\(\int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1279]

Optimal result

Integrand size = 27, antiderivative size = 76 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^3 \log (a+b \sin (c+d x))}{b^4 d}+\frac {a^2 \sin (c+d x)}{b^3 d}-\frac {a \sin ^2(c+d x)}{2 b^2 d}+\frac {\sin ^3(c+d x)}{3 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^3 \log (a+b \sin (c+d x))}{b^4 d}+\frac {a^2 \sin (c+d x)}{b^3 d}-\frac {a \sin ^2(c+d x)}{2 b^2 d}+\frac {\sin ^3(c+d x)}{3 b d} \]

[In]

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

[Out]

-((a^3*Log[a + b*Sin[c + d*x]])/(b^4*d)) + (a^2*Sin[c + d*x])/(b^3*d) - (a*Sin[c + d*x]^2)/(2*b^2*d) + Sin[c +
d*x]^3/(3*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^3}{b^3 (a+x)} \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{a+x} \, dx,x,b \sin (c+d x)\right )}{b^4 d} \\ & = \frac {\text {Subst}\left (\int \left (a^2-a x+x^2-\frac {a^3}{a+x}\right ) \, dx,x,b \sin (c+d x)\right )}{b^4 d} \\ & = -\frac {a^3 \log (a+b \sin (c+d x))}{b^4 d}+\frac {a^2 \sin (c+d x)}{b^3 d}-\frac {a \sin ^2(c+d x)}{2 b^2 d}+\frac {\sin ^3(c+d x)}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-6 a^3 \log (a+b \sin (c+d x))+6 a^2 b \sin (c+d x)-3 a b^2 \sin ^2(c+d x)+2 b^3 \sin ^3(c+d x)}{6 b^4 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sin \left (d x +c \right )}{b^{3}}-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(65\)
default	\(\frac {\frac {\frac {\left (\sin ^{3}\left (d x +c \right )\right ) b^{2}}{3}-\frac {b a \left (\sin ^{2}\left (d x +c \right )\right )}{2}+a^{2} \sin \left (d x +c \right )}{b^{3}}-\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}}{d}\)	\(65\)
parallelrisch	\(\frac {-12 a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+12 a^{3} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a \,b^{2} \cos \left (2 d x +2 c \right )-b^{3} \sin \left (3 d x +3 c \right )+3 \left (4 a^{2} b +b^{3}\right ) \sin \left (d x +c \right )-3 a \,b^{2}}{12 d \,b^{4}}\)	\(112\)
risch	\(\frac {i a^{3} x}{b^{4}}+\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{8 b d}+\frac {a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {2 i a^{3} c}{b^{4} d}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {\sin \left (3 d x +3 c \right )}{12 b d}\)	\(195\)
norman	\(\frac {\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{3} d}+\frac {2 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3} d}+\frac {2 \left (9 a^{2}+4 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}+\frac {2 \left (9 a^{2}+4 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 b^{3} d}-\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {2 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{4}}-\frac {a^{3} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{4}}\)	\(244\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 \, a b^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{6 \, b^{4} d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.38 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\begin {cases} \frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{a} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin ^{4}{\left (c + d x \right )}}{4 a d} & \text {for}\: b = 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{a + b \sin {\left (c \right )}} & \text {for}\: d = 0 \\- \frac {a^{3} \log {\left (\frac {a}{b} + \sin {\left (c + d x \right )} \right )}}{b^{4} d} + \frac {a^{2} \sin {\left (c + d x \right )}}{b^{3} d} - \frac {a \sin ^{2}{\left (c + d x \right )}}{2 b^{2} d} + \frac {\sin ^{3}{\left (c + d x \right )}}{3 b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*sin(c)**3*cos(c)/a, Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)**4/(4*a*d), Eq(b, 0)), (x*sin(c)**3*cos(c
)/(a + b*sin(c)), Eq(d, 0)), (-a**3*log(a/b + sin(c + d*x))/(b**4*d) + a**2*sin(c + d*x)/(b**3*d) - a*sin(c +
d*x)**2/(2*b**2*d) + sin(c + d*x)**3/(3*b*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {6 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}} - \frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{b^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {6 \, a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{4}} - \frac {2 \, b^{2} \sin \left (d x + c\right )^{3} - 3 \, a b \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{b^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3}{3\,b}-\frac {a^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{b^4}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2\,b^2}+\frac {a^2\,\sin \left (c+d\,x\right )}{b^3}}{d} \]

[In]

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

[Out]

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