Integrand size = 25, antiderivative size = 79 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {2 \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \]
2/3*tan(f*x+e)/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)+1/3*sec(f*x+e)^2*tan(f *x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)
Time = 5.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) (2 a+3 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \tan (e+f x)}{6 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
((a + 2*b + a*Cos[2*(e + f*x)])*(2*a + 3*b + a*Cos[2*(e + f*x)])*Sec[e + f *x]^4*Tan[e + f*x])/(6*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^(5/2))
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4634, 292, 208}
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.
\(\displaystyle \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (e+f x)^4}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)+1}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 292 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 (a+b)}+\frac {\tan (e+f x) \left (\tan ^2(e+f x)+1\right )}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {2 \tan (e+f x)}{3 (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\left (\tan ^2(e+f x)+1\right ) \tan (e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
((Tan[e + f*x]*(1 + Tan[e + f*x]^2))/(3*(a + b)*(a + b + b*Tan[e + f*x]^2) ^(3/2)) + (2*Tan[e + f*x])/(3*(a + b)^2*Sqrt[a + b + b*Tan[e + f*x]^2]))/f
3.3.90.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Si mp[(-x)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*(p + 1))), x] - Simp[c*(q/( a*(p + 1))) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1), x], x] /; FreeQ[ {a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && EqQ[2*(p + q + 1) + 1, 0] && Gt Q[q, 0] && NeQ[p, -1]
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]
Time = 3.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {\left (b +a \cos \left (f x +e \right )^{2}\right ) \left (2 a \cos \left (f x +e \right )^{2}+a +3 b \right ) \tan \left (f x +e \right ) \sec \left (f x +e \right )^{4}}{3 f \left (a^{2}+2 a b +b^{2}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(75\) |
1/3/f/(a^2+2*a*b+b^2)*(b+a*cos(f*x+e)^2)*(2*a*cos(f*x+e)^2+a+3*b)/(a+b*sec (f*x+e)^2)^(5/2)*tan(f*x+e)*sec(f*x+e)^4
Time = 0.34 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.70 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left (2 \, a \cos \left (f x + e\right )^{3} + {\left (a + 3 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}} \]
1/3*(2*a*cos(f*x + e)^3 + (a + 3*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)/((a^4 + 2*a^3*b + a^2*b^2)*f*cos(f*x + e) ^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^2*b^2 + 2*a*b^3 + b^4)*f)
\[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sec ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.48 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} + \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} - \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {\tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \]
1/3*(2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)^2) + tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*(a + b)) - tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)^(3/2)*b) + tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a + b)*(a + b)*b))/f
\[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 29.44 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.94 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {2\,\left ({\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {b}{{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}\right )}^2}}\,\left (a\,1{}\mathrm {i}+a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,4{}\mathrm {i}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}+b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,6{}\mathrm {i}\right )}{3\,f\,{\left (a+b\right )}^2\,{\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}^2} \]