3.1.22 \(\int \frac {\cos ^3(a+b x)}{(c+d x)^3} \, dx\) [22]

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

[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 = 0.22, 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 = {3395, 3384, 3380, 3383, 3393} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/(c + d*x)^3,x]

[Out]

-1/2*Cos[a + b*x]^3/(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 3380

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 3383

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 3384

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 3393

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 3395

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 = 0.53, size = 221, normalized size = 1.20 \begin {gather*} \frac {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 \left (\cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right )+3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\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 )}{16 d^3 (c+d x)^2} \end {gather*}

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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Maple [A]
time = 0.29, size = 316, normalized size = 1.72

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {9 \sinIntegral \left (-3 b x -3 a -\frac {3 \left (-d a +b c \right )}{d}\right ) \sin \left (\frac {-3 d a +3 b c}{d}\right )}{d}+\frac {9 \cosineIntegral \left (3 b x +3 a +\frac {-3 d a +3 b c}{d}\right ) \cos \left (\frac {-3 d a +3 b c}{d}\right )}{d}}{d}\right )}{2 d}\right )}{12}+\frac {3 b^{3} \left (-\frac {\cos \left (b x +a \right )}{2 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {\sinIntegral \left (-b x -a -\frac {-d a +b c}{d}\right ) \sin \left (\frac {-d a +b c}{d}\right )}{d}+\frac {\cosineIntegral \left (b x +a +\frac {-d a +b c}{d}\right ) \cos \left (\frac {-d a +b c}{d}\right )}{d}}{d}}{2 d}\right )}{4}}{b}\)	\(316\)
default	\(\frac {\frac {b^{3} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {9 \sinIntegral \left (-3 b x -3 a -\frac {3 \left (-d a +b c \right )}{d}\right ) \sin \left (\frac {-3 d a +3 b c}{d}\right )}{d}+\frac {9 \cosineIntegral \left (3 b x +3 a +\frac {-3 d a +3 b c}{d}\right ) \cos \left (\frac {-3 d a +3 b c}{d}\right )}{d}}{d}\right )}{2 d}\right )}{12}+\frac {3 b^{3} \left (-\frac {\cos \left (b x +a \right )}{2 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {\sinIntegral \left (-b x -a -\frac {-d a +b c}{d}\right ) \sin \left (\frac {-d a +b c}{d}\right )}{d}+\frac {\cosineIntegral \left (b x +a +\frac {-d a +b c}{d}\right ) \cos \left (\frac {-d a +b c}{d}\right )}{d}}{d}}{2 d}\right )}{4}}{b}\)	\(316\)
risch	\(\frac {9 b^{2} {\mathrm e}^{-\frac {3 i \left (d a -b c \right )}{d}} \expIntegral \left (1, 3 i b x +3 i a -\frac {3 i \left (d a -b c \right )}{d}\right )}{16 d^{3}}+\frac {3 b^{2} {\mathrm e}^{-\frac {i \left (d a -b c \right )}{d}} \expIntegral \left (1, i b x +i a -\frac {i \left (d a -b c \right )}{d}\right )}{16 d^{3}}+\frac {3 b^{2} {\mathrm e}^{\frac {i \left (d a -b c \right )}{d}} \expIntegral \left (1, -i b x -i a -\frac {-i a d +i b c}{d}\right )}{16 d^{3}}+\frac {9 b^{2} {\mathrm e}^{\frac {3 i \left (d a -b c \right )}{d}} \expIntegral \left (1, -3 i b x -3 i a -\frac {3 \left (-i a d +i b c \right )}{d}\right )}{16 d^{3}}+\frac {3 \left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (b x +a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {3 i \left (2 i b^{3} d^{3} x^{3}+6 i b^{3} c \,d^{2} x^{2}+6 i b^{3} c^{2} d x +2 i b^{3} c^{3}\right ) \sin \left (b x +a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}+\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (3 b x +3 a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {i \left (6 i b^{3} d^{3} x^{3}+18 i b^{3} c \,d^{2} x^{2}+18 i b^{3} c^{2} d x +6 i b^{3} c^{3}\right ) \sin \left (3 b x +3 a \right )}{16 d^{2} \left (d x +c \right )^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}\)	\(548\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/12*b^3*(-3/2*cos(3*b*x+3*a)/(-d*a+b*c+d*(b*x+a))^2/d-3/2*(-3*sin(3*b*x+3*a)/(-d*a+b*c+d*(b*x+a))/d+3*(-
3*Si(-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)/(-d*a+b*c+d*(b*x+a))^2/d-1/2*(-sin(b*x+a)/(-d*a+b*c+d*(b*x+a))/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] Result contains complex when optimal does not.
time = 0.53, size = 339, normalized size = 1.84 \begin {gather*} -\frac {3 \, 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 ) + b^{3} {\left (E_{3}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 3 \, b^{3} {\left (-i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 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 (i \, E_{3}\left (\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{3}\left (-\frac {3 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\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} \end {gather*}

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]

-1/8*(3*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) + b^3*(exp_integral_e(3, 3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_integral
_e(3, -3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-3*(b*c - a*d)/d) + 3*b^3*(-I*exp_integral_e(3, (I*b*c + I*(
b*x + a)*d - I*a*d)/d) + 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*(I
*exp_integral_e(3, 3*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) - I*exp_integral_e(3, -3*(-I*b*c - I*(b*x + a)*d + 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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (172) = 344\).
time = 0.43, size = 375, normalized size = 2.04 \begin {gather*} -\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 )}} \end {gather*}

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

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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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.71, size = 115446, normalized size = 627.42 \begin {gather*} \text {Too large to display} \end {gather*}

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]

-1/16*(9*b^2*d^2*x^2*real_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1
/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 3*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*
tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 3*b^2*d^2*x^2*real_part(cos_integ
ral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 +
9*b^2*d^2*x^2*real_part(cos_integral(-3*b*x - 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)
^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 - 6*b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1
/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) + 6*b^2*d^2*x^2*imag_part(cos_integral(-b*
x - b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) - 12*b^2*d
^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*
tan(1/2*b*c/d) - 18*b^2*d^2*x^2*imag_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2
*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)*tan(1/2*b*c/d)^2 + 18*b^2*d^2*x^2*imag_part(cos_integral(-3*b*x - 3*b*c/d))*
tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)*tan(1/2*b*c/d)^2 - 36*b^2*d^2*x^2*sin_i
ntegral(3*(b*d*x + b*c)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)*tan(1/2*b*c/
d)^2 + 6*b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a
)*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 - 6*b^2*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1
/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 12*b^2*d^2*x^2*sin_integral((b*d*x + b*c
)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 18*b^2*d^2*x^2*
imag_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)*tan(1/2*a)^2*tan(3/2*b*c/d)^
2*tan(1/2*b*c/d)^2 - 18*b^2*d^2*x^2*imag_part(cos_integral(-3*b*x - 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*ta
n(3/2*a)*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 36*b^2*d^2*x^2*sin_integral(3*(b*d*x + b*c)/d)*tan(3
/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 18*b^2*c*d*x*real_part(co
s_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*
b*c/d)^2 + 6*b^2*c*d*x*real_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2
*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 6*b^2*c*d*x*real_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan
(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 + 18*b^2*c*d*x*real_part(cos_integral(
-3*b*x - 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d)^2 +
9*b^2*d^2*x^2*real_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^
2*tan(3/2*b*c/d)^2 - 3*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*
a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2 - 3*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1/
2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2 + 9*b^2*d^2*x^2*real_part(cos_integral(-3*b*x - 3*b*c/d))*
tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2 + 12*b^2*d^2*x^2*real_part(cos_integr
al(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) + 12*b^
2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)*tan(3/2*
b*c/d)^2*tan(1/2*b*c/d) - 12*b^2*c*d*x*imag_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(
3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) + 12*b^2*c*d*x*imag_part(cos_integral(-b*x - b*c/d))*tan
(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) - 24*b^2*c*d*x*sin_integr
al((b*d*x + b*c)/d)*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)^2*tan(1/2*b*c/d) -
9*b^2*d^2*x^2*real_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2
*tan(1/2*b*c/d)^2 + 3*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a
)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 3*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(3/2*b*x)^2*tan(1/2
*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - 9*b^2*d^2*x^2*real_part(cos_integral(-3*b*x - 3*b*c/d))*t
an(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 36*b^2*d^2*x^2*real_part(cos_integra
l(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)*tan(1/2*a)^2*tan(3/2*b*c/d)*tan(1/2*b*c/d)^2 + 36
*b^2*d^2*x^2*real_part(cos_integral(-3*b*x - 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*b*x)^2*tan(3/2*a)*tan(1/2*a)^2*t
an(3/2*b*c/d)*tan(1/2*b*c/d)^2 - 36*b^2*c*d*x*imag_part(cos_integral(3*b*x + 3*b*c/d))*tan(3/2*b*x)^2*tan(1/2*
b*x)^2*tan(3/2*a)^2*tan(1/2*a)^2*tan(3/2*b*c/d)...

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

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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