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

Optimal. Leaf size=338 \[ \frac {25 b^2 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b c}{d}+5 b x\right )}{32 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^3}-\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{16 d^3}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{16 d^3}-\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^3}+\frac {25 b^2 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{32 d^3}-\frac {b \cos (a+b x)}{16 d^2 (c+d x)}-\frac {3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac {5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac {\sin (a+b x)}{16 d (c+d x)^2}-\frac {\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac {\sin (5 a+5 b x)}{32 d (c+d x)^2} \]

[Out]

-1/16*b*cos(b*x+a)/d^2/(d*x+c)-3/32*b*cos(3*b*x+3*a)/d^2/(d*x+c)+5/32*b*cos(5*b*x+5*a)/d^2/(d*x+c)-1/16*b^2*co
s(a-b*c/d)*Si(b*c/d+b*x)/d^3-9/32*b^2*cos(3*a-3*b*c/d)*Si(3*b*c/d+3*b*x)/d^3+25/32*b^2*cos(5*a-5*b*c/d)*Si(5*b
*c/d+5*b*x)/d^3+25/32*b^2*Ci(5*b*c/d+5*b*x)*sin(5*a-5*b*c/d)/d^3-9/32*b^2*Ci(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d
^3-1/16*b^2*Ci(b*c/d+b*x)*sin(a-b*c/d)/d^3-1/16*sin(b*x+a)/d/(d*x+c)^2-1/32*sin(3*b*x+3*a)/d/(d*x+c)^2+1/32*si
n(5*b*x+5*a)/d/(d*x+c)^2

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Rubi [A]  time = 0.50, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac {25 b^2 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {CosIntegral}\left (\frac {5 b c}{d}+5 b x\right )}{32 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^3}-\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{16 d^3}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{16 d^3}-\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^3}+\frac {25 b^2 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{32 d^3}-\frac {b \cos (a+b x)}{16 d^2 (c+d x)}-\frac {3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac {5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac {\sin (a+b x)}{16 d (c+d x)^2}-\frac {\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac {\sin (5 a+5 b x)}{32 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[a + b*x])/(16*d^2*(c + d*x)) - (3*b*Cos[3*a + 3*b*x])/(32*d^2*(c + d*x)) + (5*b*Cos[5*a + 5*b*x])/(32*
d^2*(c + d*x)) + (25*b^2*CosIntegral[(5*b*c)/d + 5*b*x]*Sin[5*a - (5*b*c)/d])/(32*d^3) - (9*b^2*CosIntegral[(3
*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/(32*d^3) - (b^2*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/(16*d^3) -
Sin[a + b*x]/(16*d*(c + d*x)^2) - Sin[3*a + 3*b*x]/(32*d*(c + d*x)^2) + Sin[5*a + 5*b*x]/(32*d*(c + d*x)^2) -
(b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(16*d^3) - (9*b^2*Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/
d + 3*b*x])/(32*d^3) + (25*b^2*Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x])/(32*d^3)

Rule 3297

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 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)^3} \, dx &=\int \left (\frac {\sin (a+b x)}{8 (c+d x)^3}+\frac {\sin (3 a+3 b x)}{16 (c+d x)^3}-\frac {\sin (5 a+5 b x)}{16 (c+d x)^3}\right ) \, dx\\ &=\frac {1}{16} \int \frac {\sin (3 a+3 b x)}{(c+d x)^3} \, dx-\frac {1}{16} \int \frac {\sin (5 a+5 b x)}{(c+d x)^3} \, dx+\frac {1}{8} \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx\\ &=-\frac {\sin (a+b x)}{16 d (c+d x)^2}-\frac {\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac {\sin (5 a+5 b x)}{32 d (c+d x)^2}+\frac {b \int \frac {\cos (a+b x)}{(c+d x)^2} \, dx}{16 d}+\frac {(3 b) \int \frac {\cos (3 a+3 b x)}{(c+d x)^2} \, dx}{32 d}-\frac {(5 b) \int \frac {\cos (5 a+5 b x)}{(c+d x)^2} \, dx}{32 d}\\ &=-\frac {b \cos (a+b x)}{16 d^2 (c+d x)}-\frac {3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac {5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac {\sin (a+b x)}{16 d (c+d x)^2}-\frac {\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac {\sin (5 a+5 b x)}{32 d (c+d x)^2}-\frac {b^2 \int \frac {\sin (a+b x)}{c+d x} \, dx}{16 d^2}-\frac {\left (9 b^2\right ) \int \frac {\sin (3 a+3 b x)}{c+d x} \, dx}{32 d^2}+\frac {\left (25 b^2\right ) \int \frac {\sin (5 a+5 b x)}{c+d x} \, dx}{32 d^2}\\ &=-\frac {b \cos (a+b x)}{16 d^2 (c+d x)}-\frac {3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac {5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}-\frac {\sin (a+b x)}{16 d (c+d x)^2}-\frac {\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac {\sin (5 a+5 b x)}{32 d (c+d x)^2}+\frac {\left (25 b^2 \cos \left (5 a-\frac {5 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {5 b c}{d}+5 b x\right )}{c+d x} \, dx}{32 d^2}-\frac {\left (9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{32 d^2}-\frac {\left (b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{16 d^2}+\frac {\left (25 b^2 \sin \left (5 a-\frac {5 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {5 b c}{d}+5 b x\right )}{c+d x} \, dx}{32 d^2}-\frac {\left (9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{32 d^2}-\frac {\left (b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{16 d^2}\\ &=-\frac {b \cos (a+b x)}{16 d^2 (c+d x)}-\frac {3 b \cos (3 a+3 b x)}{32 d^2 (c+d x)}+\frac {5 b \cos (5 a+5 b x)}{32 d^2 (c+d x)}+\frac {25 b^2 \text {Ci}\left (\frac {5 b c}{d}+5 b x\right ) \sin \left (5 a-\frac {5 b c}{d}\right )}{32 d^3}-\frac {9 b^2 \text {Ci}\left (\frac {3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{32 d^3}-\frac {b^2 \text {Ci}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{16 d^3}-\frac {\sin (a+b x)}{16 d (c+d x)^2}-\frac {\sin (3 a+3 b x)}{32 d (c+d x)^2}+\frac {\sin (5 a+5 b x)}{32 d (c+d x)^2}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{16 d^3}-\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^3}+\frac {25 b^2 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{32 d^3}\\ \end {align*}

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Mathematica [A]  time = 3.90, size = 279, normalized size = 0.83 \[ \frac {-2 \left (b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )+\frac {d (b (c+d x) \cos (a+b x)+d \sin (a+b x))}{(c+d x)^2}\right )+25 b^2 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b (c+d x)}{d}\right )-9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )-9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )+25 b^2 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b (c+d x)}{d}\right )-\frac {d (3 b (c+d x) \cos (3 (a+b x))+d \sin (3 (a+b x)))}{(c+d x)^2}+\frac {d (5 b (c+d x) \cos (5 (a+b x))+d \sin (5 (a+b x)))}{(c+d x)^2}}{32 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(25*b^2*CosIntegral[(5*b*(c + d*x))/d]*Sin[5*a - (5*b*c)/d] - 9*b^2*CosIntegral[(3*b*(c + d*x))/d]*Sin[3*a - (
3*b*c)/d] - (d*(3*b*(c + d*x)*Cos[3*(a + b*x)] + d*Sin[3*(a + b*x)]))/(c + d*x)^2 + (d*(5*b*(c + d*x)*Cos[5*(a
+ b*x)] + d*Sin[5*(a + b*x)]))/(c + d*x)^2 - 2*(b^2*CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + (d*(b*(c + d*
x)*Cos[a + b*x] + d*Sin[a + b*x]))/(c + d*x)^2 + b^2*Cos[a - (b*c)/d]*SinIntegral[b*(c/d + x)]) - 9*b^2*Cos[3*
a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d] + 25*b^2*Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*(c + d*x))/d])/(3
2*d^3)

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fricas [A]  time = 0.83, size = 585, normalized size = 1.73 \[ \frac {160 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{5} - 224 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} + 50 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \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 ) - 18 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \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 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 64 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) + 32 \, {\left (d^{2} \cos \left (b x + a\right )^{4} - d^{2} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right ) - 2 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - 9 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {3 \, {\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 {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 25 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {5 \, {\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 {5 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right )}{64 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/64*(160*(b*d^2*x + b*c*d)*cos(b*x + a)^5 - 224*(b*d^2*x + b*c*d)*cos(b*x + a)^3 + 50*(b^2*d^2*x^2 + 2*b^2*c*
d*x + b^2*c^2)*cos(-5*(b*c - a*d)/d)*sin_integral(5*(b*d*x + b*c)/d) - 18*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2
)*cos(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) - 4*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-(b*c -
a*d)/d)*sin_integral((b*d*x + b*c)/d) + 64*(b*d^2*x + b*c*d)*cos(b*x + a) + 32*(d^2*cos(b*x + a)^4 - d^2*cos(b
*x + a)^2)*sin(b*x + a) - 2*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral((b*d*x + b*c)/d) + (b^2*d^2*x^
2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-(b*d*x + b*c)/d))*sin(-(b*c - a*d)/d) - 9*((b^2*d^2*x^2 + 2*b^2*c*d*x
+ b^2*c^2)*cos_integral(3*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-3*(b*d*x + b
*c)/d))*sin(-3*(b*c - a*d)/d) + 25*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(5*(b*d*x + b*c)/d) + (b
^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-5*(b*d*x + b*c)/d))*sin(-5*(b*c - a*d)/d))/(d^5*x^2 + 2*c*d^
4*x + c^2*d^3)

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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)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 475, normalized size = 1.41 \[ \frac {-\frac {b^{3} \left (-\frac {5 \sin \left (5 b x +5 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {25 \cos \left (5 b x +5 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {25 \left (\frac {5 \Si \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \cos \left (\frac {-5 d a +5 c b}{d}\right )}{d}-\frac {5 \Ci \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \sin \left (\frac {-5 d a +5 c b}{d}\right )}{d}\right )}{2 d}}{d}\right )}{80}+\frac {b^{3} \left (-\frac {\sin \left (b x +a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}-\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}\right )}{8}+\frac {b^{3} \left (-\frac {3 \sin \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {9 \cos \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {9 \left (\frac {3 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}-\frac {3 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}\right )}{2 d}}{d}\right )}{48}}{b} \]

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)^3,x)

[Out]

1/b*(-1/80*b^3*(-5/2*sin(5*b*x+5*a)/((b*x+a)*d-d*a+c*b)^2/d+5/2*(-5*cos(5*b*x+5*a)/((b*x+a)*d-d*a+c*b)/d-5*(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)/d)/d)
+1/8*b^3*(-1/2*sin(b*x+a)/((b*x+a)*d-d*a+c*b)^2/d+1/2*(-cos(b*x+a)/((b*x+a)*d-d*a+c*b)/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)+1/48*b^3*(-3/2*sin(3*b*x+3*a)/((b*x+a
)*d-d*a+c*b)^2/d+3/2*(-3*cos(3*b*x+3*a)/((b*x+a)*d-d*a+c*b)/d-3*(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)/d)/d))

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maxima [C]  time = 0.78, size = 473, normalized size = 1.40 \[ \frac {b^{3} {\left (-2 i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 2 i \, E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{3} {\left (-i \, E_{3}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + i \, E_{3}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (i \, E_{3}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{3} {\left (E_{3}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{3}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (E_{3}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{3}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right )}{32 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]

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)^3,x, algorithm="maxima")

[Out]

1/32*(b^3*(-2*I*exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 2*I*exp_integral_e(3, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^3*(-I*exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) +
I*exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b^3*(I*exp_integral_e(
3, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) - I*exp_integral_e(3, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*c
os(-5*(b*c - a*d)/d) - 2*b^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(3, -(I*b*c
+ I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b^3*(exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*
d)/d) + exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d) + b^3*(exp_integral
_e(3, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) + exp_integral_e(3, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*
sin(-5*(b*c - a*d)/d))/((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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^3} \,d x \]

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)^3,x)

[Out]

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

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

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)**3,x)

[Out]

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

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