\(\int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [506]

Optimal result

Integrand size = 23, antiderivative size = 152 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}{b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac {2 (a+b \sin (c+d x))^{9/2}}{9 b^5 d} \]

[Out]

-8/3*a*(a^2-b^2)*(a+b*sin(d*x+c))^(3/2)/b^5/d+4/5*(3*a^2-b^2)*(a+b*sin(d*x+c))^(5/2)/b^5/d-8/7*a*(a+b*sin(d*x+
c))^(7/2)/b^5/d+2/9*(a+b*sin(d*x+c))^(9/2)/b^5/d+2*(a^2-b^2)^2*(a+b*sin(d*x+c))^(1/2)/b^5/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 152, 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} \[ \int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac {2 \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}{b^5 d}+\frac {2 (a+b \sin (c+d x))^{9/2}}{9 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{7/2}}{7 b^5 d} \]

[In]

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

[Out]

(2*(a^2 - b^2)^2*Sqrt[a + b*Sin[c + d*x]])/(b^5*d) - (8*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/(3*b^5*d) +
(4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(5/2))/(5*b^5*d) - (8*a*(a + b*Sin[c + d*x])^(7/2))/(7*b^5*d) + (2*(a +
b*Sin[c + d*x])^(9/2))/(9*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}{\sqrt {a+x}} \, 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}{\sqrt {a+x}}-4 \left (a^3-a b^2\right ) \sqrt {a+x}+2 \left (3 a^2-b^2\right ) (a+x)^{3/2}-4 a (a+x)^{5/2}+(a+x)^{7/2}\right ) \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {2 \left (a^2-b^2\right )^2 \sqrt {a+b \sin (c+d x)}}{b^5 d}-\frac {8 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{5/2}}{5 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{7/2}}{7 b^5 d}+\frac {2 (a+b \sin (c+d x))^{9/2}}{9 b^5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {\sqrt {a+b \sin (c+d x)} \left (1024 a^4-2496 a^2 b^2+2121 b^4-4 \left (48 a^2 b^2-91 b^4\right ) \cos (2 (c+d x))+35 b^4 \cos (4 (c+d x))-512 a^3 b \sin (c+d x)+1104 a b^3 \sin (c+d x)+80 a b^3 \sin (3 (c+d x))\right )}{1260 b^5 d} \]

[In]

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

[Out]

(Sqrt[a + b*Sin[c + d*x]]*(1024*a^4 - 2496*a^2*b^2 + 2121*b^4 - 4*(48*a^2*b^2 - 91*b^4)*Cos[2*(c + d*x)] + 35*
b^4*Cos[4*(c + d*x)] - 512*a^3*b*Sin[c + d*x] + 1104*a*b^3*Sin[c + d*x] + 80*a*b^3*Sin[3*(c + d*x)]))/(1260*b^
5*d)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a +b \sin \left (d x +c \right )}}{d \,b^{5}}\)	\(149\)
default	\(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (a +b \right )^{2}+\left (-2 a -2 b \right ) \left (-2 a +2 b \right )+\left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (a +b \right )^{2} \left (-2 a +2 b \right )+\left (-2 a -2 b \right ) \left (a -b \right )^{2}\right ) \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a +b \sin \left (d x +c \right )}}{d \,b^{5}}\)	\(149\)

[In]

int(cos(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/d/b^5*(1/9*(a+b*sin(d*x+c))^(9/2)-4/7*a*(a+b*sin(d*x+c))^(7/2)+1/5*((a+b)^2+(-2*a-2*b)*(-2*a+2*b)+(a-b)^2)*(
a+b*sin(d*x+c))^(5/2)+1/3*((a+b)^2*(-2*a+2*b)+(-2*a-2*b)*(a-b)^2)*(a+b*sin(d*x+c))^(3/2)+(a-b)^2*(a+b)^2*(a+b*
sin(d*x+c))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \, {\left (35 \, b^{4} \cos \left (d x + c\right )^{4} + 128 \, a^{4} - 288 \, a^{2} b^{2} + 224 \, b^{4} - 8 \, {\left (6 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (5 \, a b^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3} b + 16 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{315 \, b^{5} d} \]

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*b^4*cos(d*x + c)^4 + 128*a^4 - 288*a^2*b^2 + 224*b^4 - 8*(6*a^2*b^2 - 7*b^4)*cos(d*x + c)^2 + 8*(5*a
*b^3*cos(d*x + c)^2 - 8*a^3*b + 16*a*b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^5*d)

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**5/(a+b*sin(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \, {\left (315 \, \sqrt {b \sin \left (d x + c\right ) + a} - \frac {42 \, {\left (3 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{2}\right )}}{b^{2}} + \frac {35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 180 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{4}}{b^{4}}\right )}}{315 \, b d} \]

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/315*(315*sqrt(b*sin(d*x + c) + a) - 42*(3*(b*sin(d*x + c) + a)^(5/2) - 10*(b*sin(d*x + c) + a)^(3/2)*a + 15*
sqrt(b*sin(d*x + c) + a)*a^2)/b^2 + (35*(b*sin(d*x + c) + a)^(9/2) - 180*(b*sin(d*x + c) + a)^(7/2)*a + 378*(b
*sin(d*x + c) + a)^(5/2)*a^2 - 420*(b*sin(d*x + c) + a)^(3/2)*a^3 + 315*sqrt(b*sin(d*x + c) + a)*a^4)/b^4)/(b*
d)

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \, {\left (35 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 180 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{4} - 126 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} b^{2} + 420 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a b^{2} - 630 \, \sqrt {b \sin \left (d x + c\right ) + a} a^{2} b^{2} + 315 \, \sqrt {b \sin \left (d x + c\right ) + a} b^{4}\right )}}{315 \, b^{5} d} \]

[In]

integrate(cos(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

2/315*(35*(b*sin(d*x + c) + a)^(9/2) - 180*(b*sin(d*x + c) + a)^(7/2)*a + 378*(b*sin(d*x + c) + a)^(5/2)*a^2 -
420*(b*sin(d*x + c) + a)^(3/2)*a^3 + 315*sqrt(b*sin(d*x + c) + a)*a^4 - 126*(b*sin(d*x + c) + a)^(5/2)*b^2 +
420*(b*sin(d*x + c) + a)^(3/2)*a*b^2 - 630*sqrt(b*sin(d*x + c) + a)*a^2*b^2 + 315*sqrt(b*sin(d*x + c) + a)*b^4
)/(b^5*d)

Mupad [F(-1)]

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

[In]

int(cos(c + d*x)^5/(a + b*sin(c + d*x))^(1/2),x)

[Out]

int(cos(c + d*x)^5/(a + b*sin(c + d*x))^(1/2), x)