3.298 \(\int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=87 \[ \frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos (c+d x)}{a d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {3 x}{8 a} \]

[Out]

-3/8*x/a-cos(d*x+c)/a/d+1/3*cos(d*x+c)^3/a/d+3/8*cos(d*x+c)*sin(d*x+c)/a/d+1/4*cos(d*x+c)*sin(d*x+c)^3/a/d

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Rubi [A]  time = 0.13, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2839, 2633, 2635, 8} \[ \frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos (c+d x)}{a d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}-\frac {3 x}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(8*a) - Cos[c + d*x]/(a*d) + Cos[c + d*x]^3/(3*a*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (Cos[c +
d*x]*Sin[c + d*x]^3)/(4*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2839

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sin ^3(c+d x) \, dx}{a}-\frac {\int \sin ^4(c+d x) \, dx}{a}\\ &=\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a d}-\frac {3 \int \sin ^2(c+d x) \, dx}{4 a}-\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac {\cos (c+d x)}{a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a d}-\frac {3 \int 1 \, dx}{8 a}\\ &=-\frac {3 x}{8 a}-\frac {\cos (c+d x)}{a d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [B]  time = 1.72, size = 271, normalized size = 3.11 \[ \frac {-72 d x \sin \left (\frac {c}{2}\right )+72 \sin \left (\frac {c}{2}+d x\right )-72 \sin \left (\frac {3 c}{2}+d x\right )+24 \sin \left (\frac {3 c}{2}+2 d x\right )+24 \sin \left (\frac {5 c}{2}+2 d x\right )-8 \sin \left (\frac {5 c}{2}+3 d x\right )+8 \sin \left (\frac {7 c}{2}+3 d x\right )-3 \sin \left (\frac {7 c}{2}+4 d x\right )-3 \sin \left (\frac {9 c}{2}+4 d x\right )+24 \cos \left (\frac {c}{2}\right ) (c-3 d x)-72 \cos \left (\frac {c}{2}+d x\right )-72 \cos \left (\frac {3 c}{2}+d x\right )+24 \cos \left (\frac {3 c}{2}+2 d x\right )-24 \cos \left (\frac {5 c}{2}+2 d x\right )+8 \cos \left (\frac {5 c}{2}+3 d x\right )+8 \cos \left (\frac {7 c}{2}+3 d x\right )-3 \cos \left (\frac {7 c}{2}+4 d x\right )+3 \cos \left (\frac {9 c}{2}+4 d x\right )+24 c \sin \left (\frac {c}{2}\right )-48 \sin \left (\frac {c}{2}\right )}{192 a d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*(c - 3*d*x)*Cos[c/2] - 72*Cos[c/2 + d*x] - 72*Cos[(3*c)/2 + d*x] + 24*Cos[(3*c)/2 + 2*d*x] - 24*Cos[(5*c)/
2 + 2*d*x] + 8*Cos[(5*c)/2 + 3*d*x] + 8*Cos[(7*c)/2 + 3*d*x] - 3*Cos[(7*c)/2 + 4*d*x] + 3*Cos[(9*c)/2 + 4*d*x]
- 48*Sin[c/2] + 24*c*Sin[c/2] - 72*d*x*Sin[c/2] + 72*Sin[c/2 + d*x] - 72*Sin[(3*c)/2 + d*x] + 24*Sin[(3*c)/2
+ 2*d*x] + 24*Sin[(5*c)/2 + 2*d*x] - 8*Sin[(5*c)/2 + 3*d*x] + 8*Sin[(7*c)/2 + 3*d*x] - 3*Sin[(7*c)/2 + 4*d*x]
- 3*Sin[(9*c)/2 + 4*d*x])/(192*a*d*(Cos[c/2] + Sin[c/2]))

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fricas [A]  time = 0.50, size = 58, normalized size = 0.67 \[ \frac {8 \, \cos \left (d x + c\right )^{3} - 9 \, d x - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 5 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 24 \, \cos \left (d x + c\right )}{24 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*cos(d*x + c)^3 - 9*d*x - 3*(2*cos(d*x + c)^3 - 5*cos(d*x + c))*sin(d*x + c) - 24*cos(d*x + c))/(a*d)

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giac [A]  time = 0.15, size = 114, normalized size = 1.31 \[ -\frac {\frac {9 \, {\left (d x + c\right )}}{a} + \frac {2 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 33 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 64 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 16\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(9*(d*x + c)/a + 2*(9*tan(1/2*d*x + 1/2*c)^7 + 33*tan(1/2*d*x + 1/2*c)^5 + 48*tan(1/2*d*x + 1/2*c)^4 - 3
3*tan(1/2*d*x + 1/2*c)^3 + 64*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 16)/((tan(1/2*d*x + 1/2*c)^2 +
1)^4*a))/d

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maple [B]  time = 0.29, size = 245, normalized size = 2.82 \[ -\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {11 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {16 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {4}{3 a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-11/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c
)^5-4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4+11/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*
c)^3-16/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2+3/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1
/2*c)-4/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4-3/4/a/d*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.43, size = 237, normalized size = 2.72 \[ \frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {64 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {33 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {48 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {33 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {9 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 16}{a + \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((9*sin(d*x + c)/(cos(d*x + c) + 1) - 64*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 33*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 - 48*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 33*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 9*sin(d*x + c
)^7/(cos(d*x + c) + 1)^7 - 16)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*sin(d*x + c)^4/(cos(d*x + c)
+ 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 9*arctan(sin(d*x
+ c)/(cos(d*x + c) + 1))/a)/d

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mupad [B]  time = 8.61, size = 79, normalized size = 0.91 \[ \frac {{\cos \left (c+d\,x\right )}^3}{3\,a\,d}-\frac {\cos \left (c+d\,x\right )}{a\,d}-\frac {3\,x}{8\,a}-\frac {{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,a\,d}+\frac {5\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(c + d*x)^3/(3*a*d) - cos(c + d*x)/(a*d) - (3*x)/(8*a) - (cos(c + d*x)^3*sin(c + d*x))/(4*a*d) + (5*cos(c +
d*x)*sin(c + d*x))/(8*a*d)

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sympy [A]  time = 19.76, size = 1049, normalized size = 12.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*sin(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((-9*d*x*tan(c/2 + d*x/2)**8/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c
/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 36*d*x*tan(c/2 + d*x/2)**6/(24*a*d*tan(c/2 + d*x/2)**8
+ 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 54*d*x*ta
n(c/2 + d*x/2)**4/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*
a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 36*d*x*tan(c/2 + d*x/2)**2/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 +
d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 9*d*x/(24*a*d*tan(c/2 + d*x/2
)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 18*ta
n(c/2 + d*x/2)**7/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*
a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 66*tan(c/2 + d*x/2)**5/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/
2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 96*tan(c/2 + d*x/2)**4/(24*a*d*ta
n(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24
*a*d) + 66*tan(c/2 + d*x/2)**3/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*
x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) - 128*tan(c/2 + d*x/2)**2/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*
tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d) + 18*tan(c/2 + d*x/2)
/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 + 96*a*d*tan(c/2 + d*x
/2)**2 + 24*a*d) - 32/(24*a*d*tan(c/2 + d*x/2)**8 + 96*a*d*tan(c/2 + d*x/2)**6 + 144*a*d*tan(c/2 + d*x/2)**4 +
96*a*d*tan(c/2 + d*x/2)**2 + 24*a*d), Ne(d, 0)), (x*sin(c)**3*cos(c)**2/(a*sin(c) + a), True))

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