3.1012 \(\int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=51 \[ \frac {B (a-a \sin (c+d x))^4}{4 a^6 d}-\frac {(A+B) (a-a \sin (c+d x))^3}{3 a^5 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 51, 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, 43} \[ \frac {B (a-a \sin (c+d x))^4}{4 a^6 d}-\frac {(A+B) (a-a \sin (c+d x))^3}{3 a^5 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.06, size = 34, normalized size = 0.67 \[ \frac {(\sin (c+d x)-1)^3 (4 A+3 B \sin (c+d x)+B)}{12 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

((-1 + Sin[c + d*x])^3*(4*A + B + 3*B*Sin[c + d*x]))/(12*a^2*d)

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fricas [A]  time = 0.82, size = 64, normalized size = 1.25 \[ \frac {3 \, B \cos \left (d x + c\right )^{4} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, A + 2 \, B\right )} \sin \left (d x + c\right )}{12 \, a^{2} d} \]

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))^2,x, algorithm="fricas")

[Out]

1/12*(3*B*cos(d*x + c)^4 + 12*(A - B)*cos(d*x + c)^2 - 4*((A - 2*B)*cos(d*x + c)^2 - 4*A + 2*B)*sin(d*x + c))/
(a^2*d)

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giac [A]  time = 0.22, size = 73, normalized size = 1.43 \[ \frac {3 \, B \sin \left (d x + c\right )^{4} + 4 \, A \sin \left (d x + c\right )^{3} - 8 \, B \sin \left (d x + c\right )^{3} - 12 \, A \sin \left (d x + c\right )^{2} + 6 \, B \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \]

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))^2,x, algorithm="giac")

[Out]

1/12*(3*B*sin(d*x + c)^4 + 4*A*sin(d*x + c)^3 - 8*B*sin(d*x + c)^3 - 12*A*sin(d*x + c)^2 + 6*B*sin(d*x + c)^2
+ 12*A*sin(d*x + c))/(a^2*d)

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maple [A]  time = 0.59, size = 58, normalized size = 1.14 \[ \frac {\frac {B \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A -2 B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-2 A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )}{d \,a^{2}} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 61, normalized size = 1.20 \[ \frac {3 \, B \sin \left (d x + c\right )^{4} + 4 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{3} - 6 \, {\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \]

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))^2,x, algorithm="maxima")

[Out]

1/12*(3*B*sin(d*x + c)^4 + 4*(A - 2*B)*sin(d*x + c)^3 - 6*(2*A - B)*sin(d*x + c)^2 + 12*A*sin(d*x + c))/(a^2*d
)

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mupad [B]  time = 9.12, size = 68, normalized size = 1.33 \[ \frac {\frac {{\sin \left (c+d\,x\right )}^3\,\left (A-2\,B\right )}{3\,a^2}+\frac {B\,{\sin \left (c+d\,x\right )}^4}{4\,a^2}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,A-B\right )}{2\,a^2}+\frac {A\,\sin \left (c+d\,x\right )}{a^2}}{d} \]

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))^2,x)

[Out]

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

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sympy [A]  time = 53.39, size = 1182, normalized size = 23.18 \[ \text {result too large to display} \]

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))**2,x)

[Out]

Piecewise((6*A*tan(c/2 + d*x/2)**7/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 12*A*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 +
d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*
a**2*d) + 26*A*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 24*A*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 +
d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*
a**2*d) + 26*A*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*t
an(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 12*A*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 +
d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*
a**2*d) + 6*A*tan(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c
/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*B*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2
)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*
d) - 16*B*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/
2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 24*B*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2
)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*
d) - 16*B*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/
2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) + 6*B*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)
**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d
), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**5/(a*sin(c) + a)**2, True))

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