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

Optimal. Leaf size=257 \[ \frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{8 d^2}+\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{16 d^2}-\frac {5 b \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b c}{d}+5 b x\right )}{16 d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^2}-\frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{16 d^2}+\frac {5 b \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{16 d^2}-\frac {\sin (a+b x)}{8 d (c+d x)}-\frac {\sin (3 a+3 b x)}{16 d (c+d x)}+\frac {\sin (5 a+5 b x)}{16 d (c+d x)} \]

[Out]

-5/16*b*Ci(5*b*c/d+5*b*x)*cos(5*a-5*b*c/d)/d^2+3/16*b*Ci(3*b*c/d+3*b*x)*cos(3*a-3*b*c/d)/d^2+1/8*b*Ci(b*c/d+b*
x)*cos(a-b*c/d)/d^2+5/16*b*Si(5*b*c/d+5*b*x)*sin(5*a-5*b*c/d)/d^2-3/16*b*Si(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^
2-1/8*b*Si(b*c/d+b*x)*sin(a-b*c/d)/d^2-1/8*sin(b*x+a)/d/(d*x+c)-1/16*sin(3*b*x+3*a)/d/(d*x+c)+1/16*sin(5*b*x+5
*a)/d/(d*x+c)

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Rubi [A]  time = 0.42, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ \frac {b \cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{8 d^2}+\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{16 d^2}-\frac {5 b \cos \left (5 a-\frac {5 b c}{d}\right ) \text {CosIntegral}\left (\frac {5 b c}{d}+5 b x\right )}{16 d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^2}-\frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{16 d^2}+\frac {5 b \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b c}{d}+5 b x\right )}{16 d^2}-\frac {\sin (a+b x)}{8 d (c+d x)}-\frac {\sin (3 a+3 b x)}{16 d (c+d x)}+\frac {\sin (5 a+5 b x)}{16 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(8*d^2) + (3*b*Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*
b*x])/(16*d^2) - (5*b*Cos[5*a - (5*b*c)/d]*CosIntegral[(5*b*c)/d + 5*b*x])/(16*d^2) - Sin[a + b*x]/(8*d*(c + d
*x)) - Sin[3*a + 3*b*x]/(16*d*(c + d*x)) + Sin[5*a + 5*b*x]/(16*d*(c + d*x)) - (b*Sin[a - (b*c)/d]*SinIntegral
[(b*c)/d + b*x])/(8*d^2) - (3*b*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(16*d^2) + (5*b*Sin[5*a -
(5*b*c)/d]*SinIntegral[(5*b*c)/d + 5*b*x])/(16*d^2)

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

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Mathematica [A]  time = 1.47, size = 213, normalized size = 0.83 \[ \frac {2 b \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )-5 b \cos \left (5 a-\frac {5 b c}{d}\right ) \text {Ci}\left (\frac {5 b (c+d x)}{d}\right )-2 b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )-3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )+5 b \sin \left (5 a-\frac {5 b c}{d}\right ) \text {Si}\left (\frac {5 b (c+d x)}{d}\right )-\frac {2 d \sin (a+b x)}{c+d x}-\frac {d \sin (3 (a+b x))}{c+d x}+\frac {d \sin (5 (a+b x))}{c+d x}}{16 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.91, size = 347, normalized size = 1.35 \[ \frac {10 \, {\left (b d x + b c\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {5 \, {\left (b d x + b c\right )}}{d}\right ) - 6 \, {\left (b d x + b c\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - 4 \, {\left (b d x + b c\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 2 \, {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + 3 \, {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - 5 \, {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {5 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {5 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) + 32 \, {\left (d \cos \left (b x + a\right )^{4} - d \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{32 \, {\left (d^{3} x + c d^{2}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

1/32*(10*(b*d*x + b*c)*sin(-5*(b*c - a*d)/d)*sin_integral(5*(b*d*x + b*c)/d) - 6*(b*d*x + b*c)*sin(-3*(b*c - a
*d)/d)*sin_integral(3*(b*d*x + b*c)/d) - 4*(b*d*x + b*c)*sin(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) + 2
*((b*d*x + b*c)*cos_integral((b*d*x + b*c)/d) + (b*d*x + b*c)*cos_integral(-(b*d*x + b*c)/d))*cos(-(b*c - a*d)
/d) + 3*((b*d*x + b*c)*cos_integral(3*(b*d*x + b*c)/d) + (b*d*x + b*c)*cos_integral(-3*(b*d*x + b*c)/d))*cos(-
3*(b*c - a*d)/d) - 5*((b*d*x + b*c)*cos_integral(5*(b*d*x + b*c)/d) + (b*d*x + b*c)*cos_integral(-5*(b*d*x + b
*c)/d))*cos(-5*(b*c - a*d)/d) + 32*(d*cos(b*x + a)^4 - d*cos(b*x + a)^2)*sin(b*x + a))/(d^3*x + c*d^2)

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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)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 365, normalized size = 1.42 \[ \frac {-\frac {b^{2} \left (-\frac {5 \sin \left (5 b x +5 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {25 \Si \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \sin \left (\frac {-5 d a +5 c b}{d}\right )}{d}+\frac {25 \Ci \left (5 b x +5 a +\frac {-5 d a +5 c b}{d}\right ) \cos \left (\frac {-5 d a +5 c b}{d}\right )}{d}}{d}\right )}{80}+\frac {b^{2} \left (-\frac {\sin \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}+\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}}{d}\right )}{8}+\frac {b^{2} \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {9 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}+\frac {9 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}}{d}\right )}{48}}{b} \]

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)^2,x)

[Out]

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

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maxima [C]  time = 0.58, size = 438, normalized size = 1.70 \[ \frac {b^{2} {\left (-2 i \, E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 2 i \, E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{2} {\left (-i \, E_{2}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + i \, E_{2}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (i \, E_{2}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) - i \, E_{2}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \cos \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, b^{2} {\left (E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) - b^{2} {\left (E_{2}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{2}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right ) + E_{2}\left (-\frac {5 i \, b c + 5 i \, {\left (b x + a\right )} d - 5 i \, a d}{d}\right )\right )} \sin \left (-\frac {5 \, {\left (b c - a d\right )}}{d}\right )}{32 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

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)^2,x, algorithm="maxima")

[Out]

1/32*(b^2*(-2*I*exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 2*I*exp_integral_e(2, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^2*(-I*exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) +
I*exp_integral_e(2, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b^2*(I*exp_integral_e(
2, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) - I*exp_integral_e(2, -(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^2*(exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(2, -(I*b*c
+ I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d) - b^2*(exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*
d)/d) + exp_integral_e(2, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d) + b^2*(exp_integral
_e(2, (5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d) + exp_integral_e(2, -(5*I*b*c + 5*I*(b*x + a)*d - 5*I*a*d)/d))*
sin(-5*(b*c - a*d)/d))/((b*c*d + (b*x + a)*d^2 - a*d^2)*b)

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

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)^2,x)

[Out]

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

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

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)**2,x)

[Out]

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

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