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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.64, size = 128, normalized size = 0.86 \[ \frac {-30 a b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)+60 b \left (a^2-b^2\right )^2 \sin (c+d x)-60 a \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))+20 b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-15 a b^4 \sin ^4(c+d x)+12 b^5 \sin ^5(c+d x)}{60 b^6 d} \]

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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fricas [A]  time = 0.90, size = 142, normalized size = 0.96 \[ -\frac {15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{4} b - 25 \, a^{2} b^{3} + 8 \, b^{5} - {\left (5 \, a^{2} b^{3} - 4 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \]

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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giac [A]  time = 0.18, size = 165, normalized size = 1.11 \[ \frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} - 40 \, b^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b \sin \left (d x + c\right )^{2} + 60 \, a b^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right ) - 120 \, a^{2} b^{2} \sin \left (d x + c\right ) + 60 \, b^{4} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}}}{60 \, d} \]

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

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maple [A]  time = 0.22, size = 215, normalized size = 1.45 \[ \frac {\sin ^{5}\left (d x +c \right )}{5 b d}-\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{4 b^{2} d}+\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{2}}{3 d \,b^{3}}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3 b d}-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) a^{3}}{2 d \,b^{4}}+\frac {a \left (\sin ^{2}\left (d x +c \right )\right )}{b^{2} d}+\frac {a^{4} \sin \left (d x +c \right )}{d \,b^{5}}-\frac {2 a^{2} \sin \left (d x +c \right )}{b^{3} d}+\frac {\sin \left (d x +c \right )}{b d}-\frac {a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{6}}+\frac {2 a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4} d}-\frac {a \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{2}} \]

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*a^4*sin(d*x+c)-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 = 0.31, size = 139, normalized size = 0.94 \[ \frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (a^{2} b^{2} - 2 \, b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \]

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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mupad [B]  time = 0.07, size = 150, normalized size = 1.01 \[ \frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b^2}\right )-{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,b}-\frac {a^2}{3\,b^3}\right )+\frac {{\sin \left (c+d\,x\right )}^5}{5\,b}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^5-2\,a^3\,b^2+a\,b^4\right )}{b^6}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{4\,b^2}+\frac {a\,{\sin \left (c+d\,x\right )}^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{2\,b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x))/(a + b*sin(c + d*x)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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