Integrand size = 25, antiderivative size = 217 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {163 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac {17 \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {95 \sin (c+d x)}{48 a^2 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {299 \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \sec (c+d x)}} \]
163/32*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d* x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)/sec (d*x+c)^(1/2)-17/16*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)/sec(d*x+c)^(1/2) +95/48*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-299/48*sin (d*x+c)*sec(d*x+c)^(1/2)/a^2/d/(a+a*sec(d*x+c))^(1/2)
Time = 2.65 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\sec (c+d x) \left (978 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} (487+379 \sec (c+d x)+16 \cos (2 (c+d x)) (-1+5 \sec (c+d x))) \tan (c+d x)\right )}{48 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (a (1+\sec (c+d x)))^{5/2}} \]
-1/48*(Sec[c + d*x]*(978*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[ 1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*Sin[c + d*x] + Sq rt[1 - Sec[c + d*x]]*(487 + 379*Sec[c + d*x] + 16*Cos[2*(c + d*x)]*(-1 + 5 *Sec[c + d*x]))*Tan[c + d*x]))/(d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x]) ]*(a*(1 + Sec[c + d*x]))^(5/2))
Time = 1.18 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4304, 27, 3042, 4508, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}
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.
| \(\displaystyle \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4304 |
| \(\displaystyle -\frac {\int -\frac {11 a-6 a \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {11 a-6 a \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {11 a-6 a \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {95 a^2-68 a^2 \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {95 a^2-68 a^2 \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {95 a^2-68 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\frac {2 \int -\frac {299 a^3-190 a^3 \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {299 a^3-190 a^3 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {299 a^3-190 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {598 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-489 a^3 \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {598 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-489 a^3 \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {978 a^3 \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {598 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {190 a^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {598 a^3 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {489 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {17 a \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}}\) |
-1/4*Sin[c + d*x]/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)) + ((-1 7*a*Sin[c + d*x])/(2*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + (( 190*a^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-489*Sqrt[2]*a^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/( Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (598*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(4*a^2))/(8*a^2)
3.3.64.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc [e + f*x])^n/(f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ [m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d *n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
Time = 1.40 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.66
| method | result | size |
| default | \(-\frac {\left (6 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-69 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+489 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-668 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+489 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-465 \csc \left (d x +c \right )+465 \cot \left (d x +c \right )\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}{96 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {3}{2}}}\) | \(361\) |
-1/96/d/a^3*(6*(1-cos(d*x+c))^7*csc(d*x+c)^7-69*(1-cos(d*x+c))^5*csc(d*x+c )^5+489*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc (d*x+c)))*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos(d*x+c))^2*csc(d* x+c)^2-668*(1-cos(d*x+c))^3*csc(d*x+c)^3+489*arctan(1/(-(1-cos(d*x+c))^2*c sc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(-(1-cos(d*x+c))^2*csc(d*x+ c)^2-1)^(1/2)-465*csc(d*x+c)+465*cot(d*x+c))*(-2*a/((1-cos(d*x+c))^2*csc(d *x+c)^2-1))^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/(-((1-cos(d*x+c))^2*cs c(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(3/2)
Time = 0.32 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.15 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (32 \, \cos \left (d x + c\right )^{4} - 160 \, \cos \left (d x + c\right )^{3} - 503 \, \cos \left (d x + c\right )^{2} - 299 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (32 \, \cos \left (d x + c\right )^{4} - 160 \, \cos \left (d x + c\right )^{3} - 503 \, \cos \left (d x + c\right )^{2} - 299 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]
[1/192*(489*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3 *a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(32*cos(d*x + c)^4 - 160*co s(d*x + c)^3 - 503*cos(d*x + c)^2 - 299*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/96*(489*sqrt (2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)*arct an(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) - 2*(32*cos(d*x + c)^4 - 160*cos(d*x + c)^3 - 503*c os(d*x + c)^2 - 299*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))* sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
Timed out. \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 148823 vs. \(2 (180) = 360\).
Time = 2.75 (sec) , antiderivative size = 148823, normalized size of antiderivative = 685.82 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/96*(32*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin (3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*a rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(11/3*arctan2(sin( 3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 41472*(cos(3*d*x + 3*c)^2*sin (3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d* x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos (3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3 /2*c)))^4 + 8192*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3* c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*s in(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(5/3*arcta n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 288*sin(3/2*d*x + 3/2* c)^5 + 32*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*si n(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3* arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(11/3*arctan2(sin (3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 41472*(cos(3*d*x + 3*c)^2*si n(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d *x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co s(3/2*d*x + 3/2*c))))*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 8192*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3 *c)^2*sin(3/2*d*x + 3/2*c) - 15*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^...
\[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]