3.1546 \(\int \frac{\cos ^3(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=111 \[ \frac{\left (a^2 (-B)+a A b+b^2 B\right ) \sin (c+d x)}{b^3 d}-\frac{\left (a^2-b^2\right ) (A b-a B) \log (a+b \sin (c+d x))}{b^4 d}-\frac{(A b-a B) \sin ^2(c+d x)}{2 b^2 d}-\frac{B \sin ^3(c+d x)}{3 b d} \]

[Out]

-(((a^2 - b^2)*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/(b^4*d)) + ((a*A*b - a^2*B + b^2*B)*Sin[c + d*x])/(b^3*d)
- ((A*b - a*B)*Sin[c + d*x]^2)/(2*b^2*d) - (B*Sin[c + d*x]^3)/(3*b*d)

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Rubi [A]  time = 0.162742, antiderivative size = 111, 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 = {2837, 772} \[ \frac{\left (a^2 (-B)+a A b+b^2 B\right ) \sin (c+d x)}{b^3 d}-\frac{\left (a^2-b^2\right ) (A b-a B) \log (a+b \sin (c+d x))}{b^4 d}-\frac{(A b-a B) \sin ^2(c+d x)}{2 b^2 d}-\frac{B \sin ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)*(A*b - a*B)*Log[a + b*Sin[c + d*x]])/(b^4*d)) + ((a*A*b - a^2*B + b^2*B)*Sin[c + d*x])/(b^3*d)
- ((A*b - a*B)*Sin[c + d*x]^2)/(2*b^2*d) - (B*Sin[c + d*x]^3)/(3*b*d)

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.37608, size = 89, normalized size = 0.8 \[ \frac{\left (A-\frac{a B}{b}\right ) \left (\left (b^2-a^2\right ) \log (a+b \sin (c+d x))+a b \sin (c+d x)-\frac{1}{2} b^2 \sin ^2(c+d x)\right )+\frac{1}{12} b^2 B (9 \sin (c+d x)+\sin (3 (c+d x)))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A - (a*B)/b)*((-a^2 + b^2)*Log[a + b*Sin[c + d*x]] + a*b*Sin[c + d*x] - (b^2*Sin[c + d*x]^2)/2) + (b^2*B*(9*
Sin[c + d*x] + Sin[3*(c + d*x)]))/12)/(b^3*d)

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Maple [A]  time = 0.066, size = 186, normalized size = 1.7 \begin{align*} -{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,bd}}-{\frac{A \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,bd}}+{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{2}a}{2\,{b}^{2}d}}+{\frac{A\sin \left ( dx+c \right ) a}{{b}^{2}d}}-{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d{b}^{3}}}+{\frac{B\sin \left ( dx+c \right ) }{bd}}-{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) A{a}^{2}}{d{b}^{3}}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) A}{bd}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) B{a}^{3}}{d{b}^{4}}}-{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) Ba}{{b}^{2}d}} \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+b*sin(d*x+c)),x)

[Out]

-1/3*B*sin(d*x+c)^3/b/d-1/2/d/b*A*sin(d*x+c)^2+1/2/d/b^2*B*sin(d*x+c)^2*a+1/d/b^2*A*a*sin(d*x+c)-1/d/b^3*B*a^2
*sin(d*x+c)+B*sin(d*x+c)/b/d-1/d/b^3*ln(a+b*sin(d*x+c))*A*a^2+1/d/b*ln(a+b*sin(d*x+c))*A+1/d/b^4*ln(a+b*sin(d*
x+c))*B*a^3-1/d/b^2*ln(a+b*sin(d*x+c))*B*a

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Maxima [A]  time = 0.984976, size = 151, normalized size = 1.36 \begin{align*} -\frac{\frac{2 \, B b^{2} \sin \left (d x + c\right )^{3} - 3 \,{\left (B a b - A b^{2}\right )} \sin \left (d x + c\right )^{2} + 6 \,{\left (B a^{2} - A a b - B b^{2}\right )} \sin \left (d x + c\right )}{b^{3}} - \frac{6 \,{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{6 \, 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+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/6*((2*B*b^2*sin(d*x + c)^3 - 3*(B*a*b - A*b^2)*sin(d*x + c)^2 + 6*(B*a^2 - A*a*b - B*b^2)*sin(d*x + c))/b^3
- 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(b*sin(d*x + c) + a)/b^4)/d

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Fricas [A]  time = 1.53238, size = 255, normalized size = 2.3 \begin{align*} -\frac{3 \,{\left (B a b^{2} - A b^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \,{\left (B b^{3} \cos \left (d x + c\right )^{2} - 3 \, B a^{2} b + 3 \, A a b^{2} + 2 \, B b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} 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+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(3*(B*a*b^2 - A*b^3)*cos(d*x + c)^2 - 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(b*sin(d*x + c) + a) - 2*(
B*b^3*cos(d*x + c)^2 - 3*B*a^2*b + 3*A*a*b^2 + 2*B*b^3)*sin(d*x + c))/(b^4*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.21751, size = 174, normalized size = 1.57 \begin{align*} -\frac{\frac{2 \, B b^{2} \sin \left (d x + c\right )^{3} - 3 \, B a b \sin \left (d x + c\right )^{2} + 3 \, A b^{2} \sin \left (d x + c\right )^{2} + 6 \, B a^{2} \sin \left (d x + c\right ) - 6 \, A a b \sin \left (d x + c\right ) - 6 \, B b^{2} \sin \left (d x + c\right )}{b^{3}} - \frac{6 \,{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4}}}{6 \, 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+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*((2*B*b^2*sin(d*x + c)^3 - 3*B*a*b*sin(d*x + c)^2 + 3*A*b^2*sin(d*x + c)^2 + 6*B*a^2*sin(d*x + c) - 6*A*a
*b*sin(d*x + c) - 6*B*b^2*sin(d*x + c))/b^3 - 6*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)*log(abs(b*sin(d*x + c) + a
))/b^4)/d