\(\int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx\) [224]

Optimal result

Integrand size = 27, antiderivative size = 85 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\log (1+\sin (c+d x))}{a d}-\frac {\sin (c+d x)}{a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^4(c+d x)}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, 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} \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^4(c+d x)}{4 a d}-\frac {\sin ^3(c+d x)}{3 a d}+\frac {\sin ^2(c+d x)}{2 a d}-\frac {\sin (c+d x)}{a d}+\frac {\log (\sin (c+d x)+1)}{a d} \]

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {12 \log (1+\sin (c+d x))-12 \sin (c+d x)+6 \sin ^2(c+d x)-4 \sin ^3(c+d x)+3 \sin ^4(c+d x)}{12 a d} \]

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

(12*Log[1 + Sin[c + d*x]] - 12*Sin[c + d*x] + 6*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + 3*Sin[c + d*x]^4)/(12*a*d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )+\ln \left (1+\sin \left (d x +c \right )\right )}{d a}\)	\(56\)
default	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )+\ln \left (1+\sin \left (d x +c \right )\right )}{d a}\)	\(56\)
parallelrisch	\(\frac {192 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-96 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+33-36 \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right )-120 \sin \left (d x +c \right )+8 \sin \left (3 d x +3 c \right )}{96 d a}\)	\(80\)
risch	\(-\frac {i x}{a}+\frac {5 i {\mathrm e}^{i \left (d x +c \right )}}{8 d a}-\frac {5 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d a}-\frac {2 i c}{a d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\sin \left (3 d x +3 c \right )}{12 d a}-\frac {3 \cos \left (2 d x +2 c \right )}{8 a d}\)	\(127\)
norman	\(\frac {\frac {2}{a d}+\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {38 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {38 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {58 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {58 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}\)	\(253\)

[In]

int(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) + 12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{12 \, a d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.94 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin ^{4}{\left (c + d x \right )}}{4 a d} - \frac {\sin ^{3}{\left (c + d x \right )}}{3 a d} + \frac {\sin ^{2}{\left (c + d x \right )}}{2 a d} - \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{4}{\left (c \right )} \cos {\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((log(sin(c + d*x) + 1)/(a*d) + sin(c + d*x)**4/(4*a*d) - sin(c + d*x)**3/(3*a*d) + sin(c + d*x)**2/(
2*a*d) - sin(c + d*x)/(a*d), Ne(d, 0)), (x*sin(c)**4*cos(c)/(a*sin(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right )}{a} + \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a}}{12 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4} - 4 \, a^{3} \sin \left (d x + c\right )^{3} + 6 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a^{3} \sin \left (d x + c\right )}{a^{4}}}{12 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (c+d x) \sin ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a}-\frac {\sin \left (c+d\,x\right )}{a}+\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}}{d} \]

[In]

int((cos(c + d*x)*sin(c + d*x)^4)/(a + a*sin(c + d*x)),x)

[Out]

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