Integrand size = 23, antiderivative size = 150 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=-\frac {2 \left (a^2-b^2\right )^2}{b^5 d \sqrt {a+b \sin (c+d x)}}-\frac {8 a \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}}{b^5 d}+\frac {4 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}{3 b^5 d}-\frac {8 a (a+b \sin (c+d x))^{5/2}}{5 b^5 d}+\frac {2 (a+b \sin (c+d x))^{7/2}}{7 b^5 d} \]
4/3*(3*a^2-b^2)*(a+b*sin(d*x+c))^(3/2)/b^5/d-8/5*a*(a+b*sin(d*x+c))^(5/2)/ b^5/d+2/7*(a+b*sin(d*x+c))^(7/2)/b^5/d-2*(a^2-b^2)^2/b^5/d/(a+b*sin(d*x+c) )^(1/2)-8*a*(a^2-b^2)*(a+b*sin(d*x+c))^(1/2)/b^5/d
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {30 b^4 \cos ^4(c+d x)-16 \left (48 a^4-70 a^2 b^2+15 b^4+a b \left (24 a^2-35 b^2\right ) \sin (c+d x)+\left (-6 a^2 b^2+5 b^4\right ) \sin ^2(c+d x)+3 a b^3 \sin ^3(c+d x)\right )}{105 b^5 d \sqrt {a+b \sin (c+d x)}} \]
(30*b^4*Cos[c + d*x]^4 - 16*(48*a^4 - 70*a^2*b^2 + 15*b^4 + a*b*(24*a^2 - 35*b^2)*Sin[c + d*x] + (-6*a^2*b^2 + 5*b^4)*Sin[c + d*x]^2 + 3*a*b^3*Sin[c + d*x]^3))/(105*b^5*d*Sqrt[a + b*Sin[c + d*x]])
Time = 0.31 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3147, 476, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5}{(a+b \sin (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int \frac {\left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^{3/2}}d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 476 |
\(\displaystyle \frac {\int \left ((a+b \sin (c+d x))^{5/2}-4 a (a+b \sin (c+d x))^{3/2}+2 \left (3 a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}-\frac {4 \left (a^3-a b^2\right )}{\sqrt {a+b \sin (c+d x)}}+\frac {\left (a^2-b^2\right )^2}{(a+b \sin (c+d x))^{3/2}}\right )d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{3} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}-8 a \left (a^2-b^2\right ) \sqrt {a+b \sin (c+d x)}-\frac {2 \left (a^2-b^2\right )^2}{\sqrt {a+b \sin (c+d x)}}+\frac {2}{7} (a+b \sin (c+d x))^{7/2}-\frac {8}{5} a (a+b \sin (c+d x))^{5/2}}{b^5 d}\) |
((-2*(a^2 - b^2)^2)/Sqrt[a + b*Sin[c + d*x]] - 8*a*(a^2 - b^2)*Sqrt[a + b* Sin[c + d*x]] + (4*(3*a^2 - b^2)*(a + b*Sin[c + d*x])^(3/2))/3 - (8*a*(a + b*Sin[c + d*x])^(5/2))/5 + (2*(a + b*Sin[c + d*x])^(7/2))/7)/(b^5*d)
3.6.16.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[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] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+4 a^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}-\frac {4 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-8 a^{3} \sqrt {a +b \sin \left (d x +c \right )}+8 a \,b^{2} \sqrt {a +b \sin \left (d x +c \right )}-\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{\sqrt {a +b \sin \left (d x +c \right )}}}{d \,b^{5}}\) | \(137\) |
default | \(\frac {\frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {8 a \left (a +b \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+4 a^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}-\frac {4 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-8 a^{3} \sqrt {a +b \sin \left (d x +c \right )}+8 a \,b^{2} \sqrt {a +b \sin \left (d x +c \right )}-\frac {2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{\sqrt {a +b \sin \left (d x +c \right )}}}{d \,b^{5}}\) | \(137\) |
2/d/b^5*(1/7*(a+b*sin(d*x+c))^(7/2)-4/5*a*(a+b*sin(d*x+c))^(5/2)+2*a^2*(a+ b*sin(d*x+c))^(3/2)-2/3*b^2*(a+b*sin(d*x+c))^(3/2)-4*a^3*(a+b*sin(d*x+c))^ (1/2)+4*a*b^2*(a+b*sin(d*x+c))^(1/2)-(a^4-2*a^2*b^2+b^4)/(a+b*sin(d*x+c))^ (1/2))
Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (15 \, b^{4} \cos \left (d x + c\right )^{4} - 384 \, a^{4} + 608 \, a^{2} b^{2} - 160 \, b^{4} - 8 \, {\left (6 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, a b^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} b + 32 \, a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{105 \, {\left (b^{6} d \sin \left (d x + c\right ) + a b^{5} d\right )}} \]
2/105*(15*b^4*cos(d*x + c)^4 - 384*a^4 + 608*a^2*b^2 - 160*b^4 - 8*(6*a^2* b^2 - 5*b^4)*cos(d*x + c)^2 + 8*(3*a*b^3*cos(d*x + c)^2 - 24*a^3*b + 32*a* b^3)*sin(d*x + c))*sqrt(b*sin(d*x + c) + a)/(b^6*d*sin(d*x + c) + a*b^5*d)
Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\frac {2 \, {\left (\frac {15 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 84 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 70 \, {\left (3 \, a^{2} - b^{2}\right )} {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 420 \, {\left (a^{3} - a b^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{b^{4}} - \frac {105 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}{\sqrt {b \sin \left (d x + c\right ) + a} b^{4}}\right )}}{105 \, b d} \]
2/105*((15*(b*sin(d*x + c) + a)^(7/2) - 84*(b*sin(d*x + c) + a)^(5/2)*a + 70*(3*a^2 - b^2)*(b*sin(d*x + c) + a)^(3/2) - 420*(a^3 - a*b^2)*sqrt(b*sin (d*x + c) + a))/b^4 - 105*(a^4 - 2*a^2*b^2 + b^4)/(sqrt(b*sin(d*x + c) + a )*b^4))/(b*d)
Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^5(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]