3.1.15 \(\int \frac {\cos ^2(a+b x)}{(c+d x)^3} \, dx\) [15]

Optimal. Leaf size=112 \[ -\frac {\cos ^2(a+b x)}{2 d (c+d x)^2}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3395, 31, 3393, 3384, 3380, 3383} \begin {gather*} -\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {b \sin (a+b x) \cos (a+b x)}{d^2 (c+d x)}-\frac {\cos ^2(a+b x)}{2 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2/(c + d*x)^3,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 ^2(a+b x)}{(c+d x)^3} \, dx &=-\frac {\cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {b^2 \int \frac {1}{c+d x} \, dx}{d^2}-\frac {\left (2 b^2\right ) \int \frac {\cos ^2(a+b x)}{c+d x} \, dx}{d^2}\\ &=-\frac {\cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {b^2 \log (c+d x)}{d^3}+\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}-\frac {\left (2 b^2\right ) \int \left (\frac {1}{2 (c+d x)}+\frac {\cos (2 a+2 b x)}{2 (c+d x)}\right ) \, dx}{d^2}\\ &=-\frac {\cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}-\frac {b^2 \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{d^2}\\ &=-\frac {\cos ^2(a+b x)}{2 d (c+d x)^2}+\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}-\frac {\left (b^2 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}+\frac {\left (b^2 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d^2}\\ &=-\frac {\cos ^2(a+b x)}{2 d (c+d x)^2}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}\\ \end {align*}

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Mathematica [A]
time = 0.64, size = 102, normalized size = 0.91 \begin {gather*} \frac {-2 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d \left (-d \cos ^2(a+b x)+b (c+d x) \sin (2 (a+b x))\right )}{(c+d x)^2}+2 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.20, size = 193, normalized size = 1.72

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \left (-\frac {\cos \left (2 b x +2 a \right )}{\left (-d a +b c +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {4 \sinIntegral \left (-2 b x -2 a -\frac {2 \left (-d a +b c \right )}{d}\right ) \sin \left (\frac {-2 d a +2 b c}{d}\right )}{d}+\frac {4 \cosineIntegral \left (2 b x +2 a +\frac {-2 d a +2 b c}{d}\right ) \cos \left (\frac {-2 d a +2 b c}{d}\right )}{d}}{d}}{d}\right )}{4}-\frac {b^{3}}{4 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}}{b}\)	\(193\)
default	\(\frac {\frac {b^{3} \left (-\frac {\cos \left (2 b x +2 a \right )}{\left (-d a +b c +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-d a +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {4 \sinIntegral \left (-2 b x -2 a -\frac {2 \left (-d a +b c \right )}{d}\right ) \sin \left (\frac {-2 d a +2 b c}{d}\right )}{d}+\frac {4 \cosineIntegral \left (2 b x +2 a +\frac {-2 d a +2 b c}{d}\right ) \cos \left (\frac {-2 d a +2 b c}{d}\right )}{d}}{d}}{d}\right )}{4}-\frac {b^{3}}{4 \left (-d a +b c +d \left (b x +a \right )\right )^{2} d}}{b}\)	\(193\)
risch	\(-\frac {1}{4 d \left (d x +c \right )^{2}}+\frac {b^{2} {\mathrm e}^{-\frac {2 i \left (d a -b c \right )}{d}} \expIntegral \left (1, 2 i b x +2 i a -\frac {2 i \left (d a -b c \right )}{d}\right )}{2 d^{3}}+\frac {b^{2} {\mathrm e}^{\frac {2 i \left (d a -b c \right )}{d}} \expIntegral \left (1, -2 i b x -2 i a -\frac {2 \left (-i a d +i b c \right )}{d}\right )}{2 d^{3}}+\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (2 b x +2 a \right )}{8 d^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}-\frac {i \left (4 i b^{3} d^{3} x^{3}+12 i b^{3} c \,d^{2} x^{2}+12 i b^{3} c^{2} d x +4 i b^{3} c^{3}\right ) \sin \left (2 b x +2 a \right )}{8 d^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.42, size = 204, normalized size = 1.82 \begin {gather*} -\frac {b^{3} {\left (E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3}}{4 \, {\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)^2/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/4*(b^3*(exp_integral_e(3, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_integral_e(3, -2*(-I*b*c - I*(b*x + a
)*d + I*a*d)/d))*cos(-2*(b*c - a*d)/d) + b^3*(I*exp_integral_e(3, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) - I*ex
p_integral_e(3, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-2*(b*c - a*d)/d) + b^3)/((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 [A]
time = 0.37, size = 218, normalized size = 1.95 \begin {gather*} -\frac {d^{2} \cos \left (b x + a\right )^{2} - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\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 {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{2 \, {\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)^2/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/2*(d^2*cos(b*x + a)^2 - 2*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*
c^2)*sin(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) + ((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integr
al(2*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-2*(b*d*x + b*c)/d))*cos(-2*(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 ^{2}{\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)**2/(d*x+c)**3,x)

[Out]

Integral(cos(a + b*x)**2/(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 = 0.65, size = 5136, normalized size = 45.86 \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)^2/(d*x+c)^3,x, algorithm="giac")

[Out]

-1/2*(b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + b^2*d^2*x^2*real
_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 2*b^2*d^2*x^2*imag_part(cos_integral(
2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b^2*d^2*x^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan
(b*x)^2*tan(a)^2*tan(b*c/d) - 4*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2
*b^2*d^2*x^2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 - 2*b^2*d^2*x^2*imag_part
(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 4*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d
)*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 2*b^2*c*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*ta
n(b*c/d)^2 + 2*b^2*c*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b^2*d^2*
x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - b^2*d^2*x^2*real_part(cos_integral(-2*b*x -
2*b*c/d))*tan(b*x)^2*tan(a)^2 + 4*b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(
b*c/d) + 4*b^2*d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) - 4*b^2*c*d*x*im
ag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 4*b^2*c*d*x*imag_part(cos_integral(-2*
b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 8*b^2*c*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)
^2*tan(b*c/d) - b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 - b^2*d^2*x^2*rea
l_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 + 4*b^2*c*d*x*imag_part(cos_integral(2*b*x + 2*
b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 - 4*b^2*c*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(
a)*tan(b*c/d)^2 + 8*b^2*c*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + b^2*d^2*x^2*rea
l_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 + b^2*d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*
c/d))*tan(a)^2*tan(b*c/d)^2 + b^2*c^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^
2 + b^2*c^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 2*b^2*d^2*x^2*imag_pa
rt(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a) + 2*b^2*d^2*x^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*
tan(b*x)^2*tan(a) - 4*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a) - 2*b^2*c*d*x*real_part(co
s_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - 2*b^2*c*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b
*x)^2*tan(a)^2 + 2*b^2*d^2*x^2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d) - 2*b^2*d^2*x^2*
imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d) + 4*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d
)*tan(b*x)^2*tan(b*c/d) + 8*b^2*c*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) +
8*b^2*c*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) - 2*b^2*d^2*x^2*imag_part(c
os_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d) + 2*b^2*d^2*x^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*ta
n(a)^2*tan(b*c/d) - 4*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d)*tan(a)^2*tan(b*c/d) - 2*b^2*c^2*imag_part(co
s_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b^2*c^2*imag_part(cos_integral(-2*b*x - 2*b*c/
d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 4*b^2*c^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d)
- 2*b^2*c*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 - 2*b^2*c*d*x*real_part(cos_int
egral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 + 2*b^2*d^2*x^2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(
a)*tan(b*c/d)^2 - 2*b^2*d^2*x^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d)^2 + 4*b^2*d^2*x^2*
sin_integral(2*(b*d*x + b*c)/d)*tan(a)*tan(b*c/d)^2 + 2*b^2*c^2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b
*x)^2*tan(a)*tan(b*c/d)^2 - 2*b^2*c^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2
+ 4*b^2*c^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*tan(b*c/d)^2 + 2*b^2*c*d*x*real_part(cos_integr
al(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 + 2*b^2*c*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*ta
n(b*c/d)^2 + b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2 + b^2*d^2*x^2*real_part(cos_integ
ral(-2*b*x - 2*b*c/d))*tan(b*x)^2 - 4*b^2*c*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a) + 4
*b^2*c*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a) - 8*b^2*c*d*x*sin_integral(2*(b*d*x + b
*c)/d)*tan(b*x)^2*tan(a) - b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2 - b^2*d^2*x^2*real_pa
rt(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2 - b^2*c^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(
a)^2 - b^2*c^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 + 4*b^2*c*d*x*imag_part(cos_integ
ral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d) - 4*b^2*c*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2
*tan(b*c/d) + 8*b^2*c*d*x*sin_integral(2*(b*d*x...

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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 )}^2}{{\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)^2/(c + d*x)^3,x)

[Out]

int(cos(a + b*x)^2/(c + d*x)^3, x)

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