3.1011 \(\int \frac{\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=79 \[ \frac{(A+3 B) (a-a \sin (c+d x))^5}{5 a^7 d}-\frac{(A+B) (a-a \sin (c+d x))^4}{2 a^6 d}-\frac{B (a-a \sin (c+d x))^6}{6 a^8 d} \]

[Out]

-((A + B)*(a - a*Sin[c + d*x])^4)/(2*a^6*d) + ((A + 3*B)*(a - a*Sin[c + d*x])^5)/(5*a^7*d) - (B*(a - a*Sin[c +
d*x])^6)/(6*a^8*d)

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Rubi [A]  time = 0.124974, 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} \[ \frac{(A+3 B) (a-a \sin (c+d x))^5}{5 a^7 d}-\frac{(A+B) (a-a \sin (c+d x))^4}{2 a^6 d}-\frac{B (a-a \sin (c+d x))^6}{6 a^8 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

-((A + B)*(a - a*Sin[c + d*x])^4)/(2*a^6*d) + ((A + 3*B)*(a - a*Sin[c + d*x])^5)/(5*a^7*d) - (B*(a - a*Sin[c +
d*x])^6)/(6*a^8*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 ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^3 (a+x) \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a (A+B) (a-x)^3+(-A-3 B) (a-x)^4+\frac{B (a-x)^5}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=-\frac{(A+B) (a-a \sin (c+d x))^4}{2 a^6 d}+\frac{(A+3 B) (a-a \sin (c+d x))^5}{5 a^7 d}-\frac{B (a-a \sin (c+d x))^6}{6 a^8 d}\\ \end{align*}

Mathematica [A]  time = 0.16659, size = 52, normalized size = 0.66 \[ -\frac{(\sin (c+d x)-1)^4 \left ((6 A+8 B) \sin (c+d x)+9 A+5 B \sin ^2(c+d x)+2 B\right )}{30 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^7*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

-((-1 + Sin[c + d*x])^4*(9*A + 2*B + (6*A + 8*B)*Sin[c + d*x] + 5*B*Sin[c + d*x]^2))/(30*a^2*d)

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Maple [A]  time = 0.162, size = 82, normalized size = 1. \begin{align*}{\frac{1}{d{a}^{2}} \left ( -{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{6}}+{\frac{ \left ( -A+2\,B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{A \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2}}-{\frac{2\,B \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( -2\,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)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)

[Out]

1/d/a^2*(-1/6*B*sin(d*x+c)^6+1/5*(-A+2*B)*sin(d*x+c)^5+1/2*A*sin(d*x+c)^4-2/3*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 = 1.02463, size = 112, normalized size = 1.42 \begin{align*} -\frac{5 \, B \sin \left (d x + c\right )^{6} + 6 \,{\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{5} - 15 \, A \sin \left (d x + c\right )^{4} + 20 \, B \sin \left (d x + c\right )^{3} + 15 \,{\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} - 30 \, A \sin \left (d x + c\right )}{30 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(5*B*sin(d*x + c)^6 + 6*(A - 2*B)*sin(d*x + c)^5 - 15*A*sin(d*x + c)^4 + 20*B*sin(d*x + c)^3 + 15*(2*A -
B)*sin(d*x + c)^2 - 30*A*sin(d*x + c))/(a^2*d)

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Fricas [A]  time = 1.39974, size = 204, normalized size = 2.58 \begin{align*} \frac{5 \, B \cos \left (d x + c\right )^{6} + 15 \,{\left (A - B\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \,{\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} - 12 \, A + 4 \, B\right )} \sin \left (d x + c\right )}{30 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/30*(5*B*cos(d*x + c)^6 + 15*(A - B)*cos(d*x + c)^4 - 2*(3*(A - 2*B)*cos(d*x + c)^4 - 2*(3*A - B)*cos(d*x + c
)^2 - 12*A + 4*B)*sin(d*x + c))/(a^2*d)

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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)**7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.29591, size = 128, normalized size = 1.62 \begin{align*} -\frac{5 \, B \sin \left (d x + c\right )^{6} + 6 \, A \sin \left (d x + c\right )^{5} - 12 \, B \sin \left (d x + c\right )^{5} - 15 \, A \sin \left (d x + c\right )^{4} + 20 \, B \sin \left (d x + c\right )^{3} + 30 \, A \sin \left (d x + c\right )^{2} - 15 \, B \sin \left (d x + c\right )^{2} - 30 \, A \sin \left (d x + c\right )}{30 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/30*(5*B*sin(d*x + c)^6 + 6*A*sin(d*x + c)^5 - 12*B*sin(d*x + c)^5 - 15*A*sin(d*x + c)^4 + 20*B*sin(d*x + c)
^3 + 30*A*sin(d*x + c)^2 - 15*B*sin(d*x + c)^2 - 30*A*sin(d*x + c))/(a^2*d)