\(\int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1295]

Optimal result

Integrand size = 29, antiderivative size = 149 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {a \sin ^4(c+d x)}{4 b^2 d}-\frac {\sin ^5(c+d x)}{5 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 908} \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {a^3 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^6 d}+\frac {a \sin ^4(c+d x)}{4 b^2 d}-\frac {\sin ^5(c+d x)}{5 b d} \]

[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 12

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

Rule 908

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]))

Rule 2916

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/b)*x)^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*} \text {integral}& = \frac {\text {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 {\text {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 {\text {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] (verified)

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {60 a^3 (a-b) (a+b) \log (a+b \sin (c+d x))}{b^6}-\frac {60 a^2 (a-b) (a+b) \sin (c+d x)}{b^5}+\frac {30 a (a-b) (a+b) \sin ^2(c+d x)}{b^4}-\frac {20 (a-b) (a+b) \sin ^3(c+d x)}{b^3}+\frac {15 a \sin ^4(c+d x)}{b^2}-\frac {12 \sin ^5(c+d x)}{b}}{60 d} \]

[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)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {-\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right ) b^{4}}{5}-\frac {a \left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {a^{3} b \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\frac {a \,b^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{4} \sin \left (d x +c \right )-\sin \left (d x +c \right ) a^{2} b^{2}}{b^{5}}+\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\)	\(147\)
default	\(\frac {-\frac {\frac {\left (\sin ^{5}\left (d x +c \right )\right ) b^{4}}{5}-\frac {a \left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a^{2} b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {b^{4} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {a^{3} b \left (\sin ^{2}\left (d x +c \right )\right )}{2}+\frac {a \,b^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{4} \sin \left (d x +c \right )-\sin \left (d x +c \right ) a^{2} b^{2}}{b^{5}}+\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\)	\(147\)
parallelrisch	\(\frac {480 \left (a^{5}-a^{3} b^{2}\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+480 \left (-a^{5}+a^{3} b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 \left (-2 a^{3} b^{2}+a \,b^{4}\right ) \cos \left (2 d x +2 c \right )+10 \left (4 a^{2} b^{3}-b^{5}\right ) \sin \left (3 d x +3 c \right )+15 b^{4} \cos \left (4 d x +4 c \right ) a -6 b^{5} \sin \left (5 d x +5 c \right )+60 \left (-8 a^{4} b +6 a^{2} b^{3}+b^{5}\right ) \sin \left (d x +c \right )+120 a^{3} b^{2}-75 a \,b^{4}}{480 d \,b^{6}}\)	\(197\)
risch	\(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{16 b d}-\frac {2 i a^{5} c}{b^{6} d}-\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,b^{4}}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 d \,b^{2}}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {2 i a^{3} c}{b^{4} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{16 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4}}{2 b^{5} d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4}}{2 b^{5} d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,b^{4}}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 d \,b^{2}}-\frac {i x \,a^{5}}{b^{6}}+\frac {i a^{3} x}{b^{4}}+\frac {a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{6} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {\sin \left (5 d x +5 c \right )}{80 b d}+\frac {a \cos \left (4 d x +4 c \right )}{32 b^{2} d}+\frac {\sin \left (3 d x +3 c \right ) a^{2}}{12 b^{3} d}-\frac {\sin \left (3 d x +3 c \right )}{48 b d}\)	\(393\)
norman	\(\frac {\frac {\left (8 a^{3}-4 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}+\frac {\left (8 a^{3}-4 a \,b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}-\frac {4 \left (25 a^{4}-15 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 b^{5} d}-\frac {4 \left (25 a^{4}-15 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 b^{5} d}+\frac {4 \left (3 a^{3}-a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}+\frac {2 \left (a^{3}-a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}+\frac {2 \left (a^{3}-a \,b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}-\frac {2 a^{2} \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{5}}-\frac {2 a^{2} \left (a^{2}-b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{5}}-\frac {2 \left (a^{2}-b^{2}\right ) \left (15 a^{2}+4 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{5} d}-\frac {2 \left (a^{2}-b^{2}\right ) \left (15 a^{2}+4 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{5} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{6}}-\frac {a^{3} \left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{6}}\)	\(463\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {15 \, a b^{4} \cos \left (d x + c\right )^{4} - 30 \, a^{3} b^{2} \cos \left (d x + c\right )^{2} + 60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{4} b - 10 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {12 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 20 \, a^{2} b^{2} \sin \left (d x + c\right )^{3} - 20 \, b^{4} \sin \left (d x + c\right )^{3} - 30 \, a^{3} b \sin \left (d x + c\right )^{2} + 30 \, a b^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{4} \sin \left (d x + c\right ) - 60 \, a^{2} b^{2} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (a^{5} - a^{3} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6}}}{60 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {1}{3\,b}-\frac {a^2}{3\,b^3}\right )-\frac {{\sin \left (c+d\,x\right )}^5}{5\,b}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^5-a^3\,b^2\right )}{b^6}-\frac {a\,{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{2\,b}+\frac {a^2\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{b^2}}{d} \]

[In]

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

[Out]

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