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

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

[Out]

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

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Rubi [A]  time = 0.19, 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, 2637} \[ \frac {d (c+d x) \cos (a+b x)}{4 b^2}-\frac {d (c+d x) \cos (3 a+3 b x)}{72 b^2}-\frac {d (c+d x) \cos (5 a+5 b x)}{200 b^2}-\frac {d^2 \sin (a+b x)}{4 b^3}+\frac {d^2 \sin (3 a+3 b x)}{216 b^3}+\frac {d^2 \sin (5 a+5 b x)}{1000 b^3}+\frac {(c+d x)^2 \sin (a+b x)}{8 b}-\frac {(c+d x)^2 \sin (3 a+3 b x)}{48 b}-\frac {(c+d x)^2 \sin (5 a+5 b x)}{80 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2637

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

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Mathematica [A]  time = 0.96, size = 252, normalized size = 1.37 \[ -\frac {-6750 b^2 c^2 \sin (a+b x)+1125 b^2 c^2 \sin (3 (a+b x))+675 b^2 c^2 \sin (5 (a+b x))-13500 b^2 c d x \sin (a+b x)+2250 b^2 c d x \sin (3 (a+b x))+1350 b^2 c d x \sin (5 (a+b x))-6750 b^2 d^2 x^2 \sin (a+b x)+1125 b^2 d^2 x^2 \sin (3 (a+b x))+675 b^2 d^2 x^2 \sin (5 (a+b x))-13500 b d (c+d x) \cos (a+b x)+750 b d (c+d x) \cos (3 (a+b x))+270 b c d \cos (5 (a+b x))+13500 d^2 \sin (a+b x)-250 d^2 \sin (3 (a+b x))-54 d^2 \sin (5 (a+b x))+270 b d^2 x \cos (5 (a+b x))}{54000 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/54000*(-13500*b*d*(c + d*x)*Cos[a + b*x] + 750*b*d*(c + d*x)*Cos[3*(a + b*x)] + 270*b*c*d*Cos[5*(a + b*x)]
+ 270*b*d^2*x*Cos[5*(a + b*x)] - 6750*b^2*c^2*Sin[a + b*x] + 13500*d^2*Sin[a + b*x] - 13500*b^2*c*d*x*Sin[a +
b*x] - 6750*b^2*d^2*x^2*Sin[a + b*x] + 1125*b^2*c^2*Sin[3*(a + b*x)] - 250*d^2*Sin[3*(a + b*x)] + 2250*b^2*c*d
*x*Sin[3*(a + b*x)] + 1125*b^2*d^2*x^2*Sin[3*(a + b*x)] + 675*b^2*c^2*Sin[5*(a + b*x)] - 54*d^2*Sin[5*(a + b*x
)] + 1350*b^2*c*d*x*Sin[5*(a + b*x)] + 675*b^2*d^2*x^2*Sin[5*(a + b*x)])/b^3

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fricas [A]  time = 0.51, size = 193, normalized size = 1.05 \[ -\frac {270 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{5} - 150 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{3} - 900 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) - {\left (450 \, b^{2} d^{2} x^{2} + 900 \, b^{2} c d x - 27 \, {\left (25 \, b^{2} d^{2} x^{2} + 50 \, b^{2} c d x + 25 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{4} + 450 \, b^{2} c^{2} + {\left (225 \, b^{2} d^{2} x^{2} + 450 \, b^{2} c d x + 225 \, b^{2} c^{2} + 22 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - 856 \, d^{2}\right )} \sin \left (b x + a\right )}{3375 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3375*(270*(b*d^2*x + b*c*d)*cos(b*x + a)^5 - 150*(b*d^2*x + b*c*d)*cos(b*x + a)^3 - 900*(b*d^2*x + b*c*d)*c
os(b*x + a) - (450*b^2*d^2*x^2 + 900*b^2*c*d*x - 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)^4 + 450*b^2*c^2 + (225*b^2*d^2*x^2 + 450*b^2*c*d*x + 225*b^2*c^2 + 22*d^2)*cos(b*x + a)^2 - 856*d^2)*s
in(b*x + a))/b^3

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giac [A]  time = 2.11, size = 209, normalized size = 1.14 \[ -\frac {{\left (b d^{2} x + b c d\right )} \cos \left (5 \, b x + 5 \, a\right )}{200 \, b^{3}} - \frac {{\left (b d^{2} x + b c d\right )} \cos \left (3 \, b x + 3 \, a\right )}{72 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )}{4 \, b^{3}} - \frac {{\left (25 \, b^{2} d^{2} x^{2} + 50 \, b^{2} c d x + 25 \, b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (5 \, b x + 5 \, a\right )}{2000 \, b^{3}} - \frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (3 \, b x + 3 \, a\right )}{432 \, b^{3}} + \frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \sin \left (b x + a\right )}{8 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 484, normalized size = 2.63 \[ \frac {\frac {d^{2} \left (\frac {\left (b x +a \right )^{2} \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}-\frac {4 \sin \left (b x +a \right )}{15}+\frac {4 \left (b x +a \right ) \cos \left (b x +a \right )}{15}+\frac {2 \left (b x +a \right ) \left (\cos ^{3}\left (b x +a \right )\right )}{45}-\frac {2 \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{135}-\frac {\left (b x +a \right )^{2} \left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{5}-\frac {2 \left (b x +a \right ) \left (\cos ^{5}\left (b x +a \right )\right )}{25}+\frac {2 \left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{125}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (\frac {\left (b x +a \right ) \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}+\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \cos \left (b x +a \right )}{15}-\frac {\left (b x +a \right ) \left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{5}-\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{25}\right )}{b^{2}}+\frac {2 c d \left (\frac {\left (b x +a \right ) \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}+\frac {\left (\cos ^{3}\left (b x +a \right )\right )}{45}+\frac {2 \cos \left (b x +a \right )}{15}-\frac {\left (b x +a \right ) \left (\frac {8}{3}+\cos ^{4}\left (b x +a \right )+\frac {4 \left (\cos ^{2}\left (b x +a \right )\right )}{3}\right ) \sin \left (b x +a \right )}{5}-\frac {\left (\cos ^{5}\left (b x +a \right )\right )}{25}\right )}{b}+\frac {d^{2} a^{2} \left (-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{5}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{15}\right )}{b^{2}}-\frac {2 c d a \left (-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{5}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{15}\right )}{b}+c^{2} \left (-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{5}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{15}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.60, size = 375, normalized size = 2.04 \[ -\frac {3600 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} c^{2} - \frac {7200 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} a c d}{b} + \frac {3600 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )} a^{2} d^{2}}{b^{2}} + \frac {30 \, {\left (45 \, {\left (b x + a\right )} \sin \left (5 \, b x + 5 \, a\right ) + 75 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + 9 \, \cos \left (5 \, b x + 5 \, a\right ) + 25 \, \cos \left (3 \, b x + 3 \, a\right ) - 450 \, \cos \left (b x + a\right )\right )} c d}{b} - \frac {30 \, {\left (45 \, {\left (b x + a\right )} \sin \left (5 \, b x + 5 \, a\right ) + 75 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 450 \, {\left (b x + a\right )} \sin \left (b x + a\right ) + 9 \, \cos \left (5 \, b x + 5 \, a\right ) + 25 \, \cos \left (3 \, b x + 3 \, a\right ) - 450 \, \cos \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left (270 \, {\left (b x + a\right )} \cos \left (5 \, b x + 5 \, a\right ) + 750 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 13500 \, {\left (b x + a\right )} \cos \left (b x + a\right ) + 27 \, {\left (25 \, {\left (b x + a\right )}^{2} - 2\right )} \sin \left (5 \, b x + 5 \, a\right ) + 125 \, {\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \sin \left (3 \, b x + 3 \, a\right ) - 6750 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{54000 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54000*(3600*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)*c^2 - 7200*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)*a*c*d/b
+ 3600*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)*a^2*d^2/b^2 + 30*(45*(b*x + a)*sin(5*b*x + 5*a) + 75*(b*x + a)*si
n(3*b*x + 3*a) - 450*(b*x + a)*sin(b*x + a) + 9*cos(5*b*x + 5*a) + 25*cos(3*b*x + 3*a) - 450*cos(b*x + a))*c*d
/b - 30*(45*(b*x + a)*sin(5*b*x + 5*a) + 75*(b*x + a)*sin(3*b*x + 3*a) - 450*(b*x + a)*sin(b*x + a) + 9*cos(5*
b*x + 5*a) + 25*cos(3*b*x + 3*a) - 450*cos(b*x + a))*a*d^2/b^2 + (270*(b*x + a)*cos(5*b*x + 5*a) + 750*(b*x +
a)*cos(3*b*x + 3*a) - 13500*(b*x + a)*cos(b*x + a) + 27*(25*(b*x + a)^2 - 2)*sin(5*b*x + 5*a) + 125*(9*(b*x +
a)^2 - 2)*sin(3*b*x + 3*a) - 6750*((b*x + a)^2 - 2)*sin(b*x + a))*d^2/b^2)/b

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mupad [B]  time = 0.85, size = 295, normalized size = 1.60 \[ \frac {52\,d^2\,x\,{\cos \left (a+b\,x\right )}^5}{225\,b^2}-\frac {52\,d^2\,{\cos \left (a+b\,x\right )}^4\,\sin \left (a+b\,x\right )}{225\,b^3}-\frac {{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3\,\left (338\,d^2-225\,b^2\,c^2\right )}{675\,b^3}-\frac {2\,{\sin \left (a+b\,x\right )}^5\,\left (428\,d^2-225\,b^2\,c^2\right )}{3375\,b^3}+\frac {2\,d^2\,x^2\,{\sin \left (a+b\,x\right )}^5}{15\,b}+\frac {52\,c\,d\,{\cos \left (a+b\,x\right )}^5}{225\,b^2}+\frac {4\,c\,d\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^4}{15\,b^2}+\frac {4\,c\,d\,x\,{\sin \left (a+b\,x\right )}^5}{15\,b}+\frac {d^2\,x^2\,{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{3\,b}+\frac {26\,c\,d\,{\cos \left (a+b\,x\right )}^3\,{\sin \left (a+b\,x\right )}^2}{45\,b^2}+\frac {4\,d^2\,x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^4}{15\,b^2}+\frac {26\,d^2\,x\,{\cos \left (a+b\,x\right )}^3\,{\sin \left (a+b\,x\right )}^2}{45\,b^2}+\frac {2\,c\,d\,x\,{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^3}{3\,b} \]

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)^2,x)

[Out]

(52*d^2*x*cos(a + b*x)^5)/(225*b^2) - (52*d^2*cos(a + b*x)^4*sin(a + b*x))/(225*b^3) - (cos(a + b*x)^2*sin(a +
b*x)^3*(338*d^2 - 225*b^2*c^2))/(675*b^3) - (2*sin(a + b*x)^5*(428*d^2 - 225*b^2*c^2))/(3375*b^3) + (2*d^2*x^
2*sin(a + b*x)^5)/(15*b) + (52*c*d*cos(a + b*x)^5)/(225*b^2) + (4*c*d*cos(a + b*x)*sin(a + b*x)^4)/(15*b^2) +
(4*c*d*x*sin(a + b*x)^5)/(15*b) + (d^2*x^2*cos(a + b*x)^2*sin(a + b*x)^3)/(3*b) + (26*c*d*cos(a + b*x)^3*sin(a
+ b*x)^2)/(45*b^2) + (4*d^2*x*cos(a + b*x)*sin(a + b*x)^4)/(15*b^2) + (26*d^2*x*cos(a + b*x)^3*sin(a + b*x)^2
)/(45*b^2) + (2*c*d*x*cos(a + b*x)^2*sin(a + b*x)^3)/(3*b)

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sympy [A]  time = 6.23, size = 382, normalized size = 2.08 \[ \begin {cases} \frac {2 c^{2} \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {c^{2} \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {4 c d x \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {2 c d x \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {2 d^{2} x^{2} \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {d^{2} x^{2} \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {4 c d \sin ^{4}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{15 b^{2}} + \frac {26 c d \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac {52 c d \cos ^{5}{\left (a + b x \right )}}{225 b^{2}} + \frac {4 d^{2} x \sin ^{4}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{15 b^{2}} + \frac {26 d^{2} x \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac {52 d^{2} x \cos ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac {856 d^{2} \sin ^{5}{\left (a + b x \right )}}{3375 b^{3}} - \frac {338 d^{2} \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{675 b^{3}} - \frac {52 d^{2} \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{225 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin ^{2}{\relax (a )} \cos ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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