\(\int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [555]

Optimal result

Integrand size = 29, antiderivative size = 82 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \log (1+\sin (c+d x))}{a^3 d}-\frac {4 \sin (c+d x)}{a^3 d}+\frac {2 \sin ^2(c+d x)}{a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {\sin ^4(c+d x)}{4 a^3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 82, 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 = {2915, 12, 90} \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^4(c+d x)}{4 a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {2 \sin ^2(c+d x)}{a^3 d}-\frac {4 \sin (c+d x)}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(4*Log[1 + Sin[c + d*x]])/(a^3*d) - (4*Sin[c + d*x])/(a^3*d) + (2*Sin[c + d*x]^2)/(a^3*d) - Sin[c + d*x]^3/(a^
3*d) + Sin[c + d*x]^4/(4*a^3*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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^2}{a^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x^2}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {\text {Subst}\left (\int \left (-4 a^3+4 a^2 x-3 a x^2+x^3+\frac {4 a^4}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {4 \log (1+\sin (c+d x))}{a^3 d}-\frac {4 \sin (c+d x)}{a^3 d}+\frac {2 \sin ^2(c+d x)}{a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {\sin ^4(c+d x)}{4 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {16 \log (1+\sin (c+d x))-16 \sin (c+d x)+8 \sin ^2(c+d x)-4 \sin ^3(c+d x)+\sin ^4(c+d x)}{4 a^3 d} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(16*Log[1 + Sin[c + d*x]] - 16*Sin[c + d*x] + 8*Sin[c + d*x]^2 - 4*Sin[c + d*x]^3 + Sin[c + d*x]^4)/(4*a^3*d)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\left (\sin ^{3}\left (d x +c \right )\right )+2 \left (\sin ^{2}\left (d x +c \right )\right )-4 \sin \left (d x +c \right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(58\)
default	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\left (\sin ^{3}\left (d x +c \right )\right )+2 \left (\sin ^{2}\left (d x +c \right )\right )-4 \sin \left (d x +c \right )+4 \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{3}}\)	\(58\)
parallelrisch	\(\frac {256 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-128 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35-36 \cos \left (2 d x +2 c \right )+\cos \left (4 d x +4 c \right )-152 \sin \left (d x +c \right )+8 \sin \left (3 d x +3 c \right )}{32 d \,a^{3}}\)	\(78\)
risch	\(-\frac {4 i x}{a^{3}}+\frac {19 i {\mathrm e}^{i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {19 i {\mathrm e}^{-i \left (d x +c \right )}}{8 d \,a^{3}}-\frac {8 i c}{d \,a^{3}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}+\frac {\cos \left (4 d x +4 c \right )}{32 d \,a^{3}}+\frac {\sin \left (3 d x +3 c \right )}{4 d \,a^{3}}-\frac {9 \cos \left (2 d x +2 c \right )}{8 d \,a^{3}}\)	\(127\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.68 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{4 \, a^{3} d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 5)*sin(d*x + c) + 16*log(sin(d*x + c) + 1))/(a^3
*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1698 vs. \(2 (73) = 146\).

Time = 55.41 (sec) , antiderivative size = 1698, normalized size of antiderivative = 20.71 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**2/(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)**8/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*
x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 32*log(tan(c/2 + d*x/2) + 1)
*tan(c/2 + d*x/2)**6/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4
+ 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 48*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(a**3*d*tan(c/2 +
d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3
*d) + 32*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)
**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(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)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*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)**8/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*t
an(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 16*log(tan(c/2 +
d*x/2)**2 + 1)*tan(c/2 + d*x/2)**6/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c
/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 24*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)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 +
d*x/2)**2 + a**3*d) - 16*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)**8 + 4*a**
3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*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)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 +
4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 8*tan(c/2 + d*x/2)**7/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2
+ d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 8*tan(c/2 + d*x/2)**6/(
a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 +
d*x/2)**2 + a**3*d) - 32*tan(c/2 + d*x/2)**5/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a*
*3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 20*tan(c/2 + d*x/2)**4/(a**3*d*tan(c/2 + d
*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d
) - 32*tan(c/2 + d*x/2)**3/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x
/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) + 8*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d
*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c/2 + d*x/2)**2 + a**3*d) - 8*tan(c/2 + d*x
/2)/(a**3*d*tan(c/2 + d*x/2)**8 + 4*a**3*d*tan(c/2 + d*x/2)**6 + 6*a**3*d*tan(c/2 + d*x/2)**4 + 4*a**3*d*tan(c
/2 + d*x/2)**2 + a**3*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**5/(a*sin(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.74 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {\sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 8 \, \sin \left (d x + c\right )^{2} - 16 \, \sin \left (d x + c\right )}{a^{3}} + \frac {16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{4 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((sin(d*x + c)^4 - 4*sin(d*x + c)^3 + 8*sin(d*x + c)^2 - 16*sin(d*x + c))/a^3 + 16*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. 167 vs. \(2 (80) = 160\).

Time = 0.34 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.04 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {12 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac {24 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 124 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 124 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/3*(12*log(tan(1/2*d*x + 1/2*c)^2 + 1)/a^3 - 24*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - (25*tan(1/2*d*x + 1
/2*c)^8 - 24*tan(1/2*d*x + 1/2*c)^7 + 124*tan(1/2*d*x + 1/2*c)^6 - 96*tan(1/2*d*x + 1/2*c)^5 + 210*tan(1/2*d*x
+ 1/2*c)^4 - 96*tan(1/2*d*x + 1/2*c)^3 + 124*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 25)/((tan(1/2
*d*x + 1/2*c)^2 + 1)^4*a^3))/d

Mupad [B] (verification not implemented)

Time = 10.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {2\,{\sin \left (c+d\,x\right )}^2}{a^3}-\frac {{\sin \left (c+d\,x\right )}^3}{a^3}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a^3}}{d} \]

[In]

int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)

[Out]

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