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

Optimal. Leaf size=51 \[ \frac{(a+2 b) \cos ^3(c+d x)}{3 d}-\frac{(a+b) \cos (c+d x)}{d}-\frac{b \cos ^5(c+d x)}{5 d} \]

[Out]

-(((a + b)*Cos[c + d*x])/d) + ((a + 2*b)*Cos[c + d*x]^3)/(3*d) - (b*Cos[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0451092, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3013, 373} \[ \frac{(a+2 b) \cos ^3(c+d x)}{3 d}-\frac{(a+b) \cos (c+d x)}{d}-\frac{b \cos ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)*Cos[c + d*x])/d) + ((a + 2*b)*Cos[c + d*x]^3)/(3*d) - (b*Cos[c + d*x]^5)/(5*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]

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \sin ^3(c+d x) \left (a+b \sin ^2(c+d x)\right ) \, dx &=-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \left (a+b-b x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (a \left (1+\frac{b}{a}\right )-(a+2 b) x^2+b x^4\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{(a+b) \cos (c+d x)}{d}+\frac{(a+2 b) \cos ^3(c+d x)}{3 d}-\frac{b \cos ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.0285599, size = 77, normalized size = 1.51 \[ -\frac{3 a \cos (c+d x)}{4 d}+\frac{a \cos (3 (c+d x))}{12 d}-\frac{5 b \cos (c+d x)}{8 d}+\frac{5 b \cos (3 (c+d x))}{48 d}-\frac{b \cos (5 (c+d x))}{80 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Cos[c + d*x])/(4*d) - (5*b*Cos[c + d*x])/(8*d) + (a*Cos[3*(c + d*x)])/(12*d) + (5*b*Cos[3*(c + d*x)])/(4
8*d) - (b*Cos[5*(c + d*x)])/(80*d)

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Maple [A]  time = 0.029, size = 54, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( -{\frac{b\cos \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }-{\frac{a \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/5*b*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)-1/3*a*(2+sin(d*x+c)^2)*cos(d*x+c))

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Maxima [A]  time = 0.963426, size = 58, normalized size = 1.14 \begin{align*} -\frac{3 \, b \cos \left (d x + c\right )^{5} - 5 \,{\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (a + b\right )} \cos \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*b*cos(d*x + c)^5 - 5*(a + 2*b)*cos(d*x + c)^3 + 15*(a + b)*cos(d*x + c))/d

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Fricas [A]  time = 1.92667, size = 115, normalized size = 2.25 \begin{align*} -\frac{3 \, b \cos \left (d x + c\right )^{5} - 5 \,{\left (a + 2 \, b\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (a + b\right )} \cos \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*b*cos(d*x + c)^5 - 5*(a + 2*b)*cos(d*x + c)^3 + 15*(a + b)*cos(d*x + c))/d

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Sympy [A]  time = 2.33809, size = 107, normalized size = 2.1 \begin{align*} \begin{cases} - \frac{a \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{2 a \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{b \sin ^{4}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{4 b \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{8 b \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin ^{2}{\left (c \right )}\right ) \sin ^{3}{\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)**3*(a+b*sin(d*x+c)**2),x)

[Out]

Piecewise((-a*sin(c + d*x)**2*cos(c + d*x)/d - 2*a*cos(c + d*x)**3/(3*d) - b*sin(c + d*x)**4*cos(c + d*x)/d -
4*b*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 8*b*cos(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a + b*sin(c)**2)*sin(c)
**3, True))

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Giac [A]  time = 1.10308, size = 90, normalized size = 1.76 \begin{align*} -\frac{b \cos \left (d x + c\right )^{5}}{5 \, d} + \frac{a \cos \left (d x + c\right )^{3}}{3 \, d} + \frac{2 \, b \cos \left (d x + c\right )^{3}}{3 \, d} - \frac{a \cos \left (d x + c\right )}{d} - \frac{b \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)^3*(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/5*b*cos(d*x + c)^5/d + 1/3*a*cos(d*x + c)^3/d + 2/3*b*cos(d*x + c)^3/d - a*cos(d*x + c)/d - b*cos(d*x + c)/
d