Integrand size = 23, antiderivative size = 317 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (a^2+3 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (a-3 b) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}+\frac {2 a \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
2/3*(a^2+3*b^2)*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^2/(a+b)^(3/2)/d+2/3*(a-3*b)*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/(a+b)^(3/2)/d+2/3*a*tan(d*x+c) /(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*(a^2+3*b^2)*tan(d*x+c)/(a^2-b^2)^2 /d/(a+b*sec(d*x+c))^(1/2)
Time = 11.18 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.53 \[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (-\frac {2 \left (a^2+3 b^2\right ) \sin (c+d x)}{3 b \left (-a^2+b^2\right )^2}+\frac {2 b \sin (c+d x)}{3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}+\frac {4 \left (a^2 \sin (c+d x)+b^2 \sin (c+d x)\right )}{3 \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (a^3+a^2 b+3 a b^2+3 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (a^2+4 a b+3 b^2\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (a^2+3 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b \left (a^2-b^2\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-2*(a^2 + 3*b^2)*Sin[c + d*x])/(3 *b*(-a^2 + b^2)^2) + (2*b*Sin[c + d*x])/(3*(-a^2 + b^2)*(b + a*Cos[c + d*x ])^2) + (4*(a^2*Sin[c + d*x] + b^2*Sin[c + d*x]))/(3*(-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*(a^3 + a^2* b + 3*a*b^2 + 3*b^3)*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*(a^2 + 4*a*b + 3*b^2)*Sqrt[Cos[c + d*x]/(1 + Cos [c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Ellipt icF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (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]))/(3*b*(a^2 - b ^2)^2*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2))
Time = 1.15 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4323, 27, 3042, 4491, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4323 |
\(\displaystyle \frac {2 \int -\frac {\sec (c+d x) (3 b-a \sec (c+d x))}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 \left (a^2-b^2\right )}+\frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec (c+d x) (3 b-a \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}}dx}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 b-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4491 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {-\frac {2 \int -\frac {\sec (c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sec (c+d x) \left (4 a b+\left (a^2+3 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (4 a b+\left (a^2+3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\left (a^2+3 b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-3 b) (a-b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\left (a^2+3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-3 b) (a-b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {\left (a^2+3 b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-3 b) (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {2 a \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (a^2+3 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-3 b) (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {2 \left (a^2+3 b^2\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}\) |
(2*a*Tan[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (((-2*(a - b)*Sqrt[a + b]*(a^2 + 3*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*S ec[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 - 3*b)*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq rt[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))/(a^2 - b^2) - (2*(a^2 + 3*b^2)*Tan [c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*(a^2 - b^2))
3.6.73.3.1 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_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*( a^2 - b^2))), x] - Simp[1/((m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*C sc[e + f*x])^(m + 1)*(b*(m + 1) - a*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[ {a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 /((m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp [(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m , -1]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(3000\) vs. \(2(287)=574\).
Time = 4.99 (sec) , antiderivative size = 3001, normalized size of antiderivative = 9.47
2/3/d/(a-b)^2/(a+b)^2/b*((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^3*b*cos(d*x+c)^3+4*(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^2*cos(d*x+c)^3+3*(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)-cs c(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*cos(d*x+c)^3-(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*cos(d*x+c)^3-3*(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^2*cos(d*x+c)^3-3*(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^3*cos(d*x+c)^3-6 *(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^4*cos(d*x+ c)-3*(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)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^4*cos( d*x+c)^2+2*a^3*b*cos(d*x+c)^2*sin(d*x+c)-3*a^2*b^2*cos(d*x+c)^2*sin(d*x+c) +2*a*b^3*cos(d*x+c)^2*sin(d*x+c)-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...
\[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^2/(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)
\[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]