\(\int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [187]

Optimal result

Integrand size = 23, antiderivative size = 71 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {8 \sqrt {a+a \sin (c+d x)}}{a^3 d}-\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a^4 d}+\frac {2 (a+a \sin (c+d x))^{5/2}}{5 a^5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 71, 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 = {2746, 45} \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 (a \sin (c+d x)+a)^{5/2}}{5 a^5 d}-\frac {8 (a \sin (c+d x)+a)^{3/2}}{3 a^4 d}+\frac {8 \sqrt {a \sin (c+d x)+a}}{a^3 d} \]

[In]

Int[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.62 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 \sqrt {a (1+\sin (c+d x))} \left (43-14 \sin (c+d x)+3 \sin ^2(c+d x)\right )}{15 a^3 d} \]

[In]

Integrate[Cos[c + d*x]^5/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(2*Sqrt[a*(1 + Sin[c + d*x])]*(43 - 14*Sin[c + d*x] + 3*Sin[c + d*x]^2))/(15*a^3*d)

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {8 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+8 a^{2} \sqrt {a +a \sin \left (d x +c \right )}}{d \,a^{5}}\)	\(56\)
default	\(\frac {\frac {2 \left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {8 a \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+8 a^{2} \sqrt {a +a \sin \left (d x +c \right )}}{d \,a^{5}}\)	\(56\)

[In]

int(cos(d*x+c)^5/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 14 \, \sin \left (d x + c\right ) - 46\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, a^{3} d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(3*cos(d*x + c)^2 + 14*sin(d*x + c) - 46)*sqrt(a*sin(d*x + c) + a)/(a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 20 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 60 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{2}\right )}}{15 \, a^{5} d} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2/15*(3*(a*sin(d*x + c) + a)^(5/2) - 20*(a*sin(d*x + c) + a)^(3/2)*a + 60*sqrt(a*sin(d*x + c) + a)*a^2)/(a^5*d
)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {8 \, {\left (3 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \sqrt {2} \sqrt {a} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15 \, a^{3} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

integrate(cos(d*x+c)^5/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

8/15*(3*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 10*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3
+ 15*sqrt(2)*sqrt(a)*cos(-1/4*pi + 1/2*d*x + 1/2*c))/(a^3*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(cos(c + d*x)^5/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^5/(a + a*sin(c + d*x))^(5/2), x)