[next] [prev] [prev-tail] [tail] [up]

3.1004 \(\int \frac{\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=79 \[ \frac{(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac{2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d}-\frac{B (a-a \sin (c+d x))^5}{5 a^6 d} \]

[Out]

(-2*(A + B)*(a - a*Sin[c + d*x])^3)/(3*a^4*d) + ((A + 3*B)*(a - a*Sin[c + d*x])^4)/(4*a^5*d) - (B*(a - a*Sin[c
 + d*x])^5)/(5*a^6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.116178, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2836, 77} \[ \frac{(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac{2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d}-\frac{B (a-a \sin (c+d x))^5}{5 a^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A + B)*(a - a*Sin[c + d*x])^3)/(3*a^4*d) + ((A + 3*B)*(a - a*Sin[c + d*x])^4)/(4*a^5*d) - (B*(a - a*Sin[c
 + d*x])^5)/(5*a^6*d)

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]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^2 (a+x) \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a (A+B) (a-x)^2+(-A-3 B) (a-x)^3+\frac{B (a-x)^4}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=-\frac{2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d}+\frac{(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac{B (a-a \sin (c+d x))^5}{5 a^6 d}\\ \end{align*}

Mathematica [A]  time = 0.149055, size = 72, normalized size = 0.91 \[ \frac{\sin (c+d x) \left (15 (A-B) \sin ^3(c+d x)-20 (A+B) \sin ^2(c+d x)-30 (A-B) \sin (c+d x)+60 A+12 B \sin ^4(c+d x)\right )}{60 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(60*A - 30*(A - B)*Sin[c + d*x] - 20*(A + B)*Sin[c + d*x]^2 + 15*(A - B)*Sin[c + d*x]^3 + 12*B*S
in[c + d*x]^4))/(60*a*d)

________________________________________________________________________________________

Maple [A]  time = 0.095, size = 75, normalized size = 1. \begin{align*}{\frac{1}{da} \left ({\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{ \left ( A-B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}+{\frac{ \left ( -A-B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( -A+B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}}+A\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)

[Out]

1/d/a*(1/5*B*sin(d*x+c)^5+1/4*(A-B)*sin(d*x+c)^4+1/3*(-A-B)*sin(d*x+c)^3+1/2*(-A+B)*sin(d*x+c)^2+A*sin(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.10156, size = 97, normalized size = 1.23 \begin{align*} \frac{12 \, B \sin \left (d x + c\right )^{5} + 15 \,{\left (A - B\right )} \sin \left (d x + c\right )^{4} - 20 \,{\left (A + B\right )} \sin \left (d x + c\right )^{3} - 30 \,{\left (A - B\right )} \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/60*(12*B*sin(d*x + c)^5 + 15*(A - B)*sin(d*x + c)^4 - 20*(A + B)*sin(d*x + c)^3 - 30*(A - B)*sin(d*x + c)^2
+ 60*A*sin(d*x + c))/(a*d)

________________________________________________________________________________________

Fricas [A]  time = 1.66173, size = 159, normalized size = 2.01 \begin{align*} \frac{15 \,{\left (A - B\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (3 \, B \cos \left (d x + c\right )^{4} +{\left (5 \, A - B\right )} \cos \left (d x + c\right )^{2} + 10 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{60 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(15*(A - B)*cos(d*x + c)^4 + 4*(3*B*cos(d*x + c)^4 + (5*A - B)*cos(d*x + c)^2 + 10*A - 2*B)*sin(d*x + c))
/(a*d)

________________________________________________________________________________________

Sympy [A]  time = 46.5843, size = 1703, normalized size = 21.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((30*A*tan(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/
2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*tan(c/2 + d*x/2)**8/
(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*
x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 80*A*tan(c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*
d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2
 + 15*a*d) - 30*A*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(
c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 100*A*tan(c/2 + d*x/2)*
*5/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 +
 d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*tan(c/2 + d*x/2)**4/(15*a*d*tan(c/2 + d*x/2)**10 + 75
*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)
**2 + 15*a*d) + 80*A*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*t
an(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*A*tan(c/2 + d*x/2
)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2
 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*A*tan(c/2 + d*x/2)/(15*a*d*tan(c/2 + d*x/2)**10 + 75*
a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)*
*2 + 15*a*d) + 30*B*tan(c/2 + d*x/2)**8/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*ta
n(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*B*tan(c/2 + d*x/2)
**7/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2
+ d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*B*tan(c/2 + d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**10 + 7
5*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2
)**2 + 15*a*d) + 16*B*tan(c/2 + d*x/2)**5/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*
tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*B*tan(c/2 + d*x/
2)**4/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/
2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 40*B*tan(c/2 + d*x/2)**3/(15*a*d*tan(c/2 + d*x/2)**10 +
 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x
/2)**2 + 15*a*d) + 30*B*tan(c/2 + d*x/2)**2/(15*a*d*tan(c/2 + d*x/2)**10 + 75*a*d*tan(c/2 + d*x/2)**8 + 150*a*
d*tan(c/2 + d*x/2)**6 + 150*a*d*tan(c/2 + d*x/2)**4 + 75*a*d*tan(c/2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*(A +
 B*sin(c))*cos(c)**5/(a*sin(c) + a), True))

________________________________________________________________________________________

Giac [A]  time = 1.34271, size = 128, normalized size = 1.62 \begin{align*} \frac{12 \, B \sin \left (d x + c\right )^{5} + 15 \, A \sin \left (d x + c\right )^{4} - 15 \, B \sin \left (d x + c\right )^{4} - 20 \, A \sin \left (d x + c\right )^{3} - 20 \, B \sin \left (d x + c\right )^{3} - 30 \, A \sin \left (d x + c\right )^{2} + 30 \, B \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/60*(12*B*sin(d*x + c)^5 + 15*A*sin(d*x + c)^4 - 15*B*sin(d*x + c)^4 - 20*A*sin(d*x + c)^3 - 20*B*sin(d*x + c
)^3 - 30*A*sin(d*x + c)^2 + 30*B*sin(d*x + c)^2 + 60*A*sin(d*x + c))/(a*d)

[next] [prev] [prev-tail] [front] [up]