3.93 \(\int \frac {\cos ^2(a+b x) \sin ^3(a+b x)}{c+d x} \, dx\)

Optimal. Leaf size=185 \[ -\frac {\sin \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b c}{d}+5 b x\right )}{16 d}+\frac {\sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{16 d}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{8 d}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d}+\frac {\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{16 d}-\frac {\cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{16 d} \]

[Out]

1/8*cos(a-b*c/d)*Si(b*c/d+b*x)/d+1/16*cos(3*a-3*b*c/d)*Si(3*b*c/d+3*b*x)/d-1/16*cos(5*a-5*b*c/d)*Si(5*b*c/d+5*
b*x)/d-1/16*Ci(5*b*c/d+5*b*x)*sin(5*a-5*b*c/d)/d+1/16*Ci(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d+1/8*Ci(b*c/d+b*x)*s
in(a-b*c/d)/d

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Rubi [A]  time = 0.34, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4406, 3303, 3299, 3302} \[ -\frac {\sin \left (5 a-\frac {5 b c}{d}\right ) \text {CosIntegral}\left (\frac {5 b c}{d}+5 b x\right )}{16 d}+\frac {\sin \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{16 d}+\frac {\sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{8 d}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d}+\frac {\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{16 d}-\frac {\cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(CosIntegral[(5*b*c)/d + 5*b*x]*Sin[5*a - (5*b*c)/d])/(16*d) + (CosIntegral[(3*b*c)/d + 3*b*x]*Sin[3*a - (3*b
*c)/d])/(16*d) + (CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/(8*d) + (Cos[a - (b*c)/d]*SinIntegral[(b*c)/d +
b*x])/(8*d) + (Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(16*d) - (Cos[5*a - (5*b*c)/d]*SinIntegra
l[(5*b*c)/d + 5*b*x])/(16*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 ^3(a+b x)}{c+d x} \, dx &=\int \left (\frac {\sin (a+b x)}{8 (c+d x)}+\frac {\sin (3 a+3 b x)}{16 (c+d x)}-\frac {\sin (5 a+5 b x)}{16 (c+d x)}\right ) \, dx\\ &=\frac {1}{16} \int \frac {\sin (3 a+3 b x)}{c+d x} \, dx-\frac {1}{16} \int \frac {\sin (5 a+5 b x)}{c+d x} \, dx+\frac {1}{8} \int \frac {\sin (a+b x)}{c+d x} \, dx\\ &=-\left (\frac {1}{16} \cos \left (5 a-\frac {5 b c}{d}\right ) \int \frac {\sin \left (\frac {5 b c}{d}+5 b x\right )}{c+d x} \, dx\right )+\frac {1}{16} \cos \left (3 a-\frac {3 b c}{d}\right ) \int \frac {\sin \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx+\frac {1}{8} \cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx-\frac {1}{16} \sin \left (5 a-\frac {5 b c}{d}\right ) \int \frac {\cos \left (\frac {5 b c}{d}+5 b x\right )}{c+d x} \, dx+\frac {1}{16} \sin \left (3 a-\frac {3 b c}{d}\right ) \int \frac {\cos \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx+\frac {1}{8} \sin \left (a-\frac {b c}{d}\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx\\ &=-\frac {\text {Ci}\left (\frac {5 b c}{d}+5 b x\right ) \sin \left (5 a-\frac {5 b c}{d}\right )}{16 d}+\frac {\text {Ci}\left (\frac {3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{16 d}+\frac {\text {Ci}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{8 d}+\frac {\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d}+\frac {\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{16 d}-\frac {\cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{16 d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 154, normalized size = 0.83 \[ \frac {\sin \left (5 a-\frac {5 b c}{d}\right ) \left (-\text {Ci}\left (\frac {5 b (c+d x)}{d}\right )\right )+\sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )+2 \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )+\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )-\cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b (c+d x)}{d}\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(CosIntegral[(5*b*(c + d*x))/d]*Sin[5*a - (5*b*c)/d]) + CosIntegral[(3*b*(c + d*x))/d]*Sin[3*a - (3*b*c)/d]
+ 2*CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + 2*Cos[a - (b*c)/d]*SinIntegral[b*(c/d + x)] + Cos[3*a - (3*b*c
)/d]*SinIntegral[(3*b*(c + d*x))/d] - Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*(c + d*x))/d])/(16*d)

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fricas [A]  time = 0.65, size = 228, normalized size = 1.23 \[ \frac {2 \, {\left (\operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + {\left (\operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + \operatorname {Ci}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - {\left (\operatorname {Ci}\left (\frac {5 \, {\left (b d x + b c\right )}}{d}\right ) + \operatorname {Ci}\left (-\frac {5 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {5 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + 4 \, \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^3/(d*x+c),x, algorithm="fricas")

[Out]

1/32*(2*(cos_integral((b*d*x + b*c)/d) + cos_integral(-(b*d*x + b*c)/d))*sin(-(b*c - a*d)/d) + (cos_integral(3
*(b*d*x + b*c)/d) + cos_integral(-3*(b*d*x + b*c)/d))*sin(-3*(b*c - a*d)/d) - (cos_integral(5*(b*d*x + b*c)/d)
+ cos_integral(-5*(b*d*x + b*c)/d))*sin(-5*(b*c - a*d)/d) - 2*cos(-5*(b*c - a*d)/d)*sin_integral(5*(b*d*x + b
*c)/d) + 2*cos(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) + 4*cos(-(b*c - a*d)/d)*sin_integral((b*d*x +
b*c)/d))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^3/(d*x+c),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 253, normalized size = 1.37 \[ \frac {-\frac {b \left (\frac {5 \Si \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \cos \left (\frac {-5 d a +5 c b}{d}\right )}{d}-\frac {5 \Ci \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \sin \left (\frac {-5 d a +5 c b}{d}\right )}{d}\right )}{80}+\frac {b \left (\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}-\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}\right )}{8}+\frac {b \left (\frac {3 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}-\frac {3 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}\right )}{48}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/80*b*(5*Si(5*b*x+5*a+5*(-a*d+b*c)/d)*cos(5*(-a*d+b*c)/d)/d-5*Ci(5*b*x+5*a+5*(-a*d+b*c)/d)*sin(5*(-a*d+
b*c)/d)/d)+1/8*b*(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)+1/48*
b*(3*Si(3*b*x+3*a+3*(-a*d+b*c)/d)*cos(3*(-a*d+b*c)/d)/d-3*Ci(3*b*x+3*a+3*(-a*d+b*c)/d)*sin(3*(-a*d+b*c)/d)/d))

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maxima [C]  time = 0.47, size = 407, normalized size = 2.20 \[ \frac {b {\left (-2 i \, E_{1}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 2 i \, E_{1}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b {\left (-i \, E_{1}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + i \, E_{1}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b {\left (i \, E_{1}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - i \, E_{1}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b {\left (E_{1}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{1}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b {\left (E_{1}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{1}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b {\left (E_{1}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{1}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right )}{32 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(b*x+a)^3/(d*x+c),x, algorithm="maxima")

[Out]

1/32*(b*(-2*I*exp_integral_e(1, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 2*I*exp_integral_e(1, -(I*b*c + I*(b*x +
a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b*(-I*exp_integral_e(1, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) + I*e
xp_integral_e(1, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b*(I*exp_integral_e(1, (5*
I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) - I*exp_integral_e(1, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*cos(-5*
(b*c - a*d)/d) - 2*b*(exp_integral_e(1, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(1, -(I*b*c + I*(b*
x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b*(exp_integral_e(1, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) + ex
p_integral_e(1, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d) + b*(exp_integral_e(1, (5*I*b
*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) + exp_integral_e(1, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*sin(-5*(b*c
- a*d)/d))/(b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{c+d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{c + d x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2*sin(b*x+a)**3/(d*x+c),x)

[Out]

Integral(sin(a + b*x)**3*cos(a + b*x)**2/(c + d*x), x)

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