3.91 \(\int (c+d x)^2 \cos ^2(a+b x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=184 \[ \frac {d^2 \cos (a+b x)}{4 b^3}+\frac {d^2 \cos (3 a+3 b x)}{216 b^3}-\frac {d^2 \cos (5 a+5 b x)}{1000 b^3}+\frac {d (c+d x) \sin (a+b x)}{4 b^2}+\frac {d (c+d x) \sin (3 a+3 b x)}{72 b^2}-\frac {d (c+d x) \sin (5 a+5 b x)}{200 b^2}-\frac {(c+d x)^2 \cos (a+b x)}{8 b}-\frac {(c+d x)^2 \cos (3 a+3 b x)}{48 b}+\frac {(c+d x)^2 \cos (5 a+5 b x)}{80 b} \]

[Out]

1/4*d^2*cos(b*x+a)/b^3-1/8*(d*x+c)^2*cos(b*x+a)/b+1/216*d^2*cos(3*b*x+3*a)/b^3-1/48*(d*x+c)^2*cos(3*b*x+3*a)/b
-1/1000*d^2*cos(5*b*x+5*a)/b^3+1/80*(d*x+c)^2*cos(5*b*x+5*a)/b+1/4*d*(d*x+c)*sin(b*x+a)/b^2+1/72*d*(d*x+c)*sin
(3*b*x+3*a)/b^2-1/200*d*(d*x+c)*sin(5*b*x+5*a)/b^2

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Rubi [A]  time = 0.20, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4406, 3296, 2638} \[ \frac {d (c+d x) \sin (a+b x)}{4 b^2}+\frac {d (c+d x) \sin (3 a+3 b x)}{72 b^2}-\frac {d (c+d x) \sin (5 a+5 b x)}{200 b^2}+\frac {d^2 \cos (a+b x)}{4 b^3}+\frac {d^2 \cos (3 a+3 b x)}{216 b^3}-\frac {d^2 \cos (5 a+5 b x)}{1000 b^3}-\frac {(c+d x)^2 \cos (a+b x)}{8 b}-\frac {(c+d x)^2 \cos (3 a+3 b x)}{48 b}+\frac {(c+d x)^2 \cos (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*Cos[a + b*x])/(4*b^3) - ((c + d*x)^2*Cos[a + b*x])/(8*b) + (d^2*Cos[3*a + 3*b*x])/(216*b^3) - ((c + d*x)^
2*Cos[3*a + 3*b*x])/(48*b) - (d^2*Cos[5*a + 5*b*x])/(1000*b^3) + ((c + d*x)^2*Cos[5*a + 5*b*x])/(80*b) + (d*(c
+ d*x)*Sin[a + b*x])/(4*b^2) + (d*(c + d*x)*Sin[3*a + 3*b*x])/(72*b^2) - (d*(c + d*x)*Sin[5*a + 5*b*x])/(200*
b^2)

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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 (c+d x)^2 \cos ^2(a+b x) \sin ^3(a+b x) \, dx &=\int \left (\frac {1}{8} (c+d x)^2 \sin (a+b x)+\frac {1}{16} (c+d x)^2 \sin (3 a+3 b x)-\frac {1}{16} (c+d x)^2 \sin (5 a+5 b x)\right ) \, dx\\ &=\frac {1}{16} \int (c+d x)^2 \sin (3 a+3 b x) \, dx-\frac {1}{16} \int (c+d x)^2 \sin (5 a+5 b x) \, dx+\frac {1}{8} \int (c+d x)^2 \sin (a+b x) \, dx\\ &=-\frac {(c+d x)^2 \cos (a+b x)}{8 b}-\frac {(c+d x)^2 \cos (3 a+3 b x)}{48 b}+\frac {(c+d x)^2 \cos (5 a+5 b x)}{80 b}-\frac {d \int (c+d x) \cos (5 a+5 b x) \, dx}{40 b}+\frac {d \int (c+d x) \cos (3 a+3 b x) \, dx}{24 b}+\frac {d \int (c+d x) \cos (a+b x) \, dx}{4 b}\\ &=-\frac {(c+d x)^2 \cos (a+b x)}{8 b}-\frac {(c+d x)^2 \cos (3 a+3 b x)}{48 b}+\frac {(c+d x)^2 \cos (5 a+5 b x)}{80 b}+\frac {d (c+d x) \sin (a+b x)}{4 b^2}+\frac {d (c+d x) \sin (3 a+3 b x)}{72 b^2}-\frac {d (c+d x) \sin (5 a+5 b x)}{200 b^2}+\frac {d^2 \int \sin (5 a+5 b x) \, dx}{200 b^2}-\frac {d^2 \int \sin (3 a+3 b x) \, dx}{72 b^2}-\frac {d^2 \int \sin (a+b x) \, dx}{4 b^2}\\ &=\frac {d^2 \cos (a+b x)}{4 b^3}-\frac {(c+d x)^2 \cos (a+b x)}{8 b}+\frac {d^2 \cos (3 a+3 b x)}{216 b^3}-\frac {(c+d x)^2 \cos (3 a+3 b x)}{48 b}-\frac {d^2 \cos (5 a+5 b x)}{1000 b^3}+\frac {(c+d x)^2 \cos (5 a+5 b x)}{80 b}+\frac {d (c+d x) \sin (a+b x)}{4 b^2}+\frac {d (c+d x) \sin (3 a+3 b x)}{72 b^2}-\frac {d (c+d x) \sin (5 a+5 b x)}{200 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 127, normalized size = 0.69 \[ \frac {-6750 \cos (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )-125 \cos (3 (a+b x)) \left (9 b^2 (c+d x)^2-2 d^2\right )+27 \cos (5 (a+b x)) \left (25 b^2 (c+d x)^2-2 d^2\right )+30 b d (c+d x) (450 \sin (a+b x)+25 \sin (3 (a+b x))-9 \sin (5 (a+b x)))}{54000 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6750*(-2*d^2 + b^2*(c + d*x)^2)*Cos[a + b*x] - 125*(-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*(a + b*x)] + 27*(-2*d^
2 + 25*b^2*(c + d*x)^2)*Cos[5*(a + b*x)] + 30*b*d*(c + d*x)*(450*Sin[a + b*x] + 25*Sin[3*(a + b*x)] - 9*Sin[5*
(a + b*x)]))/(54000*b^3)

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fricas [A]  time = 0.47, size = 166, normalized size = 0.90 \[ \frac {27 \, {\left (25 \, b^{2} d^{2} x^{2} + 50 \, b^{2} c d x + 25 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{5} - 5 \, {\left (225 \, b^{2} d^{2} x^{2} + 450 \, b^{2} c d x + 225 \, b^{2} c^{2} - 26 \, d^{2}\right )} \cos \left (b x + a\right )^{3} + 780 \, d^{2} \cos \left (b x + a\right ) - 30 \, {\left (9 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} - 26 \, b d^{2} x - 26 \, b c d - 13 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{3375 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3375*(27*(25*b^2*d^2*x^2 + 50*b^2*c*d*x + 25*b^2*c^2 - 2*d^2)*cos(b*x + a)^5 - 5*(225*b^2*d^2*x^2 + 450*b^2*
c*d*x + 225*b^2*c^2 - 26*d^2)*cos(b*x + a)^3 + 780*d^2*cos(b*x + a) - 30*(9*(b*d^2*x + b*c*d)*cos(b*x + a)^4 -
26*b*d^2*x - 26*b*c*d - 13*(b*d^2*x + b*c*d)*cos(b*x + a)^2)*sin(b*x + a))/b^3

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giac [A]  time = 0.25, size = 209, normalized size = 1.14 \[ \frac {{\left (25 \, b^{2} d^{2} x^{2} + 50 \, b^{2} c d x + 25 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (5 \, b x + 5 \, a\right )}{2000 \, b^{3}} - \frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (3 \, b x + 3 \, a\right )}{432 \, b^{3}} - \frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{8 \, b^{3}} - \frac {{\left (b d^{2} x + b c d\right )} \sin \left (5 \, b x + 5 \, a\right )}{200 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (3 \, b x + 3 \, a\right )}{72 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{4 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2000*(25*b^2*d^2*x^2 + 50*b^2*c*d*x + 25*b^2*c^2 - 2*d^2)*cos(5*b*x + 5*a)/b^3 - 1/432*(9*b^2*d^2*x^2 + 18*b
^2*c*d*x + 9*b^2*c^2 - 2*d^2)*cos(3*b*x + 3*a)/b^3 - 1/8*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x
+ a)/b^3 - 1/200*(b*d^2*x + b*c*d)*sin(5*b*x + 5*a)/b^3 + 1/72*(b*d^2*x + b*c*d)*sin(3*b*x + 3*a)/b^3 + 1/4*(
b*d^2*x + b*c*d)*sin(b*x + a)/b^3

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maple [B]  time = 0.02, size = 466, normalized size = 2.53 \[ \frac {\frac {d^{2} \left (-\frac {\left (b x +a \right )^{2} \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {4 \cos \left (b x +a \right )}{15}+\frac {4 \left (b x +a \right ) \sin \left (b x +a \right )}{15}+\frac {2 \left (b x +a \right ) \left (\sin ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{135}+\frac {\left (b x +a \right )^{2} \left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{5}-\frac {2 \left (b x +a \right ) \left (\sin ^{5}\left (b x +a \right )\right )}{25}-\frac {2 \left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{125}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (-\frac {\left (b x +a \right ) \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \sin \left (b x +a \right )}{15}+\frac {\left (b x +a \right ) \left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{5}-\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{25}\right )}{b^{2}}+\frac {2 c d \left (-\frac {\left (b x +a \right ) \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \sin \left (b x +a \right )}{15}+\frac {\left (b x +a \right ) \left (\frac {8}{3}+\sin ^{4}\left (b x +a \right )+\frac {4 \left (\sin ^{2}\left (b x +a \right )\right )}{3}\right ) \cos \left (b x +a \right )}{5}-\frac {\left (\sin ^{5}\left (b x +a \right )\right )}{25}\right )}{b}+\frac {a^{2} d^{2} \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (b x +a \right )\right )}{15}\right )}{b^{2}}-\frac {2 a c d \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (b x +a \right )\right )}{15}\right )}{b}+c^{2} \left (-\frac {\left (\sin ^{2}\left (b x +a \right )\right ) \left (\cos ^{3}\left (b x +a \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (b x +a \right )\right )}{15}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b^2*d^2*(-1/3*(b*x+a)^2*(2+sin(b*x+a)^2)*cos(b*x+a)+4/15*cos(b*x+a)+4/15*(b*x+a)*sin(b*x+a)+2/45*(b*x+a
)*sin(b*x+a)^3+2/135*(2+sin(b*x+a)^2)*cos(b*x+a)+1/5*(b*x+a)^2*(8/3+sin(b*x+a)^4+4/3*sin(b*x+a)^2)*cos(b*x+a)-
2/25*(b*x+a)*sin(b*x+a)^5-2/125*(8/3+sin(b*x+a)^4+4/3*sin(b*x+a)^2)*cos(b*x+a))-2/b^2*a*d^2*(-1/3*(b*x+a)*(2+s
in(b*x+a)^2)*cos(b*x+a)+1/45*sin(b*x+a)^3+2/15*sin(b*x+a)+1/5*(b*x+a)*(8/3+sin(b*x+a)^4+4/3*sin(b*x+a)^2)*cos(
b*x+a)-1/25*sin(b*x+a)^5)+2/b*c*d*(-1/3*(b*x+a)*(2+sin(b*x+a)^2)*cos(b*x+a)+1/45*sin(b*x+a)^3+2/15*sin(b*x+a)+
1/5*(b*x+a)*(8/3+sin(b*x+a)^4+4/3*sin(b*x+a)^2)*cos(b*x+a)-1/25*sin(b*x+a)^5)+1/b^2*a^2*d^2*(-1/5*sin(b*x+a)^2
*cos(b*x+a)^3-2/15*cos(b*x+a)^3)-2/b*a*c*d*(-1/5*sin(b*x+a)^2*cos(b*x+a)^3-2/15*cos(b*x+a)^3)+c^2*(-1/5*sin(b*
x+a)^2*cos(b*x+a)^3-2/15*cos(b*x+a)^3))

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maxima [B]  time = 0.35, size = 375, normalized size = 2.04 \[ \frac {3600 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} c^{2} - \frac {7200 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} a c d}{b} + \frac {3600 \, {\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )} a^{2} d^{2}}{b^{2}} + \frac {30 \, {\left (45 \, {\left (b x + a\right )} \cos \left (5 \, b x + 5 \, a\right ) - 75 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - 9 \, \sin \left (5 \, b x + 5 \, a\right ) + 25 \, \sin \left (3 \, b x + 3 \, a\right ) + 450 \, \sin \left (b x + a\right )\right )} c d}{b} - \frac {30 \, {\left (45 \, {\left (b x + a\right )} \cos \left (5 \, b x + 5 \, a\right ) - 75 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - 9 \, \sin \left (5 \, b x + 5 \, a\right ) + 25 \, \sin \left (3 \, b x + 3 \, a\right ) + 450 \, \sin \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left (27 \, {\left (25 \, {\left (b x + a\right )}^{2} - 2\right )} \cos \left (5 \, b x + 5 \, a\right ) - 125 \, {\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \cos \left (3 \, b x + 3 \, a\right ) - 6750 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 270 \, {\left (b x + a\right )} \sin \left (5 \, b x + 5 \, a\right ) + 750 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 13500 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{54000 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54000*(3600*(3*cos(b*x + a)^5 - 5*cos(b*x + a)^3)*c^2 - 7200*(3*cos(b*x + a)^5 - 5*cos(b*x + a)^3)*a*c*d/b +
3600*(3*cos(b*x + a)^5 - 5*cos(b*x + a)^3)*a^2*d^2/b^2 + 30*(45*(b*x + a)*cos(5*b*x + 5*a) - 75*(b*x + a)*cos
(3*b*x + 3*a) - 450*(b*x + a)*cos(b*x + a) - 9*sin(5*b*x + 5*a) + 25*sin(3*b*x + 3*a) + 450*sin(b*x + a))*c*d/
b - 30*(45*(b*x + a)*cos(5*b*x + 5*a) - 75*(b*x + a)*cos(3*b*x + 3*a) - 450*(b*x + a)*cos(b*x + a) - 9*sin(5*b
*x + 5*a) + 25*sin(3*b*x + 3*a) + 450*sin(b*x + a))*a*d^2/b^2 + (27*(25*(b*x + a)^2 - 2)*cos(5*b*x + 5*a) - 12
5*(9*(b*x + a)^2 - 2)*cos(3*b*x + 3*a) - 6750*((b*x + a)^2 - 2)*cos(b*x + a) - 270*(b*x + a)*sin(5*b*x + 5*a)
+ 750*(b*x + a)*sin(3*b*x + 3*a) + 13500*(b*x + a)*sin(b*x + a))*d^2/b^2)/b

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mupad [B]  time = 0.81, size = 249, normalized size = 1.35 \[ \frac {780\,d^2\,\cos \left (a+b\,x\right )+130\,d^2\,{\cos \left (a+b\,x\right )}^3-54\,d^2\,{\cos \left (a+b\,x\right )}^5-1125\,b^2\,c^2\,{\cos \left (a+b\,x\right )}^3+675\,b^2\,c^2\,{\cos \left (a+b\,x\right )}^5+780\,b\,d^2\,x\,\sin \left (a+b\,x\right )-1125\,b^2\,d^2\,x^2\,{\cos \left (a+b\,x\right )}^3+675\,b^2\,d^2\,x^2\,{\cos \left (a+b\,x\right )}^5+780\,b\,c\,d\,\sin \left (a+b\,x\right )-2250\,b^2\,c\,d\,x\,{\cos \left (a+b\,x\right )}^3+1350\,b^2\,c\,d\,x\,{\cos \left (a+b\,x\right )}^5+390\,b\,d^2\,x\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )-270\,b\,d^2\,x\,{\cos \left (a+b\,x\right )}^4\,\sin \left (a+b\,x\right )+390\,b\,c\,d\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )-270\,b\,c\,d\,{\cos \left (a+b\,x\right )}^4\,\sin \left (a+b\,x\right )}{3375\,b^3} \]

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]

(780*d^2*cos(a + b*x) + 130*d^2*cos(a + b*x)^3 - 54*d^2*cos(a + b*x)^5 - 1125*b^2*c^2*cos(a + b*x)^3 + 675*b^2
*c^2*cos(a + b*x)^5 + 780*b*d^2*x*sin(a + b*x) - 1125*b^2*d^2*x^2*cos(a + b*x)^3 + 675*b^2*d^2*x^2*cos(a + b*x
)^5 + 780*b*c*d*sin(a + b*x) - 2250*b^2*c*d*x*cos(a + b*x)^3 + 1350*b^2*c*d*x*cos(a + b*x)^5 + 390*b*d^2*x*cos
(a + b*x)^2*sin(a + b*x) - 270*b*d^2*x*cos(a + b*x)^4*sin(a + b*x) + 390*b*c*d*cos(a + b*x)^2*sin(a + b*x) - 2
70*b*c*d*cos(a + b*x)^4*sin(a + b*x))/(3375*b^3)

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sympy [A]  time = 5.96, size = 382, normalized size = 2.08 \[ \begin {cases} - \frac {c^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 c^{2} \cos ^{5}{\left (a + b x \right )}}{15 b} - \frac {2 c d x \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {4 c d x \cos ^{5}{\left (a + b x \right )}}{15 b} - \frac {d^{2} x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 d^{2} x^{2} \cos ^{5}{\left (a + b x \right )}}{15 b} + \frac {52 c d \sin ^{5}{\left (a + b x \right )}}{225 b^{2}} + \frac {26 c d \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {4 c d \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{15 b^{2}} + \frac {52 d^{2} x \sin ^{5}{\left (a + b x \right )}}{225 b^{2}} + \frac {26 d^{2} x \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {4 d^{2} x \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{15 b^{2}} + \frac {52 d^{2} \sin ^{4}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{225 b^{3}} + \frac {338 d^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{675 b^{3}} + \frac {856 d^{2} \cos ^{5}{\left (a + b x \right )}}{3375 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin ^{3}{\relax (a )} \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-c**2*sin(a + b*x)**2*cos(a + b*x)**3/(3*b) - 2*c**2*cos(a + b*x)**5/(15*b) - 2*c*d*x*sin(a + b*x)*
*2*cos(a + b*x)**3/(3*b) - 4*c*d*x*cos(a + b*x)**5/(15*b) - d**2*x**2*sin(a + b*x)**2*cos(a + b*x)**3/(3*b) -
2*d**2*x**2*cos(a + b*x)**5/(15*b) + 52*c*d*sin(a + b*x)**5/(225*b**2) + 26*c*d*sin(a + b*x)**3*cos(a + b*x)**
2/(45*b**2) + 4*c*d*sin(a + b*x)*cos(a + b*x)**4/(15*b**2) + 52*d**2*x*sin(a + b*x)**5/(225*b**2) + 26*d**2*x*
sin(a + b*x)**3*cos(a + b*x)**2/(45*b**2) + 4*d**2*x*sin(a + b*x)*cos(a + b*x)**4/(15*b**2) + 52*d**2*sin(a +
b*x)**4*cos(a + b*x)/(225*b**3) + 338*d**2*sin(a + b*x)**2*cos(a + b*x)**3/(675*b**3) + 856*d**2*cos(a + b*x)*
*5/(3375*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)**3*cos(a)**2, True))

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