Integrand size = 27, antiderivative size = 310 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right ) \tan (e+f x)}{c^3 f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {a^{5/2} \left (3 c^3-15 c^2 d-20 c d^2-8 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right ) \tan (e+f x)}{4 c^3 \sqrt {d} (c+d)^{5/2} f \sqrt {a-a \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}+\frac {a^2 (c-d) \tan (e+f x)}{2 c (c+d) f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))^2}+\frac {a^2 \left (3 c^2-7 c d-4 d^2\right ) \tan (e+f x)}{4 c^2 (c+d)^2 f \sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \]
1/2*a^2*(c-d)*tan(f*x+e)/c/(c+d)/f/(c+d*sec(f*x+e))^2/(a+a*sec(f*x+e))^(1/ 2)+1/4*a^2*(3*c^2-7*c*d-4*d^2)*tan(f*x+e)/c^2/(c+d)^2/f/(c+d*sec(f*x+e))/( a+a*sec(f*x+e))^(1/2)+2*a^(5/2)*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*ta n(f*x+e)/c^3/f/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)+1/4*a^(5/2)*( 3*c^3-15*c^2*d-20*c*d^2-8*d^3)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1 /2)/(c+d)^(1/2))*tan(f*x+e)/c^3/(c+d)^(5/2)/f/d^(1/2)/(a-a*sec(f*x+e))^(1/ 2)/(a+a*sec(f*x+e))^(1/2)
Time = 5.86 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\frac {(d+c \cos (e+f x))^3 \sec ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^{\frac {3}{2}}(e+f x) (a (1+\sec (e+f x)))^{3/2} \left (\frac {\sqrt {2} \left (8 \sqrt {d} (c+d)^{5/2} \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )+\left (3 c^3-15 c^2 d-20 c d^2-8 d^3\right ) \arctan \left (\frac {\sqrt {d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right ) \sec \left (\frac {1}{2} (e+f x)\right ) \sqrt {\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right ) \sec (e+f x)}}{\sqrt {d} (c+d)^{5/2} \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}+\frac {2 c \sqrt {\sec (e+f x)} \left (c \left (5 c^2-7 c d-6 d^2\right )+d \left (3 c^2-7 c d-4 d^2\right ) \sec (e+f x)\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{(c+d)^2 (c+d \sec (e+f x))^2}\right )}{16 c^3 f (c+d \sec (e+f x))^3} \]
((d + c*Cos[e + f*x])^3*Sec[(e + f*x)/2]^3*Sec[e + f*x]^(3/2)*(a*(1 + Sec[ e + f*x]))^(3/2)*((Sqrt[2]*(8*Sqrt[d]*(c + d)^(5/2)*ArcTan[Tan[(e + f*x)/2 ]/Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]] + (3*c^3 - 15*c^2*d - 20*c*d^2 - 8*d^3)*ArcTan[(Sqrt[d]*Tan[(e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])])])*Sec[(e + f*x)/2]*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^2 ]*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]])/(Sqrt[d]*(c + d)^(5/2)*Sqrt[Sec[( e + f*x)/2]^2]) + (2*c*Sqrt[Sec[e + f*x]]*(c*(5*c^2 - 7*c*d - 6*d^2) + d*( 3*c^2 - 7*c*d - 4*d^2)*Sec[e + f*x])*Sin[(e + f*x)/2])/((c + d)^2*(c + d*S ec[e + f*x])^2)))/(16*c^3*f*(c + d*Sec[e + f*x])^3)
Time = 0.47 (sec) , antiderivative size = 288, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 4428, 27, 168, 27, 168, 27, 174, 73, 219, 221}
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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{3/2}}{(c+d \sec (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2}}{\left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4428 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {a \cos (e+f x) (\sec (e+f x)+1)}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \int \frac {\cos (e+f x) (\sec (e+f x)+1)}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^3}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\int \frac {a \cos (e+f x) (4 (c+d)+3 (c-d) \sec (e+f x))}{2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{2 a c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\int \frac {\cos (e+f x) (4 (c+d)+3 (c-d) \sec (e+f x))}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}d\sec (e+f x)}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\frac {\int \frac {a \cos (e+f x) \left (8 (c+d)^2+\left (3 c^2-7 d c-4 d^2\right ) \sec (e+f x)\right )}{2 \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{a c (c+d)}-\frac {\left (3 c^2-7 c d-4 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\frac {\int \frac {\cos (e+f x) \left (8 (c+d)^2+\left (3 c^2-7 d c-4 d^2\right ) \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{2 c (c+d)}-\frac {\left (3 c^2-7 c d-4 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\frac {\frac {\left (3 c^3-15 c^2 d-20 c d^2-8 d^3\right ) \int \frac {1}{\sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))}d\sec (e+f x)}{c}+\frac {8 (c+d)^2 \int \frac {\cos (e+f x)}{\sqrt {a-a \sec (e+f x)}}d\sec (e+f x)}{c}}{2 c (c+d)}-\frac {\left (3 c^2-7 c d-4 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\frac {-\frac {2 \left (3 c^3-15 c^2 d-20 c d^2-8 d^3\right ) \int \frac {1}{c+d-\frac {d (a-a \sec (e+f x))}{a}}d\sqrt {a-a \sec (e+f x)}}{a c}-\frac {16 (c+d)^2 \int \frac {1}{1-\frac {a-a \sec (e+f x)}{a}}d\sqrt {a-a \sec (e+f x)}}{a c}}{2 c (c+d)}-\frac {\left (3 c^2-7 c d-4 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\frac {-\frac {2 \left (3 c^3-15 c^2 d-20 c d^2-8 d^3\right ) \int \frac {1}{c+d-\frac {d (a-a \sec (e+f x))}{a}}d\sqrt {a-a \sec (e+f x)}}{a c}-\frac {16 (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c}}{2 c (c+d)}-\frac {\left (3 c^2-7 c d-4 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a^3 \tan (e+f x) \left (\frac {\frac {-\frac {2 \left (3 c^3-15 c^2 d-20 c d^2-8 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a-a \sec (e+f x)}}{\sqrt {a} \sqrt {c+d}}\right )}{\sqrt {a} c \sqrt {d} \sqrt {c+d}}-\frac {16 (c+d)^2 \text {arctanh}\left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} c}}{2 c (c+d)}-\frac {\left (3 c^2-7 c d-4 d^2\right ) \sqrt {a-a \sec (e+f x)}}{a c (c+d) (c+d \sec (e+f x))}}{4 c (c+d)}-\frac {(c-d) \sqrt {a-a \sec (e+f x)}}{2 a c (c+d) (c+d \sec (e+f x))^2}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^3*(-1/2*((c - d)*Sqrt[a - a*Sec[e + f*x]])/(a*c*(c + d)*(c + d*Sec[e + f*x])^2) + (((-16*(c + d)^2*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a]])/( Sqrt[a]*c) - (2*(3*c^3 - 15*c^2*d - 20*c*d^2 - 8*d^3)*ArcTanh[(Sqrt[d]*Sqr t[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])])/(Sqrt[a]*c*Sqrt[d]*Sqrt[c + d]))/(2*c*(c + d)) - ((3*c^2 - 7*c*d - 4*d^2)*Sqrt[a - a*Sec[e + f*x]])/( a*c*(c + d)*(c + d*Sec[e + f*x])))/(4*c*(c + d)))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.2.59.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a^2*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^n/(x*Sqrt[a - b*x])), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0 ] && IntegerQ[m - 1/2]
Leaf count of result is larger than twice the leaf count of optimal. \(55535\) vs. \(2(272)=544\).
Time = 16.91 (sec) , antiderivative size = 55536, normalized size of antiderivative = 179.15
Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (272) = 544\).
Time = 11.67 (sec) , antiderivative size = 2729, normalized size of antiderivative = 8.80 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\text {Too large to display} \]
[-1/8*((3*a*c^3*d^2 - 15*a*c^2*d^3 - 20*a*c*d^4 - 8*a*d^5 + (3*a*c^5 - 15* a*c^4*d - 20*a*c^3*d^2 - 8*a*c^2*d^3)*cos(f*x + e)^3 + (3*a*c^5 - 9*a*c^4* d - 50*a*c^3*d^2 - 48*a*c^2*d^3 - 16*a*c*d^4)*cos(f*x + e)^2 + (6*a*c^4*d - 27*a*c^3*d^2 - 55*a*c^2*d^3 - 36*a*c*d^4 - 8*a*d^5)*cos(f*x + e))*sqrt(- a/(c*d + d^2))*log((2*(c*d + d^2)*sqrt(-a/(c*d + d^2))*sqrt((a*cos(f*x + e ) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (a*c + 2*a*d)*cos(f*x + e )^2 - a*d + (a*c + a*d)*cos(f*x + e))/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) - 8*(a*c^2*d^2 + 2*a*c*d^3 + a*d^4 + (a*c^4 + 2*a*c^3*d + a*c^2 *d^2)*cos(f*x + e)^3 + (a*c^4 + 4*a*c^3*d + 5*a*c^2*d^2 + 2*a*c*d^3)*cos(f *x + e)^2 + (2*a*c^3*d + 5*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(f*x + e))*sq rt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos( f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 2*((5*a*c^4 - 7*a*c^3*d - 6*a*c^2*d^2)*cos(f*x + e)^2 + (3*a*c^3*d - 7*a*c^2*d^2 - 4*a*c*d^3)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/((c^7 + 2*c^6*d + c^5*d^2)*f*cos(f*x + e)^3 + (c^7 + 4 *c^6*d + 5*c^5*d^2 + 2*c^4*d^3)*f*cos(f*x + e)^2 + (2*c^6*d + 5*c^5*d^2 + 4*c^4*d^3 + c^3*d^4)*f*cos(f*x + e) + (c^5*d^2 + 2*c^4*d^3 + c^3*d^4)*f), -1/8*(16*(a*c^2*d^2 + 2*a*c*d^3 + a*d^4 + (a*c^4 + 2*a*c^3*d + a*c^2*d^2)* cos(f*x + e)^3 + (a*c^4 + 4*a*c^3*d + 5*a*c^2*d^2 + 2*a*c*d^3)*cos(f*x + e )^2 + (2*a*c^3*d + 5*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(f*x + e))*sqrt(...
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c+d \sec (e+f x))^3} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \]