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

Optimal. Leaf size=123 \[ -\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}+\frac {14 d^2 \cos (a+b x)}{9 b^3}+\frac {2 d (c+d x) \sin ^3(a+b x)}{9 b^2}+\frac {4 d (c+d x) \sin (a+b x)}{3 b^2}-\frac {2 (c+d x)^2 \cos (a+b x)}{3 b}-\frac {(c+d x)^2 \sin ^2(a+b x) \cos (a+b x)}{3 b} \]

[Out]

14/9*d^2*cos(b*x+a)/b^3-2/3*(d*x+c)^2*cos(b*x+a)/b-2/27*d^2*cos(b*x+a)^3/b^3+4/3*d*(d*x+c)*sin(b*x+a)/b^2-1/3*
(d*x+c)^2*cos(b*x+a)*sin(b*x+a)^2/b+2/9*d*(d*x+c)*sin(b*x+a)^3/b^2

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Rubi [A]  time = 0.10, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3311, 3296, 2638, 2633} \[ \frac {2 d (c+d x) \sin ^3(a+b x)}{9 b^2}+\frac {4 d (c+d x) \sin (a+b x)}{3 b^2}-\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}+\frac {14 d^2 \cos (a+b x)}{9 b^3}-\frac {2 (c+d x)^2 \cos (a+b x)}{3 b}-\frac {(c+d x)^2 \sin ^2(a+b x) \cos (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Sin[a + b*x]^3,x]

[Out]

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 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 \sin ^3(a+b x) \, dx &=-\frac {(c+d x)^2 \cos (a+b x) \sin ^2(a+b x)}{3 b}+\frac {2 d (c+d x) \sin ^3(a+b x)}{9 b^2}+\frac {2}{3} \int (c+d x)^2 \sin (a+b x) \, dx-\frac {\left (2 d^2\right ) \int \sin ^3(a+b x) \, dx}{9 b^2}\\ &=-\frac {2 (c+d x)^2 \cos (a+b x)}{3 b}-\frac {(c+d x)^2 \cos (a+b x) \sin ^2(a+b x)}{3 b}+\frac {2 d (c+d x) \sin ^3(a+b x)}{9 b^2}+\frac {(4 d) \int (c+d x) \cos (a+b x) \, dx}{3 b}+\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{9 b^3}\\ &=\frac {2 d^2 \cos (a+b x)}{9 b^3}-\frac {2 (c+d x)^2 \cos (a+b x)}{3 b}-\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}+\frac {4 d (c+d x) \sin (a+b x)}{3 b^2}-\frac {(c+d x)^2 \cos (a+b x) \sin ^2(a+b x)}{3 b}+\frac {2 d (c+d x) \sin ^3(a+b x)}{9 b^2}-\frac {\left (4 d^2\right ) \int \sin (a+b x) \, dx}{3 b^2}\\ &=\frac {14 d^2 \cos (a+b x)}{9 b^3}-\frac {2 (c+d x)^2 \cos (a+b x)}{3 b}-\frac {2 d^2 \cos ^3(a+b x)}{27 b^3}+\frac {4 d (c+d x) \sin (a+b x)}{3 b^2}-\frac {(c+d x)^2 \cos (a+b x) \sin ^2(a+b x)}{3 b}+\frac {2 d (c+d x) \sin ^3(a+b x)}{9 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 86, normalized size = 0.70 \[ \frac {-81 \cos (a+b x) \left (b^2 (c+d x)^2-2 d^2\right )+\cos (3 (a+b x)) \left (9 b^2 (c+d x)^2-2 d^2\right )-6 b d (c+d x) (\sin (3 (a+b x))-27 \sin (a+b x))}{108 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-81*(-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)
*(-27*Sin[a + b*x] + Sin[3*(a + b*x)]))/(108*b^3)

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fricas [A]  time = 0.61, size = 131, normalized size = 1.07 \[ \frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{3} - 3 \, {\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 14 \, d^{2}\right )} \cos \left (b x + a\right ) + 6 \, {\left (7 \, b d^{2} x + 7 \, b c d - {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{27 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sin(b*x+a)^3,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 - 3*(9*b^2*d^2*x^2 + 18*b^2*c*d*x + 9*
b^2*c^2 - 14*d^2)*cos(b*x + a) + 6*(7*b*d^2*x + 7*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 = 2.03, size = 137, normalized size = 1.11 \[ \frac {{\left (9 \, b^{2} d^{2} x^{2} + 18 \, b^{2} c d x + 9 \, b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (3 \, b x + 3 \, a\right )}{108 \, b^{3}} - \frac {3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )}{4 \, b^{3}} - \frac {{\left (b d^{2} x + b c d\right )} \sin \left (3 \, b x + 3 \, a\right )}{18 \, b^{3}} + \frac {3 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*sin(b*x+a)^3,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 - 3/4*(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 + 3/2*(b*d^2*x + b*c*d)*sin
(b*x + a)/b^3

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maple [B]  time = 0.02, size = 265, normalized size = 2.15 \[ \frac {\frac {d^{2} \left (-\frac {\left (b x +a \right )^{2} \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {4 \cos \left (b x +a \right )}{3}+\frac {4 \left (b x +a \right ) \sin \left (b x +a \right )}{3}+\frac {2 \left (b x +a \right ) \left (\sin ^{3}\left (b x +a \right )\right )}{9}+\frac {2 \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{27}\right )}{b^{2}}-\frac {2 a \,d^{2} \left (-\frac {\left (b x +a \right ) \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{9}+\frac {2 \sin \left (b x +a \right )}{3}\right )}{b^{2}}+\frac {2 c d \left (-\frac {\left (b x +a \right ) \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{9}+\frac {2 \sin \left (b x +a \right )}{3}\right )}{b}-\frac {a^{2} d^{2} \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3 b^{2}}+\frac {2 a c d \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3 b}-\frac {c^{2} \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.46, size = 270, normalized size = 2.20 \[ \frac {36 \, {\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} c^{2} - \frac {72 \, {\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} a c d}{b} + \frac {36 \, {\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} a^{2} d^{2}}{b^{2}} + \frac {6 \, {\left (3 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 27 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) + 27 \, \sin \left (b x + a\right )\right )} c d}{b} - \frac {6 \, {\left (3 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 27 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) + 27 \, \sin \left (b x + a\right )\right )} a d^{2}}{b^{2}} + \frac {{\left ({\left (9 \, {\left (b x + a\right )}^{2} - 2\right )} \cos \left (3 \, b x + 3 \, a\right ) - 81 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 6 \, {\left (b x + a\right )} \sin \left (3 \, b x + 3 \, a\right ) + 162 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} d^{2}}{b^{2}}}{108 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108*(36*(cos(b*x + a)^3 - 3*cos(b*x + a))*c^2 - 72*(cos(b*x + a)^3 - 3*cos(b*x + a))*a*c*d/b + 36*(cos(b*x +
a)^3 - 3*cos(b*x + a))*a^2*d^2/b^2 + 6*(3*(b*x + a)*cos(3*b*x + 3*a) - 27*(b*x + a)*cos(b*x + a) - sin(3*b*x
+ 3*a) + 27*sin(b*x + a))*c*d/b - 6*(3*(b*x + a)*cos(3*b*x + 3*a) - 27*(b*x + a)*cos(b*x + a) - sin(3*b*x + 3*
a) + 27*sin(b*x + a))*a*d^2/b^2 + ((9*(b*x + a)^2 - 2)*cos(3*b*x + 3*a) - 81*((b*x + a)^2 - 2)*cos(b*x + a) -
6*(b*x + a)*sin(3*b*x + 3*a) + 162*(b*x + a)*sin(b*x + a))*d^2/b^2)/b

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mupad [B]  time = 0.98, size = 174, normalized size = 1.41 \[ \frac {\frac {3\,d^2\,x\,\sin \left (a+b\,x\right )}{2}-\frac {d^2\,x\,\sin \left (3\,a+3\,b\,x\right )}{18}+\frac {3\,c\,d\,\sin \left (a+b\,x\right )}{2}-\frac {c\,d\,\sin \left (3\,a+3\,b\,x\right )}{18}}{b^2}-\frac {\frac {3\,c^2\,\cos \left (a+b\,x\right )}{4}-\frac {c^2\,\cos \left (3\,a+3\,b\,x\right )}{12}+\frac {3\,d^2\,x^2\,\cos \left (a+b\,x\right )}{4}-\frac {d^2\,x^2\,\cos \left (3\,a+3\,b\,x\right )}{12}-\frac {c\,d\,x\,\cos \left (3\,a+3\,b\,x\right )}{6}+\frac {3\,c\,d\,x\,\cos \left (a+b\,x\right )}{2}}{b}+\frac {3\,d^2\,\cos \left (a+b\,x\right )}{2\,b^3}-\frac {d^2\,\cos \left (3\,a+3\,b\,x\right )}{54\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3*(c + d*x)^2,x)

[Out]

((3*d^2*x*sin(a + b*x))/2 - (d^2*x*sin(3*a + 3*b*x))/18 + (3*c*d*sin(a + b*x))/2 - (c*d*sin(3*a + 3*b*x))/18)/
b^2 - ((3*c^2*cos(a + b*x))/4 - (c^2*cos(3*a + 3*b*x))/12 + (3*d^2*x^2*cos(a + b*x))/4 - (d^2*x^2*cos(3*a + 3*
b*x))/12 - (c*d*x*cos(3*a + 3*b*x))/6 + (3*c*d*x*cos(a + b*x))/2)/b + (3*d^2*cos(a + b*x))/(2*b^3) - (d^2*cos(
3*a + 3*b*x))/(54*b^3)

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sympy [A]  time = 3.04, size = 284, normalized size = 2.31 \[ \begin {cases} - \frac {c^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} - \frac {2 c^{2} \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 c d x \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} - \frac {4 c d x \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {d^{2} x^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} - \frac {2 d^{2} x^{2} \cos ^{3}{\left (a + b x \right )}}{3 b} + \frac {14 c d \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {4 c d \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} + \frac {14 d^{2} x \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {4 d^{2} x \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} + \frac {14 d^{2} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{9 b^{3}} + \frac {40 d^{2} \cos ^{3}{\left (a + b x \right )}}{27 b^{3}} & \text {for}\: b \neq 0 \\\left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) \sin ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-c**2*sin(a + b*x)**2*cos(a + b*x)/b - 2*c**2*cos(a + b*x)**3/(3*b) - 2*c*d*x*sin(a + b*x)**2*cos(a
+ b*x)/b - 4*c*d*x*cos(a + b*x)**3/(3*b) - d**2*x**2*sin(a + b*x)**2*cos(a + b*x)/b - 2*d**2*x**2*cos(a + b*x
)**3/(3*b) + 14*c*d*sin(a + b*x)**3/(9*b**2) + 4*c*d*sin(a + b*x)*cos(a + b*x)**2/(3*b**2) + 14*d**2*x*sin(a +
b*x)**3/(9*b**2) + 4*d**2*x*sin(a + b*x)*cos(a + b*x)**2/(3*b**2) + 14*d**2*sin(a + b*x)**2*cos(a + b*x)/(9*b
**3) + 40*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)**3, True))

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