\(\int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [233]

Optimal result

Integrand size = 27, antiderivative size = 70 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \log (1+\sin (c+d x))}{a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {\sin ^2(c+d x)}{2 a^2 d}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 70, 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} \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^2(c+d x)}{2 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d}+\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {3 \log (\sin (c+d x)+1)}{a^2 d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2+6 \log (1+\sin (c+d x))+(-4+6 \log (1+\sin (c+d x))) \sin (c+d x)-3 \sin ^2(c+d x)+\sin ^3(c+d x)}{2 a^2 d (1+\sin (c+d x))} \]

[In]

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

[Out]

(2 + 6*Log[1 + Sin[c + d*x]] + (-4 + 6*Log[1 + Sin[c + d*x]])*Sin[c + d*x] - 3*Sin[c + d*x]^2 + Sin[c + d*x]^3
)/(2*a^2*d*(1 + Sin[c + d*x]))

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-2 \sin \left (d x +c \right )+3 \ln \left (1+\sin \left (d x +c \right )\right )+\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(48\)
default	\(\frac {\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-2 \sin \left (d x +c \right )+3 \ln \left (1+\sin \left (d x +c \right )\right )+\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\)	\(48\)
parallelrisch	\(\frac {\left (-24 \sin \left (d x +c \right )-24\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (48 \sin \left (d x +c \right )+48\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \cos \left (2 d x +2 c \right )-21 \sin \left (d x +c \right )-\sin \left (3 d x +3 c \right )-6}{8 d \,a^{2} \left (1+\sin \left (d x +c \right )\right )}\)	\(97\)
risch	\(-\frac {3 i x}{a^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}-\frac {6 i c}{d \,a^{2}}+\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {\cos \left (2 d x +2 c \right )}{4 d \,a^{2}}\)	\(125\)
norman	\(\frac {-\frac {12 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {12 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {26 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {26 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {38 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {38 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {46 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {46 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {3 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\)	\(265\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} + 12 \, {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, \cos \left (d x + c\right )^{2} + 7\right )} \sin \left (d x + c\right ) - 3}{4 \, {\left (a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/4*(6*cos(d*x + c)^2 + 12*(sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - (2*cos(d*x + c)^2 + 7)*sin(d*x + c) - 3)
/(a^2*d*sin(d*x + c) + a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (61) = 122\).

Time = 0.59 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.43 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {6 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {\sin ^{3}{\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} - \frac {3 \sin ^{2}{\left (c + d x \right )}}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} + \frac {6}{2 a^{2} d \sin {\left (c + d x \right )} + 2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{3}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((6*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 6*log(sin(c + d*x) + 1)/(
2*a**2*d*sin(c + d*x) + 2*a**2*d) + sin(c + d*x)**3/(2*a**2*d*sin(c + d*x) + 2*a**2*d) - 3*sin(c + d*x)**2/(2*
a**2*d*sin(c + d*x) + 2*a**2*d) + 6/(2*a**2*d*sin(c + d*x) + 2*a**2*d), Ne(d, 0)), (x*sin(c)**3*cos(c)/(a*sin(
c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2}{a^{2} \sin \left (d x + c\right ) + a^{2}} + \frac {\sin \left (d x + c\right )^{2} - 4 \, \sin \left (d x + c\right )}{a^{2}} + \frac {6 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.29 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (\frac {6 \, a}{a \sin \left (d x + c\right ) + a} - 1\right )}}{a^{4}} + \frac {6 \, \log \left (\frac {{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left | a \right |}}\right )}{a^{2}} - \frac {2}{{\left (a \sin \left (d x + c\right ) + a\right )} a}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.16 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int \frac {\cos (c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {1}{a^2\,\sin \left (c+d\,x\right )+a^2}+\frac {3\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2}-\frac {2\,\sin \left (c+d\,x\right )}{a^2}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a^2}}{d} \]

[In]

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

[Out]

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