Integrand size = 16, antiderivative size = 125 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {b \tan (e+f x)}{3 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}-\frac {b (5 a+3 b) \tan (e+f x)}{3 a^2 (a+b)^2 f \sqrt {a+b+b \tan ^2(e+f x)}} \] Output:
arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f-1/3*b*tan( f*x+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^(3/2)-1/3*b*(5*a+3*b)*tan(f*x+e)/a^2 /(a+b)^2/f/(a+b+b*tan(f*x+e)^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 14.71 (sec) , antiderivative size = 1927, normalized size of antiderivative = 15.42 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)^(-5/2),x]
Output:
(3*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)]*Cos[e + f*x]^4*Sin[e + f*x])/(4*Sqrt[2]*f*(a + b*Sec[e + f*x]^2)^ (5/2)*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -2, 5/2, 3 /2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -2, 7 /2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[ 3/2, -1, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f* x]^2)*((15*a*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^5*Sin[e + f*x]^2)/(4*Sqrt[2]*(a + b - a*Si n[e + f*x]^2)^(7/2)*(3*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a*AppellF1[3/2, -2, 7/2, 5/2, Sin[e + f* x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)])*Sin[e + f*x]^2)) + (3*(a + b) *AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*C os[e + f*x]^5)/(4*Sqrt[2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*Appe llF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] + (5*a *AppellF1[3/2, -2, 7/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)] - 4*(a + b)*AppellF1[3/2, -1, 5/2, 5/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/ (a + b)])*Sin[e + f*x]^2)) - (3*(a + b)*AppellF1[1/2, -2, 5/2, 3/2, Sin[e + f*x]^2, (a*Sin[e + f*x]^2)/(a + b)]*Cos[e + f*x]^3*Sin[e + f*x]^2)/(Sqrt [2]*(a + b - a*Sin[e + f*x]^2)^(5/2)*(3*(a + b)*AppellF1[1/2, -2, 5/2, ...
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3042, 4616, 316, 402, 27, 291, 216}
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.
\(\displaystyle \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4616 |
\(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {\frac {\int \frac {-2 b \tan ^2(e+f x)+3 a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^{3/2}}d\tan (e+f x)}{3 a (a+b)}-\frac {b \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle \frac {\frac {\frac {\int \frac {3 (a+b)^2}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a (a+b)}-\frac {b (5 a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {3 (a+b) \int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{a}-\frac {b (5 a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\frac {\frac {3 (a+b) \int \frac {1}{\frac {a \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}+1}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{a}-\frac {b (5 a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {3 (a+b) \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2}}-\frac {b (5 a+3 b) \tan (e+f x)}{a (a+b) \sqrt {a+b \tan ^2(e+f x)+b}}}{3 a (a+b)}-\frac {b \tan (e+f x)}{3 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^{3/2}}}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)^(-5/2),x]
Output:
(-1/3*(b*Tan[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^(3/2)) + ((3* (a + b)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/a^( 3/2) - (b*(5*a + 3*b)*Tan[e + f*x])/(a*(a + b)*Sqrt[a + b + b*Tan[e + f*x] ^2]))/(3*a*(a + b)))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(860\) vs. \(2(111)=222\).
Time = 6.84 (sec) , antiderivative size = 861, normalized size of antiderivative = 6.89
Input:
int(1/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^4*ln(4*(-a)^(1/2)* ((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(-3-3*sec(f*x+e))-1/ 3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^3*b*ln(4*(-a)^(1/2)*((b+a* cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f* x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(-6-6*sec(f*x+e)-6*sec(f*x +e)^2-6*sec(f*x+e)^3)-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2* b^2*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e) +4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)* (-3-3*sec(f*x+e)-12*sec(f*x+e)^2-12*sec(f*x+e)^3-3*sec(f*x+e)^4-3*sec(f*x+ e)^5)-1/3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b^3*ln(4*(-a)^(1/2 )*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+ a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*(-6*sec(f*x+e)^2-6 *sec(f*x+e)^3-6*sec(f*x+e)^4-6*sec(f*x+e)^5)-1/3*((b+a*cos(f*x+e)^2)/(1+co s(f*x+e))^2)^(1/2)*b^4*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 2)-4*sin(f*x+e)*a)*(-3*sec(f*x+e)^4-3*sec(f*x+e)^5)-2*(-a)^(1/2)*a^3*b*tan (f*x+e)-1/3*(-a)^(1/2)*a^2*b^2*(4*tan(f*x+e)+11*tan(f*x+e)*sec(f*x+e)^2)-1 /3*(7*cos(f*x+e)^2+5)*(-a)^(1/2)*a*b^3*tan(f*x+e)*sec(f*x+e)^4-(-a)^(1/2)* b^4*tan(f*x+e)*sec(f*x+e)^4)/(a+b)^2/a^2/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^...
Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (111) = 222\).
Time = 0.56 (sec) , antiderivative size = 881, normalized size of antiderivative = 7.05 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")
Output:
[-1/24*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos(f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f*x + e)^2)*sqrt(-a)*log(128*a^4*c os(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2) *cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*s qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(2*(3*a^3*b + 2*a^2*b^2)*cos(f*x + e)^3 + (5*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*c os(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2 )*f*cos(f*x + e)^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a ^5*b^2 + 2*a^4*b^3 + a^3*b^4)*f), -1/12*(3*((a^4 + 2*a^3*b + a^2*b^2)*cos( f*x + e)^4 + a^2*b^2 + 2*a*b^3 + b^4 + 2*(a^3*b + 2*a^2*b^2 + a*b^3)*cos(f *x + e)^2)*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f* x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^ 2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a ^2*b)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(2*(3*a^3*b + 2*a^2*b^2)*cos(f*x + e)^3 + (5*a^2*b^2 + 3*a*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/c os(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a^5*b^2 + 2*a^4*b^...
\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+b*sec(f*x+e)**2)**(5/2),x)
Output:
Integral((a + b*sec(e + f*x)**2)**(-5/2), x)
Timed out. \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")
Output:
integrate((b*sec(f*x + e)^2 + a)^(-5/2), x)
Timed out. \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \] Input:
int(1/(a + b/cos(e + f*x)^2)^(5/2),x)
Output:
int(1/(a + b/cos(e + f*x)^2)^(5/2), x)
\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}}{\sec \left (f x +e \right )^{6} b^{3}+3 \sec \left (f x +e \right )^{4} a \,b^{2}+3 \sec \left (f x +e \right )^{2} a^{2} b +a^{3}}d x \] Input:
int(1/(a+b*sec(f*x+e)^2)^(5/2),x)
Output:
int(sqrt(sec(e + f*x)**2*b + a)/(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)