Integrand size = 23, antiderivative size = 362 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac {2 \left (8 a^3+6 a^2 b-9 a b^2-3 b^3\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}-\frac {2 a^2 \sec (c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
-2/3*(8*a^4-15*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)/(a-b)/b^4/(a+b)^(3/2)/d-2/3*(8*a^3+6*a^2*b-9*a*b^2-3 *b^3)*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)/(a -b)/b^3/(a+b)^(3/2)/d-2/3*a^2*sec(d*x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec (d*x+c))^(3/2)-8/3*a^2*(a^2-2*b^2)*tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d *x+c))^(1/2)
Time = 14.47 (sec) , antiderivative size = 556, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (\frac {2 \left (8 a^4-15 a^2 b^2+3 b^4\right ) \sin (c+d x)}{3 b^3 \left (-a^2+b^2\right )^2}+\frac {2 a^2 \sin (c+d x)}{3 b \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}+\frac {8 \left (-a^4 \sin (c+d x)+2 a^2 b^2 \sin (c+d x)\right )}{3 b^2 \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}-\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (8 a^5+8 a^4 b-15 a^3 b^2-15 a^2 b^3+3 a b^4+3 b^5\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (8 a^4+2 a^3 b-15 a^2 b^2-6 a b^3+3 b^4\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (8 a^4-15 a^2 b^2+3 b^4\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((2*(8*a^4 - 15*a^2*b^2 + 3*b^4)*Si n[c + d*x])/(3*b^3*(-a^2 + b^2)^2) + (2*a^2*Sin[c + d*x])/(3*b*(-a^2 + b^2 )*(b + a*Cos[c + d*x])^2) + (8*(-(a^4*Sin[c + d*x]) + 2*a^2*b^2*Sin[c + d* x]))/(3*b^2*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x]) ^(5/2)) - (2*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x)/ 2]^2*Sec[c + d*x]]*(2*(8*a^5 + 8*a^4*b - 15*a^3*b^2 - 15*a^2*b^3 + 3*a*b^4 + 3*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/ ((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/ (a + b)] - 2*b*(8*a^4 + 2*a^3*b - 15*a^2*b^2 - 6*a*b^3 + 3*b^4)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (8*a^4 - 15*a^2*b^2 + 3*b^4)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*T an[(c + d*x)/2]))/(3*b^3*(a^2 - b^2)^2*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*S ec[c + d*x])^(5/2))
Time = 1.31 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4332, 27, 3042, 4568, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4332 |
| \(\displaystyle -\frac {2 \int \frac {\sec (c+d x) \left (2 a^2-3 b \sec (c+d x) a-\left (4 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec (c+d x) \left (2 a^2-3 b \sec (c+d x) a-\left (4 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a^2-3 b \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (3 b^2-4 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4568 |
| \(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\sec (c+d x) \left (2 a b \left (a^2-3 b^2\right )+\left (8 a^4-15 b^2 a^2+3 b^4\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\sec (c+d x) \left (2 a b \left (a^2-3 b^2\right )+\left (8 a^4-15 b^2 a^2+3 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a b \left (a^2-3 b^2\right )+\left (8 a^4-15 b^2 a^2+3 b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4493 |
| \(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (8 a^3+6 a^2 b-9 a b^2-3 b^3\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (8 a^3+6 a^2 b-9 a b^2-3 b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle -\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\left (8 a^4-15 a^2 b^2+3 b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3+6 a^2 b-9 a b^2-3 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle -\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {8 a^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4-15 a^2 b^2+3 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3+6 a^2 b-9 a b^2-3 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{b \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}\) |
(-2*a^2*Sec[c + d*x]*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x]) ^(3/2)) - (-(((-2*(a - b)*Sqrt[a + b]*(8*a^4 - 15*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]*(8*a^3 + 6*a^2*b - 9*a*b^2 - 3*b ^3)*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*(a^2 - b^2))) + (8*a^2*(a^2 - 2*b^2)*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*b*(a^2 - b^2))
3.6.71.3.1 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[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] )^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(4449\) vs. \(2(332)=664\).
Time = 11.24 (sec) , antiderivative size = 4450, normalized size of antiderivative = 12.29
2/3/d/(a-b)^2/(a+b)^2/b^3*(8*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b)) ^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d *x+c)+1))^(1/2)*a^5*b*cos(d*x+c)^3-15*EllipticE(cot(d*x+c)-csc(d*x+c),((a- b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c ))/(cos(d*x+c)+1))^(1/2)*a^4*b^2*cos(d*x+c)^3-15*EllipticE(cot(d*x+c)-csc( d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+ a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b^3*cos(d*x+c)^3+3*EllipticE(cot(d *x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1 /(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^4*cos(d*x+c)^3+3*Ellip ticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1) )^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^5*cos(d*x+c)^3 -8*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d *x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^5*b*cos( d*x+c)^3-2*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c )/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^ 4*b^2*cos(d*x+c)^3+15*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)) *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 ))^(1/2)*a^3*b^3*cos(d*x+c)^3+6*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+ b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co s(d*x+c)+1))^(1/2)*a^2*b^4*cos(d*x+c)^3-3*EllipticF(cot(d*x+c)-csc(d*x+...
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^4/(b^3*sec(d*x + c)^3 + 3*a *b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]