\(\int \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx\) [229]

   Optimal result
   Rubi [A] (verified)
   Mathematica [F]
   Maple [C] (warning: unable to verify)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 288 \[ \int \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{3 b f}-\frac {(a+2 b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )}}{3 b f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {2 (a+b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right ) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}{3 f \left (a+b-a \sin ^2(e+f x)\right )}+\frac {\sec (e+f x) \sqrt {\sec ^2(e+f x) \left (a+b-a \sin ^2(e+f x)\right )} \tan (e+f x)}{3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 288, 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} \[ \int \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {2 (a+b) \sqrt {\cos ^2(e+f x)} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),\frac {a}{a+b}\right )}{3 f \left (-a \sin ^2(e+f x)+a+b\right )}-\frac {(a+2 b) \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\arcsin (\sin (e+f x))\left |\frac {a}{a+b}\right .\right )}{3 b f \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}+\frac {(a+2 b) \sin (e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{3 b f}+\frac {\tan (e+f x) \sec (e+f x) \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{3 f} \]

[In]

Int[Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

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

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

Mathematica [F]

\[ \int \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx \]

[In]

Integrate[Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

Integrate[Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2], x]

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.80 (sec) , antiderivative size = 6342, normalized size of antiderivative = 22.02

method result size
default \(\text {Expression too large to display}\) \(6342\)

[In]

int(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 782, normalized size of antiderivative = 2.72 \[ \int \sec ^3(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\frac {{\left (2 \, {\left (i \, a^{2} + 2 i \, a b\right )} \sqrt {a} \sqrt {\frac {a b + b^{2}}{a^{2}}} \cos \left (f x + e\right )^{2} - {\left (i \, a^{2} + 4 i \, a b + 4 i \, b^{2}\right )} \sqrt {a} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} E(\arcsin \left (\sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} + 2 \, a b\right )} \sqrt {\frac {a b + b^{2}}{a^{2}}}}{a^{2}}) + {\left (2 \, {\left (-i \, a^{2} - 2 i \, a b\right )} \sqrt {a} \sqrt {\frac {a b + b^{2}}{a^{2}}} \cos \left (f x + e\right )^{2} - {\left (-i \, a^{2} - 4 i \, a b - 4 i \, b^{2}\right )} \sqrt {a} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} E(\arcsin \left (\sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} + 2 \, a b\right )} \sqrt {\frac {a b + b^{2}}{a^{2}}}}{a^{2}}) - 2 \, {\left (2 i \, a^{\frac {3}{2}} b \sqrt {\frac {a b + b^{2}}{a^{2}}} \cos \left (f x + e\right )^{2} + {\left (-i \, a^{2} - 3 i \, a b - 2 i \, b^{2}\right )} \sqrt {a} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} F(\arcsin \left (\sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} + 2 \, a b\right )} \sqrt {\frac {a b + b^{2}}{a^{2}}}}{a^{2}}) - 2 \, {\left (-2 i \, a^{\frac {3}{2}} b \sqrt {\frac {a b + b^{2}}{a^{2}}} \cos \left (f x + e\right )^{2} + {\left (i \, a^{2} + 3 i \, a b + 2 i \, b^{2}\right )} \sqrt {a} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} F(\arcsin \left (\sqrt {\frac {2 \, a \sqrt {\frac {a b + b^{2}}{a^{2}}} - a - 2 \, b}{a}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} + 2 \, a b\right )} \sqrt {\frac {a b + b^{2}}{a^{2}}}}{a^{2}}) + 2 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b\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 )}{6 \, a b f \cos \left (f x + e\right )^{2}} \]

[In]

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*((2*(I*a^2 + 2*I*a*b)*sqrt(a)*sqrt((a*b + b^2)/a^2)*cos(f*x + e)^2 - (I*a^2 + 4*I*a*b + 4*I*b^2)*sqrt(a)*c
os(f*x + e)^2)*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^2 - 2*I*a*b)*sqrt(a)*sqrt((a*b + b^2)/a^2)*cos(f*x + e)^2 - (-I*a^2 - 4*I*a*b - 4*I*b^2)*sqrt
(a)*cos(f*x + e)^2)*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*(2*I*a^(3/2)*b*sqrt((a*b + b^2)/a^2)*cos(f*x + e)^2 + (-I*a^2 - 3*I*a*b - 2*I*b^2)*sqrt(a)*cos(
f*x + e)^2)*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*I*a^(3/2)*b*sqrt((a*b + b^2)/a^2)*cos(f*x + e)^2 + (I*a^2 + 3*I*a*b + 2*I*b^2)*sqrt(a)*cos(f*x + e)
^2)*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*(
(a^2 + 2*a*b)*cos(f*x + e)^2 + a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a*b*f*cos(f*x +
 e)^2)

Sympy [F]

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

[In]

integrate(sec(f*x+e)**3*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

Maxima [F]

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

[In]

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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