Integrand size = 23, antiderivative size = 157 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=-\frac {36 a^3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {36 a^3 \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a^3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d}+\frac {2 a^3 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]
2*a^3*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a^3*sec(d*x+c)^(5/2)*sin(d*x+c)/d+ 36/5*a^3*sin(d*x+c)*sec(d*x+c)^(1/2)/d-36/5*a^3*(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+4*a^3*(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 = 3.06 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.70 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=\frac {a^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^3 \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^3(c+d x) \left (9 \left (1+e^{2 i (c+d x)}\right )+9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+5 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {18 \cos (d x) \csc (c)+(5+\sec (c+d x)) \tan (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}\right )}{20 d} \]
(a^3*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Cos[c + d*x]^3*(9*(1 + E^((2*I)*(c + d *x))) + 9*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2 F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 5*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))) + (18*Cos[d*x]*Csc [c] + (5 + Sec[c + d*x])*Tan[c + d*x])/Sec[c + d*x]^(5/2)))/(20*d)
Time = 0.37 (sec) , antiderivative size = 157, 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.
\(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3dx\) |
\(\Big \downarrow \) 4278 |
\(\displaystyle \int \left (a^3 \sec ^{\frac {7}{2}}(c+d x)+3 a^3 \sec ^{\frac {5}{2}}(c+d x)+3 a^3 \sec ^{\frac {3}{2}}(c+d x)+a^3 \sqrt {\sec (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^3 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a^3 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}+\frac {36 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {36 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\) |
(-36*a^3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/ (5*d) + (4*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d *x]])/d + (36*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a^3*Sec[c + d*x]^(3/2)*Sin[c + d*x])/d + (2*a^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d)
3.2.80.3.1 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. \(385\) vs. \(2(189)=378\).
Time = 29.17 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.46
method | result | size |
default | \(-\frac {a^{3} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {56 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}-\frac {72 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {36 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{10 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{3}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(386\) |
parts | \(\text {Expression too large to display}\) | \(896\) |
-a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(56/5*(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)) -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)^2-72/5*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)-36/5*(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))-Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2)))-1/10*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)^3)/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 = 200, normalized size of antiderivative = 1.27 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (18 \, a^{3} \cos \left (d x + c\right )^{2} + 5 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{5 \, d \cos \left (d x + c\right )^{2}} \]
-2/5*(5*I*sqrt(2)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*a^3*cos(d*x + c)^2*weierstrassPInverse( -4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*I*sqrt(2)*a^3*cos(d*x + c)^2*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(2)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPI nverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (18*a^3*cos(d*x + c)^2 + 5 *a^3*cos(d*x + c) + a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^ 2)
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=\text {Timed out} \]
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3 \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]