3.1005 \(\int \frac{\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac{(A-B) \sin ^2(c+d x)}{2 a d}+\frac{A \sin (c+d x)}{a d}-\frac{B \sin ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.0954406, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2836, 43} \[ -\frac{(A-B) \sin ^2(c+d x)}{2 a d}+\frac{A \sin (c+d x)}{a d}-\frac{B \sin ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]

[Out]

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

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0941487, size = 44, normalized size = 0.77 \[ \frac{\sin (c+d x) \left (-3 (A-B) \sin (c+d x)+6 A-2 B \sin ^2(c+d x)\right )}{6 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]

[Out]

(Sin[c + d*x]*(6*A - 3*(A - B)*Sin[c + d*x] - 2*B*Sin[c + d*x]^2))/(6*a*d)

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Maple [A]  time = 0.085, size = 43, normalized size = 0.8 \begin{align*}{\frac{1}{da} \left ( -{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( -A+B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}}+A\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.04723, size = 59, normalized size = 1.04 \begin{align*} -\frac{2 \, B \sin \left (d x + c\right )^{3} + 3 \,{\left (A - B\right )} \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/6*(2*B*sin(d*x + c)^3 + 3*(A - B)*sin(d*x + c)^2 - 6*A*sin(d*x + c))/(a*d)

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Fricas [A]  time = 1.72857, size = 113, normalized size = 1.98 \begin{align*} \frac{3 \,{\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (B \cos \left (d x + c\right )^{2} + 3 \, A - B\right )} \sin \left (d x + c\right )}{6 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*(A - B)*cos(d*x + c)^2 + 2*(B*cos(d*x + c)^2 + 3*A - B)*sin(d*x + c))/(a*d)

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Sympy [A]  time = 13.6641, size = 588, normalized size = 10.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((6*A*tan(c/2 + d*x/2)**5/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*
x/2)**2 + 3*a*d) - 6*A*tan(c/2 + d*x/2)**4/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(
c/2 + d*x/2)**2 + 3*a*d) + 12*A*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9
*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*A*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2
)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*A*tan(c/2 + d*x/2)/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 +
d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*B*tan(c/2 + d*x/2)**4/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*ta
n(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 8*B*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)**6 +
9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*B*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/
2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**3
/(a*sin(c) + a), True))

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Giac [A]  time = 1.30967, size = 69, normalized size = 1.21 \begin{align*} -\frac{2 \, B \sin \left (d x + c\right )^{3} + 3 \, A \sin \left (d x + c\right )^{2} - 3 \, B \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(2*B*sin(d*x + c)^3 + 3*A*sin(d*x + c)^2 - 3*B*sin(d*x + c)^2 - 6*A*sin(d*x + c))/(a*d)