Integrand size = 28, antiderivative size = 140 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^4 f}-\frac {2 a \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}-\frac {8 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a c^4 f} \]
2*a^(5/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^4/f-2/3*a*co t(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^4/f-8/7*cot(f*x+e)^7*(a+a*sec(f*x+e))^ (7/2)/a/c^4/f+2*a^2*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2)/c^4/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 4.70 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.52 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=-\frac {2 a^3 \left (8+5 \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},1-\sec (e+f x)\right )+7 \sec (e+f x)\right ) \tan (e+f x)}{35 c^4 f (-1+\sec (e+f x))^4 \sqrt {a (1+\sec (e+f x))}} \]
(-2*a^3*(8 + 5*Hypergeometric2F1[-7/2, 1, -5/2, 1 - Sec[e + f*x]] + 7*Sec[ e + f*x])*Tan[e + f*x])/(35*c^4*f*(-1 + Sec[e + f*x])^4*Sqrt[a*(1 + Sec[e + f*x])])
Time = 0.41 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.90, 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, 364, 2009}
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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{5/2}}{(c-c \sec (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle \frac {\int \cot ^8(e+f x) (\sec (e+f x) a+a)^{13/2}dx}{a^4 c^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{13/2}}{\cot \left (e+f x+\frac {\pi }{2}\right )^8}dx}{a^4 c^4}\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 \int \frac {\cot ^8(e+f x) (\sec (e+f x) a+a)^4 \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )^2}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{a c^4 f}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle -\frac {2 \int \left (4 (\sec (e+f x) a+a)^4 \cot ^8(e+f x)+a^2 (\sec (e+f x) a+a)^2 \cot ^4(e+f x)-a^3 (\sec (e+f x) a+a) \cot ^2(e+f x)+\frac {a^4}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}\right )d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{a c^4 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \left (-a^{7/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )-a^3 \cot (e+f x) \sqrt {a \sec (e+f x)+a}+\frac {1}{3} a^2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}+\frac {4}{7} \cot ^7(e+f x) (a \sec (e+f x)+a)^{7/2}\right )}{a c^4 f}\) |
(-2*(-(a^(7/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]]) - a^3*Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]] + (a^2*Cot[e + f*x]^3*(a + a*Sec [e + f*x])^(3/2))/3 + (4*Cot[e + f*x]^7*(a + a*Sec[e + f*x])^(7/2))/7))/(a *c^4*f)
3.1.63.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
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])
Leaf count of result is larger than twice the leaf count of optimal. \(350\) vs. \(2(124)=248\).
Time = 1.64 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.51
\[\frac {2 a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (21 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \cos \left (f x +e \right )^{3}-63 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \cos \left (f x +e \right )^{2}+40 \cos \left (f x +e \right )^{3} \cot \left (f x +e \right )+63 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )-77 \cos \left (f x +e \right )^{2} \cot \left (f x +e \right )-21 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+70 \cos \left (f x +e \right ) \cot \left (f x +e \right )-21 \cot \left (f x +e \right )\right )}{21 c^{4} f \left (\cos \left (f x +e \right )-1\right )^{3}}\]
2/21/c^4/f*a^2*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)-1)^3*(21*(-cos(f*x+e)/ (cos(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)^3-63*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arcta nh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2))*cos(f*x+e )^2+40*cos(f*x+e)^3*cot(f*x+e)+63*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 )-77*cos(f*x+e)^2*cot(f*x+e)-21*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)+70*cos(f*x+ e)*cot(f*x+e)-21*cot(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (124) = 248\).
Time = 0.34 (sec) , antiderivative size = 527, normalized size of antiderivative = 3.76 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=\left [\frac {21 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}\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 (40 \, a^{2} \cos \left (f x + e\right )^{4} - 77 \, a^{2} \cos \left (f x + e\right )^{3} + 70 \, a^{2} \cos \left (f x + e\right )^{2} - 21 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{42 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}, \frac {21 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}\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 (40 \, a^{2} \cos \left (f x + e\right )^{4} - 77 \, a^{2} \cos \left (f x + e\right )^{3} + 70 \, a^{2} \cos \left (f x + e\right )^{2} - 21 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{21 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/42*(21*(a^2*cos(f*x + e)^3 - 3*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) - a^2)*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*c os(f*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e) + 4*(40*a^2*cos(f*x + e) ^4 - 77*a^2*cos(f*x + e)^3 + 70*a^2*cos(f*x + e)^2 - 21*a^2*cos(f*x + e))* sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c^4*f* cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e)), 1/21*(21*(a^ 2*cos(f*x + e)^3 - 3*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) - a^2)*sqrt(a )*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*si n(f*x + e)/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e) + 2*(40 *a^2*cos(f*x + e)^4 - 77*a^2*cos(f*x + e)^3 + 70*a^2*cos(f*x + e)^2 - 21*a ^2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e))]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (c \sec \left (f x + e\right ) - c\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^4} \,d x \]