Integrand size = 25, antiderivative size = 127 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {42 a^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {4 a^7 (e \cos (c+d x))^{7/2}}{5 d e^7 (a-a \sin (c+d x))^3}-\frac {28 a^8 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \]
4/5*a^7*(e*cos(d*x+c))^(7/2)/d/e^7/(a-a*sin(d*x+c))^3-28/5*a^8*(e*cos(d*x+ c))^(3/2)/d/e^5/(a^4-a^4*sin(d*x+c))+42/5*a^4*(cos(1/2*d*x+1/2*c)^2)^(1/2) /cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^( 1/2)/d/e^4/cos(d*x+c)^(1/2)
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 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\frac {8\ 2^{3/4} a^4 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {5}{4},-\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/4}}{5 d e (e \cos (c+d x))^{5/2}} \]
(8*2^(3/4)*a^4*Hypergeometric2F1[-7/4, -5/4, -1/4, (1 - Sin[c + d*x])/2]*( 1 + Sin[c + d*x])^(5/4))/(5*d*e*(e*Cos[c + d*x])^(5/2))
Time = 0.62 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06, 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, 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.
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^4}{(e \cos (c+d x))^{7/2}}dx\) |
| \(\Big \downarrow \) 3149 |
| \(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{9/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \int \frac {(e \cos (c+d x))^{9/2}}{(a-a \sin (c+d x))^4}dx}{e^8}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2}dx}{5 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2}dx}{5 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \int \sqrt {e \cos (c+d x)}dx}{a^2}\right )}{5 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}\right )}{5 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a^2 \sqrt {\cos (c+d x)}}\right )}{5 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2 \sqrt {\cos (c+d x)}}\right )}{5 a^2}\right )}{e^8}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a^8 \left (\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a-a \sin (c+d x))^3}-\frac {7 e^2 \left (\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a^2 d \sqrt {\cos (c+d x)}}\right )}{5 a^2}\right )}{e^8}\) |
(a^8*((4*e*(e*Cos[c + d*x])^(7/2))/(5*a*d*(a - a*Sin[c + d*x])^3) - (7*e^2 *((-6*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[Cos[ c + d*x]]) + (4*e*(e*Cos[c + d*x])^(3/2))/(d*(a^2 - a^2*Sin[c + d*x]))))/( 5*a^2)))/e^8
3.3.29.3.1 Defintions of rubi rules used
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 _)])^(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. \(331\) vs. \(2(139)=278\).
Time = 8.64 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.61
| method | result | size |
| default | \(-\frac {2 \left (128 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-84 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-128 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+84 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-80 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{3} d}\) | \(332\) |
| parts | \(\text {Expression too large to display}\) | \(1165\) |
-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*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^3*(128*sin(1/2*d*x+1/2*c)^6*cos(1/2* d*x+1/2*c)-84*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(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)^4-128*sin(1/2*d *x+1/2*c)^4*cos(1/2*d*x+1/2*c)+84*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti cE(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)^2+80*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)- 21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE (cos(1/2*d*x+1/2*c),2^(1/2))-80*sin(1/2*d*x+1/2*c)^3+12*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 = 317, normalized size of antiderivative = 2.50 \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {21 \, {\left (-i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} a^{4} + {\left (-i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, {\left (i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} a^{4} + {\left (i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 8 \, {\left (4 \, a^{4} \cos \left (d x + c\right )^{2} + 3 \, a^{4} \cos \left (d x + c\right ) - a^{4} - {\left (4 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4} \cos \left (d x + c\right ) - 2 \, d e^{4} + {\left (d e^{4} \cos \left (d x + c\right ) + 2 \, d e^{4}\right )} \sin \left (d x + c\right )\right )}} \]
-1/5*(21*(-I*sqrt(2)*a^4*cos(d*x + c)^2 + I*sqrt(2)*a^4*cos(d*x + c) + 2*I *sqrt(2)*a^4 + (-I*sqrt(2)*a^4*cos(d*x + c) - 2*I*sqrt(2)*a^4)*sin(d*x + c ))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*(I*sqrt(2)*a^4*cos(d*x + c)^2 - I*sqrt(2)*a^4*cos( d*x + c) - 2*I*sqrt(2)*a^4 + (I*sqrt(2)*a^4*cos(d*x + c) + 2*I*sqrt(2)*a^4 )*sin(d*x + c))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 8*(4*a^4*cos(d*x + c)^2 + 3*a^4*cos(d*x + c) - a^4 - (4*a^4*cos(d*x + c) + a^4)*sin(d*x + c))*sqrt(e*cos(d*x + c)) )/(d*e^4*cos(d*x + c)^2 - d*e^4*cos(d*x + c) - 2*d*e^4 + (d*e^4*cos(d*x + c) + 2*d*e^4)*sin(d*x + c))
Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]