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

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Rubi [A]  time = 0.10, antiderivative size = 57, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 0.78, size = 49, normalized size = 0.86 \[ \frac {3 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B \cos \left (d x + c\right )^{2} + 3 \, A - B\right )} \sin \left (d x + c\right )}{6 \, a d} \]

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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giac [A]  time = 0.18, size = 51, normalized size = 0.89 \[ -\frac {2 \, B \sin \left (d x + c\right )^{3} + 3 \, A \sin \left (d x + c\right )^{2} - 3 \, B \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \]

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)

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maple [A]  time = 0.44, size = 43, normalized size = 0.75 \[ \frac {-\frac {B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{d a} \]

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 = 0.33, size = 44, normalized size = 0.77 \[ -\frac {2 \, B \sin \left (d x + c\right )^{3} + 3 \, {\left (A - B\right )} \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \]

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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mupad [B]  time = 0.08, size = 47, normalized size = 0.82 \[ \frac {\sin \left (c+d\,x\right )\,\left (6\,A-3\,A\,\sin \left (c+d\,x\right )+3\,B\,\sin \left (c+d\,x\right )-2\,B\,{\sin \left (c+d\,x\right )}^2\right )}{6\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [A]  time = 9.86, size = 588, normalized size = 10.32 \[ \begin {cases} \frac {6 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 A \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {12 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 A \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 B \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {8 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 B \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \sin {\relax (c )}\right ) \cos ^{3}{\relax (c )}}{a \sin {\relax (c )} + a} & \text {otherwise} \end {cases} \]

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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