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

Optimal. Leaf size=103 \[ \frac {2 d^2 \cos ^3(a+b x)}{27 b^3}+\frac {4 d^2 \cos (a+b x)}{9 b^3}+\frac {4 d (c+d x) \sin (a+b x)}{9 b^2}+\frac {2 d (c+d x) \sin (a+b x) \cos ^2(a+b x)}{9 b^2}-\frac {(c+d x)^2 \cos ^3(a+b x)}{3 b} \]

[Out]

4/9*d^2*cos(b*x+a)/b^3+2/27*d^2*cos(b*x+a)^3/b^3-1/3*(d*x+c)^2*cos(b*x+a)^3/b+4/9*d*(d*x+c)*sin(b*x+a)/b^2+2/9
*d*(d*x+c)*cos(b*x+a)^2*sin(b*x+a)/b^2

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Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4405, 3310, 3296, 2638} \[ \frac {4 d (c+d x) \sin (a+b x)}{9 b^2}+\frac {2 d (c+d x) \sin (a+b x) \cos ^2(a+b x)}{9 b^2}+\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}+\frac {4 d^2 \cos (a+b x)}{9 b^3}-\frac {(c+d x)^2 \cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*d^2*Cos[a + b*x])/(9*b^3) + (2*d^2*Cos[a + b*x]^3)/(27*b^3) - ((c + d*x)^2*Cos[a + b*x]^3)/(3*b) + (4*d*(c
+ d*x)*Sin[a + b*x])/(9*b^2) + (2*d*(c + d*x)*Cos[a + b*x]^2*Sin[a + b*x])/(9*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 3310

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

Rule 4405

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)], x_Symbol] :> -Simp[((c +
d*x)^m*Cos[a + b*x]^(n + 1))/(b*(n + 1)), x] + Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Cos[a + b*x]^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rubi steps

\begin {align*} \int (c+d x)^2 \cos ^2(a+b x) \sin (a+b x) \, dx &=-\frac {(c+d x)^2 \cos ^3(a+b x)}{3 b}+\frac {(2 d) \int (c+d x) \cos ^3(a+b x) \, dx}{3 b}\\ &=\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}-\frac {(c+d x)^2 \cos ^3(a+b x)}{3 b}+\frac {2 d (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^2}+\frac {(4 d) \int (c+d x) \cos (a+b x) \, dx}{9 b}\\ &=\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}-\frac {(c+d x)^2 \cos ^3(a+b x)}{3 b}+\frac {4 d (c+d x) \sin (a+b x)}{9 b^2}+\frac {2 d (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^2}-\frac {\left (4 d^2\right ) \int \sin (a+b x) \, dx}{9 b^2}\\ &=\frac {4 d^2 \cos (a+b x)}{9 b^3}+\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}-\frac {(c+d x)^2 \cos ^3(a+b x)}{3 b}+\frac {4 d (c+d x) \sin (a+b x)}{9 b^2}+\frac {2 d (c+d x) \cos ^2(a+b x) \sin (a+b x)}{9 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 86, normalized size = 0.83 \[ -\frac {27 \cos (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )+\cos (3 (a+b x)) \left (9 b^2 (c+d x)^2-2 d^2\right )-6 b d (c+d x) (9 \sin (a+b x)+\sin (3 (a+b x)))}{108 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/108*(27*(-2*d^2 + b^2*(c + d*x)^2)*Cos[a + b*x] + (-2*d^2 + 9*b^2*(c + d*x)^2)*Cos[3*(a + b*x)] - 6*b*d*(c
+ d*x)*(9*Sin[a + b*x] + Sin[3*(a + b*x)]))/b^3

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fricas [A]  time = 0.72, size = 100, normalized size = 0.97 \[ -\frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{3} - 12 \, d^{2} \cos \left (b x + a\right ) - 6 \, {\left (2 \, b d^{2} x + 2 \, b c d + {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{27 \, b^{3}} \]

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),x, algorithm="fricas")

[Out]

-1/27*((9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*b^2*c^2 - 2*d^2)*cos(b*x + a)^3 - 12*d^2*cos(b*x + a) - 6*(2*b*d^2*x
+ 2*b*c*d + (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 = 4.95, size = 137, normalized size = 1.33 \[ -\frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (3 \, b x + 3 \, a\right )}{108 \, b^{3}} - \frac {{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{4 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (3 \, b x + 3 \, a\right )}{18 \, b^{3}} + \frac {{\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{2 \, b^{3}} \]

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),x, algorithm="giac")

[Out]

-1/108*(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 - 1/4*(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 + 1/18*(b*d^2*x + b*c*d)*sin(3*b*x + 3*a)/b^3 + 1/2*(b*d^2*x + b*c*d)*si
n(b*x + a)/b^3

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maple [B]  time = 0.01, size = 204, normalized size = 1.98 \[ \frac {\frac {d^{2} \left (-\frac {\left (b x +a \right )^{2} \left (\cos ^{3}\left (b x +a \right )\right )}{3}+\frac {2 \left (b x +a \right ) \left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{9}+\frac {2 \left (\cos ^{3}\left (b x +a \right )\right )}{27}+\frac {4 \cos \left (b x +a \right )}{9}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (-\frac {\left (b x +a \right ) \left (\cos ^{3}\left (b x +a \right )\right )}{3}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{9}\right )}{b^{2}}+\frac {2 c d \left (-\frac {\left (b x +a \right ) \left (\cos ^{3}\left (b x +a \right )\right )}{3}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{9}\right )}{b}-\frac {a^{2} d^{2} \left (\cos ^{3}\left (b x +a \right )\right )}{3 b^{2}}+\frac {2 a c d \left (\cos ^{3}\left (b x +a \right )\right )}{3 b}-\frac {c^{2} \left (\cos ^{3}\left (b x +a \right )\right )}{3}}{b} \]

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),x)

[Out]

1/b*(1/b^2*d^2*(-1/3*(b*x+a)^2*cos(b*x+a)^3+2/9*(b*x+a)*(2+cos(b*x+a)^2)*sin(b*x+a)+2/27*cos(b*x+a)^3+4/9*cos(
b*x+a))-2/b^2*a*d^2*(-1/3*(b*x+a)*cos(b*x+a)^3+1/9*(2+cos(b*x+a)^2)*sin(b*x+a))+2/b*c*d*(-1/3*(b*x+a)*cos(b*x+
a)^3+1/9*(2+cos(b*x+a)^2)*sin(b*x+a))-1/3/b^2*a^2*d^2*cos(b*x+a)^3+2/3/b*a*c*d*cos(b*x+a)^3-1/3*c^2*cos(b*x+a)
^3)

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maxima [B]  time = 0.37, size = 243, normalized size = 2.36 \[ -\frac {36 \, c^{2} \cos \left (b x + a\right )^{3} - \frac {72 \, a c d \cos \left (b x + a\right )^{3}}{b} + \frac {36 \, a^{2} d^{2} \cos \left (b x + a\right )^{3}}{b^{2}} + \frac {6 \, {\left (3 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) + 9 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) - 9 \, \sin \left (b x + a\right )\right )} c d}{b} - \frac {6 \, {\left (3 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) + 9 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) - 9 \, \sin \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \cos \left (3 \, b x + 3 \, a\right ) + 27 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 6 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) - 54 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{108 \, b} \]

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),x, algorithm="maxima")

[Out]

-1/108*(36*c^2*cos(b*x + a)^3 - 72*a*c*d*cos(b*x + a)^3/b + 36*a^2*d^2*cos(b*x + a)^3/b^2 + 6*(3*(b*x + a)*cos
(3*b*x + 3*a) + 9*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*a) - 9*sin(b*x + a))*c*d/b - 6*(3*(b*x + a)*cos(3*b*x
+ 3*a) + 9*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*a) - 9*sin(b*x + a))*a*d^2/b^2 + ((9*(b*x + a)^2 - 2)*cos(3
*b*x + 3*a) + 27*((b*x + a)^2 - 2)*cos(b*x + a) - 6*(b*x + a)*sin(3*b*x + 3*a) - 54*(b*x + a)*sin(b*x + a))*d^
2/b^2)/b

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mupad [B]  time = 1.13, size = 145, normalized size = 1.41 \[ \frac {12\,d^2\,\cos \left (a+b\,x\right )+2\,d^2\,{\cos \left (a+b\,x\right )}^3-9\,b^2\,c^2\,{\cos \left (a+b\,x\right )}^3+12\,b\,d^2\,x\,\sin \left (a+b\,x\right )-9\,b^2\,d^2\,x^2\,{\cos \left (a+b\,x\right )}^3+12\,b\,c\,d\,\sin \left (a+b\,x\right )-18\,b^2\,c\,d\,x\,{\cos \left (a+b\,x\right )}^3+6\,b\,d^2\,x\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )+6\,b\,c\,d\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )}{27\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12*d^2*cos(a + b*x) + 2*d^2*cos(a + b*x)^3 - 9*b^2*c^2*cos(a + b*x)^3 + 12*b*d^2*x*sin(a + b*x) - 9*b^2*d^2*x
^2*cos(a + b*x)^3 + 12*b*c*d*sin(a + b*x) - 18*b^2*c*d*x*cos(a + b*x)^3 + 6*b*d^2*x*cos(a + b*x)^2*sin(a + b*x
) + 6*b*c*d*cos(a + b*x)^2*sin(a + b*x))/(27*b^3)

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sympy [A]  time = 2.03, size = 216, normalized size = 2.10 \[ \begin {cases} - \frac {c^{2} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 c d x \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {d^{2} x^{2} \cos ^{3}{\left (a + b x \right )}}{3 b} + \frac {4 c d \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {2 c d \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} + \frac {4 d^{2} x \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {2 d^{2} x \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} + \frac {4 d^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{9 b^{3}} + \frac {14 d^{2} \cos ^{3}{\left (a + b x \right )}}{27 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin {\relax (a )} \cos ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

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),x)

[Out]

Piecewise((-c**2*cos(a + b*x)**3/(3*b) - 2*c*d*x*cos(a + b*x)**3/(3*b) - d**2*x**2*cos(a + b*x)**3/(3*b) + 4*c
*d*sin(a + b*x)**3/(9*b**2) + 2*c*d*sin(a + b*x)*cos(a + b*x)**2/(3*b**2) + 4*d**2*x*sin(a + b*x)**3/(9*b**2)
+ 2*d**2*x*sin(a + b*x)*cos(a + b*x)**2/(3*b**2) + 4*d**2*sin(a + b*x)**2*cos(a + b*x)/(9*b**3) + 14*d**2*cos(
a + b*x)**3/(27*b**3), Ne(b, 0)), ((c**2*x + c*d*x**2 + d**2*x**3/3)*sin(a)*cos(a)**2, True))

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