3.555 \(\int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)
Optimal. Leaf size=82 \[ \frac {\sin ^4(c+d x)}{4 a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {2 \sin ^2(c+d x)}{a^3 d}-\frac {4 \sin (c+d x)}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]
[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
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Rubi [A] time = 0.12, antiderivative size = 82, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used =
{2836, 12, 88} \[ \frac {\sin ^4(c+d x)}{4 a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {2 \sin ^2(c+d x)}{a^3 d}-\frac {4 \sin (c+d x)}{a^3 d}+\frac {4 \log (\sin (c+d x)+1)}{a^3 d} \]
Antiderivative was successfully verified.
[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 88
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 2836
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*x)/b
)^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*} \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {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 {\operatorname {Subst}\left (\int \frac {(a-x)^2 x^2}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {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*}
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Mathematica [A] time = 0.94, size = 59, normalized size = 0.72 \[ \frac {-152 \sin (c+d x)+8 \sin (3 (c+d x))-36 \cos (2 (c+d x))+\cos (4 (c+d x))+128 \log (\sin (c+d x)+1)+35}{32 a^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
[Out]
(35 - 36*Cos[2*(c + d*x)] + Cos[4*(c + d*x)] + 128*Log[1 + Sin[c + d*x]] - 152*Sin[c + d*x] + 8*Sin[3*(c + d*x
)])/(32*a^3*d)
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fricas [A] time = 0.61, size = 56, normalized size = 0.68 \[ \frac {\cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{4 \, a^{3} d} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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giac [B] time = 0.22, size = 167, normalized size = 2.04 \[ -\frac {\frac {12 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{a^{3}} - \frac {24 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 124 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 124 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[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
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maple [A] time = 0.44, size = 81, normalized size = 0.99 \[ \frac {4 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{3} d}-\frac {4 \sin \left (d x +c \right )}{a^{3} d}+\frac {2 \left (\sin ^{2}\left (d x +c \right )\right )}{a^{3} d}-\frac {\sin ^{3}\left (d x +c \right )}{a^{3} d}+\frac {\sin ^{4}\left (d x +c \right )}{4 a^{3} d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
[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
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maxima [A] time = 0.41, size = 61, normalized size = 0.74 \[ \frac {\frac {\sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3} + 8 \, \sin \left (d x + c\right )^{2} - 16 \, \sin \left (d x + c\right )}{a^{3}} + \frac {16 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{4 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[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
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mupad [B] time = 8.82, size = 69, normalized size = 0.84 \[ \frac {\frac {4\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3}-\frac {4\,\sin \left (c+d\,x\right )}{a^3}+\frac {2\,{\sin \left (c+d\,x\right )}^2}{a^3}-\frac {{\sin \left (c+d\,x\right )}^3}{a^3}+\frac {{\sin \left (c+d\,x\right )}^4}{4\,a^3}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[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
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sympy [A] time = 135.28, size = 1698, normalized size = 20.71 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[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))
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