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

Optimal. Leaf size=79 \[ \frac {d^2 \sin (4 a+4 b x)}{256 b^3}-\frac {d (c+d x) \cos (4 a+4 b x)}{64 b^2}-\frac {(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^3}{24 d} \]

[Out]

1/24*(d*x+c)^3/d-1/64*d*(d*x+c)*cos(4*b*x+4*a)/b^2+1/256*d^2*sin(4*b*x+4*a)/b^3-1/32*(d*x+c)^2*sin(4*b*x+4*a)/
b

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Rubi [A]  time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4406, 3296, 2637} \[ -\frac {d (c+d x) \cos (4 a+4 b x)}{64 b^2}+\frac {d^2 \sin (4 a+4 b x)}{256 b^3}-\frac {(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^3}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^3/(24*d) - (d*(c + d*x)*Cos[4*a + 4*b*x])/(64*b^2) + (d^2*Sin[4*a + 4*b*x])/(256*b^3) - ((c + d*x)^2
*Sin[4*a + 4*b*x])/(32*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 ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac {1}{8} (c+d x)^2-\frac {1}{8} (c+d x)^2 \cos (4 a+4 b x)\right ) \, dx\\ &=\frac {(c+d x)^3}{24 d}-\frac {1}{8} \int (c+d x)^2 \cos (4 a+4 b x) \, dx\\ &=\frac {(c+d x)^3}{24 d}-\frac {(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac {d \int (c+d x) \sin (4 a+4 b x) \, dx}{16 b}\\ &=\frac {(c+d x)^3}{24 d}-\frac {d (c+d x) \cos (4 a+4 b x)}{64 b^2}-\frac {(c+d x)^2 \sin (4 a+4 b x)}{32 b}+\frac {d^2 \int \cos (4 a+4 b x) \, dx}{64 b^2}\\ &=\frac {(c+d x)^3}{24 d}-\frac {d (c+d x) \cos (4 a+4 b x)}{64 b^2}+\frac {d^2 \sin (4 a+4 b x)}{256 b^3}-\frac {(c+d x)^2 \sin (4 a+4 b x)}{32 b}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 77, normalized size = 0.97 \[ \frac {-3 \sin (4 (a+b x)) \left (8 b^2 (c+d x)^2-d^2\right )-12 b d (c+d x) \cos (4 (a+b x))+32 b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )}{768 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*x*(3*c^2 + 3*c*d*x + d^2*x^2) - 12*b*d*(c + d*x)*Cos[4*(a + b*x)] - 3*(-d^2 + 8*b^2*(c + d*x)^2)*Sin[4
*(a + b*x)])/(768*b^3)

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fricas [B]  time = 0.44, size = 180, normalized size = 2.28 \[ \frac {8 \, b^{3} d^{2} x^{3} + 24 \, b^{3} c d x^{2} - 24 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} + 24 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 3 \, {\left (8 \, b^{3} c^{2} - b d^{2}\right )} x - 3 \, {\left (2 \, {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )^{3} - {\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{192 \, b^{3}} \]

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)^2,x, algorithm="fricas")

[Out]

1/192*(8*b^3*d^2*x^3 + 24*b^3*c*d*x^2 - 24*(b*d^2*x + b*c*d)*cos(b*x + a)^4 + 24*(b*d^2*x + b*c*d)*cos(b*x + a
)^2 + 3*(8*b^3*c^2 - b*d^2)*x - 3*(2*(8*b^2*d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - d^2)*cos(b*x + a)^3 - (8*b^2*
d^2*x^2 + 16*b^2*c*d*x + 8*b^2*c^2 - d^2)*cos(b*x + a))*sin(b*x + a))/b^3

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giac [A]  time = 0.20, size = 94, normalized size = 1.19 \[ \frac {1}{24} \, d^{2} x^{3} + \frac {1}{8} \, c d x^{2} + \frac {1}{8} \, c^{2} x - \frac {{\left (b d^{2} x + b c d\right )} \cos \left (4 \, b x + 4 \, a\right )}{64 \, b^{3}} - \frac {{\left (8 \, b^{2} d^{2} x^{2} + 16 \, b^{2} c d x + 8 \, b^{2} c^{2} - d^{2}\right )} \sin \left (4 \, b x + 4 \, a\right )}{256 \, b^{3}} \]

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)^2,x, algorithm="giac")

[Out]

1/24*d^2*x^3 + 1/8*c*d*x^2 + 1/8*c^2*x - 1/64*(b*d^2*x + b*c*d)*cos(4*b*x + 4*a)/b^3 - 1/256*(8*b^2*d^2*x^2 +
16*b^2*c*d*x + 8*b^2*c^2 - d^2)*sin(4*b*x + 4*a)/b^3

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maple [B]  time = 0.02, size = 519, normalized size = 6.57 \[ \frac {\frac {d^{2} \left (\left (b x +a \right )^{2} \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{8}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{16}+\frac {7 b x}{64}+\frac {7 a}{64}-\frac {\left (b x +a \right )^{3}}{12}-\left (b x +a \right )^{2} \left (-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {\left (b x +a \right ) \left (\sin ^{4}\left (b x +a \right )\right )}{8}-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{32}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (\left (b x +a \right ) \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{16}+\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{16}-\left (b x +a \right ) \left (-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {\left (\sin ^{4}\left (b x +a \right )\right )}{16}\right )}{b^{2}}+\frac {2 c d \left (\left (b x +a \right ) \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{16}+\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{16}-\left (b x +a \right ) \left (-\frac {\left (\sin ^{3}\left (b x +a \right )+\frac {3 \sin \left (b x +a \right )}{2}\right ) \cos \left (b x +a \right )}{4}+\frac {3 b x}{8}+\frac {3 a}{8}\right )-\frac {\left (\sin ^{4}\left (b x +a \right )\right )}{16}\right )}{b}+\frac {a^{2} d^{2} \left (-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{4}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}+\frac {b x}{8}+\frac {a}{8}\right )}{b^{2}}-\frac {2 a c d \left (-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{4}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}+\frac {b x}{8}+\frac {a}{8}\right )}{b}+c^{2} \left (-\frac {\left (\cos ^{3}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{4}+\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{8}+\frac {b x}{8}+\frac {a}{8}\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)^2,x)

[Out]

1/b*(1/b^2*d^2*((b*x+a)^2*(-1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-1/8*(b*x+a)*cos(b*x+a)^2+1/16*cos(b*x+a)*
sin(b*x+a)+7/64*b*x+7/64*a-1/12*(b*x+a)^3-(b*x+a)^2*(-1/4*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)+3/8*b*x+3/8
*a)-1/8*(b*x+a)*sin(b*x+a)^4-1/32*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a))-2/b^2*a*d^2*((b*x+a)*(-1/2*cos(b*x
+a)*sin(b*x+a)+1/2*b*x+1/2*a)-1/16*(b*x+a)^2+1/16*sin(b*x+a)^2-(b*x+a)*(-1/4*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos
(b*x+a)+3/8*b*x+3/8*a)-1/16*sin(b*x+a)^4)+2/b*c*d*((b*x+a)*(-1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)-1/16*(b*
x+a)^2+1/16*sin(b*x+a)^2-(b*x+a)*(-1/4*(sin(b*x+a)^3+3/2*sin(b*x+a))*cos(b*x+a)+3/8*b*x+3/8*a)-1/16*sin(b*x+a)
^4)+1/b^2*a^2*d^2*(-1/4*cos(b*x+a)^3*sin(b*x+a)+1/8*cos(b*x+a)*sin(b*x+a)+1/8*b*x+1/8*a)-2/b*a*c*d*(-1/4*cos(b
*x+a)^3*sin(b*x+a)+1/8*cos(b*x+a)*sin(b*x+a)+1/8*b*x+1/8*a)+c^2*(-1/4*cos(b*x+a)^3*sin(b*x+a)+1/8*cos(b*x+a)*s
in(b*x+a)+1/8*b*x+1/8*a))

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maxima [B]  time = 0.34, size = 232, normalized size = 2.94 \[ \frac {24 \, {\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} c^{2} - \frac {48 \, {\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} a c d}{b} + \frac {24 \, {\left (4 \, b x + 4 \, a - \sin \left (4 \, b x + 4 \, a\right )\right )} a^{2} d^{2}}{b^{2}} + \frac {12 \, {\left (8 \, {\left (b x + a\right )}^{2} - 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) - \cos \left (4 \, b x + 4 \, a\right )\right )} c d}{b} - \frac {12 \, {\left (8 \, {\left (b x + a\right )}^{2} - 4 \, {\left (b x + a\right )} \sin \left (4 \, b x + 4 \, a\right ) - \cos \left (4 \, b x + 4 \, a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left (32 \, {\left (b x + a\right )}^{3} - 12 \, {\left (b x + a\right )} \cos \left (4 \, b x + 4 \, a\right ) - 3 \, {\left (8 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} d^{2}}{b^{2}}}{768 \, b} \]

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)^2,x, algorithm="maxima")

[Out]

1/768*(24*(4*b*x + 4*a - sin(4*b*x + 4*a))*c^2 - 48*(4*b*x + 4*a - sin(4*b*x + 4*a))*a*c*d/b + 24*(4*b*x + 4*a
- sin(4*b*x + 4*a))*a^2*d^2/b^2 + 12*(8*(b*x + a)^2 - 4*(b*x + a)*sin(4*b*x + 4*a) - cos(4*b*x + 4*a))*c*d/b
- 12*(8*(b*x + a)^2 - 4*(b*x + a)*sin(4*b*x + 4*a) - cos(4*b*x + 4*a))*a*d^2/b^2 + (32*(b*x + a)^3 - 12*(b*x +
a)*cos(4*b*x + 4*a) - 3*(8*(b*x + a)^2 - 1)*sin(4*b*x + 4*a))*d^2/b^2)/b

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mupad [B]  time = 1.31, size = 179, normalized size = 2.27 \[ x\,\left (\frac {c^2}{32}+\frac {3\,d^2}{256\,b^2}\right )+x\,\left (\frac {3\,c^2}{32}-\frac {3\,d^2}{256\,b^2}\right )+\frac {d^2\,x^3}{24}+\frac {\sin \left (4\,a+4\,b\,x\right )\,\left (d^2-8\,b^2\,c^2\right )}{256\,b^3}-\frac {x\,\cos \left (4\,a+4\,b\,x\right )\,\left (\frac {c^2}{4}+\frac {3\,d^2}{32\,b^2}\right )}{8}+\frac {x\,\cos \left (4\,a+4\,b\,x\right )\,\left (\frac {c^2}{8}-\frac {d^2}{64\,b^2}\right )}{4}+\frac {c\,d\,x^2}{8}-\frac {d^2\,x^2\,\sin \left (4\,a+4\,b\,x\right )}{32\,b}-\frac {c\,d\,\cos \left (4\,a+4\,b\,x\right )}{64\,b^2}-\frac {c\,d\,x\,\sin \left (4\,a+4\,b\,x\right )}{16\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(c^2/32 + (3*d^2)/(256*b^2)) + x*((3*c^2)/32 - (3*d^2)/(256*b^2)) + (d^2*x^3)/24 + (sin(4*a + 4*b*x)*(d^2 -
8*b^2*c^2))/(256*b^3) - (x*cos(4*a + 4*b*x)*(c^2/4 + (3*d^2)/(32*b^2)))/8 + (x*cos(4*a + 4*b*x)*(c^2/8 - d^2/(
64*b^2)))/4 + (c*d*x^2)/8 - (d^2*x^2*sin(4*a + 4*b*x))/(32*b) - (c*d*cos(4*a + 4*b*x))/(64*b^2) - (c*d*x*sin(4
*a + 4*b*x))/(16*b)

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sympy [A]  time = 3.98, size = 493, normalized size = 6.24 \[ \begin {cases} \frac {c^{2} x \sin ^{4}{\left (a + b x \right )}}{8} + \frac {c^{2} x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {c^{2} x \cos ^{4}{\left (a + b x \right )}}{8} + \frac {c d x^{2} \sin ^{4}{\left (a + b x \right )}}{8} + \frac {c d x^{2} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} + \frac {c d x^{2} \cos ^{4}{\left (a + b x \right )}}{8} + \frac {d^{2} x^{3} \sin ^{4}{\left (a + b x \right )}}{24} + \frac {d^{2} x^{3} \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{12} + \frac {d^{2} x^{3} \cos ^{4}{\left (a + b x \right )}}{24} + \frac {c^{2} \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {c^{2} \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} + \frac {c d x \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b} - \frac {c d x \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{4 b} + \frac {d^{2} x^{2} \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{8 b} - \frac {d^{2} x^{2} \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{8 b} - \frac {c d \sin ^{4}{\left (a + b x \right )}}{16 b^{2}} - \frac {c d \cos ^{4}{\left (a + b x \right )}}{16 b^{2}} - \frac {d^{2} x \sin ^{4}{\left (a + b x \right )}}{64 b^{2}} + \frac {3 d^{2} x \sin ^{2}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{32 b^{2}} - \frac {d^{2} x \cos ^{4}{\left (a + b x \right )}}{64 b^{2}} - \frac {d^{2} \sin ^{3}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{64 b^{3}} + \frac {d^{2} \sin {\left (a + b x \right )} \cos ^{3}{\left (a + b x \right )}}{64 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin ^{2}{\relax (a )} \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

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)**2,x)

[Out]

Piecewise((c**2*x*sin(a + b*x)**4/8 + c**2*x*sin(a + b*x)**2*cos(a + b*x)**2/4 + c**2*x*cos(a + b*x)**4/8 + c*
d*x**2*sin(a + b*x)**4/8 + c*d*x**2*sin(a + b*x)**2*cos(a + b*x)**2/4 + c*d*x**2*cos(a + b*x)**4/8 + d**2*x**3
*sin(a + b*x)**4/24 + d**2*x**3*sin(a + b*x)**2*cos(a + b*x)**2/12 + d**2*x**3*cos(a + b*x)**4/24 + c**2*sin(a
+ b*x)**3*cos(a + b*x)/(8*b) - c**2*sin(a + b*x)*cos(a + b*x)**3/(8*b) + c*d*x*sin(a + b*x)**3*cos(a + b*x)/(
4*b) - c*d*x*sin(a + b*x)*cos(a + b*x)**3/(4*b) + d**2*x**2*sin(a + b*x)**3*cos(a + b*x)/(8*b) - d**2*x**2*sin
(a + b*x)*cos(a + b*x)**3/(8*b) - c*d*sin(a + b*x)**4/(16*b**2) - c*d*cos(a + b*x)**4/(16*b**2) - d**2*x*sin(a
+ b*x)**4/(64*b**2) + 3*d**2*x*sin(a + b*x)**2*cos(a + b*x)**2/(32*b**2) - d**2*x*cos(a + b*x)**4/(64*b**2) -
d**2*sin(a + b*x)**3*cos(a + b*x)/(64*b**3) + d**2*sin(a + b*x)*cos(a + b*x)**3/(64*b**3), Ne(b, 0)), ((c**2*
x + c*d*x**2 + d**2*x**3/3)*sin(a)**2*cos(a)**2, True))

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