3.423 \(\int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=87 \[ -\frac {2 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^2 d}-\frac {7 \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {7 x}{8 a^2} \]

[Out]

7/8*x/a^2+2*cos(d*x+c)/a^2/d-2/3*cos(d*x+c)^3/a^2/d-7/8*cos(d*x+c)*sin(d*x+c)/a^2/d-1/4*cos(d*x+c)*sin(d*x+c)^
3/a^2/d

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Rubi [A]  time = 0.20, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2869, 2757, 2635, 8, 2633} \[ -\frac {2 \cos ^3(c+d x)}{3 a^2 d}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 a^2 d}-\frac {7 \sin (c+d x) \cos (c+d x)}{8 a^2 d}+\frac {7 x}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(7*x)/(8*a^2) + (2*Cos[c + d*x])/(a^2*d) - (2*Cos[c + d*x]^3)/(3*a^2*d) - (7*Cos[c + d*x]*Sin[c + d*x])/(8*a^2
*d) - (Cos[c + d*x]*Sin[c + d*x]^3)/(4*a^2*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 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 2869

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {\int \sin ^2(c+d x) (a-a \sin (c+d x))^2 \, dx}{a^4}\\ &=\frac {\int \left (a^2 \sin ^2(c+d x)-2 a^2 \sin ^3(c+d x)+a^2 \sin ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac {\int \sin ^2(c+d x) \, dx}{a^2}+\frac {\int \sin ^4(c+d x) \, dx}{a^2}-\frac {2 \int \sin ^3(c+d x) \, dx}{a^2}\\ &=-\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}+\frac {\int 1 \, dx}{2 a^2}+\frac {3 \int \sin ^2(c+d x) \, dx}{4 a^2}+\frac {2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=\frac {x}{2 a^2}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}+\frac {3 \int 1 \, dx}{8 a^2}\\ &=\frac {7 x}{8 a^2}+\frac {2 \cos (c+d x)}{a^2 d}-\frac {2 \cos ^3(c+d x)}{3 a^2 d}-\frac {7 \cos (c+d x) \sin (c+d x)}{8 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{4 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 1.35, size = 258, normalized size = 2.97 \[ \frac {168 d x \sin \left (\frac {c}{2}\right )-144 \sin \left (\frac {c}{2}+d x\right )+144 \sin \left (\frac {3 c}{2}+d x\right )-48 \sin \left (\frac {3 c}{2}+2 d x\right )-48 \sin \left (\frac {5 c}{2}+2 d x\right )+16 \sin \left (\frac {5 c}{2}+3 d x\right )-16 \sin \left (\frac {7 c}{2}+3 d x\right )+3 \sin \left (\frac {7 c}{2}+4 d x\right )+3 \sin \left (\frac {9 c}{2}+4 d x\right )+168 d x \cos \left (\frac {c}{2}\right )+144 \cos \left (\frac {c}{2}+d x\right )+144 \cos \left (\frac {3 c}{2}+d x\right )-48 \cos \left (\frac {3 c}{2}+2 d x\right )+48 \cos \left (\frac {5 c}{2}+2 d x\right )-16 \cos \left (\frac {5 c}{2}+3 d x\right )-16 \cos \left (\frac {7 c}{2}+3 d x\right )+3 \cos \left (\frac {7 c}{2}+4 d x\right )-3 \cos \left (\frac {9 c}{2}+4 d x\right )+8 \sin \left (\frac {c}{2}\right )}{192 a^2 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(168*d*x*Cos[c/2] + 144*Cos[c/2 + d*x] + 144*Cos[(3*c)/2 + d*x] - 48*Cos[(3*c)/2 + 2*d*x] + 48*Cos[(5*c)/2 + 2
*d*x] - 16*Cos[(5*c)/2 + 3*d*x] - 16*Cos[(7*c)/2 + 3*d*x] + 3*Cos[(7*c)/2 + 4*d*x] - 3*Cos[(9*c)/2 + 4*d*x] +
8*Sin[c/2] + 168*d*x*Sin[c/2] - 144*Sin[c/2 + d*x] + 144*Sin[(3*c)/2 + d*x] - 48*Sin[(3*c)/2 + 2*d*x] - 48*Sin
[(5*c)/2 + 2*d*x] + 16*Sin[(5*c)/2 + 3*d*x] - 16*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^2*d*(Cos[c/2] + Sin[c/2]))

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fricas [A]  time = 0.44, size = 58, normalized size = 0.67 \[ -\frac {16 \, \cos \left (d x + c\right )^{3} - 21 \, d x - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} - 9 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 48 \, \cos \left (d x + c\right )}{24 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/24*(16*cos(d*x + c)^3 - 21*d*x - 3*(2*cos(d*x + c)^3 - 9*cos(d*x + c))*sin(d*x + c) - 48*cos(d*x + c))/(a^2
*d)

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giac [A]  time = 0.17, size = 114, normalized size = 1.31 \[ \frac {\frac {21 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 96 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 128 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 32\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{2}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(21*(d*x + c)/a^2 + 2*(21*tan(1/2*d*x + 1/2*c)^7 + 45*tan(1/2*d*x + 1/2*c)^5 + 96*tan(1/2*d*x + 1/2*c)^4
- 45*tan(1/2*d*x + 1/2*c)^3 + 128*tan(1/2*d*x + 1/2*c)^2 - 21*tan(1/2*d*x + 1/2*c) + 32)/((tan(1/2*d*x + 1/2*c
)^2 + 1)^4*a^2))/d

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maple [B]  time = 0.36, size = 245, normalized size = 2.82 \[ \frac {7 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {15 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {15 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {32 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8}{3 a^{2} d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x)

[Out]

7/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+15/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/
2*c)^5+8/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^4-15/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d
*x+1/2*c)^3+32/3/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^2-7/4/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4*ta
n(1/2*d*x+1/2*c)+8/3/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^4+7/4/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [B]  time = 0.47, size = 247, normalized size = 2.84 \[ -\frac {\frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {128 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {45 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {96 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {45 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {21 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - 32}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {21 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*((21*sin(d*x + c)/(cos(d*x + c) + 1) - 128*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 45*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 96*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 45*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 21*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - 32)/(a^2 + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^2*sin(d*x + c)^4/(co
s(d*x + c) + 1)^4 + 4*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 21*
arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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mupad [B]  time = 8.66, size = 79, normalized size = 0.91 \[ \frac {7\,x}{8\,a^2}+\frac {2\,\cos \left (c+d\,x\right )}{a^2\,d}-\frac {2\,{\cos \left (c+d\,x\right )}^3}{3\,a^2\,d}+\frac {{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,a^2\,d}-\frac {9\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*sin(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)

[Out]

(7*x)/(8*a^2) + (2*cos(c + d*x))/(a^2*d) - (2*cos(c + d*x)^3)/(3*a^2*d) + (cos(c + d*x)^3*sin(c + d*x))/(4*a^2
*d) - (9*cos(c + d*x)*sin(c + d*x))/(8*a^2*d)

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sympy [A]  time = 59.86, size = 1153, normalized size = 13.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((21*d*x*tan(c/2 + d*x/2)**8/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**
2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 84*d*x*tan(c/2 + d*x/2)**6/(24*a**2*d*t
an(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/
2)**2 + 24*a**2*d) + 126*d*x*tan(c/2 + d*x/2)**4/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**
6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 84*d*x*tan(c/2 + d*x/2)**2/(
24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan
(c/2 + d*x/2)**2 + 24*a**2*d) + 21*d*x/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a*
*2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 42*tan(c/2 + d*x/2)**7/(24*a**2*d*tan(
c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)*
*2 + 24*a**2*d) + 90*tan(c/2 + d*x/2)**5/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*
a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 192*tan(c/2 + d*x/2)**4/(24*a**2*d*t
an(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/
2)**2 + 24*a**2*d) - 90*tan(c/2 + d*x/2)**3/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 1
44*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 256*tan(c/2 + d*x/2)**2/(24*a**2*
d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d
*x/2)**2 + 24*a**2*d) - 42*tan(c/2 + d*x/2)/(24*a**2*d*tan(c/2 + d*x/2)**8 + 96*a**2*d*tan(c/2 + d*x/2)**6 + 1
44*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d) + 64/(24*a**2*d*tan(c/2 + d*x/2)**8
+ 96*a**2*d*tan(c/2 + d*x/2)**6 + 144*a**2*d*tan(c/2 + d*x/2)**4 + 96*a**2*d*tan(c/2 + d*x/2)**2 + 24*a**2*d)
, Ne(d, 0)), (x*sin(c)**2*cos(c)**4/(a*sin(c) + a)**2, True))

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