Integrand size = 28, antiderivative size = 70 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c f}+\frac {4 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c f} \]
2*a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c/f+4*a*cot(f* x+e)*(a+a*sec(f*x+e))^(1/2)/c/f
Time = 0.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=\frac {2 a \cot (e+f x) \sqrt {a (1+\sec (e+f x))} \left (2 \sqrt {c}-\text {arctanh}\left (\frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {c}}\right ) \sqrt {c-c \sec (e+f x)}\right )}{c^{3/2} f} \]
(2*a*Cot[e + f*x]*Sqrt[a*(1 + Sec[e + f*x])]*(2*Sqrt[c] - ArcTanh[Sqrt[c - c*Sec[e + f*x]]/Sqrt[c]]*Sqrt[c - c*Sec[e + f*x]]))/(c^(3/2)*f)
Time = 0.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4392, 3042, 4375, 359, 216}
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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{3/2}}{c-c \sec (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2}}{c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle -\frac {\int \cot ^2(e+f x) (\sec (e+f x) a+a)^{5/2}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}{\cot \left (e+f x+\frac {\pi }{2}\right )^2}dx}{a c}\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle \frac {2 a \int \frac {\cot ^2(e+f x) (\sec (e+f x) a+a) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{c f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {2 a \left (2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}-a \int \frac {1}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )\right )}{c f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 a \left (\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )+2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}\right )}{c f}\) |
(2*a*(Sqrt[a]*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]] + 2* Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]]))/(c*f)
3.1.53.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d) Subst[Int[x^m*((2 + a*x^2 )^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] ]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I ntegerQ[n - 1/2]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[Cot[e + f*x]^(2*m)*( c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !( IntegerQ[n] && GtQ[m - n, 0])
Time = 2.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {2 a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+2 \cot \left (f x +e \right )\right )}{c f}\) | \(90\) |
2/c/f*a*(a*(sec(f*x+e)+1))^(1/2)*(arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos( f*x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+2*cot(f*x +e))
Time = 0.31 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.84 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=\left [\frac {\sqrt {-a} a \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 8 \, a \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, c f \sin \left (f x + e\right )}, \frac {a^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 4 \, a \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{c f \sin \left (f x + e\right )}\right ] \]
[1/2*(sqrt(-a)*a*log(-(8*a*cos(f*x + e)^3 - 4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7*a* cos(f*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e) + 8*a*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e))/(c*f*sin(f*x + e)), (a^(3/2)*arctan(2 *sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) /(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e) + 4*a*sqrt((a*cos (f*x + e) + a)/cos(f*x + e))*cos(f*x + e))/(c*f*sin(f*x + e))]
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=- \frac {\int \frac {a \sqrt {a \sec {\left (e + f x \right )} + a}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {a \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx}{c} \]
-(Integral(a*sqrt(a*sec(e + f*x) + a)/(sec(e + f*x) - 1), x) + Integral(a* sqrt(a*sec(e + f*x) + a)*sec(e + f*x)/(sec(e + f*x) - 1), x))/c
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=\int { -\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{c \sec \left (f x + e\right ) - c} \,d x } \]
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=\int { -\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{c \sec \left (f x + e\right ) - c} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{c-c \sec (e+f x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{c-\frac {c}{\cos \left (e+f\,x\right )}} \,d x \]