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

Optimal. Leaf size=287 \[ -\frac {b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3} \]

[Out]

-4/3*b^3*Ci(4*b*c/d+4*b*x)*cos(4*a-4*b*c/d)/d^4-1/3*b^3*Ci(2*b*c/d+2*b*x)*cos(2*a-2*b*c/d)/d^4-1/12*b*cos(2*b*
x+2*a)/d^2/(d*x+c)^2-1/12*b*cos(4*b*x+4*a)/d^2/(d*x+c)^2+4/3*b^3*Si(4*b*c/d+4*b*x)*sin(4*a-4*b*c/d)/d^4+1/3*b^
3*Si(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^4-1/12*sin(2*b*x+2*a)/d/(d*x+c)^3+1/6*b^2*sin(2*b*x+2*a)/d^3/(d*x+c)-1/
24*sin(4*b*x+4*a)/d/(d*x+c)^3+1/3*b^2*sin(4*b*x+4*a)/d^3/(d*x+c)

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Rubi [A]  time = 0.45, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4406, 3297, 3303, 3299, 3302} \[ -\frac {b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[2*a + 2*b*x])/(12*d^2*(c + d*x)^2) - (b*Cos[4*a + 4*b*x])/(12*d^2*(c + d*x)^2) - (b^3*Cos[2*a - (2*b*c
)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/(3*d^4) - (4*b^3*Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*c)/d + 4*b*x])/(3*
d^4) - Sin[2*a + 2*b*x]/(12*d*(c + d*x)^3) + (b^2*Sin[2*a + 2*b*x])/(6*d^3*(c + d*x)) - Sin[4*a + 4*b*x]/(24*d
*(c + d*x)^3) + (b^2*Sin[4*a + 4*b*x])/(3*d^3*(c + d*x)) + (b^3*Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2
*b*x])/(3*d^4) + (4*b^3*Sin[4*a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/(3*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 (a+b x)}{(c+d x)^4} \, dx &=\int \left (\frac {\sin (2 a+2 b x)}{4 (c+d x)^4}+\frac {\sin (4 a+4 b x)}{8 (c+d x)^4}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sin (4 a+4 b x)}{(c+d x)^4} \, dx+\frac {1}{4} \int \frac {\sin (2 a+2 b x)}{(c+d x)^4} \, dx\\ &=-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b \int \frac {\cos (2 a+2 b x)}{(c+d x)^3} \, dx}{6 d}+\frac {b \int \frac {\cos (4 a+4 b x)}{(c+d x)^3} \, dx}{6 d}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}-\frac {b^2 \int \frac {\sin (2 a+2 b x)}{(c+d x)^2} \, dx}{6 d^2}-\frac {b^2 \int \frac {\sin (4 a+4 b x)}{(c+d x)^2} \, dx}{3 d^2}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {b^3 \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{3 d^3}-\frac {\left (4 b^3\right ) \int \frac {\cos (4 a+4 b x)}{c+d x} \, dx}{3 d^3}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}-\frac {\left (4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx}{3 d^3}-\frac {\left (b^3 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{3 d^3}+\frac {\left (4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx}{3 d^3}+\frac {\left (b^3 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{3 d^3}\\ &=-\frac {b \cos (2 a+2 b x)}{12 d^2 (c+d x)^2}-\frac {b \cos (4 a+4 b x)}{12 d^2 (c+d x)^2}-\frac {b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}-\frac {4 b^3 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}-\frac {\sin (2 a+2 b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (2 a+2 b x)}{6 d^3 (c+d x)}-\frac {\sin (4 a+4 b x)}{24 d (c+d x)^3}+\frac {b^2 \sin (4 a+4 b x)}{3 d^3 (c+d x)}+\frac {b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{3 d^4}+\frac {4 b^3 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{3 d^4}\\ \end {align*}

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Mathematica [A]  time = 2.49, size = 316, normalized size = 1.10 \[ -\frac {8 b^3 (c+d x)^3 \left (\cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )-\sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )\right )+32 b^3 (c+d x)^3 \left (\cos \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b (c+d x)}{d}\right )-\sin \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )\right )+2 d \cos (2 b x) \left (\sin (2 a) \left (d^2-2 b^2 (c+d x)^2\right )+b d \cos (2 a) (c+d x)\right )+d \cos (4 b x) \left (\sin (4 a) \left (d^2-8 b^2 (c+d x)^2\right )+2 b d \cos (4 a) (c+d x)\right )-2 d \sin (2 b x) \left (\cos (2 a) \left (2 b^2 (c+d x)^2-d^2\right )+b d \sin (2 a) (c+d x)\right )-d \sin (4 b x) \left (\cos (4 a) \left (8 b^2 (c+d x)^2-d^2\right )+2 b d \sin (4 a) (c+d x)\right )}{24 d^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(2*d*Cos[2*b*x]*(b*d*(c + d*x)*Cos[2*a] + (d^2 - 2*b^2*(c + d*x)^2)*Sin[2*a]) + d*Cos[4*b*x]*(2*b*d*(c +
d*x)*Cos[4*a] + (d^2 - 8*b^2*(c + d*x)^2)*Sin[4*a]) - 2*d*((-d^2 + 2*b^2*(c + d*x)^2)*Cos[2*a] + b*d*(c + d*x
)*Sin[2*a])*Sin[2*b*x] - d*((-d^2 + 8*b^2*(c + d*x)^2)*Cos[4*a] + 2*b*d*(c + d*x)*Sin[4*a])*Sin[4*b*x] + 8*b^3
*(c + d*x)^3*(Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d*x))/d] - Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c +
d*x))/d]) + 32*b^3*(c + d*x)^3*(Cos[4*a - (4*b*c)/d]*CosIntegral[(4*b*(c + d*x))/d] - Sin[4*a - (4*b*c)/d]*Si
nIntegral[(4*b*(c + d*x))/d]))/(d^4*(c + d*x)^3)

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fricas [B]  time = 0.70, size = 568, normalized size = 1.98 \[ -\frac {4 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{4} - 3 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2} - 8 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {4 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left ({\left (8 \, b^{2} d^{3} x^{2} + 16 \, b^{2} c d^{2} x + 8 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{3} - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{6 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(4*(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 - 8*(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)*sin(-4*(b*c - a*d)/d)*sin_integral(4*(b*d*x + b*c)/d) - 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)*sin(-2*(b*c - a*d)/d)*sin_integral(2*(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(2*(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(-2*(b*d*x + b*c)/d))*cos(-2*(b*c - a*d)/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_integral(4*(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(-4*(b*d*x + b*c)/d))*cos(-4*(b*c - a*d)/d) - 2*((8*b^2*d^3*x^2 + 16*b^2*c*d^2*x
+ 8*b^2*c^2*d - d^3)*cos(b*x + a)^3 - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cos(b*x + a))*sin(b*x + a))
/(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)/(d*x+c)^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 404, normalized size = 1.41 \[ \frac {\frac {b^{4} \left (-\frac {4 \sin \left (4 b x +4 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {8 \cos \left (4 b x +4 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {8 \left (-\frac {4 \sin \left (4 b x +4 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {16 \Si \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \sin \left (\frac {-4 d a +4 c b}{d}\right )}{d}+\frac {16 \Ci \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \cos \left (\frac {-4 d a +4 c b}{d}\right )}{d}}{d}\right )}{3 d}}{d}\right )}{32}+\frac {b^{4} \left (-\frac {2 \sin \left (2 b x +2 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {2 \cos \left (2 b x +2 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {2 \left (-\frac {2 \sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {4 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}+\frac {4 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}}{d}\right )}{3 d}}{d}\right )}{8}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/32*b^4*(-4/3*sin(4*b*x+4*a)/((b*x+a)*d-d*a+c*b)^3/d+4/3*(-2*cos(4*b*x+4*a)/((b*x+a)*d-d*a+c*b)^2/d-2*(-
4*sin(4*b*x+4*a)/((b*x+a)*d-d*a+c*b)/d+4*(4*Si(4*b*x+4*a+4*(-a*d+b*c)/d)*sin(4*(-a*d+b*c)/d)/d+4*Ci(4*b*x+4*a+
4*(-a*d+b*c)/d)*cos(4*(-a*d+b*c)/d)/d)/d)/d)/d)+1/8*b^4*(-2/3*sin(2*b*x+2*a)/((b*x+a)*d-d*a+c*b)^3/d+2/3*(-cos
(2*b*x+2*a)/((b*x+a)*d-d*a+c*b)^2/d-(-2*sin(2*b*x+2*a)/((b*x+a)*d-d*a+c*b)/d+2*(2*Si(2*b*x+2*a+2*(-a*d+b*c)/d)
*sin(2*(-a*d+b*c)/d)/d+2*Ci(2*b*x+2*a+2*(-a*d+b*c)/d)*cos(2*(-a*d+b*c)/d)/d)/d)/d)/d))

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maxima [C]  time = 1.90, size = 386, normalized size = 1.34 \[ -\frac {b^{4} {\left (2 i \, E_{4}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 2 i \, E_{4}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (i \, E_{4}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{4}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{4} {\left (E_{4}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{4}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (E_{4}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{4}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b^4*(2*I*exp_integral_e(4, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) - 2*I*exp_integral_e(4, -(2*I*b*c +
2*I*(b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) + b^4*(I*exp_integral_e(4, (4*I*b*c + 4*I*(b*x + a)*d -
4*I*a*d)/d) - I*exp_integral_e(4, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*cos(-4*(b*c - a*d)/d) + 2*b^4*(ex
p_integral_e(4, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(4, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I
*a*d)/d))*sin(-2*(b*c - a*d)/d) + b^4*(exp_integral_e(4, (4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d) + exp_integr
al_e(4, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*sin(-4*(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 \[ \int \frac {{\cos \left (a+b\,x\right )}^3\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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