Integrand size = 25, antiderivative size = 137 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\left (3 a^2-2 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{5/2} f}-\frac {3 (a-b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b^2 f}+\frac {\sec ^2(e+f x) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{4 b f} \]
1/8*(3*a^2-2*a*b+3*b^2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1 /2))/b^(5/2)/f-3/8*(a-b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b^2/f+1/4*s ec(f*x+e)^2*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f
Result contains complex when optimal does not.
Time = 8.55 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.38 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {e^{i (e+f x)} \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \sqrt {a+2 b+a \cos (2 e+2 f x)} \left (-\frac {i \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) \left (-3 a \left (1+e^{2 i (e+f x)}\right )^2+b \left (3+14 e^{2 i (e+f x)}+3 e^{4 i (e+f x)}\right )\right )}{\left (1+e^{2 i (e+f x)}\right )^4}-\frac {\left (3 a^2-2 a b+3 b^2\right ) \log \left (\frac {-4 \sqrt {b} \left (-1+e^{2 i (e+f x)}\right ) f+4 i \sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2} f}{1+e^{2 i (e+f x)}}\right )}{\sqrt {4 b e^{2 i (e+f x)}+a \left (1+e^{2 i (e+f x)}\right )^2}}\right ) \sec (e+f x)}{8 \sqrt {2} b^{5/2} f \sqrt {a+b \sec ^2(e+f x)}} \]
(E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]*(((-I)*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*(-3*a*(1 + E^((2*I)*(e + f*x)))^2 + b*(3 + 14*E^((2*I)*(e + f*x) ) + 3*E^((4*I)*(e + f*x)))))/(1 + E^((2*I)*(e + f*x)))^4 - ((3*a^2 - 2*a*b + 3*b^2)*Log[(-4*Sqrt[b]*(-1 + E^((2*I)*(e + f*x)))*f + (4*I)*Sqrt[4*b*E^ ((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]*f)/(1 + E^((2*I)*(e + f *x)))])/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*Sec [e + f*x])/(8*Sqrt[2]*b^(5/2)*f*Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4634, 318, 25, 299, 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.
| \(\displaystyle \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (e+f x)^6}{\sqrt {a+b \sec (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {\left (\tan ^2(e+f x)+1\right )^2}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {\frac {\int -\frac {3 (a-b) \tan ^2(e+f x)+a-3 b}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 b}+\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {\int \frac {3 (a-b) \tan ^2(e+f x)+a-3 b}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{4 b}}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {\frac {3 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (3 a^2-2 a b+3 b^2\right ) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{2 b}}{4 b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {\frac {3 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (3 a^2-2 a b+3 b^2\right ) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{2 b}}{4 b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right ) \sqrt {a+b \tan ^2(e+f x)+b}}{4 b}-\frac {\frac {3 (a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 b}-\frac {\left (3 a^2-2 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b^{3/2}}}{4 b}}{f}\) |
((Tan[e + f*x]*(1 + Tan[e + f*x]^2)*Sqrt[a + b + b*Tan[e + f*x]^2])/(4*b) - (-1/2*((3*a^2 - 2*a*b + 3*b^2)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/b^(3/2) + (3*(a - b)*Tan[e + f*x]*Sqrt[a + b + b*Ta n[e + f*x]^2])/(2*b))/(4*b))/f
3.3.63.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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. \(1611\) vs. \(2(121)=242\).
Time = 16.83 (sec) , antiderivative size = 1612, normalized size of antiderivative = 11.77
1/16/f/b^(9/2)/(a+b*sec(f*x+e)^2)^(1/2)*(6*b^(7/2)*a*tan(f*x+e)+6*b^(9/2)* tan(f*x+e)*sec(f*x+e)^2-6*b^(5/2)*a^2*tan(f*x+e)-2*b^(7/2)*a*tan(f*x+e)*se c(f*x+e)^2+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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^2-2 *((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*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+co s(f*x+e))^2)^(1/2)-sin(f*x+e)*a+a+b)/(sin(f*x+e)-1))*a*b^3+3*((b+a*cos(f*x +e)^2)/(1+cos(f*x+e))^2)^(1/2)*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))*b^4+3*((b+a*cos(f*x+e)^2)/(1+cos(f* x+e))^2)^(1/2)*ln(4*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*c os(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^2-2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )*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*b^3+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(((b+a*c os(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))*b^4+4* b^(9/2)*tan(f*x+e)*sec(f*x+e)^4+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)...
Time = 0.44 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.89 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \sqrt {b} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left (3 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, b^{3} f \cos \left (f x + e\right )^{3}}, \frac {{\left (3 \, a^{2} - 2 \, a b + 3 \, b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, b^{3} f \cos \left (f x + e\right )^{3}}\right ] \]
[1/32*((3*a^2 - 2*a*b + 3*b^2)*sqrt(b)*cos(f*x + e)^3*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*b - b^2)*cos(f*x + e)^2 - 2*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(b^3*f *cos(f*x + e)^3), 1/16*((3*a^2 - 2*a*b + 3*b^2)*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)^ 3 - 2*(3*(a*b - b^2)*cos(f*x + e)^2 - 2*b^2)*sqrt((a*cos(f*x + e)^2 + b)/c os(f*x + e)^2)*sin(f*x + e))/(b^3*f*cos(f*x + e)^3)]
\[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{3}}{b} + \frac {3 \, {\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {3 \, {\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {8 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {3 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b^{2}} + \frac {8 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )}{b}}{8 \, f} \]
1/8*(2*sqrt(b*tan(f*x + e)^2 + a + b)*tan(f*x + e)^3/b + 3*(a + b)*a*arcsi nh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(5/2) + 3*(a + b)*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) - 8*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b)) /b^(3/2) - 3*sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*tan(f*x + e)/b^2 + 8*s qrt(b*tan(f*x + e)^2 + a + b)*tan(f*x + e)/b)/f
\[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{6}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\sec ^6(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]