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

Optimal. Leaf size=270 \[ -\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{24 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^4}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{24 d^4}+\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^4}+\frac {b^2 \sin (a+b x)}{24 d^3 (c+d x)}+\frac {3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac {b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{12 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3} \]

[Out]

-9/8*b^3*Ci(3*b*c/d+3*b*x)*cos(3*a-3*b*c/d)/d^4-1/24*b^3*Ci(b*c/d+b*x)*cos(a-b*c/d)/d^4-1/24*b*cos(b*x+a)/d^2/
(d*x+c)^2-1/8*b*cos(3*b*x+3*a)/d^2/(d*x+c)^2+9/8*b^3*Si(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^4+1/24*b^3*Si(b*c/d+
b*x)*sin(a-b*c/d)/d^4-1/12*sin(b*x+a)/d/(d*x+c)^3+1/24*b^2*sin(b*x+a)/d^3/(d*x+c)-1/12*sin(3*b*x+3*a)/d/(d*x+c
)^3+3/8*b^2*sin(3*b*x+3*a)/d^3/(d*x+c)

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Rubi [A]  time = 0.38, antiderivative size = 270, 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} \[ -\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{24 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^4}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{24 d^4}+\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^4}+\frac {b^2 \sin (a+b x)}{24 d^3 (c+d x)}+\frac {3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac {b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{12 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[a + b*x])/(24*d^2*(c + d*x)^2) - (b*Cos[3*a + 3*b*x])/(8*d^2*(c + d*x)^2) - (b^3*Cos[a - (b*c)/d]*CosI
ntegral[(b*c)/d + b*x])/(24*d^4) - (9*b^3*Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*b*x])/(8*d^4) - Sin[a
+ b*x]/(12*d*(c + d*x)^3) + (b^2*Sin[a + b*x])/(24*d^3*(c + d*x)) - Sin[3*a + 3*b*x]/(12*d*(c + d*x)^3) + (3*
b^2*Sin[3*a + 3*b*x])/(8*d^3*(c + d*x)) + (b^3*Sin[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(24*d^4) + (9*b^3*
Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(8*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 ^2(a+b x) \sin (a+b x)}{(c+d x)^4} \, dx &=\int \left (\frac {\sin (a+b x)}{4 (c+d x)^4}+\frac {\sin (3 a+3 b x)}{4 (c+d x)^4}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\sin (a+b x)}{(c+d x)^4} \, dx+\frac {1}{4} \int \frac {\sin (3 a+3 b x)}{(c+d x)^4} \, dx\\ &=-\frac {\sin (a+b x)}{12 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac {b \int \frac {\cos (a+b x)}{(c+d x)^3} \, dx}{12 d}+\frac {b \int \frac {\cos (3 a+3 b x)}{(c+d x)^3} \, dx}{4 d}\\ &=-\frac {b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{12 d (c+d x)^3}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3}-\frac {b^2 \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx}{24 d^2}-\frac {\left (3 b^2\right ) \int \frac {\sin (3 a+3 b x)}{(c+d x)^2} \, dx}{8 d^2}\\ &=-\frac {b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (a+b x)}{24 d^3 (c+d x)}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac {3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac {b^3 \int \frac {\cos (a+b x)}{c+d x} \, dx}{24 d^3}-\frac {\left (9 b^3\right ) \int \frac {\cos (3 a+3 b x)}{c+d x} \, dx}{8 d^3}\\ &=-\frac {b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac {\sin (a+b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (a+b x)}{24 d^3 (c+d x)}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac {3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}-\frac {\left (9 b^3 \cos \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}{8 d^3}-\frac {\left (b^3 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{24 d^3}+\frac {\left (9 b^3 \sin \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}{8 d^3}+\frac {\left (b^3 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{24 d^3}\\ &=-\frac {b \cos (a+b x)}{24 d^2 (c+d x)^2}-\frac {b \cos (3 a+3 b x)}{8 d^2 (c+d x)^2}-\frac {b^3 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{24 d^4}-\frac {9 b^3 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^4}-\frac {\sin (a+b x)}{12 d (c+d x)^3}+\frac {b^2 \sin (a+b x)}{24 d^3 (c+d x)}-\frac {\sin (3 a+3 b x)}{12 d (c+d x)^3}+\frac {3 b^2 \sin (3 a+3 b x)}{8 d^3 (c+d x)}+\frac {b^3 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{24 d^4}+\frac {9 b^3 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^4}\\ \end {align*}

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Mathematica [A]  time = 1.82, size = 300, normalized size = 1.11 \[ -\frac {b^3 (c+d x)^3 \left (\cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )-\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )\right )+27 b^3 (c+d x)^3 \left (\cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )-\sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )\right )+d \cos (b x) \left (b d \cos (a) (c+d x)-\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )\right )+d \cos (3 b x) \left (3 b d \cos (3 a) (c+d x)-\sin (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )\right )-d \sin (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )+b d \sin (a) (c+d x)\right )-d \sin (3 b x) \left (\cos (3 a) \left (9 b^2 (c+d x)^2-2 d^2\right )+3 b d \sin (3 a) (c+d x)\right )}{24 d^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(d*Cos[b*x]*(b*d*(c + d*x)*Cos[a] - (-2*d^2 + b^2*(c + d*x)^2)*Sin[a]) + d*Cos[3*b*x]*(3*b*d*(c + d*x)*C
os[3*a] - (-2*d^2 + 9*b^2*(c + d*x)^2)*Sin[3*a]) - d*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] + b*d*(c + d*x)*Sin[a]
)*Sin[b*x] - d*((-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*a] + 3*b*d*(c + d*x)*Sin[3*a])*Sin[3*b*x] + b^3*(c + d*x)^3
*(Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)] - Sin[a - (b*c)/d]*SinIntegral[b*(c/d + x)]) + 27*b^3*(c + d*x)^3*
(Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*(c + d*x))/d] - Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d]))/(
d^4*(c + d*x)^3)

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fricas [B]  time = 0.67, size = 558, normalized size = 2.07 \[ -\frac {24 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{3} - 54 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - 16 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) + {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + 27 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 8 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - {\left (9 \, b^{2} d^{3} x^{2} + 18 \, b^{2} c d^{2} x + 9 \, b^{2} c^{2} d - 2 \, d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{48 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(24*(b*d^3*x + b*c*d^2)*cos(b*x + a)^3 - 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)*si
n(-3*(b*c - a*d)/d)*sin_integral(3*(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(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) - 16*(b*d^3*x + b*c*d^2)*cos(b*x + a) + ((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))*cos(-(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))*cos(-3*(b*c - a*d)/d) + 8*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*
c^2*d - (9*b^2*d^3*x^2 + 18*b^2*c*d^2*x + 9*b^2*c^2*d - 2*d^3)*cos(b*x + a)^2)*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)^2*sin(b*x+a)/(d*x+c)^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 381, normalized size = 1.41 \[ \frac {\frac {b^{4} \left (-\frac {\sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {9 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}+\frac {9 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}}{d}\right )}{2 d}}{d}\right )}{12}+\frac {b^{4} \left (-\frac {\sin \left (b x +a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}+\frac {-\frac {\cos \left (b x +a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}+\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}}{3 d}\right )}{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.88, size = 385, normalized size = 1.43 \[ -\frac {b^{4} {\left (i \, E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{4}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{4} {\left (i \, E_{4}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - i \, E_{4}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{4} {\left (E_{4}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{4}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + b^{4} {\left (E_{4}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{4}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} + {\left (b x + a\right )}^{3} d^{4} - a^{3} d^{4} + 3 \, {\left (b c d^{3} - a d^{4}\right )} {\left (b x + a\right )}^{2} + 3 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^4*(I*exp_integral_e(4, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_integral_e(4, -(I*b*c + I*(b*x + a)*
d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^4*(I*exp_integral_e(4, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - I*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) + 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))*sin(-(b*c - a*d)/d) + 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))*sin(-3*(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 \[ \int \frac {{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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