Integrand size = 25, antiderivative size = 188 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{21 d e^5}+\frac {2 a \left (5 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}+\frac {2 (a+b \sin (c+d x)) \left (a b+\left (5 a^2-4 b^2\right ) \sin (c+d x)\right )}{21 d e^3 (e \cos (c+d x))^{3/2}} \]
2/7*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^2/d/e/(e*cos(d*x+c))^(7/2)+2/21*(a+b *sin(d*x+c))*(a*b+(5*a^2-4*b^2)*sin(d*x+c))/d/e^3/(e*cos(d*x+c))^(3/2)+2/2 1*a*(5*a^2-6*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elliptic F(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/e^4/(e*cos(d*x+c))^(1/2)+ 2/21*b*(5*a^2-4*b^2)*(e*cos(d*x+c))^(1/2)/d/e^5
Time = 1.47 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\frac {\sqrt {e \cos (c+d x)} \sec ^4(c+d x) \left (36 a^2 b-2 b^3-14 b^3 \cos (2 (c+d x))+4 a \left (5 a^2-6 b^2\right ) \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+17 a^3 \sin (c+d x)+30 a b^2 \sin (c+d x)+5 a^3 \sin (3 (c+d x))-6 a b^2 \sin (3 (c+d x))\right )}{42 d e^5} \]
(Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^4*(36*a^2*b - 2*b^3 - 14*b^3*Cos[2*(c + d*x)] + 4*a*(5*a^2 - 6*b^2)*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 17*a^3*Sin[c + d*x] + 30*a*b^2*Sin[c + d*x] + 5*a^3*Sin[3*(c + d*x)] - 6 *a*b^2*Sin[3*(c + d*x)]))/(42*d*e^5)
Time = 0.87 (sec) , antiderivative size = 197, 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.
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3170 |
| \(\displaystyle \frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}-\frac {2 \int -\frac {(a+b \sin (c+d x)) \left (5 a^2+b \sin (c+d x) a-4 b^2\right )}{2 (e \cos (c+d x))^{5/2}}dx}{7 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \sin (c+d x)) \left (5 a^2+b \sin (c+d x) a-4 b^2\right )}{(e \cos (c+d x))^{5/2}}dx}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(a+b \sin (c+d x)) \left (5 a^2+b \sin (c+d x) a-4 b^2\right )}{(e \cos (c+d x))^{5/2}}dx}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3340 |
| \(\displaystyle \frac {\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}-\frac {2 \int -\frac {a \left (5 a^2-6 b^2\right )-b \left (5 a^2-4 b^2\right ) \sin (c+d x)}{2 \sqrt {e \cos (c+d x)}}dx}{3 e^2}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (5 a^2-6 b^2\right )-b \left (5 a^2-4 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)}}dx}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (5 a^2-6 b^2\right )-b \left (5 a^2-4 b^2\right ) \sin (c+d x)}{\sqrt {e \cos (c+d x)}}dx}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {\frac {a \left (5 a^2-6 b^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (5 a^2-6 b^2\right ) \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {\frac {\frac {a \left (5 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{\sqrt {e \cos (c+d x)}}+\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a \left (5 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt {e \cos (c+d x)}}+\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {2 b \left (5 a^2-4 b^2\right ) \sqrt {e \cos (c+d x)}}{d e}+\frac {2 a \left (5 a^2-6 b^2\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}}{3 e^2}+\frac {2 (a+b \sin (c+d x)) \left (\left (5 a^2-4 b^2\right ) \sin (c+d x)+a b\right )}{3 d e (e \cos (c+d x))^{3/2}}}{7 e^2}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{7 d e (e \cos (c+d x))^{7/2}}\) |
(2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(7*d*e*(e*Cos[c + d*x])^(7 /2)) + (((2*b*(5*a^2 - 4*b^2)*Sqrt[e*Cos[c + d*x]])/(d*e) + (2*a*(5*a^2 - 6*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d*x ]]))/(3*e^2) + (2*(a + b*Sin[c + d*x])*(a*b + (5*a^2 - 4*b^2)*Sin[c + d*x] ))/(3*d*e*(e*Cos[c + d*x])^(3/2)))/(7*e^2)
3.6.63.3.1 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[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])
Leaf count of result is larger than twice the leaf count of optimal. \(749\) vs. \(2(196)=392\).
Time = 9.54 (sec) , antiderivative size = 750, normalized size of antiderivative = 3.99
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(750\) |
| parts | \(\text {Expression too large to display}\) | \(862\) |
-2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c) ^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^4*(40*(2*si n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6*a^3-48*(2*sin(1/2*d*x+1/2*c)^2-1 )^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2) *sin(1/2*d*x+1/2*c)^6*a*b^2+40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a^3 -48*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*a*b^2-60*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*sin(1/2*d*x+1/2*c)^4*a^3+72*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2 *c)^4*a*b^2-40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^3+48*cos(1/2*d*x+ 1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+30*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) *(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x +1/2*c)^2*a^3-36*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c) ^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2*a*b^2-28*s in(1/2*d*x+1/2*c)^5*b^3+16*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3+6*c os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2-5*(sin(1/2*d*x+1/2*c)^2)^(1/2 )*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a ^3+6*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti cF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+28*sin(1/2*d*x+1/2*c)^3*b^3+9*sin(...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\frac {\sqrt {2} {\left (-5 i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (5 i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {e} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (7 \, b^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{2} b - 3 \, b^{3} - {\left (3 \, a^{3} + 9 \, a b^{2} + {\left (5 \, a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{21 \, d e^{5} \cos \left (d x + c\right )^{4}} \]
1/21*(sqrt(2)*(-5*I*a^3 + 6*I*a*b^2)*sqrt(e)*cos(d*x + c)^4*weierstrassPIn verse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(5*I*a^3 - 6*I*a*b^2 )*sqrt(e)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c)) - 2*(7*b^3*cos(d*x + c)^2 - 9*a^2*b - 3*b^3 - (3*a^3 + 9*a*b^2 + (5*a^3 - 6*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(d*e ^5*cos(d*x + c)^4)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}} \,d x \]