3.153 \(\int \frac {\cos ^3(a+b x) \sin ^2(a+b x)}{(c+d x)^4} \, dx\)

Optimal. Leaf size=413 \[ -\frac {125 b^3 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}-\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{48 d^4}+\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{48 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}-\frac {125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}+\frac {b^2 \cos (a+b x)}{48 d^3 (c+d x)}-\frac {3 b^2 \cos (3 a+3 b x)}{32 d^3 (c+d x)}-\frac {25 b^2 \cos (5 a+5 b x)}{96 d^3 (c+d x)}+\frac {b \sin (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \sin (3 a+3 b x)}{32 d^2 (c+d x)^2}-\frac {5 b \sin (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3} \]

[Out]

-1/24*cos(b*x+a)/d/(d*x+c)^3+1/48*b^2*cos(b*x+a)/d^3/(d*x+c)+1/48*cos(3*b*x+3*a)/d/(d*x+c)^3-3/32*b^2*cos(3*b*
x+3*a)/d^3/(d*x+c)+1/48*cos(5*b*x+5*a)/d/(d*x+c)^3-25/96*b^2*cos(5*b*x+5*a)/d^3/(d*x+c)+1/48*b^3*cos(a-b*c/d)*
Si(b*c/d+b*x)/d^4-9/32*b^3*cos(3*a-3*b*c/d)*Si(3*b*c/d+3*b*x)/d^4-125/96*b^3*cos(5*a-5*b*c/d)*Si(5*b*c/d+5*b*x
)/d^4-125/96*b^3*Ci(5*b*c/d+5*b*x)*sin(5*a-5*b*c/d)/d^4-9/32*b^3*Ci(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^4+1/48*b
^3*Ci(b*c/d+b*x)*sin(a-b*c/d)/d^4+1/48*b*sin(b*x+a)/d^2/(d*x+c)^2-1/32*b*sin(3*b*x+3*a)/d^2/(d*x+c)^2-5/96*b*s
in(5*b*x+5*a)/d^2/(d*x+c)^2

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Rubi [A]  time = 0.54, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 20, 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} \[ -\frac {125 b^3 \sin \left (5 a-\frac {5 b c}{d}\right ) \text {CosIntegral}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}-\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{48 d^4}+\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{48 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}-\frac {125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}+\frac {b^2 \cos (a+b x)}{48 d^3 (c+d x)}-\frac {3 b^2 \cos (3 a+3 b x)}{32 d^3 (c+d x)}-\frac {25 b^2 \cos (5 a+5 b x)}{96 d^3 (c+d x)}+\frac {b \sin (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \sin (3 a+3 b x)}{32 d^2 (c+d x)^2}-\frac {5 b \sin (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]^3*Sin[a + b*x]^2)/(c + d*x)^4,x]

[Out]

-Cos[a + b*x]/(24*d*(c + d*x)^3) + (b^2*Cos[a + b*x])/(48*d^3*(c + d*x)) + Cos[3*a + 3*b*x]/(48*d*(c + d*x)^3)
- (3*b^2*Cos[3*a + 3*b*x])/(32*d^3*(c + d*x)) + Cos[5*a + 5*b*x]/(48*d*(c + d*x)^3) - (25*b^2*Cos[5*a + 5*b*x
])/(96*d^3*(c + d*x)) - (125*b^3*CosIntegral[(5*b*c)/d + 5*b*x]*Sin[5*a - (5*b*c)/d])/(96*d^4) - (9*b^3*CosInt
egral[(3*b*c)/d + 3*b*x]*Sin[3*a - (3*b*c)/d])/(32*d^4) + (b^3*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/(4
8*d^4) + (b*Sin[a + b*x])/(48*d^2*(c + d*x)^2) - (b*Sin[3*a + 3*b*x])/(32*d^2*(c + d*x)^2) - (5*b*Sin[5*a + 5*
b*x])/(96*d^2*(c + d*x)^2) + (b^3*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(48*d^4) - (9*b^3*Cos[3*a - (3*
b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(32*d^4) - (125*b^3*Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x
])/(96*d^4)

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 ^3(a+b x) \sin ^2(a+b x)}{(c+d x)^4} \, dx &=\int \left (\frac {\cos (a+b x)}{8 (c+d x)^4}-\frac {\cos (3 a+3 b x)}{16 (c+d x)^4}-\frac {\cos (5 a+5 b x)}{16 (c+d x)^4}\right ) \, dx\\ &=-\left (\frac {1}{16} \int \frac {\cos (3 a+3 b x)}{(c+d x)^4} \, dx\right )-\frac {1}{16} \int \frac {\cos (5 a+5 b x)}{(c+d x)^4} \, dx+\frac {1}{8} \int \frac {\cos (a+b x)}{(c+d x)^4} \, dx\\ &=-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3}-\frac {b \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx}{24 d}+\frac {b \int \frac {\sin (3 a+3 b x)}{(c+d x)^3} \, dx}{16 d}+\frac {(5 b) \int \frac {\sin (5 a+5 b x)}{(c+d x)^3} \, dx}{48 d}\\ &=-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3}+\frac {b \sin (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \sin (3 a+3 b x)}{32 d^2 (c+d x)^2}-\frac {5 b \sin (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {b^2 \int \frac {\cos (a+b x)}{(c+d x)^2} \, dx}{48 d^2}+\frac {\left (3 b^2\right ) \int \frac {\cos (3 a+3 b x)}{(c+d x)^2} \, dx}{32 d^2}+\frac {\left (25 b^2\right ) \int \frac {\cos (5 a+5 b x)}{(c+d x)^2} \, dx}{96 d^2}\\ &=-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{48 d^3 (c+d x)}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}-\frac {3 b^2 \cos (3 a+3 b x)}{32 d^3 (c+d x)}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3}-\frac {25 b^2 \cos (5 a+5 b x)}{96 d^3 (c+d x)}+\frac {b \sin (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \sin (3 a+3 b x)}{32 d^2 (c+d x)^2}-\frac {5 b \sin (5 a+5 b x)}{96 d^2 (c+d x)^2}+\frac {b^3 \int \frac {\sin (a+b x)}{c+d x} \, dx}{48 d^3}-\frac {\left (9 b^3\right ) \int \frac {\sin (3 a+3 b x)}{c+d x} \, dx}{32 d^3}-\frac {\left (125 b^3\right ) \int \frac {\sin (5 a+5 b x)}{c+d x} \, dx}{96 d^3}\\ &=-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{48 d^3 (c+d x)}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}-\frac {3 b^2 \cos (3 a+3 b x)}{32 d^3 (c+d x)}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3}-\frac {25 b^2 \cos (5 a+5 b x)}{96 d^3 (c+d x)}+\frac {b \sin (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \sin (3 a+3 b x)}{32 d^2 (c+d x)^2}-\frac {5 b \sin (5 a+5 b x)}{96 d^2 (c+d x)^2}-\frac {\left (125 b^3 \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}{96 d^3}-\frac {\left (9 b^3 \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^3}+\frac {\left (b^3 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{48 d^3}-\frac {\left (125 b^3 \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}{96 d^3}-\frac {\left (9 b^3 \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^3}+\frac {\left (b^3 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{48 d^3}\\ &=-\frac {\cos (a+b x)}{24 d (c+d x)^3}+\frac {b^2 \cos (a+b x)}{48 d^3 (c+d x)}+\frac {\cos (3 a+3 b x)}{48 d (c+d x)^3}-\frac {3 b^2 \cos (3 a+3 b x)}{32 d^3 (c+d x)}+\frac {\cos (5 a+5 b x)}{48 d (c+d x)^3}-\frac {25 b^2 \cos (5 a+5 b x)}{96 d^3 (c+d x)}-\frac {125 b^3 \text {Ci}\left (\frac {5 b c}{d}+5 b x\right ) \sin \left (5 a-\frac {5 b c}{d}\right )}{96 d^4}-\frac {9 b^3 \text {Ci}\left (\frac {3 b c}{d}+3 b x\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{32 d^4}+\frac {b^3 \text {Ci}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{48 d^4}+\frac {b \sin (a+b x)}{48 d^2 (c+d x)^2}-\frac {b \sin (3 a+3 b x)}{32 d^2 (c+d x)^2}-\frac {5 b \sin (5 a+5 b x)}{96 d^2 (c+d x)^2}+\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{48 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{32 d^4}-\frac {125 b^3 \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{96 d^4}\\ \end {align*}

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Mathematica [A]  time = 3.40, size = 451, normalized size = 1.09 \[ -\frac {27 b^3 (c+d x)^3 \left (\sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )+\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )\right )+125 b^3 (c+d x)^3 \left (\sin \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b (c+d x)}{d}\right )+\cos \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b (c+d x)}{d}\right )\right )+d \cos (3 b x) \left (\cos (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )+3 b d \sin (3 a) (c+d x)\right )+d \cos (5 b x) \left (\cos (5 a) \left (25 b^2 (c+d x)^2-2 d^2\right )+5 b d \sin (5 a) (c+d x)\right )+d \sin (3 b x) \left (3 b d \cos (3 a) (c+d x)-\sin (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )\right )+d \sin (5 b x) \left (5 b d \cos (5 a) (c+d x)-\sin (5 a) \left (25 b^2 (c+d x)^2-2 d^2\right )\right )-2 \left (b^3 (c+d x)^3 \left (\sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+\cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )\right )+d \cos (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )+b d \sin (a) (c+d x)\right )+d \sin (b x) \left (b d \cos (a) (c+d x)-\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )\right )\right )}{96 d^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]^3*Sin[a + b*x]^2)/(c + d*x)^4,x]

[Out]

-1/96*(d*Cos[3*b*x]*((-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*a] + 3*b*d*(c + d*x)*Sin[3*a]) + d*Cos[5*b*x]*((-2*d^2
+ 25*b^2*(c + d*x)^2)*Cos[5*a] + 5*b*d*(c + d*x)*Sin[5*a]) + d*(3*b*d*(c + d*x)*Cos[3*a] - (-2*d^2 + 9*b^2*(c
+ d*x)^2)*Sin[3*a])*Sin[3*b*x] + d*(5*b*d*(c + d*x)*Cos[5*a] - (-2*d^2 + 25*b^2*(c + d*x)^2)*Sin[5*a])*Sin[5*
b*x] - 2*(d*Cos[b*x]*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] + b*d*(c + d*x)*Sin[a]) + d*(b*d*(c + d*x)*Cos[a] - (-
2*d^2 + b^2*(c + d*x)^2)*Sin[a])*Sin[b*x] + b^3*(c + d*x)^3*(CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + Cos[a
- (b*c)/d]*SinIntegral[b*(c/d + x)])) + 27*b^3*(c + d*x)^3*(CosIntegral[(3*b*(c + d*x))/d]*Sin[3*a - (3*b*c)/
d] + Cos[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d]) + 125*b^3*(c + d*x)^3*(CosIntegral[(5*b*(c + d*x))/d
]*Sin[5*a - (5*b*c)/d] + Cos[5*a - (5*b*c)/d]*SinIntegral[(5*b*(c + d*x))/d]))/(d^4*(c + d*x)^3)

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fricas [B]  time = 0.63, size = 811, normalized size = 1.96 \[ -\frac {32 \, {\left (25 \, b^{2} d^{3} x^{2} + 50 \, b^{2} c d^{2} x + 25 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{5} - 32 \, {\left (29 \, b^{2} d^{3} x^{2} + 58 \, b^{2} c d^{2} x + 29 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{3} + 250 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\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 ) + 54 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\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^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 192 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right ) + 32 \, {\left (5 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{4} - 3 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right ) - 2 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + 27 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\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 ) + 125 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {5 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\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 )}{192 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^2/(d*x+c)^4,x, algorithm="fricas")

[Out]

-1/192*(32*(25*b^2*d^3*x^2 + 50*b^2*c*d^2*x + 25*b^2*c^2*d - 2*d^3)*cos(b*x + a)^5 - 32*(29*b^2*d^3*x^2 + 58*b
^2*c*d^2*x + 29*b^2*c^2*d - 2*d^3)*cos(b*x + a)^3 + 250*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c
^3)*cos(-5*(b*c - a*d)/d)*sin_integral(5*(b*d*x + b*c)/d) + 54*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x
+ b^3*c^3)*cos(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d) - 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^
2*d*x + b^3*c^3)*cos(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) + 192*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^
2*d)*cos(b*x + a) + 32*(5*(b*d^3*x + b*c*d^2)*cos(b*x + a)^4 - 3*(b*d^3*x + b*c*d^2)*cos(b*x + a)^2)*sin(b*x +
a) - 2*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral((b*d*x + b*c)/d) + (b^3*d^3*x^
3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-(b*d*x + b*c)/d))*sin(-(b*c - a*d)/d) + 27*((b^3*
d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(3*(b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*
d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-3*(b*d*x + b*c)/d))*sin(-3*(b*c - a*d)/d) + 125*((b^3*d^3*x^3
+ 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(5*(b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2
+ 3*b^3*c^2*d*x + b^3*c^3)*cos_integral(-5*(b*d*x + b*c)/d))*sin(-5*(b*c - a*d)/d))/(d^7*x^3 + 3*c*d^6*x^2 +
3*c^2*d^5*x + c^3*d^4)

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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)^3*sin(b*x+a)^2/(d*x+c)^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 583, normalized size = 1.41 \[ \frac {\frac {b^{4} \left (-\frac {\cos \left (b x +a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}-\frac {-\frac {\sin \left (b x +a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}-\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}}{3 d}\right )}{8}-\frac {b^{4} \left (-\frac {5 \cos \left (5 b x +5 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}-\frac {5 \left (-\frac {5 \sin \left (5 b x +5 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {25 \cos \left (5 b x +5 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {25 \left (\frac {5 \Si \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \cos \left (\frac {-5 d a +5 c b}{d}\right )}{d}-\frac {5 \Ci \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \sin \left (\frac {-5 d a +5 c b}{d}\right )}{d}\right )}{2 d}}{d}\right )}{3 d}\right )}{80}-\frac {b^{4} \left (-\frac {\cos \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{3} d}-\frac {-\frac {3 \sin \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {9 \cos \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {9 \left (\frac {3 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}-\frac {3 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}\right )}{2 d}}{d}}{d}\right )}{48}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(b*x+a)^2/(d*x+c)^4,x)

[Out]

1/b*(1/8*b^4*(-1/3*cos(b*x+a)/((b*x+a)*d-d*a+c*b)^3/d-1/3*(-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)/d)-1/80*b^4*(-5/3*cos(5*b*x+5*a)/((b*x+a)*d-d*a+c*b)^3/d-5/3*(-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)/d)-1/48*b^4*(-cos(3*b*x+3*a)/((b*x+a)*d-d*a+c*b)^3/
d-(-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)/d))

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maxima [C]  time = 1.27, size = 524, normalized size = 1.27 \[ -\frac {1073741824 \, b^{4} {\left (E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{4}\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 ) - 536870912 \, b^{4} {\left (E_{4}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{4}\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 ) - 536870912 \, b^{4} {\left (E_{4}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{4}\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 ) + b^{4} {\left (-1073741824 i \, E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 1073741824 i \, E_{4}\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^{4} {\left (536870912 i \, E_{4}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - 536870912 i \, E_{4}\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^{4} {\left (536870912 i \, E_{4}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - 536870912 i \, E_{4}\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 )}{17179869184 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(b*x+a)^2/(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/17179869184*(1073741824*b^4*(exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(4, -(I*b
*c + I*(b*x + a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) - 536870912*b^4*(exp_integral_e(4, (3*I*b*c + 3*I*(b*x + a
)*d - 3*I*a*d)/d) + exp_integral_e(4, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) - 53687
0912*b^4*(exp_integral_e(4, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) + exp_integral_e(4, -(5*I*b*c + 5*I*(b*x
+ a)*d - 5*I*a*d)/d))*cos(-5*(b*c - a*d)/d) + b^4*(-1073741824*I*exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*
a*d)/d) + 1073741824*I*exp_integral_e(4, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) + b^4*(53687
0912*I*exp_integral_e(4, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - 536870912*I*exp_integral_e(4, -(3*I*b*c +
3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d) + b^4*(536870912*I*exp_integral_e(4, (5*I*b*c + 5*I*(b*x
+ a)*d - 5*I*a*d)/d) - 536870912*I*exp_integral_e(4, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*sin(-5*(b*c -
a*d)/d))/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 + (b*x + a)^3*d^4 - a^3*d^4 + 3*(b*c*d^3 - a*d^4)*(b*x
+ a)^2 + 3*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)*(b*x + a))*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (a+b\,x\right )}^3\,{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(a + b*x)^3*sin(a + b*x)^2)/(c + d*x)^4,x)

[Out]

int((cos(a + b*x)^3*sin(a + b*x)^2)/(c + d*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*sin(b*x+a)**2/(d*x+c)**4,x)

[Out]

Integral(sin(a + b*x)**2*cos(a + b*x)**3/(c + d*x)**4, x)

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