3.1227 \(\int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=193 \[ -\frac {\left (2-\frac {3 a^2}{b^2}\right ) \sin ^3(c+d x)}{3 b^2 d}-\frac {a^2 \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))}-\frac {2 a \left (a^2-b^2\right ) \sin ^2(c+d x)}{b^5 d}-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \sin (c+d x)}{b^6 d}-\frac {a \sin ^4(c+d x)}{2 b^3 d}+\frac {\sin ^5(c+d x)}{5 b^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 193, normalized size of antiderivative = 1.00, 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, 948} \[ -\frac {\left (2-\frac {3 a^2}{b^2}\right ) \sin ^3(c+d x)}{3 b^2 d}-\frac {2 a \left (a^2-b^2\right ) \sin ^2(c+d x)}{b^5 d}+\frac {\left (-6 a^2 b^2+5 a^4+b^4\right ) \sin (c+d x)}{b^6 d}-\frac {a^2 \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))}-\frac {2 a \left (-4 a^2 b^2+3 a^4+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \sin ^4(c+d x)}{2 b^3 d}+\frac {\sin ^5(c+d x)}{5 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(3*a^4 - 4*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]])/(b^7*d) + ((5*a^4 - 6*a^2*b^2 + b^4)*Sin[c + d*x])/(b
^6*d) - (2*a*(a^2 - b^2)*Sin[c + d*x]^2)/(b^5*d) - ((2 - (3*a^2)/b^2)*Sin[c + d*x]^3)/(3*b^2*d) - (a*Sin[c + d
*x]^4)/(2*b^3*d) + Sin[c + d*x]^5/(5*b^2*d) - (a^2*(a^2 - b^2)^2)/(b^7*d*(a + b*Sin[c + d*x]))

Rule 12

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

Rule 948

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] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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 ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (b^2-x^2\right )^2}{b^2 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (b^2-x^2\right )^2}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (5 a^4 \left (1+\frac {-6 a^2 b^2+b^4}{5 a^4}\right )-4 a \left (a^2-b^2\right ) x+\left (3 a^2-2 b^2\right ) x^2-2 a x^3+x^4+\frac {\left (a^3-a b^2\right )^2}{(a+x)^2}-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right )}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \sin (c+d x)}{b^6 d}-\frac {2 a \left (a^2-b^2\right ) \sin ^2(c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}-\frac {a \sin ^4(c+d x)}{2 b^3 d}+\frac {\sin ^5(c+d x)}{5 b^2 d}-\frac {a^2 \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.49, size = 225, normalized size = 1.17 \[ \frac {\left (40 a b^5-30 a^3 b^3\right ) \sin ^3(c+d x)-30 a b \left (a^2-b^2\right ) \sin (c+d x) \left (\left (6 a^2-2 b^2\right ) \log (a+b \sin (c+d x))-5 a^2+b^2\right )-30 a^2 \left (a^2-b^2\right ) \left (\left (6 a^2-2 b^2\right ) \log (a+b \sin (c+d x))+a^2-b^2\right )+5 b^4 \left (3 a^2-4 b^2\right ) \sin ^4(c+d x)+30 b^2 \left (3 a^4-4 a^2 b^2+b^4\right ) \sin ^2(c+d x)-9 a b^5 \sin ^5(c+d x)+6 b^6 \sin ^6(c+d x)}{30 b^7 d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*a^2*(a^2 - b^2)*(a^2 - b^2 + (6*a^2 - 2*b^2)*Log[a + b*Sin[c + d*x]]) - 30*a*b*(a^2 - b^2)*(-5*a^2 + b^2
+ (6*a^2 - 2*b^2)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] + 30*b^2*(3*a^4 - 4*a^2*b^2 + b^4)*Sin[c + d*x]^2 + (-
30*a^3*b^3 + 40*a*b^5)*Sin[c + d*x]^3 + 5*b^4*(3*a^2 - 4*b^2)*Sin[c + d*x]^4 - 9*a*b^5*Sin[c + d*x]^5 + 6*b^6*
Sin[c + d*x]^6)/(30*b^7*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 1.05, size = 246, normalized size = 1.27 \[ -\frac {48 \, b^{6} \cos \left (d x + c\right )^{6} + 240 \, a^{6} - 1440 \, a^{4} b^{2} + 1275 \, a^{2} b^{4} - 128 \, b^{6} - 8 \, {\left (15 \, a^{2} b^{4} - 2 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (45 \, a^{4} b^{2} - 45 \, a^{2} b^{4} + 4 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 480 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + a^{2} b^{4} + {\left (3 \, a^{5} b - 4 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (72 \, a b^{5} \cos \left (d x + c\right )^{4} - 1200 \, a^{5} b + 1440 \, a^{3} b^{3} - 293 \, a b^{5} - 16 \, {\left (15 \, a^{3} b^{3} - 11 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, {\left (b^{8} d \sin \left (d x + c\right ) + a b^{7} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/240*(48*b^6*cos(d*x + c)^6 + 240*a^6 - 1440*a^4*b^2 + 1275*a^2*b^4 - 128*b^6 - 8*(15*a^2*b^4 - 2*b^6)*cos(d
*x + c)^4 + 16*(45*a^4*b^2 - 45*a^2*b^4 + 4*b^6)*cos(d*x + c)^2 + 480*(3*a^6 - 4*a^4*b^2 + a^2*b^4 + (3*a^5*b
- 4*a^3*b^3 + a*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (72*a*b^5*cos(d*x + c)^4 - 1200*a^5*b + 1440*a^3*
b^3 - 293*a*b^5 - 16*(15*a^3*b^3 - 11*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^8*d*sin(d*x + c) + a*b^7*d)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 249, normalized size = 1.29 \[ -\frac {\frac {60 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7}} - \frac {30 \, {\left (6 \, a^{5} b \sin \left (d x + c\right ) - 8 \, a^{3} b^{3} \sin \left (d x + c\right ) + 2 \, a b^{5} \sin \left (d x + c\right ) + 5 \, a^{6} - 6 \, a^{4} b^{2} + a^{2} b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{7}} - \frac {6 \, b^{8} \sin \left (d x + c\right )^{5} - 15 \, a b^{7} \sin \left (d x + c\right )^{4} + 30 \, a^{2} b^{6} \sin \left (d x + c\right )^{3} - 20 \, b^{8} \sin \left (d x + c\right )^{3} - 60 \, a^{3} b^{5} \sin \left (d x + c\right )^{2} + 60 \, a b^{7} \sin \left (d x + c\right )^{2} + 150 \, a^{4} b^{4} \sin \left (d x + c\right ) - 180 \, a^{2} b^{6} \sin \left (d x + c\right ) + 30 \, b^{8} \sin \left (d x + c\right )}{b^{10}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(60*(3*a^5 - 4*a^3*b^2 + a*b^4)*log(abs(b*sin(d*x + c) + a))/b^7 - 30*(6*a^5*b*sin(d*x + c) - 8*a^3*b^3*
sin(d*x + c) + 2*a*b^5*sin(d*x + c) + 5*a^6 - 6*a^4*b^2 + a^2*b^4)/((b*sin(d*x + c) + a)*b^7) - (6*b^8*sin(d*x
+ c)^5 - 15*a*b^7*sin(d*x + c)^4 + 30*a^2*b^6*sin(d*x + c)^3 - 20*b^8*sin(d*x + c)^3 - 60*a^3*b^5*sin(d*x + c
)^2 + 60*a*b^7*sin(d*x + c)^2 + 150*a^4*b^4*sin(d*x + c) - 180*a^2*b^6*sin(d*x + c) + 30*b^8*sin(d*x + c))/b^1
0)/d

________________________________________________________________________________________

maple [A]  time = 0.52, size = 285, normalized size = 1.48 \[ \frac {\sin ^{5}\left (d x +c \right )}{5 b^{2} d}-\frac {a \left (\sin ^{4}\left (d x +c \right )\right )}{2 b^{3} d}+\frac {\left (\sin ^{3}\left (d x +c \right )\right ) a^{2}}{d \,b^{4}}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3 b^{2} d}-\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) a^{3}}{d \,b^{5}}+\frac {2 a \left (\sin ^{2}\left (d x +c \right )\right )}{b^{3} d}+\frac {5 a^{4} \sin \left (d x +c \right )}{d \,b^{6}}-\frac {6 \sin \left (d x +c \right ) a^{2}}{d \,b^{4}}+\frac {\sin \left (d x +c \right )}{b^{2} d}-\frac {6 a^{5} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}+\frac {8 a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}-\frac {2 a \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}-\frac {a^{6}}{d \,b^{7} \left (a +b \sin \left (d x +c \right )\right )}+\frac {2 a^{4}}{d \,b^{5} \left (a +b \sin \left (d x +c \right )\right )}-\frac {a^{2}}{d \,b^{3} \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x)

[Out]

1/5*sin(d*x+c)^5/b^2/d-1/2*a*sin(d*x+c)^4/b^3/d+1/d/b^4*sin(d*x+c)^3*a^2-2/3*sin(d*x+c)^3/b^2/d-2/d/b^5*sin(d*
x+c)^2*a^3+2*a*sin(d*x+c)^2/b^3/d+5/d/b^6*a^4*sin(d*x+c)-6/d/b^4*sin(d*x+c)*a^2+sin(d*x+c)/b^2/d-6/d*a^5/b^7*l
n(a+b*sin(d*x+c))+8/d*a^3/b^5*ln(a+b*sin(d*x+c))-2*a*ln(a+b*sin(d*x+c))/b^3/d-1/d*a^6/b^7/(a+b*sin(d*x+c))+2/d
*a^4/b^5/(a+b*sin(d*x+c))-1/d*a^2/b^3/(a+b*sin(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 184, normalized size = 0.95 \[ -\frac {\frac {30 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}}{b^{8} \sin \left (d x + c\right ) + a b^{7}} - \frac {6 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 10 \, {\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} \sin \left (d x + c\right )^{3} - 60 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2} + 30 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(30*(a^6 - 2*a^4*b^2 + a^2*b^4)/(b^8*sin(d*x + c) + a*b^7) - (6*b^4*sin(d*x + c)^5 - 15*a*b^3*sin(d*x +
c)^4 + 10*(3*a^2*b^2 - 2*b^4)*sin(d*x + c)^3 - 60*(a^3*b - a*b^3)*sin(d*x + c)^2 + 30*(5*a^4 - 6*a^2*b^2 + b^4
)*sin(d*x + c))/b^6 + 60*(3*a^5 - 4*a^3*b^2 + a*b^4)*log(b*sin(d*x + c) + a)/b^7)/d

________________________________________________________________________________________

mupad [B]  time = 11.46, size = 254, normalized size = 1.32 \[ \frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {a^3}{b^5}+\frac {a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,b^2}-\frac {a^2}{b^4}\right )}{d}+\frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b^2}+\frac {a^2\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {2\,a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,b^2\,d}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{2\,b^3\,d}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (6\,a^5-8\,a^3\,b^2+2\,a\,b^4\right )}{b^7\,d}-\frac {a^6-2\,a^4\,b^2+a^2\,b^4}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^7+a\,b^6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + b*sin(c + d*x))^2,x)

[Out]

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

________________________________________________________________________________________

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)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________