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

Optimal. Leaf size=31 \[ \frac{b \cos ^3(c+d x)}{3 d}-\frac{(a+b) \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0224598, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {3013} \[ \frac{b \cos ^3(c+d x)}{3 d}-\frac{(a+b) \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

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

Mathematica [A]  time = 0.0194513, size = 54, normalized size = 1.74 \[ \frac{a \sin (c) \sin (d x)}{d}-\frac{a \cos (c) \cos (d x)}{d}-\frac{3 b \cos (c+d x)}{4 d}+\frac{b \cos (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cos[c]*Cos[d*x])/d) - (3*b*Cos[c + d*x])/(4*d) + (b*Cos[3*(c + d*x)])/(12*d) + (a*Sin[c]*Sin[d*x])/d

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Maple [A]  time = 0.028, size = 34, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -{\frac{b \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}-\cos \left ( dx+c \right ) a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.936997, size = 46, normalized size = 1.48 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b - 3 \, a \cos \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70839, size = 69, normalized size = 2.23 \begin{align*} \frac{b \cos \left (d x + c\right )^{3} - 3 \,{\left (a + b\right )} \cos \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.901478, size = 58, normalized size = 1.87 \begin{align*} \begin{cases} - \frac{a \cos{\left (c + d x \right )}}{d} - \frac{b \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 b \cos ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin ^{2}{\left (c \right )}\right ) \sin{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13028, size = 54, normalized size = 1.74 \begin{align*} \frac{1}{3} \,{\left (\frac{\cos \left (d x + c\right )^{3}}{d} - \frac{3 \, \cos \left (d x + c\right )}{d}\right )} b - \frac{a \cos \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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