3.432 \(\int \frac{\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=109 \[ -\frac{\cos ^3(c+d x)}{a^3 d}+\frac{7 \cos (c+d x)}{a^3 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{4 a^3 d}-\frac{19 \sin (c+d x) \cos (c+d x)}{8 a^3 d}+\frac{4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}+\frac{51 x}{8 a^3} \]

[Out]

(51*x)/(8*a^3) + (7*Cos[c + d*x])/(a^3*d) - Cos[c + d*x]^3/(a^3*d) - (19*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d)
- (Cos[c + d*x]*Sin[c + d*x]^3)/(4*a^3*d) + (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.259208, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2875, 2872, 2638, 2635, 8, 2633, 2648} \[ -\frac{\cos ^3(c+d x)}{a^3 d}+\frac{7 \cos (c+d x)}{a^3 d}-\frac{\sin ^3(c+d x) \cos (c+d x)}{4 a^3 d}-\frac{19 \sin (c+d x) \cos (c+d x)}{8 a^3 d}+\frac{4 \cos (c+d x)}{a^3 d (\sin (c+d x)+1)}+\frac{51 x}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(51*x)/(8*a^3) + (7*Cos[c + d*x])/(a^3*d) - Cos[c + d*x]^3/(a^3*d) - (19*Cos[c + d*x]*Sin[c + d*x])/(8*a^3*d)
- (Cos[c + d*x]*Sin[c + d*x]^3)/(4*a^3*d) + (4*Cos[c + d*x])/(a^3*d*(1 + Sin[c + d*x]))

Rule 2875

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

Rule 2872

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \sin ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac{\int \sin (c+d x) (a-a \sin (c+d x))^3 \tan ^2(c+d x) \, dx}{a^6}\\ &=\frac{\int \left (4 a-4 a \sin (c+d x)+4 a \sin ^2(c+d x)-3 a \sin ^3(c+d x)+a \sin ^4(c+d x)-\frac{4 a}{1+\sin (c+d x)}\right ) \, dx}{a^4}\\ &=\frac{4 x}{a^3}+\frac{\int \sin ^4(c+d x) \, dx}{a^3}-\frac{3 \int \sin ^3(c+d x) \, dx}{a^3}-\frac{4 \int \sin (c+d x) \, dx}{a^3}+\frac{4 \int \sin ^2(c+d x) \, dx}{a^3}-\frac{4 \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=\frac{4 x}{a^3}+\frac{4 \cos (c+d x)}{a^3 d}-\frac{2 \cos (c+d x) \sin (c+d x)}{a^3 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac{3 \int \sin ^2(c+d x) \, dx}{4 a^3}+\frac{2 \int 1 \, dx}{a^3}+\frac{3 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}\\ &=\frac{6 x}{a^3}+\frac{7 \cos (c+d x)}{a^3 d}-\frac{\cos ^3(c+d x)}{a^3 d}-\frac{19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac{3 \int 1 \, dx}{8 a^3}\\ &=\frac{51 x}{8 a^3}+\frac{7 \cos (c+d x)}{a^3 d}-\frac{\cos ^3(c+d x)}{a^3 d}-\frac{19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{\cos (c+d x) \sin ^3(c+d x)}{4 a^3 d}+\frac{4 \cos (c+d x)}{a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.3382, size = 195, normalized size = 1.79 \[ \frac{2040 d x \sin \left (c+\frac{d x}{2}\right )+800 \sin \left (2 c+\frac{3 d x}{2}\right )-160 \sin \left (2 c+\frac{5 d x}{2}\right )-35 \sin \left (4 c+\frac{7 d x}{2}\right )+5 \sin \left (4 c+\frac{9 d x}{2}\right )+997 \cos \left (c+\frac{d x}{2}\right )+800 \cos \left (c+\frac{3 d x}{2}\right )+160 \cos \left (3 c+\frac{5 d x}{2}\right )-35 \cos \left (3 c+\frac{7 d x}{2}\right )-5 \cos \left (5 c+\frac{9 d x}{2}\right )-3563 \sin \left (\frac{d x}{2}\right )+2040 d x \cos \left (\frac{d x}{2}\right )}{320 a^3 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2040*d*x*Cos[(d*x)/2] + 997*Cos[c + (d*x)/2] + 800*Cos[c + (3*d*x)/2] + 160*Cos[3*c + (5*d*x)/2] - 35*Cos[3*c
+ (7*d*x)/2] - 5*Cos[5*c + (9*d*x)/2] - 3563*Sin[(d*x)/2] + 2040*d*x*Sin[c + (d*x)/2] + 800*Sin[2*c + (3*d*x)
/2] - 160*Sin[2*c + (5*d*x)/2] - 35*Sin[4*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*x)/2])/(320*a^3*d*(Cos[c/2] + Sin[
c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

________________________________________________________________________________________

Maple [B]  time = 0.119, size = 300, normalized size = 2.8 \begin{align*}{\frac{19}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+8\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{27}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+36\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{27}{4\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+40\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}-{\frac{19}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+12\,{\frac{1}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{4}}}+{\frac{51}{4\,d{a}^{3}}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+8\,{\frac{1}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x)

[Out]

19/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7+8/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*
c)^6+27/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5+36/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*
x+1/2*c)^4-27/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+40/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(
1/2*d*x+1/2*c)^2-19/4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+12/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4+
51/4/d/a^3*arctan(tan(1/2*d*x+1/2*c))+8/d/a^3/(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

Maxima [B]  time = 1.7116, size = 537, normalized size = 4.93 \begin{align*} \frac{\frac{\frac{29 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{269 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{133 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{309 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{171 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{187 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{51 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{51 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 80}{a^{3} + \frac{a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{6 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{4 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}} + \frac{51 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((29*sin(d*x + c)/(cos(d*x + c) + 1) + 269*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 133*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 309*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 171*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 187*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 51*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 51*sin(d*x + c)^8/(cos(d*x + c) + 1)
^8 + 80)/(a^3 + a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^3*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + 6*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 6*a^3*sin(d*x + c)^5/(cos(d*x + c)
+ 1)^5 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^3*sin(d*x
+ c)^8/(cos(d*x + c) + 1)^8 + a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9) + 51*arctan(sin(d*x + c)/(cos(d*x + c)
+ 1))/a^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.15889, size = 381, normalized size = 3.5 \begin{align*} -\frac{2 \, \cos \left (d x + c\right )^{5} + 8 \, \cos \left (d x + c\right )^{4} - 15 \, \cos \left (d x + c\right )^{3} - 51 \, d x -{\left (51 \, d x + 67\right )} \cos \left (d x + c\right ) - 56 \, \cos \left (d x + c\right )^{2} -{\left (2 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{3} + 51 \, d x - 21 \, \cos \left (d x + c\right )^{2} + 35 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right ) - 32}{8 \,{\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/8*(2*cos(d*x + c)^5 + 8*cos(d*x + c)^4 - 15*cos(d*x + c)^3 - 51*d*x - (51*d*x + 67)*cos(d*x + c) - 56*cos(d
*x + c)^2 - (2*cos(d*x + c)^4 - 6*cos(d*x + c)^3 + 51*d*x - 21*cos(d*x + c)^2 + 35*cos(d*x + c) - 32)*sin(d*x
+ c) - 32)/(a^3*d*cos(d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**3/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.35192, size = 196, normalized size = 1.8 \begin{align*} \frac{\frac{51 \,{\left (d x + c\right )}}{a^{3}} + \frac{64}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}} + \frac{2 \,{\left (19 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 32 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 27 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 144 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 27 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 160 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 19 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(51*(d*x + c)/a^3 + 64/(a^3*(tan(1/2*d*x + 1/2*c) + 1)) + 2*(19*tan(1/2*d*x + 1/2*c)^7 + 32*tan(1/2*d*x +
1/2*c)^6 + 27*tan(1/2*d*x + 1/2*c)^5 + 144*tan(1/2*d*x + 1/2*c)^4 - 27*tan(1/2*d*x + 1/2*c)^3 + 160*tan(1/2*d*
x + 1/2*c)^2 - 19*tan(1/2*d*x + 1/2*c) + 48)/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d