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

Optimal. Leaf size=78 \[ -\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\log (c+d x)}{8 d} \]

[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

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Rubi [A]  time = 0.14, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4406, 3303, 3299, 3302} \[ -\frac {\cos \left (4 a-\frac {4 b c}{d}\right ) \text {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{8 d}+\frac {\log (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(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 + 4*b*x])/(8*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 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 4406

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*} \int \frac {\cos ^2(a+b x) \sin ^2(a+b x)}{c+d x} \, dx &=\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 ) \text {Ci}\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*}

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Mathematica [A]  time = 0.16, size = 65, normalized size = 0.83 \[ \frac {-\cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b (c+d x)}{d}\right )+\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )+\log (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.49, size = 88, normalized size = 1.13 \[ -\frac {{\left (\operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + \operatorname {Ci}\left (-\frac {4 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, \log \left (d x + c\right )}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((cos_integral(4*(b*d*x + b*c)/d) + cos_integral(-4*(b*d*x + b*c)/d))*cos(-4*(b*c - a*d)/d) - 2*sin(-4*(
b*c - a*d)/d)*sin_integral(4*(b*d*x + b*c)/d) - 2*log(d*x + c))/d

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giac [C]  time = 0.22, size = 669, normalized size = 8.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 105, normalized size = 1.35 \[ -\frac {\Si \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \sin \left (\frac {-4 d a +4 c b}{d}\right )}{8 d}-\frac {\Ci \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \cos \left (\frac {-4 d a +4 c b}{d}\right )}{8 d}+\frac {\ln \left (\left (b x +a \right ) d -d a +c b \right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.41, size = 160, normalized size = 2.05 \[ \frac {b {\left (E_{1}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{1}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + b {\left (-i \, E_{1}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + i \, E_{1}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b \log \left (b c + {\left (b x + a\right )} d - a d\right )}{16 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2}{c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{c + d x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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