Integrand size = 25, antiderivative size = 71 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\tan (e+f x)}{3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {2 \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:
1/3*tan(f*x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)+2/3*tan(f*x+e)/(a+b)^2/f /(a+b+b*tan(f*x+e)^2)^(1/2)
Time = 5.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) (3 a+5 b+(3 a+b) \cos (2 (e+f x))) \sec ^4(e+f x) \tan (e+f x)}{12 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \] Input:
Integrate[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*(3*a + 5*b + (3*a + b)*Cos[2*(e + f*x)])*S ec[e + f*x]^4*Tan[e + f*x])/(12*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^(5/2))
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4634, 209, 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.
\(\displaystyle \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {\frac {2 \int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 (a+b)}+\frac {\tan (e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {2 \tan (e+f x)}{3 (a+b)^2 \sqrt {a+b \tan ^2(e+f x)+b}}+\frac {\tan (e+f x)}{3 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
Input:
Int[Sec[e + f*x]^2/(a + b*Sec[e + f*x]^2)^(5/2),x]
Output:
(Tan[e + f*x]/(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
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_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 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]
Time = 5.87 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18
method | result | size |
default | \(\frac {\left (3 a \cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )^{2} b +2 b \right ) \left (b +a \cos \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right ) \sec \left (f x +e \right )^{4}}{3 f \left (a^{2}+2 a b +b^{2}\right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(84\) |
Input:
int(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3/f/(a^2+2*a*b+b^2)*(3*a*cos(f*x+e)^2+cos(f*x+e)^2*b+2*b)*(b+a*cos(f*x+e )^2)/(a+b*sec(f*x+e)^2)^(5/2)*tan(f*x+e)*sec(f*x+e)^4
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (63) = 126\).
Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.89 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left ({\left (3 \, a + b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}} \] Input:
integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
Output:
1/3*((3*a + b)*cos(f*x + e)^3 + 2*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)
\[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(sec(f*x+e)**2/(a+b*sec(f*x+e)**2)**(5/2),x)
Output:
Integral(sec(e + f*x)**2/(a + b*sec(e + f*x)**2)**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {2 \, \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} + \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}}}{3 \, f} \] Input:
integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
Output:
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)))/f
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (63) = 126\).
Time = 0.72 (sec) , antiderivative size = 337, normalized size of antiderivative = 4.75 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left (\frac {3 \, {\left (a^{6} b^{4} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 2 \, a^{5} b^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + a^{4} b^{6} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{7} b^{4} + 3 \, a^{6} b^{5} + 3 \, a^{5} b^{6} + a^{4} b^{7}} - \frac {2 \, {\left (3 \, a^{6} b^{4} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 2 \, a^{5} b^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - a^{4} b^{6} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )}}{a^{7} b^{4} + 3 \, a^{6} b^{5} + 3 \, a^{5} b^{6} + a^{4} b^{7}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {3 \, {\left (a^{6} b^{4} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 2 \, a^{5} b^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + a^{4} b^{6} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )\right )}}{a^{7} b^{4} + 3 \, a^{6} b^{5} + 3 \, a^{5} b^{6} + a^{4} b^{7}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{3 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a + b\right )}^{\frac {3}{2}} f} \] Input:
integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
Output:
2/3*((3*(a^6*b^4*sgn(cos(f*x + e)) + 2*a^5*b^5*sgn(cos(f*x + e)) + a^4*b^6 *sgn(cos(f*x + e)))*tan(1/2*f*x + 1/2*e)^2/(a^7*b^4 + 3*a^6*b^5 + 3*a^5*b^ 6 + a^4*b^7) - 2*(3*a^6*b^4*sgn(cos(f*x + e)) + 2*a^5*b^5*sgn(cos(f*x + e) ) - a^4*b^6*sgn(cos(f*x + e)))/(a^7*b^4 + 3*a^6*b^5 + 3*a^5*b^6 + a^4*b^7) )*tan(1/2*f*x + 1/2*e)^2 + 3*(a^6*b^4*sgn(cos(f*x + e)) + 2*a^5*b^5*sgn(co s(f*x + e)) + a^4*b^6*sgn(cos(f*x + e)))/(a^7*b^4 + 3*a^6*b^5 + 3*a^5*b^6 + a^4*b^7))*tan(1/2*f*x + 1/2*e)/((a*tan(1/2*f*x + 1/2*e)^4 + b*tan(1/2*f* x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e)^2 + a + b)^(3/2)*f)
Time = 26.72 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.42 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\left ({\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}-1\right )\,\sqrt {a+\frac {b}{{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}\right )}^2}}\,\left (a\,3{}\mathrm {i}+b\,1{}\mathrm {i}+a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,6{}\mathrm {i}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,3{}\mathrm {i}+b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,10{}\mathrm {i}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}\right )}{3\,f\,{\left (a+b\right )}^2\,{\left (a+2\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}^2} \] Input:
int(1/(cos(e + f*x)^2*(a + b/cos(e + f*x)^2)^(5/2)),x)
Output:
-((exp(e*4i + f*x*4i) - 1)*(a + b/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x *1i)/2)^2)^(1/2)*(a*3i + b*1i + a*exp(e*2i + f*x*2i)*6i + a*exp(e*4i + f*x *4i)*3i + b*exp(e*2i + f*x*2i)*10i + b*exp(e*4i + f*x*4i)*1i))/(3*f*(a + b )^2*(a + 2*a*exp(e*2i + f*x*2i) + a*exp(e*4i + f*x*4i) + 4*b*exp(e*2i + f* x*2i))^2)
\[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:
int(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(5/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2)/(sec(e + f*x)**6*b**3 + 3*sec(e + f*x)**4*a*b**2 + 3*sec(e + f*x)**2*a**2*b + a**3),x)