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

Optimal. Leaf size=95 \[ -\frac {d^2 \sin (a+b x) \cos (a+b x)}{4 b^3}+\frac {d (c+d x) \cos ^2(a+b x)}{2 b^2}+\frac {(c+d x)^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac {d^2 x}{4 b^2}+\frac {(c+d x)^3}{6 d} \]

[Out]

-1/4*d^2*x/b^2+1/6*(d*x+c)^3/d+1/2*d*(d*x+c)*cos(b*x+a)^2/b^2-1/4*d^2*cos(b*x+a)*sin(b*x+a)/b^3+1/2*(d*x+c)^2*
cos(b*x+a)*sin(b*x+a)/b

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Rubi [A]  time = 0.05, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3311, 32, 2635, 8} \[ \frac {d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac {d^2 \sin (a+b x) \cos (a+b x)}{4 b^3}+\frac {(c+d x)^2 \sin (a+b x) \cos (a+b x)}{2 b}-\frac {d^2 x}{4 b^2}+\frac {(c+d x)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cos[a + b*x]^2,x]

[Out]

-(d^2*x)/(4*b^2) + (c + d*x)^3/(6*d) + (d*(c + d*x)*Cos[a + b*x]^2)/(2*b^2) - (d^2*Cos[a + b*x]*Sin[a + b*x])/
(4*b^3) + ((c + d*x)^2*Cos[a + b*x]*Sin[a + b*x])/(2*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rubi steps

\begin {align*} \int (c+d x)^2 \cos ^2(a+b x) \, dx &=\frac {d (c+d x) \cos ^2(a+b x)}{2 b^2}+\frac {(c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b}+\frac {1}{2} \int (c+d x)^2 \, dx-\frac {d^2 \int \cos ^2(a+b x) \, dx}{2 b^2}\\ &=\frac {(c+d x)^3}{6 d}+\frac {d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac {d^2 \cos (a+b x) \sin (a+b x)}{4 b^3}+\frac {(c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b}-\frac {d^2 \int 1 \, dx}{4 b^2}\\ &=-\frac {d^2 x}{4 b^2}+\frac {(c+d x)^3}{6 d}+\frac {d (c+d x) \cos ^2(a+b x)}{2 b^2}-\frac {d^2 \cos (a+b x) \sin (a+b x)}{4 b^3}+\frac {(c+d x)^2 \cos (a+b x) \sin (a+b x)}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 77, normalized size = 0.81 \[ \frac {3 \sin (2 (a+b x)) \left (2 b^2 (c+d x)^2-d^2\right )+6 b d (c+d x) \cos (2 (a+b x))+4 b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^3*x*(3*c^2 + 3*c*d*x + d^2*x^2) + 6*b*d*(c + d*x)*Cos[2*(a + b*x)] + 3*(-d^2 + 2*b^2*(c + d*x)^2)*Sin[2*(
a + b*x)])/(24*b^3)

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fricas [A]  time = 0.60, size = 113, normalized size = 1.19 \[ \frac {2 \, b^{3} d^{2} x^{3} + 6 \, b^{3} c d x^{2} + 6 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 3 \, {\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 3 \, {\left (2 \, b^{3} c^{2} - b d^{2}\right )} x}{12 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*b^3*d^2*x^3 + 6*b^3*c*d*x^2 + 6*(b*d^2*x + b*c*d)*cos(b*x + a)^2 + 3*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*
b^2*c^2 - d^2)*cos(b*x + a)*sin(b*x + a) + 3*(2*b^3*c^2 - b*d^2)*x)/b^3

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giac [A]  time = 0.53, size = 94, normalized size = 0.99 \[ \frac {1}{6} \, d^{2} x^{3} + \frac {1}{2} \, c d x^{2} + \frac {1}{2} \, c^{2} x + \frac {{\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right )}{4 \, b^{3}} + \frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - d^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )}{8 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*d^2*x^3 + 1/2*c*d*x^2 + 1/2*c^2*x + 1/4*(b*d^2*x + b*c*d)*cos(2*b*x + 2*a)/b^3 + 1/8*(2*b^2*d^2*x^2 + 4*b^
2*c*d*x + 2*b^2*c^2 - d^2)*sin(2*b*x + 2*a)/b^3

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maple [B]  time = 0.02, size = 289, normalized size = 3.04 \[ \frac {\frac {d^{2} \left (\left (b x +a \right )^{2} \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )+\frac {\left (b x +a \right ) \left (\cos ^{2}\left (b x +a \right )\right )}{2}-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{4}-\frac {b x}{4}-\frac {a}{4}-\frac {\left (b x +a \right )^{3}}{3}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (\left (b x +a \right ) \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}\right )}{b^{2}}+\frac {2 c d \left (\left (b x +a \right ) \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}\right )}{b}+\frac {a^{2} d^{2} \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b^{2}}-\frac {2 a c d \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c^{2} \left (\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.72, size = 222, normalized size = 2.34 \[ \frac {6 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} c^{2} - \frac {12 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a c d}{b} + \frac {6 \, {\left (2 \, b x + 2 \, a + \sin \left (2 \, b x + 2 \, a\right )\right )} a^{2} d^{2}}{b^{2}} + \frac {6 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} c d}{b} - \frac {6 \, {\left (2 \, {\left (b x + a\right )}^{2} + 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left (4 \, {\left (b x + a\right )}^{3} + 6 \, {\left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b x + a\right )}^{2} - 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} d^{2}}{b^{2}}}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*(2*b*x + 2*a + sin(2*b*x + 2*a))*c^2 - 12*(2*b*x + 2*a + sin(2*b*x + 2*a))*a*c*d/b + 6*(2*b*x + 2*a +
sin(2*b*x + 2*a))*a^2*d^2/b^2 + 6*(2*(b*x + a)^2 + 2*(b*x + a)*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*c*d/b - 6*
(2*(b*x + a)^2 + 2*(b*x + a)*sin(2*b*x + 2*a) + cos(2*b*x + 2*a))*a*d^2/b^2 + (4*(b*x + a)^3 + 6*(b*x + a)*cos
(2*b*x + 2*a) + 3*(2*(b*x + a)^2 - 1)*sin(2*b*x + 2*a))*d^2/b^2)/b

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mupad [B]  time = 0.21, size = 179, normalized size = 1.88 \[ x\,\left (\frac {c^2}{4}-\frac {d^2}{8\,b^2}\right )+x\,\left (\frac {c^2}{4}+\frac {d^2}{8\,b^2}\right )+\frac {d^2\,x^3}{6}-\frac {\sin \left (2\,a+2\,b\,x\right )\,\left (d^2-2\,b^2\,c^2\right )}{8\,b^3}-\frac {x\,\cos \left (2\,a+2\,b\,x\right )\,\left (\frac {c^2}{2}-\frac {d^2}{4\,b^2}\right )}{2}+\frac {x\,\cos \left (2\,a+2\,b\,x\right )\,\left (\frac {c^2}{2}+\frac {d^2}{4\,b^2}\right )}{2}+\frac {c\,d\,x^2}{2}+\frac {d^2\,x^2\,\sin \left (2\,a+2\,b\,x\right )}{4\,b}+\frac {c\,d\,\cos \left (2\,a+2\,b\,x\right )}{4\,b^2}+\frac {c\,d\,x\,\sin \left (2\,a+2\,b\,x\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^2*(c + d*x)^2,x)

[Out]

x*(c^2/4 - d^2/(8*b^2)) + x*(c^2/4 + d^2/(8*b^2)) + (d^2*x^3)/6 - (sin(2*a + 2*b*x)*(d^2 - 2*b^2*c^2))/(8*b^3)
- (x*cos(2*a + 2*b*x)*(c^2/2 - d^2/(4*b^2)))/2 + (x*cos(2*a + 2*b*x)*(c^2/2 + d^2/(4*b^2)))/2 + (c*d*x^2)/2 +
(d^2*x^2*sin(2*a + 2*b*x))/(4*b) + (c*d*cos(2*a + 2*b*x))/(4*b^2) + (c*d*x*sin(2*a + 2*b*x))/(2*b)

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sympy [A]  time = 1.18, size = 264, normalized size = 2.78 \[ \begin {cases} \frac {c^{2} x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {c^{2} x \cos ^{2}{\left (a + b x \right )}}{2} + \frac {c d x^{2} \sin ^{2}{\left (a + b x \right )}}{2} + \frac {c d x^{2} \cos ^{2}{\left (a + b x \right )}}{2} + \frac {d^{2} x^{3} \sin ^{2}{\left (a + b x \right )}}{6} + \frac {d^{2} x^{3} \cos ^{2}{\left (a + b x \right )}}{6} + \frac {c^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} + \frac {c d x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} + \frac {d^{2} x^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} - \frac {c d \sin ^{2}{\left (a + b x \right )}}{2 b^{2}} - \frac {d^{2} x \sin ^{2}{\left (a + b x \right )}}{4 b^{2}} + \frac {d^{2} x \cos ^{2}{\left (a + b x \right )}}{4 b^{2}} - \frac {d^{2} \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{4 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c**2*x*sin(a + b*x)**2/2 + c**2*x*cos(a + b*x)**2/2 + c*d*x**2*sin(a + b*x)**2/2 + c*d*x**2*cos(a +
b*x)**2/2 + d**2*x**3*sin(a + b*x)**2/6 + d**2*x**3*cos(a + b*x)**2/6 + c**2*sin(a + b*x)*cos(a + b*x)/(2*b)
+ c*d*x*sin(a + b*x)*cos(a + b*x)/b + d**2*x**2*sin(a + b*x)*cos(a + b*x)/(2*b) - c*d*sin(a + b*x)**2/(2*b**2)
- d**2*x*sin(a + b*x)**2/(4*b**2) + d**2*x*cos(a + b*x)**2/(4*b**2) - d**2*sin(a + b*x)*cos(a + b*x)/(4*b**3)
, Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*cos(a)**2, True))

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