Integrand size = 25, antiderivative size = 165 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a-5 b) (a+b)^2 \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}-\frac {(a-5 b) (a+b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{16 b f}-\frac {(a-5 b) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{24 b f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{5/2}}{6 b f} \]
-1/16*(a-5*b)*(a+b)^2*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2 ))/b^(3/2)/f-1/16*(a-5*b)*(a+b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f- 1/24*(a-5*b)*tan(f*x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/b/f+1/6*tan(f*x+e)*(a+b +b*tan(f*x+e)^2)^(5/2)/b/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 8.97 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.73 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {(a+b)^3 \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right )^2 \tan (e+f x) \left (5 \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \left (-9+\frac {6 a \sin ^2(e+f x)}{a+b}\right )+\frac {5 \sec ^4(e+f x) \left (-3 a^3 \cos ^4(e+f x) \left (-3+2 \sin ^2(e+f x)\right )+3 b^3 \left (3+8 \sin ^2(e+f x)-3 \sin ^4(e+f x)\right )+a b^2 \left (27+24 \sin ^2(e+f x)-49 \sin ^4(e+f x)+6 \sin ^6(e+f x)\right )+a^2 b \left (27-24 \sin ^2(e+f x)-19 \sin ^4(e+f x)+16 \sin ^6(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}{(a+b)^3}-\frac {256 \operatorname {Hypergeometric2F1}\left (2,5,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (a+b-a \sin ^2(e+f x)\right ) \left (-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}\right )^{5/2}}{a+b}\right )}{60 \sqrt {2} f (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \left (a+b-a \sin ^2(e+f x)\right )^{3/2} \left (-\frac {b \tan ^2(e+f x)}{a+b}\right )^{3/2}} \]
-1/60*((a + b)^3*(a + b*Sec[e + f*x]^2)^(3/2)*(1 - (a*Sin[e + f*x]^2)/(a + b))^2*Tan[e + f*x]*(5*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*(-9 + ( 6*a*Sin[e + f*x]^2)/(a + b)) + (5*Sec[e + f*x]^4*(-3*a^3*Cos[e + f*x]^4*(- 3 + 2*Sin[e + f*x]^2) + 3*b^3*(3 + 8*Sin[e + f*x]^2 - 3*Sin[e + f*x]^4) + a*b^2*(27 + 24*Sin[e + f*x]^2 - 49*Sin[e + f*x]^4 + 6*Sin[e + f*x]^6) + a^ 2*b*(27 - 24*Sin[e + f*x]^2 - 19*Sin[e + f*x]^4 + 16*Sin[e + f*x]^6))*Sqrt [-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2) ])/(a + b)^3 - (256*Hypergeometric2F1[2, 5, 7/2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + b - a*Sin[e + f*x]^2)*(-((b*Sec[e + f*x]^2*(a + b - a*Sin[e + f *x]^2)*Tan[e + f*x]^2)/(a + b)^2))^(5/2))/(a + b)))/(Sqrt[2]*f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*(a + b - a* Sin[e + f*x]^2)^(3/2)*(-((b*Tan[e + f*x]^2)/(a + b)))^(3/2))
Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4634, 299, 211, 211, 224, 219}
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.
| \(\displaystyle \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (e+f x)^4 \left (a+b \sec (e+f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \int \left (b \tan ^2(e+f x)+a+b\right )^{3/2}d\tan (e+f x)}{6 b}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{5/2}}{6 b}-\frac {(a-5 b) \left (\frac {3}{4} (a+b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )+\frac {1}{4} \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}\right )}{6 b}}{f}\) |
((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(5/2))/(6*b) - ((a - 5*b)*((Tan[ e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/4 + (3*(a + b)*(((a + b)*ArcTan h[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*Sqrt[b]) + (T an[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2))/4))/(6*b))/f
3.3.48.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(1416\) vs. \(2(145)=290\).
Time = 16.72 (sec) , antiderivative size = 1417, normalized size of antiderivative = 8.59
-1/96/f/b^(9/2)*(a+b*sec(f*x+e)^2)^(3/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) )^2)^(1/2)/(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))*(-30*b^(11/2)*cos(f*x+e)^2*si n(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-30*b^(11/2)*cos(f*x+e )*sin(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-44*cos(f*x+e)^2*s in(f*x+e)*a*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-20*b^(11/2 )*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e)-44*sin(f*x+e)*cos (f*x+e)*a*b^(9/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-6*sin(f*x+e) *cos(f*x+e)^2*a^2*b^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-20*b ^(11/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*tan(f*x+e)-28*b^(9/2)* ((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*sin(f*x+e)-6*sin(f*x+e)*cos( f*x+e)*a^2*b^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-16*b^(11/2) *((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*tan(f*x+e)*sec(f*x+e)-28*b^(9 /2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*tan(f*x+e)+3*cos(f*x+e)^ 3*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^( 1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f* x+e)+1))*a^3*b^3-9*cos(f*x+e)^3*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2 )^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^( 1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a^2*b^4-27*cos(f*x+e)^3*ln(4*(((b+a *cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*cos(f*x+e)+b^(1/2)*((b+a*co s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-sin(f*x+e)*a-a-b)/(sin(f*x+e)+1))*a...
Time = 1.07 (sec) , antiderivative size = 470, normalized size of antiderivative = 2.85 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt {b} \cos \left (f x + e\right )^{5} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (3 \, a^{2} b + 22 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, b^{3} + 2 \, {\left (7 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{192 \, b^{2} f \cos \left (f x + e\right )^{5}}, -\frac {3 \, {\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - 2 \, {\left ({\left (3 \, a^{2} b + 22 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 8 \, b^{3} + 2 \, {\left (7 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{96 \, b^{2} f \cos \left (f x + e\right )^{5}}\right ] \]
[-1/192*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*sqrt(b)*cos(f*x + e)^5*log((( a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b) /cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) - 4*((3*a^2*b + 22* a*b^2 + 15*b^3)*cos(f*x + e)^4 + 8*b^3 + 2*(7*a*b^2 + 5*b^3)*cos(f*x + e)^ 2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^2*f*cos(f* x + e)^5), -1/96*(3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*sqrt(-b)*arctan(-1/2 *((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e) ^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*cos(f*x + e)^5 - 2*((3*a^2*b + 22*a*b^2 + 15*b^3)*cos(f*x + e)^4 + 8*b^3 + 2*(7*a *b^2 + 5*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)* sin(f*x + e))/(b^2*f*cos(f*x + e)^5)]
\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \sec ^{4}{\left (e + f x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.47 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {3 \, {\left (a + b\right )}^{2} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, {\left (a + b\right )}^{2} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {18 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 18 \, {\left (a + b\right )} \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 12 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right ) - 18 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right ) - \frac {8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {5}{2}} \tan \left (f x + e\right )}{b} + \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )} \tan \left (f x + e\right )}{b} + \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2} \tan \left (f x + e\right )}{b}}{48 \, f} \]
-1/48*(3*(a + b)^2*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) + 3*( a + b)^2*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*a*ar csinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 18*(a + b)*sqrt(b)*arcsinh (b*tan(f*x + e)/sqrt((a + b)*b)) - 12*(b*tan(f*x + e)^2 + a + b)^(3/2)*tan (f*x + e) - 18*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e) - 8*(b* tan(f*x + e)^2 + a + b)^(5/2)*tan(f*x + e)/b + 2*(b*tan(f*x + e)^2 + a + b )^(3/2)*(a + b)*tan(f*x + e)/b + 3*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^ 2*tan(f*x + e)/b)/f
\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^4} \,d x \]