3.533 \(\int \frac{\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)
Optimal. Leaf size=73 \[ \frac{\sin ^7(c+d x)}{7 a d}-\frac{\sin ^6(c+d x)}{6 a d}-\frac{\sin ^5(c+d x)}{5 a d}+\frac{\sin ^4(c+d x)}{4 a d} \]
[Out]
Sin[c + d*x]^4/(4*a*d) - Sin[c + d*x]^5/(5*a*d) - Sin[c + d*x]^6/(6*a*d) + Sin[c + d*x]^7/(7*a*d)
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Rubi [A] time = 0.113463, antiderivative size = 73, normalized size of antiderivative = 1.,
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, 75} \[ \frac{\sin ^7(c+d x)}{7 a d}-\frac{\sin ^6(c+d x)}{6 a d}-\frac{\sin ^5(c+d x)}{5 a d}+\frac{\sin ^4(c+d x)}{4 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
[Out]
Sin[c + d*x]^4/(4*a*d) - Sin[c + d*x]^5/(5*a*d) - Sin[c + d*x]^6/(6*a*d) + Sin[c + d*x]^7/(7*a*d)
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]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 75
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])
Rubi steps
\begin{align*} \int \frac{\cos ^5(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^3 (a+x)}{a^3} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int (a-x)^2 x^3 (a+x) \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^3 x^3-a^2 x^4-a x^5+x^6\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac{\sin ^4(c+d x)}{4 a d}-\frac{\sin ^5(c+d x)}{5 a d}-\frac{\sin ^6(c+d x)}{6 a d}+\frac{\sin ^7(c+d x)}{7 a d}\\ \end{align*}
Mathematica [A] time = 0.334576, size = 48, normalized size = 0.66 \[ \frac{\sin ^4(c+d x) \left (60 \sin ^3(c+d x)-70 \sin ^2(c+d x)-84 \sin (c+d x)+105\right )}{420 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + a*Sin[c + d*x]),x]
[Out]
(Sin[c + d*x]^4*(105 - 84*Sin[c + d*x] - 70*Sin[c + d*x]^2 + 60*Sin[c + d*x]^3))/(420*a*d)
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Maple [A] time = 0.074, size = 49, normalized size = 0.7 \begin{align*}{\frac{1}{da} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{7}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{6}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x)
[Out]
1/d/a*(1/7*sin(d*x+c)^7-1/6*sin(d*x+c)^6-1/5*sin(d*x+c)^5+1/4*sin(d*x+c)^4)
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Maxima [A] time = 1.08004, size = 66, normalized size = 0.9 \begin{align*} \frac{60 \, \sin \left (d x + c\right )^{7} - 70 \, \sin \left (d x + c\right )^{6} - 84 \, \sin \left (d x + c\right )^{5} + 105 \, \sin \left (d x + c\right )^{4}}{420 \, a d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="maxima")
[Out]
1/420*(60*sin(d*x + c)^7 - 70*sin(d*x + c)^6 - 84*sin(d*x + c)^5 + 105*sin(d*x + c)^4)/(a*d)
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Fricas [A] time = 1.4254, size = 177, normalized size = 2.42 \begin{align*} \frac{70 \, \cos \left (d x + c\right )^{6} - 105 \, \cos \left (d x + c\right )^{4} - 12 \,{\left (5 \, \cos \left (d x + c\right )^{6} - 8 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{420 \, a d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")
[Out]
1/420*(70*cos(d*x + c)^6 - 105*cos(d*x + c)^4 - 12*(5*cos(d*x + c)^6 - 8*cos(d*x + c)^4 + cos(d*x + c)^2 + 2)*
sin(d*x + c))/(a*d)
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Sympy [A] time = 129.939, size = 1520, normalized size = 20.82 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**5*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)
[Out]
Piecewise((-15*tan(c/2 + d*x/2)**14/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*ta
n(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**
4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) - 105*tan(c/2 + d*x/2)**12/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*
tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2
)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 105*tan(c/2 + d*x/2)**10/(105*a
*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*
x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d)
- 672*tan(c/2 + d*x/2)**9/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d
*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a
*d*tan(c/2 + d*x/2)**2 + 105*a*d) - 385*tan(c/2 + d*x/2)**8/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 +
d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 220
5*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 576*tan(c/2 + d*x/2)**7/(105*a*d*tan(c/2
+ d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3
675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) - 385*tan(
c/2 + d*x/2)**6/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 +
3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2
+ d*x/2)**2 + 105*a*d) - 672*tan(c/2 + d*x/2)**5/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12
+ 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c
/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) + 105*tan(c/2 + d*x/2)**4/(105*a*d*tan(c/2 + d*x/2)**1
4 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan
(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2 + 105*a*d) - 105*tan(c/2 + d*x/2
)**2/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10 + 3675*a*d*t
an(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2 + d*x/2)**2
+ 105*a*d) - 15/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*tan(c/2 + d*x/2)**12 + 2205*a*d*tan(c/2 + d*x/2)**10
+ 3675*a*d*tan(c/2 + d*x/2)**8 + 3675*a*d*tan(c/2 + d*x/2)**6 + 2205*a*d*tan(c/2 + d*x/2)**4 + 735*a*d*tan(c/2
+ d*x/2)**2 + 105*a*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**5/(a*sin(c) + a), True))
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Giac [A] time = 1.15126, size = 66, normalized size = 0.9 \begin{align*} \frac{60 \, \sin \left (d x + c\right )^{7} - 70 \, \sin \left (d x + c\right )^{6} - 84 \, \sin \left (d x + c\right )^{5} + 105 \, \sin \left (d x + c\right )^{4}}{420 \, a d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")
[Out]
1/420*(60*sin(d*x + c)^7 - 70*sin(d*x + c)^6 - 84*sin(d*x + c)^5 + 105*sin(d*x + c)^4)/(a*d)