\(\int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [157]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {64 a^3 \cos ^7(c+d x)}{693 d (a+a \sin (c+d x))^{7/2}}-\frac {16 a^2 \cos ^7(c+d x)}{99 d (a+a \sin (c+d x))^{5/2}}-\frac {2 a \cos ^7(c+d x)}{11 d (a+a \sin (c+d x))^{3/2}} \]

[Out]

-64/693*a^3*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(7/2)-16/99*a^2*cos(d*x+c)^7/d/(a+a*sin(d*x+c))^(5/2)-2/11*a*cos(d
*x+c)^7/d/(a+a*sin(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2752} \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {64 a^3 \cos ^7(c+d x)}{693 d (a \sin (c+d x)+a)^{7/2}}-\frac {16 a^2 \cos ^7(c+d x)}{99 d (a \sin (c+d x)+a)^{5/2}}-\frac {2 a \cos ^7(c+d x)}{11 d (a \sin (c+d x)+a)^{3/2}} \]

[In]

Int[Cos[c + d*x]^6/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-64*a^3*Cos[c + d*x]^7)/(693*d*(a + a*Sin[c + d*x])^(7/2)) - (16*a^2*Cos[c + d*x]^7)/(99*d*(a + a*Sin[c + d*x
])^(5/2)) - (2*a*Cos[c + d*x]^7)/(11*d*(a + a*Sin[c + d*x])^(3/2))

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)}{11 d (a+a \sin (c+d x))^{3/2}}+\frac {1}{11} (8 a) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {16 a^2 \cos ^7(c+d x)}{99 d (a+a \sin (c+d x))^{5/2}}-\frac {2 a \cos ^7(c+d x)}{11 d (a+a \sin (c+d x))^{3/2}}+\frac {1}{99} \left (32 a^2\right ) \int \frac {\cos ^6(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx \\ & = -\frac {64 a^3 \cos ^7(c+d x)}{693 d (a+a \sin (c+d x))^{7/2}}-\frac {16 a^2 \cos ^7(c+d x)}{99 d (a+a \sin (c+d x))^{5/2}}-\frac {2 a \cos ^7(c+d x)}{11 d (a+a \sin (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \cos ^7(c+d x) \left (151+182 \sin (c+d x)+63 \sin ^2(c+d x)\right )}{693 d (1+\sin (c+d x))^3 \sqrt {a (1+\sin (c+d x))}} \]

[In]

Integrate[Cos[c + d*x]^6/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-2*Cos[c + d*x]^7*(151 + 182*Sin[c + d*x] + 63*Sin[c + d*x]^2))/(693*d*(1 + Sin[c + d*x])^3*Sqrt[a*(1 + Sin[c
 + d*x])])

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67

method result size
default \(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{4} \left (63 \left (\sin ^{2}\left (d x +c \right )\right )+182 \sin \left (d x +c \right )+151\right )}{693 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) \(64\)

[In]

int(cos(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/693*(1+sin(d*x+c))*(sin(d*x+c)-1)^4*(63*sin(d*x+c)^2+182*sin(d*x+c)+151)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/
d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \, {\left (63 \, \cos \left (d x + c\right )^{6} - 7 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + {\left (63 \, \cos \left (d x + c\right )^{5} + 70 \, \cos \left (d x + c\right )^{4} + 80 \, \cos \left (d x + c\right )^{3} + 96 \, \cos \left (d x + c\right )^{2} + 128 \, \cos \left (d x + c\right ) + 256\right )} \sin \left (d x + c\right ) - 128 \, \cos \left (d x + c\right ) - 256\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{693 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/693*(63*cos(d*x + c)^6 - 7*cos(d*x + c)^5 + 10*cos(d*x + c)^4 - 16*cos(d*x + c)^3 + 32*cos(d*x + c)^2 + (63*
cos(d*x + c)^5 + 70*cos(d*x + c)^4 + 80*cos(d*x + c)^3 + 96*cos(d*x + c)^2 + 128*cos(d*x + c) + 256)*sin(d*x +
 c) - 128*cos(d*x + c) - 256)*sqrt(a*sin(d*x + c) + a)/(a*d*cos(d*x + c) + a*d*sin(d*x + c) + a*d)

Sympy [F]

\[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\cos ^{6}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate(cos(d*x+c)**6/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)**6/sqrt(a*(sin(c + d*x) + 1)), x)

Maxima [F]

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

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {64 \, \sqrt {2} {\left (63 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 154 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 99 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}\right )}}{693 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

integrate(cos(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

64/693*sqrt(2)*(63*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^11 - 154*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 +
99*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7)/(a*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

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

[In]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^6/(a + a*sin(c + d*x))^(1/2), x)