Integrand size = 23, antiderivative size = 149 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=-\frac {6 b \left (5 a^2+b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {8 a b^2 \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {6 b \left (5 a^2+b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 (a+b \sec (c+d x)) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]
-6/5*b*(5*a^2+b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2*a*(a^2+b^2)*(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))/d+8/5*a*b^2*sin (d*x+c)/d/cos(d*x+c)^(3/2)+2/5*b^2*(a+b*sec(d*x+c))*sin(d*x+c)/d/cos(d*x+c )^(3/2)+6/5*b*(5*a^2+b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
Time = 1.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\frac {-6 b \left (5 a^2+b^2\right ) \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 a \left (a^2+b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+10 a b^2 \sin (c+d x)+3 \left (5 a^2 b+b^3\right ) \sin (2 (c+d x))+2 b^3 \tan (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x)} \]
(-6*b*(5*a^2 + b^2)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*a*(a ^2 + b^2)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 10*a*b^2*Sin[c + d*x] + 3*(5*a^2*b + b^3)*Sin[2*(c + d*x)] + 2*b^3*Tan[c + d*x])/(5*d*Cos[c + d*x]^(3/2))
Time = 1.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.36, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4752, 3042, 4329, 27, 3042, 4535, 3042, 4255, 3042, 4258, 3042, 3119, 4534, 3042, 4258, 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.
| \(\displaystyle \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4752 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3dx\) |
| \(\Big \downarrow \) 4329 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\sec (c+d x)} \left (12 a b^2 \sec ^2(c+d x)+3 b \left (5 a^2+b^2\right ) \sec (c+d x)+a \left (5 a^2+b^2\right )\right )dx+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \sqrt {\sec (c+d x)} \left (12 a b^2 \sec ^2(c+d x)+3 b \left (5 a^2+b^2\right ) \sec (c+d x)+a \left (5 a^2+b^2\right )\right )dx+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (5 a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (5 a^2+b^2\right )\right )dx+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (3 b \left (5 a^2+b^2\right ) \int \sec ^{\frac {3}{2}}(c+d x)dx+\int \sqrt {\sec (c+d x)} \left (12 a b^2 \sec ^2(c+d x)+a \left (5 a^2+b^2\right )\right )dx\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (3 b \left (5 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (5 a^2+b^2\right )\right )dx\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (5 a^2+b^2\right )\right )dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (5 a^2+b^2\right )\right )dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (5 a^2+b^2\right )\right )dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (5 a^2+b^2\right )\right )dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (12 a b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (5 a^2+b^2\right )\right )dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (5 a \left (a^2+b^2\right ) \int \sqrt {\sec (c+d x)}dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (5 a \left (a^2+b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (5 a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (5 a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {10 a \left (a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+3 b \left (5 a^2+b^2\right ) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {8 a b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d}\right )+\frac {2 b^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}{5 d}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*b^2*Sec[c + d*x]^(3/2)*(a + b*Se c[c + d*x])*Sin[c + d*x])/(5*d) + ((10*a*(a^2 + b^2)*Sqrt[Cos[c + d*x]]*El lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (8*a*b^2*Sec[c + d*x]^(3/2 )*Sin[c + d*x])/d + 3*b*(5*a^2 + b^2)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/ d))/5)
3.9.15.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[ (a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a* b^2*d*n + b*(b^2*d*(m + n - 2) + 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2* d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, n}, x ] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && !IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(187)=374\).
Time = 17.43 (sec) , antiderivative size = 711, normalized size of antiderivative = 4.77
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^3*(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))+2/ 5*b^3/(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*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12* (sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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)^4-24*sin(1/2*d*x+1/2*c)^4*cos (1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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+8*cos(1/ 2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE (cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(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)+6*a*b^2*(-1/6*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+1/3*(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)))+6*a^2*b/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1 )*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2* c)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(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* 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 = 0.10 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=-\frac {5 \, \sqrt {2} {\left (i \, a^{3} + i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} - i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, a b^{2} \cos \left (d x + c\right ) + b^{3} + 3 \, {\left (5 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{5 \, d \cos \left (d x + c\right )^{3}} \]
-1/5*(5*sqrt(2)*(I*a^3 + I*a*b^2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*a^3 - I*a*b^2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2) *(5*I*a^2*b + I*b^3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*a^2*b - I*b^ 3)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) - I*sin(d*x + c))) - 2*(5*a*b^2*cos(d*x + c) + b^3 + 3*(5*a^2*b + b ^3)*cos(d*x + c)^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\left (a + b \sec {\left (c + d x \right )}\right )^{3}}{\sqrt {\cos {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
\[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \]
Time = 14.90 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.05 \[ \int \frac {(a+b \sec (c+d x))^3}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {6\,a^2\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]
(2*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + (2*b^3*sin(c + d*x)*hypergeom([-5/ 4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^( 1/2)) + (6*a^2*b*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2)) /(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*a*b^2*sin(c + d*x)*hyp ergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d *x)^2)^(1/2))