Integrand size = 25, antiderivative size = 122 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {(a-3 b) (a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{8 b^{3/2} f}-\frac {(a-3 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{8 b f}+\frac {\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}{4 b f} \]
-1/8*(a-3*b)*(a+b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/ b^(3/2)/f-1/8*(a-3*b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f+1/4*tan(f* x+e)*(a+b+b*tan(f*x+e)^2)^(3/2)/b/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 7.73 (sec) , antiderivative size = 380, normalized size of antiderivative = 3.11 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {b \sec ^2(e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \tan ^3(e+f x) \left (5 \arcsin \left (\sqrt {-\frac {b \tan ^2(e+f x)}{a+b}}\right ) \left (3-\frac {2 a \sin ^2(e+f x)}{a+b}\right )-\frac {5 \sec ^2(e+f x) \left (3 b^2 \left (1+\sin ^2(e+f x)\right )-2 a b \left (-3+\sin ^2(e+f x)+\sin ^4(e+f x)\right )+a^2 \left (3-5 \sin ^2(e+f x)+2 \sin ^4(e+f x)\right )\right ) \sqrt {-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}}}{(a+b)^2}+32 \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (2,4,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (-\frac {b \sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan ^2(e+f x)}{(a+b)^2}\right )^{5/2}\right )}{20 \sqrt {2} f \sqrt {a+2 b+a \cos (2 e+2 f x)} \sqrt {\frac {a+b \sec ^2(e+f x)}{a+b}} \sqrt {a+b-a \sin ^2(e+f x)} \left (-\frac {b \tan ^2(e+f x)}{a+b}\right )^{5/2}} \]
-1/20*(b*Sec[e + f*x]^2*Sqrt[a + b*Sec[e + f*x]^2]*(1 - (a*Sin[e + f*x]^2) /(a + b))*Tan[e + f*x]^3*(5*ArcSin[Sqrt[-((b*Tan[e + f*x]^2)/(a + b))]]*(3 - (2*a*Sin[e + f*x]^2)/(a + b)) - (5*Sec[e + f*x]^2*(3*b^2*(1 + Sin[e + f *x]^2) - 2*a*b*(-3 + Sin[e + f*x]^2 + Sin[e + f*x]^4) + a^2*(3 - 5*Sin[e + f*x]^2 + 2*Sin[e + f*x]^4))*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)^2 + 32*Cos[e + f*x]^2*Hypergeo metric2F1[2, 4, 7/2, -((b*Tan[e + f*x]^2)/(a + b))]*(-((b*Sec[e + f*x]^2*( a + b - a*Sin[e + f*x]^2)*Tan[e + f*x]^2)/(a + b)^2))^(5/2)))/(Sqrt[2]*f*S qrt[a + 2*b + a*Cos[2*e + 2*f*x]]*Sqrt[(a + b*Sec[e + f*x]^2)/(a + b)]*Sqr t[a + b - a*Sin[e + f*x]^2]*(-((b*Tan[e + f*x]^2)/(a + b)))^(5/2))
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4634, 299, 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.
| \(\displaystyle \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (e+f x)^4 \sqrt {a+b \sec (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \int \sqrt {b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{4 b}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {1}{2} (a+b) \int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}\right )}{4 b}}{f}\) |
((Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(3/2))/(4*b) - ((a - 3*b)*(((a + b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*Sqr t[b]) + (Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/2))/(4*b))/f
3.3.35.3.1 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. \(906\) vs. \(2(106)=212\).
Time = 11.17 (sec) , antiderivative size = 907, normalized size of antiderivative = 7.43
1/16/f/b^2*(a+b*sec(f*x+e)^2)^(1/2)/(1+cos(f*x+e))/((b+a*cos(f*x+e)^2)/(1+ cos(f*x+e))^2)^(1/2)*(3*cos(f*x+e)*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^(5/2)+3*cos(f*x+e)*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*c os(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^( 5/2)+2*cos(f*x+e)*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^(3/2)*a+2*cos(f*x+e)*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^(3/2)*a-cos(f*x+ e)*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^(1/2)*a^2-cos(f*x+e)*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^(1/2)*a^2+2*((b+a*cos(f*x+e)^2) /(1+cos(f*x+e))^2)^(1/2)*a*b*sin(f*x+e)+6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e ))^2)^(1/2)*b^2*sin(f*x+e)+2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a *b*tan(f*x+e)+6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*tan(f*x+e) +4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2*tan(f*x+e)*sec(f*x+e...
Time = 0.42 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.20 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [-\frac {{\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} \sqrt {b} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) - 4 \, {\left ({\left (a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, b^{2} f \cos \left (f x + e\right )^{3}}, -\frac {{\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} \sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - 2 \, {\left ({\left (a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{16 \, b^{2} f \cos \left (f x + e\right )^{3}}\right ] \]
[-1/32*((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*((a*b + 3*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^2*f* cos(f*x + e)^3), -1/16*((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*((a*b + 3*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^2*f*cos(f*x + e)^3)]
\[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \sec ^{4}{\left (e + f x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.42 \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (a + b\right )} a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {{\left (a + b\right )} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {4 \, a \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 4 \, \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right ) - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \tan \left (f x + e\right )}{b} + \frac {\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \tan \left (f x + e\right )}{b}}{8 \, f} \]
-1/8*((a + b)*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/b^(3/2) + (a + b)* arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 4*a*arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 4*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt((a + b )*b)) - 4*sqrt(b*tan(f*x + e)^2 + a + b)*tan(f*x + e) - 2*(b*tan(f*x + e)^ 2 + a + b)^(3/2)*tan(f*x + e)/b + sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*t an(f*x + e)/b)/f
\[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \sec ^4(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{{\cos \left (e+f\,x\right )}^4} \,d x \]