Integrand size = 34, antiderivative size = 174 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {15 a^3 \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}-\frac {a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac {15 a^3 \tan (e+f x)}{4 c^2 f \sqrt {c-c \sec (e+f x)}} \]
-15/4*a^3*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/c^ (5/2)/f*2^(1/2)-1/2*a*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(5/ 2)+5/4*(a^3+a^3*sec(f*x+e))*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(3/2)+15/4*a^3 *tan(f*x+e)/c^2/f/(c-c*sec(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.62 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.37 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=-\frac {a^3 \operatorname {Hypergeometric2F1}\left (3,\frac {7}{2},\frac {9}{2},\frac {1}{2} (1+\sec (e+f x))\right ) (1+\sec (e+f x))^3 \tan (e+f x)}{28 c^2 f \sqrt {c-c \sec (e+f x)}} \]
-1/28*(a^3*Hypergeometric2F1[3, 7/2, 9/2, (1 + Sec[e + f*x])/2]*(1 + Sec[e + f*x])^3*Tan[e + f*x])/(c^2*f*Sqrt[c - c*Sec[e + f*x]])
Time = 0.85 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4445, 3042, 4445, 3042, 4444, 3042, 4282, 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.
| \(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^3}{(c-c \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {5 a \int \frac {\sec (e+f x) (\sec (e+f x) a+a)^2}{(c-c \sec (e+f x))^{3/2}}dx}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \int \frac {\sec (e+f x) (\sec (e+f x) a+a)}{\sqrt {c-c \sec (e+f x)}}dx}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\right )}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\right )}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4444 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (2 a \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}}dx+\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\right )}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\right )}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (2 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}\right )}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\right )}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}-\frac {4 a \int \frac {1}{\frac {c^2 \tan ^2(e+f x)}{c-c \sec (e+f x)}+2 c}d\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}}{f}\right )}{2 c}-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}\right )}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {5 a \left (-\frac {\tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))^{3/2}}-\frac {3 a \left (\frac {2 a \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}}-\frac {2 \sqrt {2} a \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {c} f}\right )}{2 c}\right )}{4 c}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}}\) |
-1/2*(a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^(5/2) ) - (5*a*(-(((a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(f*(c - c*Sec[e + f*x] )^(3/2))) - (3*a*((-2*Sqrt[2]*a*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqr t[c - c*Sec[e + f*x]])])/(Sqrt[c]*f) + (2*a*Tan[e + f*x])/(f*Sqrt[c - c*Se c[e + f*x]])))/(2*c)))/(4*c)
3.1.85.3.1 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[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.))/ Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*d*Cot[e + f*x]*((c + d*Csc[e + f*x])^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*c*((2*n - 1)/(2*n - 1)) Int[Csc[e + f*x]*((c + d*Csc[e + f*x ])^(n - 1)/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x ] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), x] - Simp[d*((2*n - 1)/(b*(2*m + 1))) Int[Csc[e + f*x]*(a + b*Csc[e + f* x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && LtQ[m, -2^ (-1)] && IntegerQ[2*m]
Time = 6.29 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.28
| method | result | size |
| default | \(\frac {a^{3} \sqrt {2}\, \left (15 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \csc \left (f x +e \right )-15 \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )+5 \left (1-\cos \left (f x +e \right )\right )^{2} \sin \left (f x +e \right )+2 \sin \left (f x +e \right )^{3}\right )}{4 c^{2} f \sqrt {\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right )^{3} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )}\) | \(223\) |
| parts | \(\text {Expression too large to display}\) | \(1034\) |
1/4*a^3/c^2/f*2^(1/2)/(c*(1-cos(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e)^2-1 )*csc(f*x+e)^2)^(1/2)/(1-cos(f*x+e))^3/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)*( 15*arctan(1/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2))*(1-cos(f*x+e))^4*((1- cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)*csc(f*x+e)-15*(1-cos(f*x+e))^4*csc(f*x +e)+5*(1-cos(f*x+e))^2*sin(f*x+e)+2*sin(f*x+e)^3)
Time = 0.32 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.53 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \sqrt {-c} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (9 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (9 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{4 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]
[-1/8*(15*sqrt(2)*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*sqrt(-c) *log((2*sqrt(2)*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) + (3*c*cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x + e) - 1)*sin(f*x + e)))*sin(f*x + e) + 4*(9*a^3*cos(f*x + e)^3 - 8*a^3*cos( f*x + e)^2 - 13*a^3*cos(f*x + e) + 4*a^3)*sqrt((c*cos(f*x + e) - c)/cos(f* x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e)), 1/4*(15*sqrt(2)*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*sqrt( c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sq rt(c)*sin(f*x + e)))*sin(f*x + e) - 2*(9*a^3*cos(f*x + e)^3 - 8*a^3*cos(f* x + e)^2 - 13*a^3*cos(f*x + e) + 4*a^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e) )]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx\right ) \]
a**3*(Integral(sec(e + f*x)/(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)** 2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c**2*sqrt(-c*sec(e + f *x) + c)), x) + Integral(3*sec(e + f*x)**2/(c**2*sqrt(-c*sec(e + f*x) + c) *sec(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c**2*sq rt(-c*sec(e + f*x) + c)), x) + Integral(3*sec(e + f*x)**3/(c**2*sqrt(-c*se c(e + f*x) + c)*sec(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x) + Integral(sec(e + f*x)**4/(c* *2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x))
Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 1.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.76 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=\frac {a^{3} {\left (\frac {15 \, \sqrt {2} \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}} + \frac {8 \, \sqrt {2}}{\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{2}} + \frac {7 \, \sqrt {2} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} + 9 \, \sqrt {2} \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c}{c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}\right )}}{4 \, f} \]
1/4*a^3*(15*sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/c^( 5/2) + 8*sqrt(2)/(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2) + (7*sqrt(2)*(c* tan(1/2*f*x + 1/2*e)^2 - c)^(3/2) + 9*sqrt(2)*sqrt(c*tan(1/2*f*x + 1/2*e)^ 2 - c)*c)/(c^4*tan(1/2*f*x + 1/2*e)^4))/f
Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^3}{\cos \left (e+f\,x\right )\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]