\(\int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx\) [527]

Optimal result

Integrand size = 23, antiderivative size = 150 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \left (a^2-b^2\right )^2}{3 b^5 d (a+b \sin (c+d x))^{3/2}}+\frac {8 a \left (a^2-b^2\right )}{b^5 d \sqrt {a+b \sin (c+d x)}}+\frac {4 \left (3 a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}-\frac {8 a (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac {2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 150, 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 = {2747, 711} \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {4 \left (3 a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}+\frac {8 a \left (a^2-b^2\right )}{b^5 d \sqrt {a+b \sin (c+d x)}}-\frac {2 \left (a^2-b^2\right )^2}{3 b^5 d (a+b \sin (c+d x))^{3/2}}+\frac {2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{3/2}}{3 b^5 d} \]

[In]

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

[Out]

(-2*(a^2 - b^2)^2)/(3*b^5*d*(a + b*Sin[c + d*x])^(3/2)) + (8*a*(a^2 - b^2))/(b^5*d*Sqrt[a + b*Sin[c + d*x]]) +
(4*(3*a^2 - b^2)*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) - (8*a*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) + (2*(a + b*S
in[c + d*x])^(5/2))/(5*b^5*d)

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2747

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{(a+x)^{5/2}} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {\left (a^2-b^2\right )^2}{(a+x)^{5/2}}-\frac {4 \left (a^3-a b^2\right )}{(a+x)^{3/2}}+\frac {2 \left (3 a^2-b^2\right )}{\sqrt {a+x}}-4 a \sqrt {a+x}+(a+x)^{3/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = -\frac {2 \left (a^2-b^2\right )^2}{3 b^5 d (a+b \sin (c+d x))^{3/2}}+\frac {8 a \left (a^2-b^2\right )}{b^5 d \sqrt {a+b \sin (c+d x)}}+\frac {4 \left (3 a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}-\frac {8 a (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac {2 (a+b \sin (c+d x))^{5/2}}{5 b^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=\frac {6 b^4 \cos ^4(c+d x)+16 \left (16 a^4-10 a^2 b^2-b^4+3 a b \left (8 a^2-5 b^2\right ) \sin (c+d x)+\left (6 a^2 b^2-3 b^4\right ) \sin ^2(c+d x)-a b^3 \sin ^3(c+d x)\right )}{15 b^5 d (a+b \sin (c+d x))^{3/2}} \]

[In]

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

[Out]

(6*b^4*Cos[c + d*x]^4 + 16*(16*a^4 - 10*a^2*b^2 - b^4 + 3*a*b*(8*a^2 - 5*b^2)*Sin[c + d*x] + (6*a^2*b^2 - 3*b^
4)*Sin[c + d*x]^2 - a*b^3*Sin[c + d*x]^3))/(15*b^5*d*(a + b*Sin[c + d*x])^(3/2))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+12 a^{2} \sqrt {a +b \sin \left (d x +c \right )}-4 b^{2} \sqrt {a +b \sin \left (d x +c \right )}-\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{3 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {8 a \left (a^{2}-b^{2}\right )}{\sqrt {a +b \sin \left (d x +c \right )}}}{d \,b^{5}}\)	\(126\)
default	\(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+12 a^{2} \sqrt {a +b \sin \left (d x +c \right )}-4 b^{2} \sqrt {a +b \sin \left (d x +c \right )}-\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{3 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {8 a \left (a^{2}-b^{2}\right )}{\sqrt {a +b \sin \left (d x +c \right )}}}{d \,b^{5}}\)	\(126\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, b^{4} \cos \left (d x + c\right )^{4} + 128 \, a^{4} - 32 \, a^{2} b^{2} - 32 \, b^{4} - 24 \, {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} + 24 \, a^{3} b - 16 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{15 \, {\left (b^{7} d \cos \left (d x + c\right )^{2} - 2 \, a b^{6} d \sin \left (d x + c\right ) - {\left (a^{2} b^{5} + b^{7}\right )} d\right )}} \]

[In]

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

[Out]

-2/15*(3*b^4*cos(d*x + c)^4 + 128*a^4 - 32*a^2*b^2 - 32*b^4 - 24*(2*a^2*b^2 - b^4)*cos(d*x + c)^2 + 8*(a*b^3*c
os(d*x + c)^2 + 24*a^3*b - 16*a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^7*d*cos(d*x + c)^2 - 2*a*b^6*d*
sin(d*x + c) - (a^2*b^5 + b^7)*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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