\(\int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [1296]

Optimal result

Integrand size = 29, antiderivative size = 119 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin (c+d x)}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^3 d}+\frac {a \sin ^3(c+d x)}{3 b^2 d}-\frac {\sin ^4(c+d x)}{4 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 119, 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} \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin (c+d x)}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^3 d}+\frac {a \sin ^3(c+d x)}{3 b^2 d}-\frac {\sin ^4(c+d x)}{4 b d} \]

[In]

Int[(Cos[c + d*x]^3*Sin[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

-((a^2*(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/(b^5*d)) + (a*(a^2 - b^2)*Sin[c + d*x])/(b^4*d) - ((a^2 - b^2)*Sin
[c + d*x]^2)/(2*b^3*d) + (a*Sin[c + d*x]^3)/(3*b^2*d) - Sin[c + d*x]^4/(4*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^2 \left (b^2-x^2\right )}{b^2 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b^3 d} \\ & = \frac {\text {Subst}\left (\int \frac {x^2 \left (b^2-x^2\right )}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \left (a^3 \left (1-\frac {b^2}{a^2}\right )-\left (a^2-b^2\right ) x+a x^2-x^3+\frac {a^2 \left (-a^2+b^2\right )}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = -\frac {a^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{b^5 d}+\frac {a \left (a^2-b^2\right ) \sin (c+d x)}{b^4 d}-\frac {\left (a^2-b^2\right ) \sin ^2(c+d x)}{2 b^3 d}+\frac {a \sin ^3(c+d x)}{3 b^2 d}-\frac {\sin ^4(c+d x)}{4 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {12 a^2 \left (-a^2+b^2\right ) \log (a+b \sin (c+d x))+12 a b \left (a^2-b^2\right ) \sin (c+d x)+6 b^2 \left (-a^2+b^2\right ) \sin ^2(c+d x)+4 a b^3 \sin ^3(c+d x)-3 b^4 \sin ^4(c+d x)}{12 b^5 d} \]

[In]

Integrate[(Cos[c + d*x]^3*Sin[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {\left (a^{2}-b^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+\sin \left (d x +c \right ) a \left (a^{2}-b^{2}\right )}{b^{4}}-\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(103\)
default	\(\frac {\frac {-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) b^{3}}{4}+\frac {a \left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {\left (a^{2}-b^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+\sin \left (d x +c \right ) a \left (a^{2}-b^{2}\right )}{b^{4}}-\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(103\)
parallelrisch	\(\frac {96 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}-96 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} b^{2}-96 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a^{4}+96 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) a^{2} b^{2}+24 a^{2} b^{2} \cos \left (2 d x +2 c \right )-12 b^{4} \cos \left (2 d x +2 c \right )+96 a^{3} b \sin \left (d x +c \right )-72 a \,b^{3} \sin \left (d x +c \right )-3 \cos \left (4 d x +4 c \right ) b^{4}-8 a \sin \left (3 d x +3 c \right ) b^{3}-24 a^{2} b^{2}+15 b^{4}}{96 b^{5} d}\)	\(211\)
risch	\(-\frac {i a^{2} x}{b^{3}}-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{4}}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )}}{16 b d}-\frac {2 i a^{2} c}{d \,b^{3}}+\frac {3 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 d \,b^{2}}+\frac {2 i a^{4} c}{d \,b^{5}}+\frac {i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,b^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b d}+\frac {i x \,a^{4}}{b^{5}}-\frac {3 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d \,b^{2}}-\frac {a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,b^{5}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,b^{3}}-\frac {\cos \left (4 d x +4 c \right )}{32 b d}-\frac {a \sin \left (3 d x +3 c \right )}{12 b^{2} d}\)	\(314\)
norman	\(\frac {-\frac {2 \left (3 a^{2}-b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3}}-\frac {2 \left (3 a^{2}-b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3}}-\frac {\left (2 a^{2}-2 b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}-\frac {\left (2 a^{2}-2 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 a \left (a^{2}-b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4} d}+\frac {2 a \left (a^{2}-b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}+\frac {8 a \left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}+\frac {8 a \left (3 a^{2}-2 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}+\frac {4 a \left (9 a^{2}-5 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{4} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{5}}-\frac {a^{2} \left (a^{2}-b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{5}}\)	\(369\)

[In]

int(cos(d*x+c)^3*sin(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {3 \, b^{4} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} + 12 \, {\left (a^{4} - a^{2} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{3} b + 2 \, a b^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*sin(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - 12 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 4 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b \sin \left (d x + c\right )^{2} - 6 \, b^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right ) + 12 \, a b^{2} \sin \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} - a^{2} b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{5}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.48 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b}-\frac {a^2}{2\,b^3}\right )-\frac {a\,{\sin \left (c+d\,x\right )}^3}{3\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^4-a^2\,b^2\right )}{b^5}+\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2}{b^3}\right )}{b}}{d} \]

[In]

int((cos(c + d*x)^3*sin(c + d*x)^2)/(a + b*sin(c + d*x)),x)

[Out]

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