\(\int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx\) [683]

Optimal result

Integrand size = 21, antiderivative size = 68 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {(a-a \sin (c+d x))^4}{a^5 d}+\frac {4 (a-a \sin (c+d x))^5}{5 a^6 d}-\frac {(a-a \sin (c+d x))^6}{6 a^7 d} \]

[Out]

-(a-a*sin(d*x+c))^4/a^5/d+4/5*(a-a*sin(d*x+c))^5/a^6/d-1/6*(a-a*sin(d*x+c))^6/a^7/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 45} \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {(a-a \sin (c+d x))^6}{6 a^7 d}+\frac {4 (a-a \sin (c+d x))^5}{5 a^6 d}-\frac {(a-a \sin (c+d x))^4}{a^5 d} \]

[In]

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

[Out]

-((a - a*Sin[c + d*x])^4/(a^5*d)) + (4*(a - a*Sin[c + d*x])^5)/(5*a^6*d) - (a - a*Sin[c + d*x])^6/(6*a^7*d)

Rule 45

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 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x)^3 (a+x)^2 \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = \frac {\text {Subst}\left (\int \left (4 a^2 (a-x)^3-4 a (a-x)^4+(a-x)^5\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d} \\ & = -\frac {(a-a \sin (c+d x))^4}{a^5 d}+\frac {4 (a-a \sin (c+d x))^5}{5 a^6 d}-\frac {(a-a \sin (c+d x))^6}{6 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sin (c+d x) \left (-30+15 \sin (c+d x)+20 \sin ^2(c+d x)-15 \sin ^3(c+d x)-6 \sin ^4(c+d x)+5 \sin ^5(c+d x)\right )}{30 a d} \]

[In]

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

[Out]

-1/30*(Sin[c + d*x]*(-30 + 15*Sin[c + d*x] + 20*Sin[c + d*x]^2 - 15*Sin[c + d*x]^3 - 6*Sin[c + d*x]^4 + 5*Sin[
c + d*x]^5))/(a*d)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(-\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )}{d a}\)	\(68\)
default	\(-\frac {\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\sin \left (d x +c \right )}{d a}\)	\(68\)
risch	\(\frac {5 \sin \left (d x +c \right )}{8 a d}+\frac {\cos \left (6 d x +6 c \right )}{192 a d}+\frac {\sin \left (5 d x +5 c \right )}{80 d a}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {5 \sin \left (3 d x +3 c \right )}{48 d a}+\frac {5 \cos \left (2 d x +2 c \right )}{64 a d}\)	\(101\)
parallelrisch	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+78 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-50 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+78 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+15\right )}{15 d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} a}\)	\(137\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {14 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {14 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {42 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {42 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {196 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {196 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {212 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {212 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(257\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {5 \, \cos \left (d x + c\right )^{6} + 2 \, {\left (3 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right )}{30 \, a d} \]

[In]

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

[Out]

1/30*(5*cos(d*x + c)^6 + 2*(3*cos(d*x + c)^4 + 4*cos(d*x + c)^2 + 8)*sin(d*x + c))/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1096 vs. \(2 (54) = 108\).

Time = 19.93 (sec) , antiderivative size = 1096, normalized size of antiderivative = 16.12 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((30*tan(c/2 + d*x/2)**11/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/
2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*
d) - 30*tan(c/2 + d*x/2)**10/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*
x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 7
0*tan(c/2 + d*x/2)**9/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8
+ 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 156*tan(
c/2 + d*x/2)**7/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300
*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 100*tan(c/2 +
d*x/2)**6/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*t
an(c/2 + d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 156*tan(c/2 + d*x/2)
**5/(15*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2
+ d*x/2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 70*tan(c/2 + d*x/2)**3/(15
*a*d*tan(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/
2)**6 + 225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) - 30*tan(c/2 + d*x/2)**2/(15*a*d*ta
n(c/2 + d*x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 +
225*a*d*tan(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d) + 30*tan(c/2 + d*x/2)/(15*a*d*tan(c/2 + d*
x/2)**12 + 90*a*d*tan(c/2 + d*x/2)**10 + 225*a*d*tan(c/2 + d*x/2)**8 + 300*a*d*tan(c/2 + d*x/2)**6 + 225*a*d*t
an(c/2 + d*x/2)**4 + 90*a*d*tan(c/2 + d*x/2)**2 + 15*a*d), Ne(d, 0)), (x*cos(c)**7/(a*sin(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \, \sin \left (d x + c\right )^{6} - 6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right )}{30 \, a d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {5 \, \sin \left (d x + c\right )^{6} - 6 \, \sin \left (d x + c\right )^{5} - 15 \, \sin \left (d x + c\right )^{4} + 20 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )^{2} - 30 \, \sin \left (d x + c\right )}{30 \, a d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^7(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {\sin \left (c+d\,x\right )}{a}-\frac {{\sin \left (c+d\,x\right )}^2}{2\,a}-\frac {2\,{\sin \left (c+d\,x\right )}^3}{3\,a}+\frac {{\sin \left (c+d\,x\right )}^4}{2\,a}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,a}-\frac {{\sin \left (c+d\,x\right )}^6}{6\,a}}{d} \]

[In]

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

[Out]

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