Integrand size = 40, antiderivative size = 116 \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=-\frac {g^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {2} \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f}+\frac {g \cot (e+f x) \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}{a c f} \]
-1/2*g^(3/2)*arctanh(1/2*a^(1/2)*g^(1/2)*tan(f*x+e)*2^(1/2)/(g*sec(f*x+e)) ^(1/2)/(a+a*sec(f*x+e))^(1/2))/c/f*2^(1/2)/a^(1/2)+g*cot(f*x+e)*(g*sec(f*x +e))^(1/2)*(a+a*sec(f*x+e))^(1/2)/a/c/f
Leaf count is larger than twice the leaf count of optimal. \(236\) vs. \(2(116)=232\).
Time = 3.92 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.03 \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=-\frac {a \cos \left (\frac {1}{2} (e+f x)\right ) (g \sec (e+f x))^{5/2} \sin ^3\left (\frac {1}{2} (e+f x)\right ) \left (-4-4 \sec (e+f x)+\frac {\left (\log \left (1-2 \sec (e+f x)-3 \sec ^2(e+f x)-2 \sqrt {2} \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \sqrt {\tan ^2(e+f x)}\right )-\log \left (1-2 \sec (e+f x)-3 \sec ^2(e+f x)+2 \sqrt {2} \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \sqrt {\tan ^2(e+f x)}\right )\right ) \sqrt {\tan ^2(e+f x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}\right )}{c f g (-1+\sec (e+f x))^2 (a (1+\sec (e+f x)))^{3/2}} \]
-((a*Cos[(e + f*x)/2]*(g*Sec[e + f*x])^(5/2)*Sin[(e + f*x)/2]^3*(-4 - 4*Se c[e + f*x] + ((Log[1 - 2*Sec[e + f*x] - 3*Sec[e + f*x]^2 - 2*Sqrt[2]*Sqrt[ Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]]*Sqrt[Tan[e + f*x]^2]] - Log[1 - 2*Sec [e + f*x] - 3*Sec[e + f*x]^2 + 2*Sqrt[2]*Sqrt[Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]]*Sqrt[Tan[e + f*x]^2]])*Sqrt[Tan[e + f*x]^2])/Sqrt[Sec[(e + f*x)/2 ]^2]))/(c*f*g*(-1 + Sec[e + f*x])^2*(a*(1 + Sec[e + f*x]))^(3/2)))
Time = 0.48 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4452, 27, 105, 104, 218}
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.
| \(\displaystyle \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (g \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4452 |
| \(\displaystyle -\frac {a c g \tan (e+f x) \int \frac {\sqrt {g \sec (e+f x)}}{a (\sec (e+f x)+1) (c-c \sec (e+f x))^{3/2}}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {c g \tan (e+f x) \int \frac {\sqrt {g \sec (e+f x)}}{(\sec (e+f x)+1) (c-c \sec (e+f x))^{3/2}}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 105 |
| \(\displaystyle -\frac {c g \tan (e+f x) \left (\frac {\sqrt {g \sec (e+f x)}}{c \sqrt {c-c \sec (e+f x)}}-\frac {g \int \frac {1}{\sqrt {g \sec (e+f x)} (\sec (e+f x)+1) \sqrt {c-c \sec (e+f x)}}d\sec (e+f x)}{2 c}\right )}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {c g \tan (e+f x) \left (\frac {\sqrt {g \sec (e+f x)}}{c \sqrt {c-c \sec (e+f x)}}-\frac {g \int \frac {1}{\frac {2 c \sec (e+f x) g}{c-c \sec (e+f x)}+g}d\frac {\sqrt {g \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}}}{c}\right )}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {c g \tan (e+f x) \left (\frac {\sqrt {g \sec (e+f x)}}{c \sqrt {c-c \sec (e+f x)}}-\frac {\sqrt {g} \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} c^{3/2}}\right )}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((c*g*(-((Sqrt[g]*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[g*Sec[e + f*x]])/(Sqrt[g]* Sqrt[c - c*Sec[e + f*x]])])/(Sqrt[2]*c^(3/2))) + Sqrt[g*Sec[e + f*x]]/(c*S qrt[c - c*Sec[e + f*x]]))*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))
3.2.81.3.1 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_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[a *c*g*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2), x], x, Cs c[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0]
Time = 3.07 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.07
| method | result | size |
| default | \(\frac {g \left (\operatorname {arcsinh}\left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ) \sqrt {2}\, \sin \left (f x +e \right )+2 \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+2 \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\right ) \sqrt {g \sec \left (f x +e \right )}\, \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \cot \left (f x +e \right )}{2 c f a \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}}\) | \(124\) |
1/2*g/c/f/a*(arcsinh(cot(f*x+e)-csc(f*x+e))*2^(1/2)*sin(f*x+e)+2*(1/(cos(f *x+e)+1))^(1/2)*cos(f*x+e)+2*(1/(cos(f*x+e)+1))^(1/2))*(g*sec(f*x+e))^(1/2 )*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)+1)/(1/(cos(f*x+e)+1))^(1/2)*cot(f*x +e)
Time = 0.28 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.84 \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=\left [\frac {\sqrt {2} a g \sqrt {\frac {g}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + g \cos \left (f x + e\right )^{2} - 2 \, g \cos \left (f x + e\right ) - 3 \, g}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, g \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, a c f \sin \left (f x + e\right )}, \frac {\sqrt {2} a g \sqrt {-\frac {g}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{g \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, g \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, a c f \sin \left (f x + e\right )}\right ] \]
[1/4*(sqrt(2)*a*g*sqrt(g/a)*log(-(2*sqrt(2)*sqrt(g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + g*cos (f*x + e)^2 - 2*g*cos(f*x + e) - 3*g)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1 ))*sin(f*x + e) + 4*g*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f *x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e)), 1/2*(sqrt(2)*a*g*sqrt(-g/a)*a rctan(sqrt(2)*sqrt(-g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/co s(f*x + e))*cos(f*x + e)/(g*sin(f*x + e)))*sin(f*x + e) + 2*g*sqrt((a*cos( f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin( f*x + e))]
\[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=- \frac {\int \frac {\left (g \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} - \sqrt {a \sec {\left (e + f x \right )} + a}}\, dx}{c} \]
-Integral((g*sec(e + f*x))**(3/2)/(sqrt(a*sec(e + f*x) + a)*sec(e + f*x) - sqrt(a*sec(e + f*x) + a)), x)/c
Leaf count of result is larger than twice the leaf count of optimal. 536 vs. \(2 (97) = 194\).
Time = 0.42 (sec) , antiderivative size = 536, normalized size of antiderivative = 4.62 \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=\frac {{\left (4 \, g \cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 4 \, g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - {\left (g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + g \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + g\right )} \log \left (\cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + {\left (g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + g \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + g\right )} \log \left (\cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 4 \, g \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {g}}{2 \, {\left (\sqrt {2} c \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sqrt {2} c \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sqrt {2} c \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + \sqrt {2} c\right )} \sqrt {a} f} \]
1/2*(4*g*cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/2*arct an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*g*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2* e))) - (g*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + g*sin(1 /2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*g*cos(1/2*arctan2(si n(2*f*x + 2*e), cos(2*f*x + 2*e))) + g)*log(cos(1/4*arctan2(sin(2*f*x + 2* e), cos(2*f*x + 2*e)))^2 + sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 *e)))^2 + 2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + (g *cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + g*sin(1/2*arctan 2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*g*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + g)*log(cos(1/4*arctan2(sin(2*f*x + 2*e), cos(2 *f*x + 2*e)))^2 + sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/4*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 4*g*sin(1/4 *arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(g)/((sqrt(2)*c*cos(1/2 *arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sqrt(2)*c*sin(1/2*arctan 2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sqrt(2)*c*cos(1/2*arctan2(sin (2*f*x + 2*e), cos(2*f*x + 2*e))) + sqrt(2)*c)*sqrt(a)*f)
\[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=\int { -\frac {\left (g \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {a \sec \left (f x + e\right ) + a} {\left (c \sec \left (f x + e\right ) - c\right )}} \,d x } \]
Timed out. \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx=\int \frac {{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )} \,d x \]