,x]

[Out]

(6*d*Cos[b*x]*(-(d*Cos[a]) + b*(c + d*x)*Sin[a]) + 2*d*Cos[3*b*x]*(-(d*Cos[3*a]) + 3*b*(c + d*x)*Sin[3*a]) + 6
*d*(b*(c + d*x)*Cos[a] + d*Sin[a])*Sin[b*x] + 2*d*(3*b*(c + d*x)*Cos[3*a] + d*Sin[3*a])*Sin[3*b*x] - 6*b^2*(c
+ d*x)^2*(Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)] + 3*Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*(c + d*x))/d] -
Sin[a - (b*c)/d]*SinIntegral[b*(c/d + x)] - 3*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d]))/(16*d^3*(c
+ d*x)^2)

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fricas [B]  time = 0.75, size = 375, normalized size = 2.04 \[ -\frac {8 \, d^{2} \cos \left (b x + a\right )^{3} - 24 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - 18 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - 6 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 3 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + 9 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/16*(8*d^2*cos(b*x + a)^3 - 24*(b*d^2*x + b*c*d)*cos(b*x + a)^2*sin(b*x + a) - 18*(b^2*d^2*x^2 + 2*b^2*c*d*x
+ b^2*c^2)*sin(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) - 6*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*si
n(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) + 3*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral((b*d*x
+ b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-(b*d*x + b*c)/d))*cos(-(b*c - a*d)/d) + 9*((b
^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(3*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*co
s_integral(-3*(b*d*x + b*c)/d))*cos(-3*(b*c - a*d)/d))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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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(b*x+a)^3/(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 311, normalized size = 1.69 \[ \frac {\frac {b^{3} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {9 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}+\frac {9 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}}{d}\right )}{2 d}\right )}{12}+\frac {3 b^{3} \left (-\frac {\cos \left (b x +a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}+\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}\right )}{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/(d*x+c)^3,x)

[Out]

1/b*(1/12*b^3*(-3/2*cos(3*b*x+3*a)/((b*x+a)*d-d*a+c*b)^2/d-3/2*(-3*sin(3*b*x+3*a)/((b*x+a)*d-d*a+c*b)/d+3*(3*S
i(3*b*x+3*a+3*(-a*d+b*c)/d)*sin(3*(-a*d+b*c)/d)/d+3*Ci(3*b*x+3*a+3*(-a*d+b*c)/d)*cos(3*(-a*d+b*c)/d)/d)/d)/d)+
3/4*b^3*(-1/2*cos(b*x+a)/((b*x+a)*d-d*a+c*b)^2/d-1/2*(-sin(b*x+a)/((b*x+a)*d-d*a+c*b)/d+(Si(b*x+a+(-a*d+b*c)/d
)*sin((-a*d+b*c)/d)/d+Ci(b*x+a+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d)/d)/d))

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maxima [C]  time = 1.01, size = 339, normalized size = 1.84 \[ -\frac {24576 \, b^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + 8192 \, b^{3} {\left (E_{3}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{3}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - b^{3} {\left (24576 i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - 24576 i \, E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{3} {\left (8192 i \, E_{3}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - 8192 i \, E_{3}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{65536 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/65536*(24576*b^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(3, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + 8192*b^3*(exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d
) + exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) - b^3*(24576*I*exp_inte
gral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - 24576*I*exp_integral_e(3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*
sin(-(b*c - a*d)/d) - b^3*(8192*I*exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - 8192*I*exp_inte
gral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d))/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x +
a)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3/(c + d*x)^3,x)

[Out]

int(cos(a + b*x)^3/(c + d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3/(d*x+c)**3,x)

[Out]

Integral(cos(a + b*x)**3/(c + d*x)**3, x)

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