Integrand size = 23, antiderivative size = 399 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \left (16 a^4-8 a^2 b^2-3 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^5 \sqrt {a+b} d}-\frac {2 (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{5 b^4 \sqrt {a+b} d}-\frac {2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 a \left (8 a^2-3 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]
-2/5*(16*a^4-8*a^2*b^2-3*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/ (a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec (d*x+c))/(a-b))^(1/2)/b^5/d/(a+b)^(1/2)-2/5*(4*a+3*b)*(4*a^2+b^2)*cot(d*x+ c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1 -sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d/(a+b)^(1/2 )-2*a^2*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)-2/5*a *(8*a^2-3*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(6*a^ 2-b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2-b^2)/d
Time = 11.78 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {(b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \left (\frac {2 \left (4 a^2+b^2\right ) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (4 a^3+4 a^2 b-3 a b^2-3 b^3\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+2 b \left (-4 a^2-a b+3 b^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+\left (4 a^2-3 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 \left (-a^2+b^2\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {2 \sqrt {\sec (c+d x)} \left (\left (-8 a^4 b+5 a^2 b^3+3 b^5\right ) \sin (c+d x)+\left (-8 a^5+4 a^3 b^2+\frac {3 a b^4}{2}\right ) \sin (2 (c+d x))+b^2 \left (-a^2+b^2\right ) (b-2 a \cos (c+d x)) \sec (c+d x) \tan (c+d x)\right )}{-a^2 b^4+b^6}\right )}{5 d (a+b \sec (c+d x))^{3/2}} \]
((b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*((2*(4*a^2 + b^2)*Sqrt[Cos[(c + d *x)/2]^2*Sec[c + d*x]]*(2*(4*a^3 + 4*a^2*b - 3*a*b^2 - 3*b^3)*EllipticE[Ar cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqr t[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + 2*b*(-4*a^2 - a*b + 3*b^2)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec [c + d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + (4*a^2 - 3*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[ (c + d*x)/2]))/(b^4*(-a^2 + b^2)*Sqrt[Sec[(c + d*x)/2]^2]) + (2*Sqrt[Sec[c + d*x]]*((-8*a^4*b + 5*a^2*b^3 + 3*b^5)*Sin[c + d*x] + (-8*a^5 + 4*a^3*b^ 2 + (3*a*b^4)/2)*Sin[2*(c + d*x)] + b^2*(-a^2 + b^2)*(b - 2*a*Cos[c + d*x] )*Sec[c + d*x]*Tan[c + d*x]))/(-(a^2*b^4) + b^6)))/(5*d*(a + b*Sec[c + d*x ])^(3/2))
Time = 1.75 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3042, 4332, 27, 3042, 4580, 27, 3042, 4570, 27, 3042, 4493, 3042, 4319, 4492}
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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4332 |
| \(\displaystyle -\frac {2 \int \frac {\sec ^2(c+d x) \left (4 a^2-b \sec (c+d x) a-\left (6 a^2-b^2\right ) \sec ^2(c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) \left (4 a^2-b \sec (c+d x) a-\left (6 a^2-b^2\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^2-b \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (b^2-6 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (-3 a \left (8 a^2-3 b^2\right ) \sec ^2(c+d x)-b \left (2 a^2+3 b^2\right ) \sec (c+d x)+2 a \left (6 a^2-b^2\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sec (c+d x) \left (-3 a \left (8 a^2-3 b^2\right ) \sec ^2(c+d x)-b \left (2 a^2+3 b^2\right ) \sec (c+d x)+2 a \left (6 a^2-b^2\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-3 a \left (8 a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (2 a^2+3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (6 a^2-b^2\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle -\frac {-\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (a b \left (4 a^2+b^2\right )+\left (16 a^4-8 b^2 a^2-3 b^4\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (a b \left (4 a^2+b^2\right )+\left (16 a^4-8 b^2 a^2-3 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (4 a^2+b^2\right )+\left (16 a^4-8 b^2 a^2-3 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle -\frac {-\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) (4 a+3 b) \left (4 a^2+b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) (4 a+3 b) \left (4 a^2+b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle -\frac {-\frac {\frac {\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (16 a^4-8 a^2 b^2-3 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{b}-\frac {2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{b d}}{5 b}}{b \left (a^2-b^2\right )}\) |
(-2*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d *x]]) - ((-2*(6*a^2 - b^2)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d *x])/(5*b*d) - (((-2*(a - b)*Sqrt[a + b]*(16*a^4 - 8*a^2*b^2 - 3*b^4)*Cot[ c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/( a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x])) /(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(4*a + 3*b)*(4*a^2 + b^2)*Cot [c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/ (a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]) )/(a - b))])/(b*d))/b - (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[ c + d*x])/(b*d))/(5*b))/(b*(a^2 - b^2))
3.6.62.3.1 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[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-a^2)*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^( m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[d^3/ (b*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x]) ^(n - 3)*Simp[a^2*(n - 3) + a*b*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*( m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n , 2]))
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B) Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B Int[Csc[e + f*x]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} , x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3170\) vs. \(2(367)=734\).
Time = 15.94 (sec) , antiderivative size = 3171, normalized size of antiderivative = 7.95
-2/5/d/(a-b)/(a+b)/b^4*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c) +1)*(3*b^5*sin(d*x+c)+5*a^2*b^3*sin(d*x+c)+b^5*tan(d*x+c)+2*a^3*b^2*sin(d* x+c)-8*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^ 3*cos(d*x+c)^2+(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c) /(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2) )*a*b^4*cos(d*x+c)^2-32*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+ b))^(1/2))*a^4*b*cos(d*x+c)+16*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^( 1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a -b)/(a+b))^(1/2))*a^3*b^2*cos(d*x+c)+16*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x +c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d *x+c),((a-b)/(a+b))^(1/2))*a^2*b^3*cos(d*x+c)-16*(1/(a+b)*(b+a*cos(d*x+c)) /(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x +c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b*cos(d*x+c)^2+6*(1/(a+b)*(b+a*cos (d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE (cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^4*cos(d*x+c)+32*(1/(a+b)*( b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*El lipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b*cos(d*x+c)+8*(1/( a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))...
\[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{5}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^5/(b^2*sec(d*x + c)^2 + 2*a *b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{5}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]