Integrand size = 23, antiderivative size = 187 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=-\frac {64 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {136 a^4 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {64 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {94 a^4 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {8 a^4 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a^4 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]
94/21*a^4*sec(d*x+c)^(3/2)*sin(d*x+c)/d+8/5*a^4*sec(d*x+c)^(5/2)*sin(d*x+c )/d+2/7*a^4*sec(d*x+c)^(7/2)*sin(d*x+c)/d+64/5*a^4*sin(d*x+c)*sec(d*x+c)^( 1/2)/d-64/5*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+136/21*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)*sec(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.84 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.49 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=\frac {a^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 \left (-\frac {4 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^4(c+d x) \left (168 \left (1+e^{2 i (c+d x)}\right )+168 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+85 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {672 \cos (d x) \csc (c)+\left (235+84 \sec (c+d x)+15 \sec ^2(c+d x)\right ) \tan (c+d x)}{\sec ^{\frac {7}{2}}(c+d x)}\right )}{840 d} \]
(a^4*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Cos[c + d*x]^4*(168*(1 + E^((2*I)*(c + d*x))) + 168*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeomet ric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 85*E^(I*(c + d*x))*(-1 + E^ ((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (672*Cos[d* x]*Csc[c] + (235 + 84*Sec[c + d*x] + 15*Sec[c + d*x]^2)*Tan[c + d*x])/Sec[ c + d*x]^(7/2)))/(840*d)
Time = 0.42 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4278, 2009}
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.
| \(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4dx\) |
| \(\Big \downarrow \) 4278 |
| \(\displaystyle \int \left (a^4 \sec ^{\frac {9}{2}}(c+d x)+4 a^4 \sec ^{\frac {7}{2}}(c+d x)+6 a^4 \sec ^{\frac {5}{2}}(c+d x)+4 a^4 \sec ^{\frac {3}{2}}(c+d x)+a^4 \sqrt {\sec (c+d x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^4 \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d}+\frac {8 a^4 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {94 a^4 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{21 d}+\frac {64 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {136 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {64 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\) |
(-64*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/ (5*d) + (136*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (64*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (94*a^4* Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (8*a^4*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a^4*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d)
3.2.87.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f *x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I GtQ[m, 0] && RationalQ[n]
Leaf count of result is larger than twice the leaf count of optimal. \(438\) vs. \(2(211)=422\).
Time = 38.79 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.35
| method | result | size |
| default | \(-\frac {a^{4} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2024 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {47 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{21 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}-\frac {128 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {64 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{5 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{28 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{4}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(439\) |
| parts | \(\text {Expression too large to display}\) | \(1141\) |
-a^4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2024/105*( sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2* d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1 /2))-47/21*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2-128/5*sin(1/2*d*x+1/2*c)^2*cos(1/2*d *x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)-64/5*( sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2* d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^( 1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2/5*cos(1/2*d*x+1/2*c)*(-2*si n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^ 3-1/28*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c )^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.15 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=-\frac {2 \, {\left (170 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 170 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 336 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 336 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (672 \, a^{4} \cos \left (d x + c\right )^{3} + 235 \, a^{4} \cos \left (d x + c\right )^{2} + 84 \, a^{4} \cos \left (d x + c\right ) + 15 \, a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{105 \, d \cos \left (d x + c\right )^{3}} \]
-2/105*(170*I*sqrt(2)*a^4*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d* x + c) + I*sin(d*x + c)) - 170*I*sqrt(2)*a^4*cos(d*x + c)^3*weierstrassPIn verse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 336*I*sqrt(2)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si n(d*x + c))) - 336*I*sqrt(2)*a^4*cos(d*x + c)^3*weierstrassZeta(-4, 0, wei erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (672*a^4*cos(d*x + c)^3 + 235*a^4*cos(d*x + c)^2 + 84*a^4*cos(d*x + c) + 15*a^4)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=\text {Timed out} \]
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{4} \sqrt {\sec \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{4} \sqrt {\sec \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^4 \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]