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

Optimal. Leaf size=55 \[ \frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {c x}{2}+\frac {d x^2}{4} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3310} \[ \frac {d \sin ^2(a+b x)}{4 b^2}-\frac {(c+d x) \sin (a+b x) \cos (a+b x)}{2 b}+\frac {c x}{2}+\frac {d x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sin[a + b*x]^2,x]

[Out]

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

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

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Mathematica [A]  time = 0.16, size = 52, normalized size = 0.95 \[ \frac {2 b (-(c+d x) \sin (2 (a+b x))+2 a c+b x (2 c+d x))-d \cos (2 (a+b x))}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sin[a + b*x]^2,x]

[Out]

(-(d*Cos[2*(a + b*x)]) + 2*b*(2*a*c + b*x*(2*c + d*x) - (c + d*x)*Sin[2*(a + b*x)]))/(8*b^2)

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fricas [A]  time = 0.82, size = 54, normalized size = 0.98 \[ \frac {b^{2} d x^{2} + 2 \, b^{2} c x - d \cos \left (b x + a\right )^{2} - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 48, normalized size = 0.87 \[ \frac {1}{4} \, d x^{2} + \frac {1}{2} \, c x - \frac {d \cos \left (2 \, b x + 2 \, a\right )}{8 \, b^{2}} - \frac {{\left (b d x + b c\right )} \sin \left (2 \, b x + 2 \, a\right )}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 112, normalized size = 2.04 \[ \frac {\frac {d \left (\left (b x +a \right ) \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )-\frac {\left (b x +a \right )^{2}}{4}+\frac {\left (\sin ^{2}\left (b x +a \right )\right )}{4}\right )}{b}-\frac {d a \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b}+c \left (-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.30, size = 96, normalized size = 1.75 \[ \frac {2 \, {\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} c - \frac {2 \, {\left (2 \, b x + 2 \, a - \sin \left (2 \, b x + 2 \, a\right )\right )} a d}{b} + \frac {{\left (2 \, {\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right )\right )} d}{b}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 57, normalized size = 1.04 \[ \frac {c\,x}{2}+\frac {d\,x^2}{4}-\frac {d\,\cos \left (2\,a+2\,b\,x\right )}{8\,b^2}-\frac {c\,\sin \left (2\,a+2\,b\,x\right )}{4\,b}-\frac {d\,x\,\sin \left (2\,a+2\,b\,x\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2*(c + d*x),x)

[Out]

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

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sympy [A]  time = 0.67, size = 126, normalized size = 2.29 \[ \begin {cases} \frac {c x \sin ^{2}{\left (a + b x \right )}}{2} + \frac {c x \cos ^{2}{\left (a + b x \right )}}{2} + \frac {d x^{2} \sin ^{2}{\left (a + b x \right )}}{4} + \frac {d x^{2} \cos ^{2}{\left (a + b x \right )}}{4} - \frac {c \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} - \frac {d x \sin {\left (a + b x \right )} \cos {\left (a + b x \right )}}{2 b} - \frac {d \cos ^{2}{\left (a + b x \right )}}{4 b^{2}} & \text {for}\: b \neq 0 \\\left (c x + \frac {d x^{2}}{2}\right ) \sin ^{2}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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