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

Optimal. Leaf size=75 \[ \frac {d \sin ^3(a+b x)}{9 b^2}+\frac {2 d \sin (a+b x)}{3 b^2}-\frac {2 (c+d x) \cos (a+b x)}{3 b}-\frac {(c+d x) \sin ^2(a+b x) \cos (a+b x)}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3310, 3296, 2637} \[ \frac {d \sin ^3(a+b x)}{9 b^2}+\frac {2 d \sin (a+b x)}{3 b^2}-\frac {2 (c+d x) \cos (a+b x)}{3 b}-\frac {(c+d x) \sin ^2(a+b x) \cos (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sin[a + b*x]^3,x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 59, normalized size = 0.79 \[ \frac {-27 b (c+d x) \cos (a+b x)+3 b (c+d x) \cos (3 (a+b x))+d (27 \sin (a+b x)-\sin (3 (a+b x)))}{36 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sin[a + b*x]^3,x]

[Out]

(-27*b*(c + d*x)*Cos[a + b*x] + 3*b*(c + d*x)*Cos[3*(a + b*x)] + d*(27*Sin[a + b*x] - Sin[3*(a + b*x)]))/(36*b
^2)

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fricas [A]  time = 0.59, size = 62, normalized size = 0.83 \[ \frac {3 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )^{3} - 9 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) - {\left (d \cos \left (b x + a\right )^{2} - 7 \, d\right )} \sin \left (b x + a\right )}{9 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*(b*d*x + b*c)*cos(b*x + a)^3 - 9*(b*d*x + b*c)*cos(b*x + a) - (d*cos(b*x + a)^2 - 7*d)*sin(b*x + a))/b^
2

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giac [A]  time = 1.21, size = 69, normalized size = 0.92 \[ \frac {{\left (b d x + b c\right )} \cos \left (3 \, b x + 3 \, a\right )}{12 \, b^{2}} - \frac {3 \, {\left (b d x + b c\right )} \cos \left (b x + a\right )}{4 \, b^{2}} - \frac {d \sin \left (3 \, b x + 3 \, a\right )}{36 \, b^{2}} + \frac {3 \, d \sin \left (b x + a\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(b*d*x + b*c)*cos(3*b*x + 3*a)/b^2 - 3/4*(b*d*x + b*c)*cos(b*x + a)/b^2 - 1/36*d*sin(3*b*x + 3*a)/b^2 + 3
/4*d*sin(b*x + a)/b^2

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maple [A]  time = 0.02, size = 95, normalized size = 1.27 \[ \frac {\frac {d \left (-\frac {\left (b x +a \right ) \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{9}+\frac {2 \sin \left (b x +a \right )}{3}\right )}{b}+\frac {d a \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3 b}-\frac {c \left (2+\sin ^{2}\left (b x +a \right )\right ) \cos \left (b x +a \right )}{3}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 104, normalized size = 1.39 \[ \frac {12 \, {\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} c - \frac {12 \, {\left (\cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} a d}{b} + \frac {{\left (3 \, {\left (b x + a\right )} \cos \left (3 \, b x + 3 \, a\right ) - 27 \, {\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (3 \, b x + 3 \, a\right ) + 27 \, \sin \left (b x + a\right )\right )} d}{b}}{36 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(12*(cos(b*x + a)^3 - 3*cos(b*x + a))*c - 12*(cos(b*x + a)^3 - 3*cos(b*x + a))*a*d/b + (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))*d/b)/b

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mupad [B]  time = 0.63, size = 79, normalized size = 1.05 \[ \frac {7\,d\,\sin \left (a+b\,x\right )}{9\,b^2}-\frac {c\,\cos \left (a+b\,x\right )-\frac {c\,{\cos \left (a+b\,x\right )}^3}{3}+d\,x\,\cos \left (a+b\,x\right )-\frac {d\,x\,{\cos \left (a+b\,x\right )}^3}{3}}{b}-\frac {d\,{\cos \left (a+b\,x\right )}^2\,\sin \left (a+b\,x\right )}{9\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3*(c + d*x),x)

[Out]

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

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sympy [A]  time = 1.25, size = 126, normalized size = 1.68 \[ \begin {cases} - \frac {c \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} - \frac {2 c \cos ^{3}{\left (a + b x \right )}}{3 b} - \frac {d x \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{b} - \frac {2 d x \cos ^{3}{\left (a + b x \right )}}{3 b} + \frac {7 d \sin ^{3}{\left (a + b x \right )}}{9 b^{2}} + \frac {2 d \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-c*sin(a + b*x)**2*cos(a + b*x)/b - 2*c*cos(a + b*x)**3/(3*b) - d*x*sin(a + b*x)**2*cos(a + b*x)/b
- 2*d*x*cos(a + b*x)**3/(3*b) + 7*d*sin(a + b*x)**3/(9*b**2) + 2*d*sin(a + b*x)*cos(a + b*x)**2/(3*b**2), Ne(b
, 0)), ((c*x + d*x**2/2)*sin(a)**3, True))

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