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]

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+1/7*sin(d*x+c)^7/a/d

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Rubi [A]  time = 0.11, antiderivative size = 73, 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, 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 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])

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 ^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*}

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Mathematica [A]  time = 0.35, 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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fricas [A]  time = 0.69, size = 67, normalized size = 0.92 \[ \frac {70 \, \cos \left (d x + c\right )^{6} - 105 \, \cos \left (d x + c\right )^{4} - 12 \, {\left (5 \, \cos \left (d x + c\right )^{6} - 8 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{420 \, a d} \]

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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giac [A]  time = 0.17, size = 49, normalized size = 0.67 \[ \frac {60 \, \sin \left (d x + c\right )^{7} - 70 \, \sin \left (d x + c\right )^{6} - 84 \, \sin \left (d x + c\right )^{5} + 105 \, \sin \left (d x + c\right )^{4}}{420 \, a d} \]

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)

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maple [A]  time = 0.25, size = 49, normalized size = 0.67 \[ \frac {\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}}{d a} \]

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 = 0.67, size = 49, normalized size = 0.67 \[ \frac {60 \, \sin \left (d x + c\right )^{7} - 70 \, \sin \left (d x + c\right )^{6} - 84 \, \sin \left (d x + c\right )^{5} + 105 \, \sin \left (d x + c\right )^{4}}{420 \, a d} \]

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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mupad [B]  time = 0.07, size = 57, normalized size = 0.78 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}-\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}+\frac {{\sin \left (c+d\,x\right )}^7}{7\,a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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) - sin(c + d*x)^5/(5*a) - sin(c + d*x)^6/(6*a) + sin(c + d*x)^7/(7*a))/d

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sympy [A]  time = 81.79, size = 981, normalized size = 13.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((420*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*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) - 672*tan(c/2 + d*x/2)**9/(105*a*d*tan(c/2 + d*x/2)**14 + 735*a*d*t
an(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) + 140*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 + 2205*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 + 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) + 140*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 + 367
5*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) + 420*tan(c/
2 + d*x/2)**4/(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 + 3
675*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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