Integrand size = 25, antiderivative size = 127 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}+\frac {4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}-\frac {20 a^8 \sqrt {e \cos (c+d x)}}{21 d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \]
4/7*a^7*(e*cos(d*x+c))^(5/2)/d/e^7/(a-a*sin(d*x+c))^3+10/21*a^4*(cos(1/2*d *x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(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)-20/21*a^8*(e*cos(d*x+c))^(1/ 2)/d/e^5/(a^4-a^4*sin(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=\frac {8 \sqrt [4]{2} a^4 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {5}{4},-\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{7/4}}{7 d e (e \cos (c+d x))^{7/2}} \]
(8*2^(1/4)*a^4*Hypergeometric2F1[-7/4, -5/4, -3/4, (1 - Sin[c + d*x])/2]*( 1 + Sin[c + d*x])^(7/4))/(7*d*e*(e*Cos[c + d*x])^(7/2))
Time = 0.61 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3149, 3042, 3159, 3042, 3159, 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.
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3149 |
| \(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{(a-a \sin (c+d x))^2}dx}{7 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \int \frac {(e \cos (c+d x))^{3/2}}{(a-a \sin (c+d x))^2}dx}{7 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \left (\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{3 a^2}\right )}{7 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \left (\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a^2}\right )}{7 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \left (\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 a^2 \sqrt {e \cos (c+d x)}}\right )}{7 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \left (\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a^2 \sqrt {e \cos (c+d x)}}\right )}{7 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{5/2}}{7 a d (a-a \sin (c+d x))^3}-\frac {5 e^2 \left (\frac {4 e \sqrt {e \cos (c+d x)}}{3 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d \sqrt {e \cos (c+d x)}}\right )}{7 a^2}\right )}{e^8}\) |
(a^8*((4*e*(e*Cos[c + d*x])^(5/2))/(7*a*d*(a - a*Sin[c + d*x])^3) - (5*e^2 *((-2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*Sqrt[e*Co s[c + d*x]]) + (4*e*Sqrt[e*Cos[c + d*x]])/(3*d*(a^2 - a^2*Sin[c + d*x])))) /(7*a^2)))/e^8
3.3.30.3.1 Defintions of rubi rules used
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 _)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m) Int[(g*Cos[e + f*x])^(2*m + p)/( a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 , 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(139)=278\).
Time = 8.72 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.16
| method | result | size |
| default | \(\frac {2 \left (-40 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+60 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-128 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+112 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-112 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{4} d}\) | \(401\) |
| parts | \(\text {Expression too large to display}\) | \(1260\) |
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*(sin( 1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(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)^6+128*sin(1/2*d*x+1/2*c)^6*cos(1/2 *d*x+1/2*c)+60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(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)^4-128*sin(1/2* d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-30*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF( cos(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+112*sin(1/2*d*x+1/2*c)^5-16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c )+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)*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))-112*sin(1/2*d*x+1/2*c)^3+4*sin(1/2*d*x+1/2*c ))*a^4/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.50 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=-\frac {5 \, {\left (i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 8 \, {\left (4 \, a^{4} \sin \left (d x + c\right ) - a^{4}\right )} \sqrt {e \cos \left (d x + c\right )}}{21 \, {\left (d e^{5} \cos \left (d x + c\right )^{2} + 2 \, d e^{5} \sin \left (d x + c\right ) - 2 \, d e^{5}\right )}} \]
-1/21*(5*(I*sqrt(2)*a^4*cos(d*x + c)^2 + 2*I*sqrt(2)*a^4*sin(d*x + c) - 2* I*sqrt(2)*a^4)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(-I*sqrt(2)*a^4*cos(d*x + c)^2 - 2*I*sqrt(2)*a^4*sin(d*x + c) + 2*I*sqrt(2)*a^4)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin( d*x + c)) + 8*(4*a^4*sin(d*x + c) - a^4)*sqrt(e*cos(d*x + c)))/(d*e^5*cos( d*x + c)^2 + 2*d*e^5*sin(d*x + c) - 2*d*e^5)
Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}} \,d x \]