\(\int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx\) [84]

Optimal result

Integrand size = 24, antiderivative size = 78 \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=-\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\log (c+d x)}{8 d}+\frac {\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{8 d} \]

[Out]

-1/8*Ci(4*b*c/d+4*b*x)*cos(4*a-4*b*c/d)/d+1/8*ln(d*x+c)/d+1/8*Si(4*b*c/d+4*b*x)*sin(4*a-4*b*c/d)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4491, 3384, 3380, 3383} \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=-\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\log (c+d x)}{8 d} \]

[In]

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

[Out]

-1/8*(Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*c)/d + 4*b*x])/d + Log[c + d*x]/(8*d) + (Sin[4*a - (4*b*c)/d]*SinI
ntegral[(4*b*c)/d + 4*b*x])/(8*d)

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 4491

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{8 (c+d x)}-\frac {\cos (4 a+4 b x)}{8 (c+d x)}\right ) \, dx \\ & = \frac {\log (c+d x)}{8 d}-\frac {1}{8} \int \frac {\cos (4 a+4 b x)}{c+d x} \, dx \\ & = \frac {\log (c+d x)}{8 d}-\frac {1}{8} \cos \left (4 a-\frac {4 b c}{d}\right ) \int \frac {\cos \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx+\frac {1}{8} \sin \left (4 a-\frac {4 b c}{d}\right ) \int \frac {\sin \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx \\ & = -\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\log (c+d x)}{8 d}+\frac {\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=\frac {-\cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 b (c+d x)}{d}\right )+\log (c+d x)+\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )}{8 d} \]

[In]

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

[Out]

(-(Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*(c + d*x))/d]) + Log[c + d*x] + Sin[4*a - (4*b*c)/d]*SinIntegral[(4*b
*(c + d*x))/d])/(8*d)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.92 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.37

method	result	size
risch	\(\frac {\ln \left (d x +c \right )}{8 d}+\frac {{\mathrm e}^{-\frac {4 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (4 i b x +4 i a -\frac {4 i \left (a d -c b \right )}{d}\right )}{16 d}+\frac {{\mathrm e}^{\frac {4 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-4 i b x -4 i a -\frac {4 \left (-i a d +i c b \right )}{d}\right )}{16 d}\)	\(107\)
derivativedivides	\(\frac {-\frac {b \left (-\frac {4 \,\operatorname {Si}\left (-4 x b -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (4 x b +4 a +\frac {-4 a d +4 c b}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}\right )}{32}+\frac {b \ln \left (-a d +c b +d \left (x b +a \right )\right )}{8 d}}{b}\)	\(114\)
default	\(\frac {-\frac {b \left (-\frac {4 \,\operatorname {Si}\left (-4 x b -4 a -\frac {4 \left (-a d +c b \right )}{d}\right ) \sin \left (\frac {-4 a d +4 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (4 x b +4 a +\frac {-4 a d +4 c b}{d}\right ) \cos \left (\frac {-4 a d +4 c b}{d}\right )}{d}\right )}{32}+\frac {b \ln \left (-a d +c b +d \left (x b +a \right )\right )}{8 d}}{b}\)	\(114\)

[In]

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

[Out]

1/8*ln(d*x+c)/d+1/16/d*exp(-4*I*(a*d-b*c)/d)*Ei(1,4*I*b*x+4*I*a-4*I*(a*d-b*c)/d)+1/16/d*exp(4*I*(a*d-b*c)/d)*E
i(1,-4*I*x*b-4*I*a-4*(-I*a*d+I*c*b)/d)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=-\frac {\cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - \log \left (d x + c\right )}{8 \, d} \]

[In]

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

[Out]

-1/8*(cos(-4*(b*c - a*d)/d)*cos_integral(4*(b*d*x + b*c)/d) - sin(-4*(b*c - a*d)/d)*sin_integral(4*(b*d*x + b*
c)/d) - log(d*x + c))/d

Sympy [F]

\[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=\int \frac {\sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{c + d x}\, dx \]

[In]

integrate(cos(b*x+a)**2*sin(b*x+a)**2/(d*x+c),x)

[Out]

Integral(sin(a + b*x)**2*cos(a + b*x)**2/(c + d*x), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.08 \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=\frac {b {\left (E_{1}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{1}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + b {\left (i \, E_{1}\left (\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{1}\left (-\frac {4 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b \log \left (b c + {\left (b x + a\right )} d - a d\right )}{16 \, b d} \]

[In]

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

[Out]

1/16*(b*(exp_integral_e(1, 4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_integral_e(1, -4*(-I*b*c - I*(b*x + a)*
d + I*a*d)/d))*cos(-4*(b*c - a*d)/d) + b*(I*exp_integral_e(1, 4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) - I*exp_in
tegral_e(1, -4*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-4*(b*c - a*d)/d) + 2*b*log(b*c + (b*x + a)*d - a*d))/
(b*d)

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.33 (sec) , antiderivative size = 669, normalized size of antiderivative = 8.58 \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/16*(2*log(abs(d*x + c))*tan(2*a)^2*tan(2*b*c/d)^2 - real_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*a)^2*tan(
2*b*c/d)^2 - real_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)^2*tan(2*b*c/d)^2 + 2*imag_part(cos_integral(4*
b*x + 4*b*c/d))*tan(2*a)^2*tan(2*b*c/d) - 2*imag_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)^2*tan(2*b*c/d)
+ 4*sin_integral(4*(b*d*x + b*c)/d)*tan(2*a)^2*tan(2*b*c/d) - 2*imag_part(cos_integral(4*b*x + 4*b*c/d))*tan(2
*a)*tan(2*b*c/d)^2 + 2*imag_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)*tan(2*b*c/d)^2 - 4*sin_integral(4*(b
*d*x + b*c)/d)*tan(2*a)*tan(2*b*c/d)^2 + 2*log(abs(d*x + c))*tan(2*a)^2 + real_part(cos_integral(4*b*x + 4*b*c
/d))*tan(2*a)^2 + real_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)^2 - 4*real_part(cos_integral(4*b*x + 4*b*
c/d))*tan(2*a)*tan(2*b*c/d) - 4*real_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*a)*tan(2*b*c/d) + 2*log(abs(d*
x + c))*tan(2*b*c/d)^2 + real_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*b*c/d)^2 + real_part(cos_integral(-4*b
*x - 4*b*c/d))*tan(2*b*c/d)^2 + 2*imag_part(cos_integral(4*b*x + 4*b*c/d))*tan(2*a) - 2*imag_part(cos_integral
(-4*b*x - 4*b*c/d))*tan(2*a) + 4*sin_integral(4*(b*d*x + b*c)/d)*tan(2*a) - 2*imag_part(cos_integral(4*b*x + 4
*b*c/d))*tan(2*b*c/d) + 2*imag_part(cos_integral(-4*b*x - 4*b*c/d))*tan(2*b*c/d) - 4*sin_integral(4*(b*d*x + b
*c)/d)*tan(2*b*c/d) + 2*log(abs(d*x + c)) - real_part(cos_integral(4*b*x + 4*b*c/d)) - real_part(cos_integral(
-4*b*x - 4*b*c/d)))/(d*tan(2*a)^2*tan(2*b*c/d)^2 + d*tan(2*a)^2 + d*tan(2*b*c/d)^2 + d)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2}{c+d\,x} \,d x \]

[In]

int((cos(a + b*x)^2*sin(a + b*x)^2)/(c + d*x),x)

[Out]

int((cos(a + b*x)^2*sin(a + b*x)^2)/(c + d*x), x)