Integrand size = 25, antiderivative size = 192 \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}+\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {6 a \left (a^2-2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}}-\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{5 d e^5} \]
-2/5*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))^2/d/e/(e*sin(d*x+c))^(5/2)-2/5*b*(3 *a^2-4*b^2)*(e*sin(d*x+c))^(3/2)/d/e^5+2/5*(a+b*cos(d*x+c))*(a*b-(3*a^2-4* b^2)*cos(d*x+c))/d/e^3/(e*sin(d*x+c))^(1/2)+6/5*a*(a^2-2*b^2)*(sin(1/2*c+1 /4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4* Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/e^4/sin(d*x+c)^(1/2)
Time = 1.68 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.68 \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {12 a^2 b-6 b^3+a \left (7 a^2+6 b^2\right ) \cos (c+d x)+10 b^3 \cos (2 (c+d x))-3 a^3 \cos (3 (c+d x))+6 a b^2 \cos (3 (c+d x))-12 a \left (a^2-2 b^2\right ) E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {5}{2}}(c+d x)}{10 d e (e \sin (c+d x))^{5/2}} \]
-1/10*(12*a^2*b - 6*b^3 + a*(7*a^2 + 6*b^2)*Cos[c + d*x] + 10*b^3*Cos[2*(c + d*x)] - 3*a^3*Cos[3*(c + d*x)] + 6*a*b^2*Cos[3*(c + d*x)] - 12*a*(a^2 - 2*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(5/2))/(d*e*(e*Si n[c + d*x])^(5/2))
Time = 0.88 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3170, 27, 3042, 3340, 27, 3042, 3148, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2}}dx\) |
\(\Big \downarrow \) 3170 |
\(\displaystyle -\frac {2 \int -\frac {(a+b \cos (c+d x)) \left (3 a^2-b \cos (c+d x) a-4 b^2\right )}{2 (e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(a+b \cos (c+d x)) \left (3 a^2-b \cos (c+d x) a-4 b^2\right )}{(e \sin (c+d x))^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right ) \left (3 a^2+b \sin \left (c+d x-\frac {\pi }{2}\right ) a-4 b^2\right )}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3340 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {2 \int \frac {3}{2} \left (a \left (a^2-2 b^2\right )+b \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}dx}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \int \left (a \left (a^2-2 b^2\right )+b \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}dx}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \int \sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a \left (a^2-2 b^2\right )-b \left (3 a^2-4 b^2\right ) \sin \left (c+d x-\frac {\pi }{2}\right )\right )dx}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \left (a \left (a^2-2 b^2\right ) \int \sqrt {e \sin (c+d x)}dx+\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e}\right )}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \left (a \left (a^2-2 b^2\right ) \int \sqrt {e \sin (c+d x)}dx+\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e}\right )}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \left (\frac {a \left (a^2-2 b^2\right ) \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}+\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e}\right )}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \left (\frac {a \left (a^2-2 b^2\right ) \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{\sqrt {\sin (c+d x)}}+\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e}\right )}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{d e \sqrt {e \sin (c+d x)}}-\frac {3 \left (\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{3 d e}+\frac {2 a \left (a^2-2 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\right )}{e^2}}{5 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}\) |
(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(5*d*e*(e*Sin[c + d*x])^( 5/2)) + ((2*(a + b*Cos[c + d*x])*(a*b - (3*a^2 - 4*b^2)*Cos[c + d*x]))/(d* e*Sqrt[e*Sin[c + d*x]]) - (3*((2*a*(a^2 - 2*b^2)*EllipticE[(c - Pi/2 + d*x )/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*Sqrt[Sin[c + d*x]]) + (2*b*(3*a^2 - 4*b^2 )*(e*Sin[c + d*x])^(3/2))/(3*d*e)))/e^2)/(5*e^2)
3.1.56.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x ])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g }, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2* p] || IntegerQ[m])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(g* Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Si n[e + f*x])^(m - 1)*Simp[a*c*(p + 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ [m, 0] && LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])
Time = 4.02 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.95
method | result | size |
default | \(\frac {-\frac {2 b \left (5 b^{2} \left (\cos ^{2}\left (d x +c \right )\right )+3 a^{2}-4 b^{2}\right )}{5 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {a \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-12 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}+6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}+6 a^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-12 b^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(375\) |
parts | \(\frac {a^{3} \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{5}\left (d x +c \right )\right )-4 \left (\sin ^{3}\left (d x +c \right )\right )-2 \sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b^{3} \left (\frac {e^{2}}{5 \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {1}{\sqrt {e \sin \left (d x +c \right )}}\right )}{d \,e^{3}}-\frac {6 a \,b^{2} \left (2 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )+2 \left (\sin ^{5}\left (d x +c \right )\right )-3 \left (\sin ^{3}\left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {6 a^{2} b}{5 \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}} e d}\) | \(402\) |
(-2/5*b/e/(e*sin(d*x+c))^(5/2)*(5*b^2*cos(d*x+c)^2+3*a^2-4*b^2)+1/5*a/e^3* (6*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE( (1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-12*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c )+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^ 2-3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF ((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+6*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c )+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^ 2+6*a^2*cos(d*x+c)^4*sin(d*x+c)-12*b^2*cos(d*x+c)^4*sin(d*x+c)-8*a^2*cos(d *x+c)^2*sin(d*x+c)+6*b^2*cos(d*x+c)^2*sin(d*x+c))/sin(d*x+c)^3/cos(d*x+c)/ (e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.35 \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=-\frac {3 \, {\left (\sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )}\right )} \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )}\right )} \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, a^{2} b - 4 \, b^{3} + {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4}\right )} \sin \left (d x + c\right )} \]
-1/5*(3*(sqrt(2)*(I*a^3 - 2*I*a*b^2)*cos(d*x + c)^2 + sqrt(2)*(-I*a^3 + 2* I*a*b^2))*sqrt(-I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInvers e(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-I*a^3 + 2*I*a*b^2)* cos(d*x + c)^2 + sqrt(2)*(I*a^3 - 2*I*a*b^2))*sqrt(I*e)*sin(d*x + c)*weier strassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(5*b^3*cos(d*x + c)^2 - 3*(a^3 - 2*a*b^2)*cos(d*x + c)^3 + 3*a^2*b - 4*b^3 + (4*a^3 - 3*a*b^2)*cos(d*x + c))*sqrt(e*sin(d*x + c)))/((d*e^4*cos( d*x + c)^2 - d*e^4)*sin(d*x + c))
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]