Integrand size = 28, antiderivative size = 102 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^2 f}+\frac {2 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^2 f}-\frac {4 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^2 f} \]
2*a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^2/f-4/3*cot( f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^2/f+2*a*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2 )/c^2/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.64 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.72 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\frac {2 a^2 \left (-2+3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\sec (e+f x)\right ) (-1+\sec (e+f x))\right ) \tan (e+f x)}{3 c^2 f (-1+\sec (e+f x))^2 \sqrt {a (1+\sec (e+f x))}} \]
(2*a^2*(-2 + 3*Hypergeometric2F1[-1/2, 1, 1/2, 1 - Sec[e + f*x]]*(-1 + Sec [e + f*x]))*Tan[e + f*x])/(3*c^2*f*(-1 + Sec[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])
Time = 0.39 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4392, 3042, 4375, 359, 264, 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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{3/2}}{(c-c \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2}}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle \frac {\int \cot ^4(e+f x) (\sec (e+f x) a+a)^{7/2}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{7/2}}{\cot \left (e+f x+\frac {\pi }{2}\right )^4}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 \int \frac {\cot ^4(e+f x) (\sec (e+f x) a+a)^2 \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{c^2 f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle -\frac {2 \left (\frac {2}{3} \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}-a \int \frac {\cot ^2(e+f x) (\sec (e+f x) a+a)}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )\right )}{c^2 f}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle -\frac {2 \left (\frac {2}{3} \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}-a \left (\cot (e+f x) \sqrt {a \sec (e+f x)+a}-a \int \frac {1}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )\right )\right )}{c^2 f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 \left (\frac {2}{3} \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}-a \left (\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )+\cot (e+f x) \sqrt {a \sec (e+f x)+a}\right )\right )}{c^2 f}\) |
(-2*((2*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/3 - a*(Sqrt[a]*ArcTan[( Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]] + Cot[e + f*x]*Sqrt[a + a* Sec[e + f*x]])))/(c^2*f)
3.1.54.3.1 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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 = 2.07 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.77
| method | result | size |
| default | \(\frac {2 a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )-3 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+5 \cos \left (f x +e \right ) \cot \left (f x +e \right )-3 \cot \left (f x +e \right )\right )}{3 c^{2} f \left (\cos \left (f x +e \right )-1\right )}\) | \(181\) |
2/3/c^2/f*a*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)-1)*(3*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)*cos(f*x+e)-3*arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(c os(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+5*cos(f*x+e)*cot(f *x+e)-3*cot(f*x+e))
Time = 0.30 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.44 \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\left [\frac {3 \, {\left (a \cos \left (f x + e\right ) - a\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (5 \, a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{6 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, \frac {3 \, {\left (a \cos \left (f x + e\right ) - a\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (5 \, a \cos \left (f x + e\right )^{2} - 3 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{3 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/6*(3*(a*cos(f*x + e) - a)*sqrt(-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) + 4*(5*a*cos(f*x + e)^2 - 3*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f* x + e)))/((c^2*f*cos(f*x + e) - c^2*f)*sin(f*x + e)), 1/3*(3*(a*cos(f*x + e) - a)*sqrt(a)*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*c os(f*x + e)*sin(f*x + e)/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f* x + e) + 2*(5*a*cos(f*x + e)^2 - 3*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^2*f*cos(f*x + e) - c^2*f)*sin(f*x + e))]
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\frac {\int \frac {a \sqrt {a \sec {\left (e + f x \right )} + a}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {a \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx}{c^{2}} \]
(Integral(a*sqrt(a*sec(e + f*x) + a)/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1 ), x) + Integral(a*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x))/c**2
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (c \sec \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^2} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \]