\(\int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\) [1011]

Optimal result

Integrand size = 31, antiderivative size = 79 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {(A+B) (a-a \sin (c+d x))^4}{2 a^6 d}+\frac {(A+3 B) (a-a \sin (c+d x))^5}{5 a^7 d}-\frac {B (a-a \sin (c+d x))^6}{6 a^8 d} \]

[Out]

-1/2*(A+B)*(a-a*sin(d*x+c))^4/a^6/d+1/5*(A+3*B)*(a-a*sin(d*x+c))^5/a^7/d-1/6*B*(a-a*sin(d*x+c))^6/a^8/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2915, 78} \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {B (a-a \sin (c+d x))^6}{6 a^8 d}+\frac {(A+3 B) (a-a \sin (c+d x))^5}{5 a^7 d}-\frac {(A+B) (a-a \sin (c+d x))^4}{2 a^6 d} \]

[In]

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

[Out]

-1/2*((A + B)*(a - a*Sin[c + d*x])^4)/(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 78

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 2915

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/b)*x
)^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*} \text {integral}& = \frac {\text {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 {\text {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] (verified)

Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {(-1+\sin (c+d x))^4 \left (9 A+2 B+(6 A+8 B) \sin (c+d x)+5 B \sin ^2(c+d x)\right )}{30 a^2 d} \]

[In]

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

[Out]

-1/30*((-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))/(a^2*d)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(-\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right ) B}{6}+\frac {\left (A -2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) A}{2}+\frac {2 B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (2 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )}{a^{2} d}\)	\(84\)
default	\(-\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right ) B}{6}+\frac {\left (A -2 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) A}{2}+\frac {2 B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (2 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )}{a^{2} d}\)	\(84\)
parallelrisch	\(\frac {\left (240 A -165 B \right ) \cos \left (2 d x +2 c \right )+\left (60 A -30 B \right ) \cos \left (4 d x +4 c \right )+\left (60 A +40 B \right ) \sin \left (3 d x +3 c \right )+\left (-12 A +24 B \right ) \sin \left (5 d x +5 c \right )+5 B \cos \left (6 d x +6 c \right )+\left (840 A -240 B \right ) \sin \left (d x +c \right )-300 A +190 B}{960 a^{2} d}\)	\(110\)
risch	\(\frac {7 \sin \left (d x +c \right ) A}{8 a^{2} d}-\frac {\sin \left (d x +c \right ) B}{4 a^{2} d}+\frac {B \cos \left (6 d x +6 c \right )}{192 a^{2} d}-\frac {\sin \left (5 d x +5 c \right ) A}{80 a^{2} d}+\frac {\sin \left (5 d x +5 c \right ) B}{40 a^{2} d}+\frac {\cos \left (4 d x +4 c \right ) A}{16 a^{2} d}-\frac {\cos \left (4 d x +4 c \right ) B}{32 a^{2} d}+\frac {\sin \left (3 d x +3 c \right ) A}{16 a^{2} d}+\frac {\sin \left (3 d x +3 c \right ) B}{24 a^{2} d}+\frac {\cos \left (2 d x +2 c \right ) A}{4 a^{2} d}-\frac {11 \cos \left (2 d x +2 c \right ) B}{64 a^{2} d}\)	\(194\)

[In]

int(cos(d*x+c)^7*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {5 \, B \cos \left (d x + c\right )^{6} + 15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, {\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} - 12 \, A + 4 \, B\right )} \sin \left (d x + c\right )}{30 \, a^{2} d} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2705 vs. \(2 (70) = 140\).

Time = 52.94 (sec) , antiderivative size = 2705, normalized size of antiderivative = 34.24 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((30*A*tan(c/2 + d*x/2)**11/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a*
*2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2
+ d*x/2)**2 + 15*a**2*d) - 60*A*tan(c/2 + d*x/2)**10/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*
x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 +
90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 150*A*tan(c/2 + d*x/2)**9/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a*
*2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c
/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 120*A*tan(c/2 + d*x/2)**8/(15*a**2*d*tan(c/2 + d
*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 +
225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 204*A*tan(c/2 + d*x/2)**7/(15*a
**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(
c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 120*A*tan(c/2
+ d*x/2)**6/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8
+ 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d)
+ 204*A*tan(c/2 + d*x/2)**5/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan
(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2
)**2 + 15*a**2*d) - 120*A*tan(c/2 + d*x/2)**4/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10
+ 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*
d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 150*A*tan(c/2 + d*x/2)**3/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan
(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x
/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 60*A*tan(c/2 + d*x/2)**2/(15*a**2*d*tan(c/2 + d*x/2)**12
+ 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2
*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 30*A*tan(c/2 + d*x/2)/(15*a**2*d*tan(c/2
+ d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)*
*6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 30*B*tan(c/2 + d*x/2)**10/(
15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*
tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 80*B*tan(c
/2 + d*x/2)**9/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)*
*8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2
*d) + 120*B*tan(c/2 + d*x/2)**8/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*
tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*
x/2)**2 + 15*a**2*d) - 48*B*tan(c/2 + d*x/2)**7/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**
10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**
2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 20*B*tan(c/2 + d*x/2)**6/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*ta
n(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*
x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 48*B*tan(c/2 + d*x/2)**5/(15*a**2*d*tan(c/2 + d*x/2)**1
2 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**
2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 120*B*tan(c/2 + d*x/2)**4/(15*a**2*d*ta
n(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**2*d*tan(c/2 + d*
x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) - 80*B*tan(c/2 + d*x/2)*
*3/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x/2)**8 + 300*a**
2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*a**2*d) + 30*B*t
an(c/2 + d*x/2)**2/(15*a**2*d*tan(c/2 + d*x/2)**12 + 90*a**2*d*tan(c/2 + d*x/2)**10 + 225*a**2*d*tan(c/2 + d*x
/2)**8 + 300*a**2*d*tan(c/2 + d*x/2)**6 + 225*a**2*d*tan(c/2 + d*x/2)**4 + 90*a**2*d*tan(c/2 + d*x/2)**2 + 15*
a**2*d), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**7/(a*sin(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 \, B \sin \left (d x + c\right )^{6} + 6 \, {\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{5} - 15 \, A \sin \left (d x + c\right )^{4} + 20 \, B \sin \left (d x + c\right )^{3} + 15 \, {\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} - 30 \, A \sin \left (d x + c\right )}{30 \, a^{2} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 \, B \sin \left (d x + c\right )^{6} + 6 \, A \sin \left (d x + c\right )^{5} - 12 \, B \sin \left (d x + c\right )^{5} - 15 \, A \sin \left (d x + c\right )^{4} + 20 \, B \sin \left (d x + c\right )^{3} + 30 \, A \sin \left (d x + c\right )^{2} - 15 \, B \sin \left (d x + c\right )^{2} - 30 \, A \sin \left (d x + c\right )}{30 \, a^{2} d} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^7(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {{\sin \left (c+d\,x\right )}^5\,\left (A-2\,B\right )}{5\,a^2}-\frac {A\,{\sin \left (c+d\,x\right )}^4}{2\,a^2}+\frac {2\,B\,{\sin \left (c+d\,x\right )}^3}{3\,a^2}+\frac {B\,{\sin \left (c+d\,x\right )}^6}{6\,a^2}+\frac {{\sin \left (c+d\,x\right )}^2\,\left (2\,A-B\right )}{2\,a^2}-\frac {A\,\sin \left (c+d\,x\right )}{a^2}}{d} \]

[In]

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

[Out]

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