3.534 \(\int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=73 \[ \frac {\sin ^6(c+d x)}{6 a d}-\frac {\sin ^5(c+d x)}{5 a d}-\frac {\sin ^4(c+d x)}{4 a d}+\frac {\sin ^3(c+d x)}{3 a d} \]

[Out]

1/3*sin(d*x+c)^3/a/d-1/4*sin(d*x+c)^4/a/d-1/5*sin(d*x+c)^5/a/d+1/6*sin(d*x+c)^6/a/d

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Rubi [A]  time = 0.16, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2835, 2564, 14} \[ \frac {\sin ^6(c+d x)}{6 a d}-\frac {\sin ^5(c+d x)}{5 a d}-\frac {\sin ^4(c+d x)}{4 a d}+\frac {\sin ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]^3/(3*a*d) - Sin[c + d*x]^4/(4*a*d) - Sin[c + d*x]^5/(5*a*d) + Sin[c + d*x]^6/(6*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 48, normalized size = 0.66 \[ \frac {\sin ^3(c+d x) \left (10 \sin ^3(c+d x)-12 \sin ^2(c+d x)-15 \sin (c+d x)+20\right )}{60 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]^3*(20 - 15*Sin[c + d*x] - 12*Sin[c + d*x]^2 + 10*Sin[c + d*x]^3))/(60*a*d)

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fricas [A]  time = 0.64, size = 59, normalized size = 0.81 \[ -\frac {10 \, \cos \left (d x + c\right )^{6} - 15 \, \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right )}{60 \, a d} \]

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)),x, algorithm="fricas")

[Out]

-1/60*(10*cos(d*x + c)^6 - 15*cos(d*x + c)^4 + 4*(3*cos(d*x + c)^4 - cos(d*x + c)^2 - 2)*sin(d*x + c))/(a*d)

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giac [A]  time = 0.18, size = 49, normalized size = 0.67 \[ \frac {10 \, \sin \left (d x + c\right )^{6} - 12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3}}{60 \, a d} \]

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)),x, algorithm="giac")

[Out]

1/60*(10*sin(d*x + c)^6 - 12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d*x + c)^3)/(a*d)

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maple [A]  time = 0.24, size = 49, normalized size = 0.67 \[ \frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d a} \]

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)),x)

[Out]

1/d/a*(1/6*sin(d*x+c)^6-1/5*sin(d*x+c)^5-1/4*sin(d*x+c)^4+1/3*sin(d*x+c)^3)

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maxima [A]  time = 0.58, size = 49, normalized size = 0.67 \[ \frac {10 \, \sin \left (d x + c\right )^{6} - 12 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3}}{60 \, a d} \]

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)),x, algorithm="maxima")

[Out]

1/60*(10*sin(d*x + c)^6 - 12*sin(d*x + c)^5 - 15*sin(d*x + c)^4 + 20*sin(d*x + c)^3)/(a*d)

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mupad [B]  time = 0.06, size = 57, normalized size = 0.78 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^3}{3\,a}-\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}+\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}}{d} \]

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)),x)

[Out]

(sin(c + d*x)^3/(3*a) - sin(c + d*x)^4/(4*a) - sin(c + d*x)^5/(5*a) + sin(c + d*x)^6/(6*a))/d

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sympy [A]  time = 51.02, size = 862, normalized size = 11.81 \[ \text {result too large to display} \]

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)),x)

[Out]

Piecewise((40*tan(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2
+ d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d
) - 60*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/
2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 24*
tan(c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 +
300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 40*tan(c/2
+ d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*
d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 24*tan(c/2 + d*x/
2)**5/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c
/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 60*tan(c/2 + d*x/2)**4/(
15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*
x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 40*tan(c/2 + d*x/2)**3/(15*a*d*
tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6
+ 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*sin(c)**2*cos(c)**5/(a*si
n(c) + a), True))

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