Integrand size = 25, antiderivative size = 193 \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac {2 (a+b \cos (c+d x)) \left (a b+\left (5 a^2-4 b^2\right ) \cos (c+d x)\right )}{21 d e^3 (e \sin (c+d x))^{3/2}}+\frac {2 a \left (5 a^2-6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 d e^4 \sqrt {e \sin (c+d x)}}-\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \sin (c+d x)}}{21 d e^5} \] Output:
-2/7*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))^2/d/e/(e*sin(d*x+c))^(7/2)-2/21*(a+ b*cos(d*x+c))*(a*b+(5*a^2-4*b^2)*cos(d*x+c))/d/e^3/(e*sin(d*x+c))^(3/2)+2/ 21*a*(5*a^2-6*b^2)*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))*sin(d*x+c )^(1/2)/d/e^4/(e*sin(d*x+c))^(1/2)-2/21*b*(5*a^2-4*b^2)*(e*sin(d*x+c))^(1/ 2)/d/e^5
Time = 2.26 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.75 \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=-\frac {2 \csc ^4(c+d x) \sqrt {e \sin (c+d x)} \left (\frac {1}{4} \left (36 a^2 b-2 b^3+a \left (17 a^2+30 b^2\right ) \cos (c+d x)+14 b^3 \cos (2 (c+d x))-5 a^3 \cos (3 (c+d x))+6 a b^2 \cos (3 (c+d x))\right )+a \left (5 a^2-6 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{21 d e^5} \] Input:
Integrate[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(9/2),x]
Output:
(-2*Csc[c + d*x]^4*Sqrt[e*Sin[c + d*x]]*((36*a^2*b - 2*b^3 + a*(17*a^2 + 3 0*b^2)*Cos[c + d*x] + 14*b^3*Cos[2*(c + d*x)] - 5*a^3*Cos[3*(c + d*x)] + 6 *a*b^2*Cos[3*(c + d*x)])/4 + a*(5*a^2 - 6*b^2)*EllipticF[(-2*c + Pi - 2*d* x)/4, 2]*Sin[c + d*x]^(7/2)))/(21*d*e^5)
Time = 0.94 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.05, 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, 3120}
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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3170 |
| \(\displaystyle -\frac {2 \int -\frac {(a+b \cos (c+d x)) \left (5 a^2+b \cos (c+d x) a-4 b^2\right )}{2 (e \sin (c+d x))^{5/2}}dx}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \cos (c+d x)) \left (5 a^2+b \cos (c+d x) a-4 b^2\right )}{(e \sin (c+d x))^{5/2}}dx}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right ) \left (5 a^2-b \sin \left (c+d x-\frac {\pi }{2}\right ) a-4 b^2\right )}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3340 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {a \left (5 a^2-6 b^2\right )-b \left (5 a^2-4 b^2\right ) \cos (c+d x)}{2 \sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (5 a^2-6 b^2\right )-b \left (5 a^2-4 b^2\right ) \cos (c+d x)}{\sqrt {e \sin (c+d x)}}dx}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (5 a^2-6 b^2\right )+b \left (5 a^2-4 b^2\right ) \sin \left (c+d x-\frac {\pi }{2}\right )}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {\frac {a \left (5 a^2-6 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (5 a^2-6 b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {\frac {\frac {a \left (5 a^2-6 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}-\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a \left (5 a^2-6 b^2\right ) \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{\sqrt {e \sin (c+d x)}}-\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {2 a \left (5 a^2-6 b^2\right ) \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {e \sin (c+d x)}}-\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \sin (c+d x)}}{d e}}{3 e^2}-\frac {2 (a+b \cos (c+d x)) \left (\left (5 a^2-4 b^2\right ) \cos (c+d x)+a b\right )}{3 d e (e \sin (c+d x))^{3/2}}}{7 e^2}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{7 d e (e \sin (c+d x))^{7/2}}\) |
Input:
Int[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(9/2),x]
Output:
(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(7*d*e*(e*Sin[c + d*x])^( 7/2)) + ((-2*(a + b*Cos[c + d*x])*(a*b + (5*a^2 - 4*b^2)*Cos[c + d*x]))/(3 *d*e*(e*Sin[c + d*x])^(3/2)) + ((2*a*(5*a^2 - 6*b^2)*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(d*Sqrt[e*Sin[c + d*x]]) - (2*b*(5*a^2 - 4 *b^2)*Sqrt[e*Sin[c + d*x]])/(d*e))/(3*e^2))/(7*e^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 = 4.95 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.37
| method | result | size |
| default | \(\frac {-\frac {2 b \left (7 \cos \left (d x +c \right )^{2} b^{2}+9 a^{2}-4 b^{2}\right )}{21 e \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {a \left (5 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {9}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {9}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right ) b^{2}-10 a^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+12 b^{2} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+16 a^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+6 b^{2} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )\right )}{21 e^{4} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(265\) |
| parts | \(-\frac {a^{3} \left (5 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {9}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-10 \sin \left (d x +c \right )^{5}+4 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )\right )}{21 e^{4} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}-\frac {2 b^{3} \left (\frac {e^{2}}{7 \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {1}{3 \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,e^{3}}-\frac {6 a^{2} b}{7 \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}} e d}+\frac {2 b^{2} a \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{\frac {9}{2}} \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{5}+5 \sin \left (d x +c \right )^{3}-3 \sin \left (d x +c \right )\right )}{7 e^{4} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\) | \(299\) |
Input:
int((a+cos(d*x+c)*b)^3/(e*sin(d*x+c))^(9/2),x,method=_RETURNVERBOSE)
Output:
(-2/21*b/e/(e*sin(d*x+c))^(7/2)*(7*cos(d*x+c)^2*b^2+9*a^2-4*b^2)-1/21*a/e^ 4*(5*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(9/2)*Elliptic F((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-6*(1-sin(d*x+c))^(1/2)*(2+2*sin(d* x+c))^(1/2)*sin(d*x+c)^(9/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b ^2-10*a^2*cos(d*x+c)^4*sin(d*x+c)+12*b^2*cos(d*x+c)^4*sin(d*x+c)+16*a^2*co s(d*x+c)^2*sin(d*x+c)+6*b^2*cos(d*x+c)^2*sin(d*x+c))/sin(d*x+c)^4/cos(d*x+ c)/(e*sin(d*x+c))^(1/2))/d
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.42 \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=\frac {2 \, {\left ({\left ({\left (5 \, a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 5 \, a^{3} - 6 \, a b^{2} - 2 \, {\left (5 \, a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left ({\left (5 \, a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 5 \, a^{3} - 6 \, a b^{2} - 2 \, {\left (5 \, a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {1}{2} i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (7 \, b^{3} \cos \left (d x + c\right )^{2} - {\left (5 \, a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 9 \, a^{2} b - 4 \, b^{3} + {\left (8 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{21 \, {\left (d e^{5} \cos \left (d x + c\right )^{4} - 2 \, d e^{5} \cos \left (d x + c\right )^{2} + d e^{5}\right )}} \] Input:
integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(9/2),x, algorithm="fricas")
Output:
2/21*(((5*a^3 - 6*a*b^2)*cos(d*x + c)^4 + 5*a^3 - 6*a*b^2 - 2*(5*a^3 - 6*a *b^2)*cos(d*x + c)^2)*sqrt(-1/2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c ) + I*sin(d*x + c)) + ((5*a^3 - 6*a*b^2)*cos(d*x + c)^4 + 5*a^3 - 6*a*b^2 - 2*(5*a^3 - 6*a*b^2)*cos(d*x + c)^2)*sqrt(1/2*I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - (7*b^3*cos(d*x + c)^2 - (5*a^3 - 6*a* b^2)*cos(d*x + c)^3 + 9*a^2*b - 4*b^3 + (8*a^3 + 3*a*b^2)*cos(d*x + c))*sq rt(e*sin(d*x + c)))/(d*e^5*cos(d*x + c)^4 - 2*d*e^5*cos(d*x + c)^2 + d*e^5 )
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3/(e*sin(d*x+c))**(9/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(9/2),x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c) + a)^3/(e*sin(d*x + c))^(9/2), x)
\[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=\int { \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\left (e \sin \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(9/2),x, algorithm="giac")
Output:
integrate((b*cos(d*x + c) + a)^3/(e*sin(d*x + c))^(9/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{9/2}} \,d x \] Input:
int((a + b*cos(c + d*x))^3/(e*sin(c + d*x))^(9/2),x)
Output:
int((a + b*cos(c + d*x))^3/(e*sin(c + d*x))^(9/2), x)
\[ \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{9/2}} \, dx=\frac {\sqrt {e}\, \left (-6 \sqrt {\sin \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} b^{3}+8 \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{2} b^{3}-18 \sqrt {\sin \left (d x +c \right )}\, a^{2} b +21 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}}{\sin \left (d x +c \right )^{5}}d x \right ) \sin \left (d x +c \right )^{4} a^{3} d +63 \left (\int \frac {\sqrt {\sin \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{5}}d x \right ) \sin \left (d x +c \right )^{4} a \,b^{2} d \right )}{21 \sin \left (d x +c \right )^{4} d \,e^{5}} \] Input:
int((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(9/2),x)
Output:
(sqrt(e)*( - 6*sqrt(sin(c + d*x))*cos(c + d*x)**2*b**3 + 8*sqrt(sin(c + d* x))*sin(c + d*x)**2*b**3 - 18*sqrt(sin(c + d*x))*a**2*b + 21*int(sqrt(sin( c + d*x))/sin(c + d*x)**5,x)*sin(c + d*x)**4*a**3*d + 63*int((sqrt(sin(c + d*x))*cos(c + d*x)**2)/sin(c + d*x)**5,x)*sin(c + d*x)**4*a*b**2*d))/(21* sin(c + d*x)**4*d*e**5)