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 \[ -\frac {B (a-a \sin (c+d x))^5}{5 a^6 d}+\frac {(A+3 B) (a-a \sin (c+d x))^4}{4 a^5 d}-\frac {2 (A+B) (a-a \sin (c+d x))^3}{3 a^4 d} \]

[Out]

-2/3*(A+B)*(a-a*sin(d*x+c))^3/a^4/d+1/4*(A+3*B)*(a-a*sin(d*x+c))^4/a^5/d-1/5*B*(a-a*sin(d*x+c))^5/a^6/d

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Rubi [A]  time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 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]))))

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) (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*}

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Mathematica [A]  time = 0.15, 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)

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fricas [A]  time = 0.72, size = 66, normalized size = 0.84 \[ \frac {15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, B \cos \left (d x + c\right )^{4} + {\left (5 \, A - B\right )} \cos \left (d x + c\right )^{2} + 10 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{60 \, a d} \]

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)

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giac [A]  time = 0.19, size = 95, normalized size = 1.20 \[ \frac {12 \, B \sin \left (d x + c\right )^{5} + 15 \, A \sin \left (d x + c\right )^{4} - 15 \, B \sin \left (d x + c\right )^{4} - 20 \, A \sin \left (d x + c\right )^{3} - 20 \, B \sin \left (d x + c\right )^{3} - 30 \, A \sin \left (d x + c\right )^{2} + 30 \, B \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \]

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)

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maple [A]  time = 0.45, size = 75, normalized size = 0.95 \[ \frac {\frac {B \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (A -B \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-A -B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{d a} \]

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))

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maxima [A]  time = 0.32, size = 72, normalized size = 0.91 \[ \frac {12 \, B \sin \left (d x + c\right )^{5} + 15 \, {\left (A - B\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (A + B\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (A - B\right )} \sin \left (d x + c\right )^{2} + 60 \, A \sin \left (d x + c\right )}{60 \, a d} \]

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)

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mupad [B]  time = 9.20, size = 82, normalized size = 1.04 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^4\,\left (A-B\right )}{4\,a}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (A-B\right )}{2\,a}+\frac {B\,{\sin \left (c+d\,x\right )}^5}{5\,a}+\frac {A\,\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (A+B\right )}{3\,a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 30.97, size = 1703, normalized size = 21.56 \[ \text {result too large to display} \]

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))

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