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

Optimal. Leaf size=45 \[ -\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2644} \[ -\frac {2 a^2 \cos (c+d x)}{d}-\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 34, normalized size = 0.76 \[ -\frac {a^2 (-6 (c+d x)+\sin (2 (c+d x))+8 \cos (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2,x]

[Out]

-1/4*(a^2*(-6*(c + d*x) + 8*Cos[c + d*x] + Sin[2*(c + d*x)]))/d

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fricas [A]  time = 0.42, size = 41, normalized size = 0.91 \[ \frac {3 \, a^{2} d x - a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, a^{2} \cos \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 38, normalized size = 0.84 \[ \frac {3}{2} \, a^{2} x - \frac {2 \, a^{2} \cos \left (d x + c\right )}{d} - \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 52, normalized size = 1.16 \[ \frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-2 a^{2} \cos \left (d x +c \right )+a^{2} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 47, normalized size = 1.04 \[ a^{2} x + \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} - \frac {2 \, a^{2} \cos \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.58, size = 123, normalized size = 2.73 \[ \frac {3\,a^2\,x}{2}-\frac {a^2\,\left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )-a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-a^2\,\left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}-4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2\,\left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )-a^2\,\left (3\,c+3\,d\,x-4\right )\right )+a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^2,x)

[Out]

(3*a^2*x)/2 - (a^2*((3*c)/2 + (3*d*x)/2) - a^2*tan(c/2 + (d*x)/2)^3 - a^2*((3*c)/2 + (3*d*x)/2 - 4) + tan(c/2
+ (d*x)/2)^2*(2*a^2*((3*c)/2 + (3*d*x)/2) - a^2*(3*c + 3*d*x - 4)) + a^2*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*
x)/2)^2 + 1)^2)

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sympy [A]  time = 0.42, size = 78, normalized size = 1.73 \[ \begin {cases} \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + a^{2} x - \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {2 a^{2} \cos {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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