\(\int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\) [170]

Optimal result

Integrand size = 23, antiderivative size = 63 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {8 a^2 \cos ^7(c+d x)}{63 d (a+a \sin (c+d x))^{7/2}}-\frac {2 a \cos ^7(c+d x)}{9 d (a+a \sin (c+d x))^{5/2}} \]

[Out]

-8/63*a^2*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(7/2)-2/9*a*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2752} \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {8 a^2 \cos ^7(c+d x)}{63 d (a \sin (c+d x)+a)^{7/2}}-\frac {2 a \cos ^7(c+d x)}{9 d (a \sin (c+d x)+a)^{5/2}} \]

[In]

Int[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-8*a^2*Cos[c + d*x]^7)/(63*d*(a + a*Sin[c + d*x])^(7/2)) - (2*a*Cos[c + d*x]^7)/(9*d*(a + a*Sin[c + d*x])^(5/
2))

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-b)*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), Int
[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2,
0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a \cos ^7(c+d x)}{9 d (a+a \sin (c+d x))^{5/2}}+\frac {1}{9} (4 a) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx \\ & = -\frac {8 a^2 \cos ^7(c+d x)}{63 d (a+a \sin (c+d x))^{7/2}}-\frac {2 a \cos ^7(c+d x)}{9 d (a+a \sin (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \cos ^7(c+d x) \sqrt {a (1+\sin (c+d x))} (11+7 \sin (c+d x))}{63 a^2 d (1+\sin (c+d x))^4} \]

[In]

Integrate[Cos[c + d*x]^6/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(-2*Cos[c + d*x]^7*Sqrt[a*(1 + Sin[c + d*x])]*(11 + 7*Sin[c + d*x]))/(63*a^2*d*(1 + Sin[c + d*x])^4)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{4} \left (7 \sin \left (d x +c \right )+11\right )}{63 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(57\)

[In]

int(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/63/a*(1+sin(d*x+c))*(sin(d*x+c)-1)^4*(7*sin(d*x+c)+11)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (55) = 110\).

Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.25 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{5} + 17 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (7 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{3} - 12 \, \cos \left (d x + c\right )^{2} - 16 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right ) - 16 \, \cos \left (d x + c\right ) - 32\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/63*(7*cos(d*x + c)^5 + 17*cos(d*x + c)^4 - 2*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - (7*cos(d*x + c)^4 - 10*cos(
d*x + c)^3 - 12*cos(d*x + c)^2 - 16*cos(d*x + c) - 32)*sin(d*x + c) - 16*cos(d*x + c) - 32)*sqrt(a*sin(d*x + c
) + a)/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\cos \left (d x + c\right )^{6}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^6/(a*sin(d*x + c) + a)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {32 \, {\left (7 \, \sqrt {2} \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 9 \, \sqrt {2} \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}\right )}}{63 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-32/63*(7*sqrt(2)*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 9*sqrt(2)*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^
7)/(a^2*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^6}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^(3/2), x)