Integrand size = 25, antiderivative size = 152 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {10 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d e^2 \sqrt {e \cos (c+d x)}}-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \]
4/3*a^7*(e*cos(d*x+c))^(9/2)/d/e^7/(a-a*sin(d*x+c))^3+12*a^8*(e*cos(d*x+c) )^(5/2)/d/e^5/(a^4-a^4*sin(d*x+c))-10*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 ^2/(e*cos(d*x+c))^(1/2)-10*a^4*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d/e^3
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\frac {16 \sqrt [4]{2} a^4 \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},-\frac {3}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{3 d e (e \cos (c+d x))^{3/2}} \]
(16*2^(1/4)*a^4*Hypergeometric2F1[-9/4, -3/4, 1/4, (1 - Sin[c + d*x])/2]*( 1 + Sin[c + d*x])^(3/4))/(3*d*e*(e*Cos[c + d*x])^(3/2))
Time = 0.82 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3149, 3042, 3159, 3042, 3159, 3042, 3115, 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.
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3149 |
| \(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{11/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{11/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^2}dx}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^2}dx}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \int (e \cos (c+d x))^{3/2}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{9/2}}{3 a d (a-a \sin (c+d x))^3}-\frac {3 e^2 \left (\frac {5 e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a^2}-\frac {4 e (e \cos (c+d x))^{5/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}\right )}{a^2}\right )}{e^8}\) |
(a^8*((4*e*(e*Cos[c + d*x])^(9/2))/(3*a*d*(a - a*Sin[c + d*x])^3) - (3*e^2 *((-4*e*(e*Cos[c + d*x])^(5/2))/(d*(a^2 - a^2*Sin[c + d*x])) + (5*e^2*((2* e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[e*Cos[c + d*x] ]) + (2*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/a^2))/a^2))/e^8
3.3.28.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
Time = 6.01 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.73
| method | result | size |
| default | \(-\frac {2 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+18 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-48 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\) | \(263\) |
| parts | \(\text {Expression too large to display}\) | \(835\) |
-2/3/(2*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^2*(8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-8*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+48*sin(1/2*d*x+1/2*c)^5+18*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15* (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(co s(1/2*d*x+1/2*c),2^(1/2))-48*sin(1/2*d*x+1/2*c)^3+20*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 = 156, normalized size of antiderivative = 1.03 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=-\frac {15 \, {\left (-i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) + i \, \sqrt {2} a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, {\left (i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) - i \, \sqrt {2} a^{4}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - 11 \, a^{4} \sin \left (d x + c\right ) + 19 \, a^{4}\right )} \sqrt {e \cos \left (d x + c\right )}}{3 \, {\left (d e^{3} \sin \left (d x + c\right ) - d e^{3}\right )}} \]
-1/3*(15*(-I*sqrt(2)*a^4*sin(d*x + c) + I*sqrt(2)*a^4)*sqrt(e)*weierstrass PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*(I*sqrt(2)*a^4*sin(d*x + c) - I*sqrt(2)*a^4)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I *sin(d*x + c)) + 2*(a^4*cos(d*x + c)^2 - 11*a^4*sin(d*x + c) + 19*a^4)*sqr t(e*cos(d*x + c)))/(d*e^3*sin(d*x + c) - d*e^3)
Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]