3.1312 \(\int \frac{\cos ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=148 \[ \frac{\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^3 d}-\frac{a \left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac{\left (a^2-b^2\right )^2 \sin (c+d x)}{b^5 d}-\frac{a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^6 d}-\frac{a \sin ^4(c+d x)}{4 b^2 d}+\frac{\sin ^5(c+d x)}{5 b d} \]

[Out]

-((a*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(b^6*d)) + ((a^2 - b^2)^2*Sin[c + d*x])/(b^5*d) - (a*(a^2 - 2*b^2)
*Sin[c + d*x]^2)/(2*b^4*d) + ((a^2 - 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 = 0.140724, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 772} \[ \frac{\left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^3 d}-\frac{a \left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac{\left (a^2-b^2\right )^2 \sin (c+d x)}{b^5 d}-\frac{a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^6 d}-\frac{a \sin ^4(c+d x)}{4 b^2 d}+\frac{\sin ^5(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

-((a*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/(b^6*d)) + ((a^2 - b^2)^2*Sin[c + d*x])/(b^5*d) - (a*(a^2 - 2*b^2)
*Sin[c + d*x]^2)/(2*b^4*d) + ((a^2 - 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 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 ^5(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x \left (b^2-x^2\right )^2}{b (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x \left (b^2-x^2\right )^2}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\left (a^2-b^2\right )^2-a \left (a^2-2 b^2\right ) x+\left (a^2-2 b^2\right ) x^2-a x^3+x^4-\frac{a \left (a^2-b^2\right )^2}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^6 d}\\ &=-\frac{a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^6 d}+\frac{\left (a^2-b^2\right )^2 \sin (c+d x)}{b^5 d}-\frac{a \left (a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}+\frac{\left (a^2-2 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 = 0.611066, size = 128, normalized size = 0.86 \[ \frac{20 b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-30 a b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)+60 b \left (a^2-b^2\right )^2 \sin (c+d x)-60 a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-15 a b^4 \sin ^4(c+d x)+12 b^5 \sin ^5(c+d x)}{60 b^6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.043, size = 215, normalized size = 1.5 \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{2\, \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}{{b}^{2}d}}+{\frac{{a}^{4}\sin \left ( dx+c \right ) }{d{b}^{5}}}-2\,{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d{b}^{3}}}+{\frac{\sin \left ( dx+c \right ) }{bd}}-{\frac{{a}^{5}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{6}}}+2\,{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{4}}}-{\frac{a\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.00154, size = 188, normalized size = 1.27 \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} - 2 \, b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \,{\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac{60 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\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)^5*sin(d*x+c)/(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 - 2*b^4)*sin(d*x + c)^3 - 30*(a^3*b - 2*a
*b^3)*sin(d*x + c)^2 + 60*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c))/b^5 - 60*(a^5 - 2*a^3*b^2 + a*b^4)*log(b*sin(d
*x + c) + a)/b^6)/d

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Fricas [A]  time = 1.50313, size = 328, normalized size = 2.22 \begin{align*} -\frac{15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + 60 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\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 - 25 \, a^{2} b^{3} + 8 \, b^{5} -{\left (5 \, a^{2} b^{3} - 4 \, 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)^5*sin(d*x+c)/(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 - a*b^4)*cos(d*x + c)^2 + 60*(a^5 - 2*a^3*b^2 + a*b^4)*log(b*sin(
d*x + c) + a) - 4*(3*b^5*cos(d*x + c)^4 + 15*a^4*b - 25*a^2*b^3 + 8*b^5 - (5*a^2*b^3 - 4*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)**5*sin(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.17555, size = 223, normalized size = 1.51 \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} - 40 \, b^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b \sin \left (d x + c\right )^{2} + 60 \, a b^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right ) - 120 \, a^{2} b^{2} \sin \left (d x + c\right ) + 60 \, b^{4} \sin \left (d x + c\right )}{b^{5}} - \frac{60 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\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)^5*sin(d*x+c)/(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 - 40*b^4*sin(d*x + c)^3 - 3
0*a^3*b*sin(d*x + c)^2 + 60*a*b^3*sin(d*x + c)^2 + 60*a^4*sin(d*x + c) - 120*a^2*b^2*sin(d*x + c) + 60*b^4*sin
(d*x + c))/b^5 - 60*(a^5 - 2*a^3*b^2 + a*b^4)*log(abs(b*sin(d*x + c) + a))/b^6)/d