3.1.8 \(\int \frac {\cos (a+b x)}{(c+d x)^4} \, dx\) [8]

Optimal. Leaf size=127 \[ -\frac {\cos (a+b x)}{3 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{6 d^3 (c+d x)}+\frac {b^3 \text {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{6 d^4}+\frac {b \sin (a+b x)}{6 d^2 (c+d x)^2}+\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{6 d^4} \]

[Out]

-1/3*cos(b*x+a)/d/(d*x+c)^3+1/6*b^2*cos(b*x+a)/d^3/(d*x+c)+1/6*b^3*cos(a-b*c/d)*Si(b*c/d+b*x)/d^4+1/6*b^3*Ci(b
*c/d+b*x)*sin(a-b*c/d)/d^4+1/6*b*sin(b*x+a)/d^2/(d*x+c)^2

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Rubi [A]
time = 0.11, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3378, 3384, 3380, 3383} \begin {gather*} \frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{6 d^4}+\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{6 d^4}+\frac {b^2 \cos (a+b x)}{6 d^3 (c+d x)}+\frac {b \sin (a+b x)}{6 d^2 (c+d x)^2}-\frac {\cos (a+b x)}{3 d (c+d x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/(c + d*x)^4,x]

[Out]

-1/3*Cos[a + b*x]/(d*(c + d*x)^3) + (b^2*Cos[a + b*x])/(6*d^3*(c + d*x)) + (b^3*CosIntegral[(b*c)/d + b*x]*Sin
[a - (b*c)/d])/(6*d^4) + (b*Sin[a + b*x])/(6*d^2*(c + d*x)^2) + (b^3*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*
x])/(6*d^4)

Rule 3378

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 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]

Rubi steps

\begin {align*} \int \frac {\cos (a+b x)}{(c+d x)^4} \, dx &=-\frac {\cos (a+b x)}{3 d (c+d x)^3}-\frac {b \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx}{3 d}\\ &=-\frac {\cos (a+b x)}{3 d (c+d x)^3}+\frac {b \sin (a+b x)}{6 d^2 (c+d x)^2}-\frac {b^2 \int \frac {\cos (a+b x)}{(c+d x)^2} \, dx}{6 d^2}\\ &=-\frac {\cos (a+b x)}{3 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{6 d^3 (c+d x)}+\frac {b \sin (a+b x)}{6 d^2 (c+d x)^2}+\frac {b^3 \int \frac {\sin (a+b x)}{c+d x} \, dx}{6 d^3}\\ &=-\frac {\cos (a+b x)}{3 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{6 d^3 (c+d x)}+\frac {b \sin (a+b x)}{6 d^2 (c+d x)^2}+\frac {\left (b^3 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{6 d^3}+\frac {\left (b^3 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{6 d^3}\\ &=-\frac {\cos (a+b x)}{3 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{6 d^3 (c+d x)}+\frac {b^3 \text {Ci}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{6 d^4}+\frac {b \sin (a+b x)}{6 d^2 (c+d x)^2}+\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{6 d^4}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 144, normalized size = 1.13 \begin {gather*} \frac {d \cos (b x) \left (\left (-2 d^2+b^2 (c+d x)^2\right ) \cos (a)+b d (c+d x) \sin (a)\right )+d \left (b d (c+d x) \cos (a)-\left (-2 d^2+b^2 (c+d x)^2\right ) \sin (a)\right ) \sin (b x)+b^3 (c+d x)^3 \left (\text {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right ) \sin \left (a-\frac {b c}{d}\right )+\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )\right )}{6 d^4 (c+d x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]/(c + d*x)^4,x]

[Out]

(d*Cos[b*x]*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] + b*d*(c + d*x)*Sin[a]) + d*(b*d*(c + d*x)*Cos[a] - (-2*d^2 + b
^2*(c + d*x)^2)*Sin[a])*Sin[b*x] + b^3*(c + d*x)^3*(CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + Cos[a - (b*c)/
d]*SinIntegral[b*(c/d + x)]))/(6*d^4*(c + d*x)^3)

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Maple [A]
time = 0.19, size = 184, normalized size = 1.45

method	result	size
derivativedivides	\(b^{3} \left (-\frac {\cos \left (b x +a \right )}{3 \left (-d a +b c +d \left (b x +a \right )\right )^{3} d}-\frac {-\frac {\sin \left (b x +a \right )}{2 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}-\frac {-\frac {\sinIntegral \left (-b x -a -\frac {-d a +b c}{d}\right ) \cos \left (\frac {-d a +b c}{d}\right )}{d}-\frac {\cosineIntegral \left (b x +a +\frac {-d a +b c}{d}\right ) \sin \left (\frac {-d a +b c}{d}\right )}{d}}{d}}{2 d}}{3 d}\right )\)	\(184\)
default	\(b^{3} \left (-\frac {\cos \left (b x +a \right )}{3 \left (-d a +b c +d \left (b x +a \right )\right )^{3} d}-\frac {-\frac {\sin \left (b x +a \right )}{2 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}-\frac {-\frac {\sinIntegral \left (-b x -a -\frac {-d a +b c}{d}\right ) \cos \left (\frac {-d a +b c}{d}\right )}{d}-\frac {\cosineIntegral \left (b x +a +\frac {-d a +b c}{d}\right ) \sin \left (\frac {-d a +b c}{d}\right )}{d}}{d}}{2 d}}{3 d}\right )\)	\(184\)
risch	\(-\frac {i b^{3} {\mathrm e}^{-\frac {i \left (d a -b c \right )}{d}} \expIntegral \left (1, i b x +i a -\frac {i \left (d a -b c \right )}{d}\right )}{12 d^{4}}+\frac {i b^{3} {\mathrm e}^{\frac {i \left (d a -b c \right )}{d}} \expIntegral \left (1, -i b x -i a -\frac {-i a d +i b c}{d}\right )}{12 d^{4}}-\frac {\left (-2 b^{5} d^{5} x^{5}-10 b^{5} c \,d^{4} x^{4}-20 b^{5} c^{2} d^{3} x^{3}-20 b^{5} c^{3} d^{2} x^{2}-10 b^{5} c^{4} d x +4 b^{3} d^{5} x^{3}-2 b^{5} c^{5}+12 b^{3} c \,d^{4} x^{2}+12 b^{3} c^{2} d^{3} x +4 b^{3} c^{3} d^{2}\right ) \cos \left (b x +a \right )}{12 d^{3} \left (d x +c \right )^{3} \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 b^{3} c^{2} d x +b^{3} c^{3}\right )}-\frac {i \left (2 i b^{4} d^{5} x^{4}+8 i b^{4} c \,d^{4} x^{3}+12 i b^{4} c^{2} d^{3} x^{2}+8 i b^{4} c^{3} d^{2} x +2 i b^{4} c^{4} d \right ) \sin \left (b x +a \right )}{12 d^{3} \left (d x +c \right )^{3} \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 b^{3} c^{2} d x +b^{3} c^{3}\right )}\)	\(405\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.49, size = 249, normalized size = 1.96 \begin {gather*} -\frac {b^{4} {\left (E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{4}\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^{4} {\left (-i \, E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + i \, E_{4}\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 )}{2 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\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)/(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/2*(b^4*(exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(4, -(I*b*c + I*(b*x + a)*d -
I*a*d)/d))*cos(-(b*c - a*d)/d) + b^4*(-I*exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + I*exp_integral
_e(4, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d))/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3
+ (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b*x + a)^2 + 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x +
a))*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (117) = 234\).
time = 0.38, size = 295, normalized size = 2.32 \begin {gather*} \frac {2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 2 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right ) + 2 \, {\left (b d^{3} x + b c d^{2}\right )} \sin \left (b x + a\right ) + {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )}{12 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos(-(b*c - a*d)/d)*sin_integral((b*d*x + b*
c)/d) + 2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cos(b*x + a) + 2*(b*d^3*x + b*c*d^2)*sin(b*x + a)
+ ((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral((b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*
b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-(b*d*x + b*c)/d))*sin(-(b*c - a*d)/d))/(d^7*x^3 + 3*c*d
^6*x^2 + 3*c^2*d^5*x + c^3*d^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)**4,x)

[Out]

Integral(cos(a + b*x)/(c + d*x)**4, x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.70, size = 8378, normalized size = 65.97 \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)/(d*x+c)^4,x, algorithm="giac")

[Out]

1/12*(b^3*d^3*x^3*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b^3*d^3*
x^3*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^3*d^3*x^3*sin_int
egral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^3*d^3*x^3*real_part(cos_integral(b*x
+ b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) + 2*b^3*d^3*x^3*real_part(cos_integral(-b*x - b*c/d))*ta
n(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) - 2*b^3*d^3*x^3*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*t
an(1/2*a)*tan(1/2*b*c/d)^2 - 2*b^3*d^3*x^3*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan
(1/2*b*c/d)^2 + 3*b^3*c*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d
)^2 - 3*b^3*c*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 6*b
^3*c*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b^3*d^3*x^3*imag_par
t(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2 + b^3*d^3*x^3*imag_part(cos_integral(-b*x - b*c/d))*t
an(1/2*b*x)^2*tan(1/2*a)^2 - 2*b^3*d^3*x^3*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2 + 4*b^3*d
^3*x^3*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) - 4*b^3*d^3*x^3*imag_part
(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 8*b^3*d^3*x^3*sin_integral((b*d*x + b*
c)/d)*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 6*b^3*c*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(1/2*
b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) + 6*b^3*c*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(
1/2*a)^2*tan(1/2*b*c/d) - b^3*d^3*x^3*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 + b
^3*d^3*x^3*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 2*b^3*d^3*x^3*sin_integral(
(b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 6*b^3*c*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(1/
2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 - 6*b^3*c*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*ta
n(1/2*a)*tan(1/2*b*c/d)^2 + b^3*d^3*x^3*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b
^3*d^3*x^3*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^3*d^3*x^3*sin_integral((b
*d*x + b*c)/d)*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 3*b^3*c^2*d*x*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)
^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - 3*b^3*c^2*d*x*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*
a)^2*tan(1/2*b*c/d)^2 + 6*b^3*c^2*d*x*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)
^2 + 2*b^3*d^3*x^3*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a) + 2*b^3*d^3*x^3*real_part(co
s_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a) - 3*b^3*c*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan
(1/2*b*x)^2*tan(1/2*a)^2 + 3*b^3*c*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2 -
6*b^3*c*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*b^3*d^3*x^3*real_part(cos_integ
ral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d) - 2*b^3*d^3*x^3*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*
b*x)^2*tan(1/2*b*c/d) + 12*b^3*c*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/
2*b*c/d) - 12*b^3*c*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 2
4*b^3*c*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 2*b^3*d^3*x^3*real_pa
rt(cos_integral(b*x + b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d) + 2*b^3*d^3*x^3*real_part(cos_integral(-b*x - b*c/d)
)*tan(1/2*a)^2*tan(1/2*b*c/d) + 6*b^3*c^2*d*x*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2
*tan(1/2*b*c/d) + 6*b^3*c^2*d*x*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/
d) - 3*b^3*c*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 + 3*b^3*c*d^2*x^2*im
ag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 6*b^3*c*d^2*x^2*sin_integral((b*d*x + b*
c)/d)*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 2*b^3*d^3*x^3*real_part(cos_integral(b*x + b*c/d))*tan(1/2*a)*tan(1/2*
b*c/d)^2 - 2*b^3*d^3*x^3*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*a)*tan(1/2*b*c/d)^2 - 6*b^3*c^2*d*x*rea
l_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 - 6*b^3*c^2*d*x*real_part(cos_int
egral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 + 3*b^3*c*d^2*x^2*imag_part(cos_integral(b*x +
b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - 3*b^3*c*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*a)^2*ta
n(1/2*b*c/d)^2 + 6*b^3*c*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^3*x^2*t
an(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + b^3*c^3*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(
1/2*a)^2*tan(1/2*b*c/d)^2 - b^3*c^3*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*
b*c/d)^2 + 2*b^3*c^3*sin_integral((b*d*x + b*c)...

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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 )}{{\left (c+d\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/(c + d*x)^4,x)

[Out]

int(cos(a + b*x)/(c + d*x)^4, x)

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