3.1295 \(\int \frac{\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=149 \[ -\frac{\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac{a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac{a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac{a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}+\frac{a \sin ^4(c+d x)}{4 b^2 d}-\frac{\sin ^5(c+d x)}{5 b d} \]

[Out]

(a^3*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^6*d) - (a^2*(a^2 - b^2)*Sin[c + d*x])/(b^5*d) + (a*(a^2 - b^2)*Si
n[c + d*x]^2)/(2*b^4*d) - ((a^2 - b^2)*Sin[c + d*x]^3)/(3*b^3*d) + (a*Sin[c + d*x]^4)/(4*b^2*d) - Sin[c + d*x]
^5/(5*b*d)

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Rubi [A]  time = 0.198837, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2837, 12, 894} \[ -\frac{\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac{a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac{a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac{a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}+\frac{a \sin ^4(c+d x)}{4 b^2 d}-\frac{\sin ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^6*d) - (a^2*(a^2 - b^2)*Sin[c + d*x])/(b^5*d) + (a*(a^2 - b^2)*Si
n[c + d*x]^2)/(2*b^4*d) - ((a^2 - b^2)*Sin[c + d*x]^3)/(3*b^3*d) + (a*Sin[c + d*x]^4)/(4*b^2*d) - Sin[c + d*x]
^5/(5*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (b^2-x^2\right )}{b^3 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (b^2-x^2\right )}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a^4 \left (1-\frac{b^2}{a^2}\right )+a \left (a^2-b^2\right ) x-\left (a^2-b^2\right ) x^2+a x^3-x^4+\frac{a^5-a^3 b^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac{a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac{a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac{a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac{\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac{a \sin ^4(c+d x)}{4 b^2 d}-\frac{\sin ^5(c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.279319, size = 127, normalized size = 0.85 \[ \frac{-\frac{60 a^2 (a-b) (a+b) \sin (c+d x)}{b^5}+\frac{60 a^3 (a-b) (a+b) \log (a+b \sin (c+d x))}{b^6}+\frac{15 a \sin ^4(c+d x)}{b^2}-\frac{20 (a-b) (a+b) \sin ^3(c+d x)}{b^3}+\frac{30 a (a-b) (a+b) \sin ^2(c+d x)}{b^4}-\frac{12 \sin ^5(c+d x)}{b}}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((60*a^3*(a - b)*(a + b)*Log[a + b*Sin[c + d*x]])/b^6 - (60*a^2*(a - b)*(a + b)*Sin[c + d*x])/b^5 + (30*a*(a -
b)*(a + b)*Sin[c + d*x]^2)/b^4 - (20*(a - b)*(a + b)*Sin[c + d*x]^3)/b^3 + (15*a*Sin[c + d*x]^4)/b^2 - (12*Si
n[c + d*x]^5)/b)/(60*d)

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Maple [A]  time = 0.052, size = 182, normalized size = 1.2 \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5\,bd}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,{b}^{2}d}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d{b}^{3}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,bd}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{2\,d{b}^{4}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}a}{2\,{b}^{2}d}}-{\frac{{a}^{4}\sin \left ( dx+c \right ) }{d{b}^{5}}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d{b}^{3}}}+{\frac{{a}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{6}}}-{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sin(d*x+c)^5/b/d+1/4*a*sin(d*x+c)^4/b^2/d-1/3/d/b^3*sin(d*x+c)^3*a^2+1/3*sin(d*x+c)^3/b/d+1/2/d/b^4*sin(d
*x+c)^2*a^3-1/2*a*sin(d*x+c)^2/b^2/d-1/d/b^5*sin(d*x+c)*a^4+a^2*sin(d*x+c)/b^3/d+1/d*a^5/b^6*ln(a+b*sin(d*x+c)
)-a^3*ln(a+b*sin(d*x+c))/b^4/d

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Maxima [A]  time = 1.0133, size = 177, normalized size = 1.19 \begin{align*} -\frac{\frac{12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \,{\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \,{\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \,{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac{60 \,{\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*((12*b^4*sin(d*x + c)^5 - 15*a*b^3*sin(d*x + c)^4 + 20*(a^2*b^2 - b^4)*sin(d*x + c)^3 - 30*(a^3*b - a*b^
3)*sin(d*x + c)^2 + 60*(a^4 - a^2*b^2)*sin(d*x + c))/b^5 - 60*(a^5 - a^3*b^2)*log(b*sin(d*x + c) + a)/b^6)/d

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Fricas [A]  time = 1.62657, size = 297, normalized size = 1.99 \begin{align*} \frac{15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \, a^{3} b^{2} \cos \left (d x + c\right )^{2} + 60 \,{\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \,{\left (3 \, b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{4} b - 10 \, a^{2} b^{3} - 2 \, b^{5} -{\left (5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*a*b^4*cos(d*x + c)^4 - 30*a^3*b^2*cos(d*x + c)^2 + 60*(a^5 - a^3*b^2)*log(b*sin(d*x + c) + a) - 4*(3*
b^5*cos(d*x + c)^4 + 15*a^4*b - 10*a^2*b^3 - 2*b^5 - (5*a^2*b^3 + b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^6*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*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.17064, size = 201, normalized size = 1.35 \begin{align*} -\frac{\frac{12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} - 20 \, b^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b \sin \left (d x + c\right )^{2} + 30 \, a b^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right ) - 60 \, a^{2} b^{2} \sin \left (d x + c\right )}{b^{5}} - \frac{60 \,{\left (a^{5} - a^{3} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*((12*b^4*sin(d*x + c)^5 - 15*a*b^3*sin(d*x + c)^4 + 20*a^2*b^2*sin(d*x + c)^3 - 20*b^4*sin(d*x + c)^3 -
30*a^3*b*sin(d*x + c)^2 + 30*a*b^3*sin(d*x + c)^2 + 60*a^4*sin(d*x + c) - 60*a^2*b^2*sin(d*x + c))/b^5 - 60*(a
^5 - a^3*b^2)*log(abs(b*sin(d*x + c) + a))/b^6)/d