\(\int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [78]

Optimal result

Integrand size = 21, antiderivative size = 50 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \log (1+\sin (c+d x))}{a^3 d}-\frac {3 \sin (c+d x)}{a^3 d}+\frac {\sin ^2(c+d x)}{2 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 45} \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^2(c+d x)}{2 a^3 d}-\frac {3 \sin (c+d x)}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

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

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 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (-3 a+x+\frac {4 a^2}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {4 \log (1+\sin (c+d x))}{a^3 d}-\frac {3 \sin (c+d x)}{a^3 d}+\frac {\sin ^2(c+d x)}{2 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8 \log (1+\sin (c+d x))-6 \sin (c+d x)+\sin ^2(c+d x)}{2 a^3 d} \]

[In]

Integrate[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \sin \left (d x +c \right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}\)	\(38\)
default	\(\frac {\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \sin \left (d x +c \right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}\)	\(38\)
parallelrisch	\(\frac {32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-16 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1-\cos \left (2 d x +2 c \right )-12 \sin \left (d x +c \right )}{4 a^{3} d}\)	\(58\)
risch	\(-\frac {4 i x}{a^{3}}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}-\frac {8 i c}{a^{3} d}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {\cos \left (2 d x +2 c \right )}{4 a^{3} d}\)	\(93\)
norman	\(\frac {-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {6 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {28 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {28 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {74 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {74 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {154 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {154 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {256 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {256 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {412 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {412 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {350 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {350 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {4 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\)	\(341\)

[In]

int(cos(d*x+c)^5/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\cos \left (d x + c\right )^{2} - 8 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 6 \, \sin \left (d x + c\right )}{2 \, a^{3} d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(cos(d*x + c)^2 - 8*log(sin(d*x + c) + 1) + 6*sin(d*x + c))/(a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (44) = 88\).

Time = 23.00 (sec) , antiderivative size = 564, normalized size of antiderivative = 11.28 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} \frac {8 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} + \frac {16 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} + \frac {8 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {4 \log {\left (\tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {8 \log {\left (\tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {4 \log {\left (\tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {6 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} + \frac {2 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{5}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((8*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*tan(c/2 + d*
x/2)**2 + a**3*d) + 16*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*ta
n(c/2 + d*x/2)**2 + a**3*d) + 8*log(tan(c/2 + d*x/2) + 1)/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*tan(c/2 + d*x
/2)**2 + a**3*d) - 4*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*t
an(c/2 + d*x/2)**2 + a**3*d) - 8*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**4
+ 2*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 4*log(tan(c/2 + d*x/2)**2 + 1)/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3
*d*tan(c/2 + d*x/2)**2 + a**3*d) - 6*tan(c/2 + d*x/2)**3/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*tan(c/2 + d*x/
2)**2 + a**3*d) + 2*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) -
6*tan(c/2 + d*x/2)/(a**3*d*tan(c/2 + d*x/2)**4 + 2*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d), Ne(d, 0)), (x*cos(c)
**5/(a*sin(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\sin \left (d x + c\right )^{2} - 6 \, \sin \left (d x + c\right )}{a^{3}} + \frac {8 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*((sin(d*x + c)^2 - 6*sin(d*x + c))/a^3 + 8*log(sin(d*x + c) + 1)/a^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (48) = 96\).

Time = 0.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 2.30 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {2 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac {4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}\right )}}{d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-2*(2*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 4*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (3*tan(1/2*d*x + 1/2*c)
^4 - 3*tan(1/2*d*x + 1/2*c)^3 + 7*tan(1/2*d*x + 1/2*c)^2 - 3*tan(1/2*d*x + 1/2*c) + 3)/((tan(1/2*d*x + 1/2*c)^
2 + 1)^2*a^3))/d

Mupad [B] (verification not implemented)

Time = 5.91 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8\,\ln \left (\sin \left (c+d\,x\right )+1\right )-6\,\sin \left (c+d\,x\right )+{\sin \left (c+d\,x\right )}^2}{2\,a^3\,d} \]

[In]

int(cos(c + d*x)^5/(a + a*sin(c + d*x))^3,x)

[Out]

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