3.1228 \(\int \frac{\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=157 \[ \frac{\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac{4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac{a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))}+\frac{\left (-6 a^2 b^2+5 a^4+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac{2 a \sin ^3(c+d x)}{3 b^3 d}+\frac{\sin ^4(c+d x)}{4 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.154681, antiderivative size = 157, 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} \[ \frac{\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac{4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac{a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))}+\frac{\left (-6 a^2 b^2+5 a^4+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac{2 a \sin ^3(c+d x)}{3 b^3 d}+\frac{\sin ^4(c+d x)}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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

Mathematica [A]  time = 1.00147, size = 188, normalized size = 1.2 \[ \frac{2 b^3 \left (5 a^2-6 b^2\right ) \sin ^3(c+d x)-6 a b^2 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)+12 b \left (b^2-a^2\right ) \sin (c+d x) \left (\left (b^2-5 a^2\right ) \log (a+b \sin (c+d x))+4 a^2\right )+12 a \left (a^2-b^2\right ) \left (\left (5 a^2-b^2\right ) \log (a+b \sin (c+d x))+a^2-b^2\right )-5 a b^4 \sin ^4(c+d x)+3 b^5 \sin ^5(c+d x)}{12 b^6 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.078, size = 229, normalized size = 1.5 \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,{b}^{2}d}}-{\frac{2\,a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,{b}^{3}d}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}}{2\,d{b}^{4}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{{b}^{2}d}}-4\,{\frac{{a}^{3}\sin \left ( dx+c \right ) }{d{b}^{5}}}+4\,{\frac{a\sin \left ( dx+c \right ) }{{b}^{3}d}}+5\,{\frac{{a}^{4}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{6}}}-6\,{\frac{{a}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{4}}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{{b}^{2}d}}+{\frac{{a}^{5}}{d{b}^{6} \left ( a+b\sin \left ( dx+c \right ) \right ) }}-2\,{\frac{{a}^{3}}{d{b}^{4} \left ( a+b\sin \left ( dx+c \right ) \right ) }}+{\frac{a}{{b}^{2}d \left ( a+b\sin \left ( dx+c \right ) \right ) }} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.02555, size = 200, normalized size = 1.27 \begin{align*} \frac{\frac{12 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{b^{7} \sin \left (d x + c\right ) + a b^{6}} + \frac{3 \, b^{3} \sin \left (d x + c\right )^{4} - 8 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \,{\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - 48 \,{\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} + \frac{12 \,{\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{12 \, 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))^2,x, algorithm="maxima")

[Out]

1/12*(12*(a^5 - 2*a^3*b^2 + a*b^4)/(b^7*sin(d*x + c) + a*b^6) + (3*b^3*sin(d*x + c)^4 - 8*a*b^2*sin(d*x + c)^3
+ 6*(3*a^2*b - 2*b^3)*sin(d*x + c)^2 - 48*(a^3 - a*b^2)*sin(d*x + c))/b^5 + 12*(5*a^4 - 6*a^2*b^2 + b^4)*log(
b*sin(d*x + c) + a)/b^6)/d

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Fricas [A]  time = 1.94763, size = 482, normalized size = 3.07 \begin{align*} -\frac{40 \, a b^{4} \cos \left (d x + c\right )^{4} - 96 \, a^{5} + 504 \, a^{3} b^{2} - 383 \, a b^{4} - 16 \,{\left (15 \, a^{3} b^{2} - 13 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 96 \,{\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4} +{\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left (24 \, b^{5} \cos \left (d x + c\right )^{4} - 384 \, a^{4} b + 392 \, a^{2} b^{3} - 33 \, b^{5} - 16 \,{\left (5 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \,{\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}} \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))^2,x, algorithm="fricas")

[Out]

-1/96*(40*a*b^4*cos(d*x + c)^4 - 96*a^5 + 504*a^3*b^2 - 383*a*b^4 - 16*(15*a^3*b^2 - 13*a*b^4)*cos(d*x + c)^2
- 96*(5*a^5 - 6*a^3*b^2 + a*b^4 + (5*a^4*b - 6*a^2*b^3 + b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (24*b^5*
cos(d*x + c)^4 - 384*a^4*b + 392*a^2*b^3 - 33*b^5 - 16*(5*a^2*b^3 - 3*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*
d*sin(d*x + c) + a*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20231, size = 262, normalized size = 1.67 \begin{align*} \frac{\frac{12 \,{\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}} - \frac{12 \,{\left (5 \, a^{4} b \sin \left (d x + c\right ) - 6 \, a^{2} b^{3} \sin \left (d x + c\right ) + b^{5} \sin \left (d x + c\right ) + 4 \, a^{5} - 4 \, a^{3} b^{2}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{6}} + \frac{3 \, b^{6} \sin \left (d x + c\right )^{4} - 8 \, a b^{5} \sin \left (d x + c\right )^{3} + 18 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} - 12 \, b^{6} \sin \left (d x + c\right )^{2} - 48 \, a^{3} b^{3} \sin \left (d x + c\right ) + 48 \, a b^{5} \sin \left (d x + c\right )}{b^{8}}}{12 \, 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))^2,x, algorithm="giac")

[Out]

1/12*(12*(5*a^4 - 6*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/b^6 - 12*(5*a^4*b*sin(d*x + c) - 6*a^2*b^3*sin
(d*x + c) + b^5*sin(d*x + c) + 4*a^5 - 4*a^3*b^2)/((b*sin(d*x + c) + a)*b^6) + (3*b^6*sin(d*x + c)^4 - 8*a*b^5
*sin(d*x + c)^3 + 18*a^2*b^4*sin(d*x + c)^2 - 12*b^6*sin(d*x + c)^2 - 48*a^3*b^3*sin(d*x + c) + 48*a*b^5*sin(d
*x + c))/b^8)/d