\(\int \frac {\sec ^3(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [284]

Optimal result

Integrand size = 25, antiderivative size = 319 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a-b) \sin (e+f x)}{3 b (a+b)^2 f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sin (e+f x)}{3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {(a-b) E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \left (a+b-a \sin ^2(e+f x)\right )}{3 a b (a+b)^2 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {\operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{3 a (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \]

[Out]

-1/3*(a-b)*sin(f*x+e)/b/(a+b)^2/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+1/3*sin(f*x+e)/(a+b)/f/(a+b-a*sin(
f*x+e)^2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+1/3*(a-b)*EllipticE(sin(f*x+e),(a/(a+b))^(1/2))*(a+b-a*sin
(f*x+e)^2)/a/b/(a+b)^2/f/(cos(f*x+e)^2)^(1/2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)/(1-a*sin(f*x+e)^2/(a+b
))^(1/2)+1/3*EllipticF(sin(f*x+e),(a/(a+b))^(1/2))*(1-a*sin(f*x+e)^2/(a+b))^(1/2)/a/(a+b)/f/(cos(f*x+e)^2)^(1/
2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4233, 1985, 1986, 423, 541, 538, 437, 435, 432, 430} \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right )}{3 a f (a+b) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac {(a-b) \left (-a \sin ^2(e+f x)+a+b\right ) E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right )}{3 a b f (a+b)^2 \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac {(a-b) \sin (e+f x)}{3 b f (a+b)^2 \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}+\frac {\sin (e+f x)}{3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right ) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]

[In]

Int[Sec[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-1/3*((a - b)*Sin[e + f*x])/(b*(a + b)^2*f*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)]) + Sin[e + f*x]/(3*
(a + b)*f*(a + b - a*Sin[e + f*x]^2)*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)]) + ((a - b)*EllipticE[Arc
Sin[Sin[e + f*x]], a/(a + b)]*(a + b - a*Sin[e + f*x]^2))/(3*a*b*(a + b)^2*f*Sqrt[Cos[e + f*x]^2]*Sqrt[Sec[e +
f*x]^2*(a + b - a*Sin[e + f*x]^2)]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)]) + (EllipticF[ArcSin[Sin[e + f*x]], a
/(a + b)]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)])/(3*a*(a + b)*f*Sqrt[Cos[e + f*x]^2]*Sqrt[Sec[e + f*x]^2*(a + b
- a*Sin[e + f*x]^2)])

Rule 423

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*((
c + d*x^n)^q/(a*n*(p + 1))), x] + Dist[1/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(n*
(p + 1) + 1) + d*(n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[
p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 1985

Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u*((b + a*c + a*d*x^n)/(c + d*x^n))^p
, x] /; FreeQ[{a, b, c, d, n, p}, x]

Rule 1986

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_.)*((c_) + (d_.)*(x_)^(n_))^(r_.))^(p_), x_Symbol] :> Dist[Simp
[(e*(a + b*x^n)^q*(c + d*x^n)^r)^p/((a + b*x^n)^(p*q)*(c + d*x^n)^(p*r))], Int[u*(a + b*x^n)^(p*q)*(c + d*x^n)
^(p*r), x], x] /; FreeQ[{a, b, c, d, e, n, p, q, r}, x]

Rule 4233

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b/(1 - ff^2*x^2)^(n/2))^p/(1 - ff^2*x^2)^((m + 1)/2), x
], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] &&  !IntegerQ
[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (a+\frac {b}{1-x^2}\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (\frac {a+b-a x^2}{1-x^2}\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sqrt {a+b-a \sin ^2(e+f x)} \text {Subst}\left (\int \frac {\sqrt {1-x^2}}{\left (a+b-a x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \\ & = \frac {\sin (e+f x)}{3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}-\frac {\sqrt {a+b-a \sin ^2(e+f x)} \text {Subst}\left (\int \frac {-2+x^2}{\sqrt {1-x^2} \left (a+b-a x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \\ & = -\frac {(a-b) \sin (e+f x)}{3 b (a+b)^2 f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sin (e+f x)}{3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sqrt {a+b-a \sin ^2(e+f x)} \text {Subst}\left (\int \frac {a+b+(-a+b) x^2}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{3 b (a+b)^2 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \\ & = -\frac {(a-b) \sin (e+f x)}{3 b (a+b)^2 f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sin (e+f x)}{3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}-\frac {\left ((-a+b) \sqrt {a+b-a \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {a+b-a x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 a b (a+b)^2 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sqrt {a+b-a \sin ^2(e+f x)} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \\ & = -\frac {(a-b) \sin (e+f x)}{3 b (a+b)^2 f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sin (e+f x)}{3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}-\frac {\left ((-a+b) \left (a+b-a \sin ^2(e+f x)\right )\right ) \text {Subst}\left (\int \frac {\sqrt {1-\frac {a x^2}{a+b}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 a b (a+b)^2 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1-\frac {a x^2}{a+b}}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \\ & = -\frac {(a-b) \sin (e+f x)}{3 b (a+b)^2 f \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {\sin (e+f x)}{3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}+\frac {(a-b) E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \left (a+b-a \sin ^2(e+f x)\right )}{3 a b (a+b)^2 f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {\operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{3 a (a+b) f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 22.18 (sec) , antiderivative size = 1204, normalized size of antiderivative = 3.77 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 e+2 f x))^{5/2} \sec ^5(e+f x) \left (-\frac {\left (-2 \sqrt {-\frac {1}{b}} (-a-a \cos (2 e+2 f x)) \left (2 a^2 (a+3 b+a \cos (2 e+2 f x))+b \left (2 b^2+3 b (a+2 b+a \cos (2 e+2 f x))-2 (a+2 b+a \cos (2 e+2 f x))^2\right )+a \left (4 b^2+5 b (a+2 b+a \cos (2 e+2 f x))-(a+2 b+a \cos (2 e+2 f x))^2\right )\right )+2 i \left (a^2+3 a b+2 b^2\right ) \sqrt {\frac {a-a \cos (2 e+2 f x)}{a+b}} (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {4-\frac {2 (a+2 b+a \cos (2 e+2 f x))}{b}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 e+2 f x)}}{\sqrt {2}}\right )|\frac {b}{a+b}\right )-i \left (2 a^2+5 a b+3 b^2\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {\frac {4 a+4 b-2 (a+2 b+a \cos (2 e+2 f x))}{a+b}} \sqrt {2-\frac {a+2 b+a \cos (2 e+2 f x)}{b}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 e+2 f x)}}{\sqrt {2}}\right ),\frac {b}{a+b}\right )\right ) \sin (2 e+2 f x)}{24 a \sqrt {-\frac {1}{b}} b^2 (a+b)^2 f \sqrt {\frac {(a-a \cos (2 e+2 f x)) (a+a \cos (2 e+2 f x))}{a^2}} (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {1-\cos ^2(2 e+2 f x)}}+\frac {\cos (2 (e+f x)) \left (-2 \sqrt {-\frac {1}{b}} (-a-a \cos (2 e+2 f x)) \left (4 b^4-b^2 (a+2 b+a \cos (2 e+2 f x))^2+2 a^3 (a+3 b+a \cos (2 e+2 f x))+a b \left (10 b^2+b (a+2 b+a \cos (2 e+2 f x))-(a+2 b+a \cos (2 e+2 f x))^2\right )+a^2 \left (8 b^2+3 b (a+2 b+a \cos (2 e+2 f x))-(a+2 b+a \cos (2 e+2 f x))^2\right )\right )+2 i \left (a^3+2 a^2 b+2 a b^2+b^3\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {\frac {4 a+4 b-2 (a+2 b+a \cos (2 e+2 f x))}{a+b}} \sqrt {2-\frac {a+2 b+a \cos (2 e+2 f x)}{b}} E\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 e+2 f x)}}{\sqrt {2}}\right )|\frac {b}{a+b}\right )-i a \left (2 a^2+3 a b+b^2\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {\frac {4 a+4 b-2 (a+2 b+a \cos (2 e+2 f x))}{a+b}} \sqrt {2-\frac {a+2 b+a \cos (2 e+2 f x)}{b}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a+2 b+a \cos (2 e+2 f x)}}{\sqrt {2}}\right ),\frac {b}{a+b}\right )\right ) \sec \left (2 \left (e+\frac {1}{2} (-2 e+\arccos (\cos (2 e+2 f x)))\right )\right ) \sin (2 e+2 f x)}{24 a^2 \sqrt {-\frac {1}{b}} b^2 (a+b)^2 f \sqrt {\frac {(a-a \cos (2 e+2 f x)) (a+a \cos (2 e+2 f x))}{a^2}} (a+2 b+a \cos (2 e+2 f x))^{3/2} \sqrt {1-\cos ^2(2 e+2 f x)}}\right )}{2 \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

[In]

Integrate[Sec[e + f*x]^3/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 2*b + a*Cos[2*e + 2*f*x])^(5/2)*Sec[e + f*x]^5*(-1/24*((-2*Sqrt[-b^(-1)]*(-a - a*Cos[2*e + 2*f*x])*(2*a^
2*(a + 3*b + a*Cos[2*e + 2*f*x]) + b*(2*b^2 + 3*b*(a + 2*b + a*Cos[2*e + 2*f*x]) - 2*(a + 2*b + a*Cos[2*e + 2*
f*x])^2) + a*(4*b^2 + 5*b*(a + 2*b + a*Cos[2*e + 2*f*x]) - (a + 2*b + a*Cos[2*e + 2*f*x])^2)) + (2*I)*(a^2 + 3
*a*b + 2*b^2)*Sqrt[(a - a*Cos[2*e + 2*f*x])/(a + b)]*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sqrt[4 - (2*(a + 2*b
+ a*Cos[2*e + 2*f*x]))/b]*EllipticE[I*ArcSinh[(Sqrt[-b^(-1)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]], b/
(a + b)] - I*(2*a^2 + 5*a*b + 3*b^2)*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sqrt[(4*a + 4*b - 2*(a + 2*b + a*Cos
[2*e + 2*f*x]))/(a + b)]*Sqrt[2 - (a + 2*b + a*Cos[2*e + 2*f*x])/b]*EllipticF[I*ArcSinh[(Sqrt[-b^(-1)]*Sqrt[a
+ 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]], b/(a + b)])*Sin[2*e + 2*f*x])/(a*Sqrt[-b^(-1)]*b^2*(a + b)^2*f*Sqrt[((a
- a*Cos[2*e + 2*f*x])*(a + a*Cos[2*e + 2*f*x]))/a^2]*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sqrt[1 - Cos[2*e +
2*f*x]^2]) + (Cos[2*(e + f*x)]*(-2*Sqrt[-b^(-1)]*(-a - a*Cos[2*e + 2*f*x])*(4*b^4 - b^2*(a + 2*b + a*Cos[2*e +
2*f*x])^2 + 2*a^3*(a + 3*b + a*Cos[2*e + 2*f*x]) + a*b*(10*b^2 + b*(a + 2*b + a*Cos[2*e + 2*f*x]) - (a + 2*b
+ a*Cos[2*e + 2*f*x])^2) + a^2*(8*b^2 + 3*b*(a + 2*b + a*Cos[2*e + 2*f*x]) - (a + 2*b + a*Cos[2*e + 2*f*x])^2)
) + (2*I)*(a^3 + 2*a^2*b + 2*a*b^2 + b^3)*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sqrt[(4*a + 4*b - 2*(a + 2*b +
a*Cos[2*e + 2*f*x]))/(a + b)]*Sqrt[2 - (a + 2*b + a*Cos[2*e + 2*f*x])/b]*EllipticE[I*ArcSinh[(Sqrt[-b^(-1)]*Sq
rt[a + 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]], b/(a + b)] - I*a*(2*a^2 + 3*a*b + b^2)*(a + 2*b + a*Cos[2*e + 2*f*
x])^(3/2)*Sqrt[(4*a + 4*b - 2*(a + 2*b + a*Cos[2*e + 2*f*x]))/(a + b)]*Sqrt[2 - (a + 2*b + a*Cos[2*e + 2*f*x])
/b]*EllipticF[I*ArcSinh[(Sqrt[-b^(-1)]*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]])/Sqrt[2]], b/(a + b)])*Sec[2*(e + (-
2*e + ArcCos[Cos[2*e + 2*f*x]])/2)]*Sin[2*e + 2*f*x])/(24*a^2*Sqrt[-b^(-1)]*b^2*(a + b)^2*f*Sqrt[((a - a*Cos[2
*e + 2*f*x])*(a + a*Cos[2*e + 2*f*x]))/a^2]*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sqrt[1 - Cos[2*e + 2*f*x]^2])
))/(2*(a + b*Sec[e + f*x]^2)^(5/2))

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.09 (sec) , antiderivative size = 12063, normalized size of antiderivative = 37.82

method	result	size
default	\(\text {Expression too large to display}\)	\(12063\)

[In]

int(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 1244, normalized size of antiderivative = 3.90 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/6*((2*((-I*a^4 + I*a^3*b)*cos(f*x + e)^4 - I*a^2*b^2 + I*a*b^3 - 2*(I*a^3*b - I*a^2*b^2)*cos(f*x + e)^2)*sqr
t(a)*sqrt((a*b + b^2)/a^2) - ((-I*a^4 - I*a^3*b + 2*I*a^2*b^2)*cos(f*x + e)^4 - I*a^2*b^2 - I*a*b^3 + 2*I*b^4
+ 2*(-I*a^3*b - I*a^2*b^2 + 2*I*a*b^3)*cos(f*x + e)^2)*sqrt(a))*sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*
elliptic_e(arcsin(sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*(cos(f*x + e) + I*sin(f*x + e))), (a^2 + 8*a*b
+ 8*b^2 + 4*(a^2 + 2*a*b)*sqrt((a*b + b^2)/a^2))/a^2) + (2*((I*a^4 - I*a^3*b)*cos(f*x + e)^4 + I*a^2*b^2 - I*
a*b^3 - 2*(-I*a^3*b + I*a^2*b^2)*cos(f*x + e)^2)*sqrt(a)*sqrt((a*b + b^2)/a^2) - ((I*a^4 + I*a^3*b - 2*I*a^2*b
^2)*cos(f*x + e)^4 + I*a^2*b^2 + I*a*b^3 - 2*I*b^4 + 2*(I*a^3*b + I*a^2*b^2 - 2*I*a*b^3)*cos(f*x + e)^2)*sqrt(
a))*sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*elliptic_e(arcsin(sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)
/a)*(cos(f*x + e) - I*sin(f*x + e))), (a^2 + 8*a*b + 8*b^2 + 4*(a^2 + 2*a*b)*sqrt((a*b + b^2)/a^2))/a^2) - 2*(
4*(I*a^3*b*cos(f*x + e)^4 + 2*I*a^2*b^2*cos(f*x + e)^2 + I*a*b^3)*sqrt(a)*sqrt((a*b + b^2)/a^2) + ((I*a^4 + 3*
I*a^3*b + 2*I*a^2*b^2)*cos(f*x + e)^4 + I*a^2*b^2 + 3*I*a*b^3 + 2*I*b^4 + 2*(I*a^3*b + 3*I*a^2*b^2 + 2*I*a*b^3
)*cos(f*x + e)^2)*sqrt(a))*sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*elliptic_f(arcsin(sqrt((2*a*sqrt((a*b
+ b^2)/a^2) - a - 2*b)/a)*(cos(f*x + e) + I*sin(f*x + e))), (a^2 + 8*a*b + 8*b^2 + 4*(a^2 + 2*a*b)*sqrt((a*b
+ b^2)/a^2))/a^2) - 2*(4*(-I*a^3*b*cos(f*x + e)^4 - 2*I*a^2*b^2*cos(f*x + e)^2 - I*a*b^3)*sqrt(a)*sqrt((a*b +
b^2)/a^2) + ((-I*a^4 - 3*I*a^3*b - 2*I*a^2*b^2)*cos(f*x + e)^4 - I*a^2*b^2 - 3*I*a*b^3 - 2*I*b^4 + 2*(-I*a^3*b
- 3*I*a^2*b^2 - 2*I*a*b^3)*cos(f*x + e)^2)*sqrt(a))*sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*elliptic_f(
arcsin(sqrt((2*a*sqrt((a*b + b^2)/a^2) - a - 2*b)/a)*(cos(f*x + e) - I*sin(f*x + e))), (a^2 + 8*a*b + 8*b^2 +
4*(a^2 + 2*a*b)*sqrt((a*b + b^2)/a^2))/a^2) + 2*(2*a^2*b^2*cos(f*x + e)^2 - (a^4 - a^3*b)*cos(f*x + e)^4)*sqrt
((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/((a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^4 + 2*(a^5
*b^2 + 2*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^2 + (a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f)

Sympy [F]

\[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\sec ^{3}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(sec(f*x+e)**3/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Integral(sec(e + f*x)**3/(a + b*sec(e + f*x)**2)**(5/2), x)

Maxima [F]

\[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{3}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^3/(b*sec(f*x + e)^2 + a)^(5/2), x)

Giac [F]

\[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sec \left (f x + e\right )^{3}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]

[In]

int(1/(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^(5/2)),x)

[Out]

int(1/(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^(5/2)), x)