\(\int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [333]

Optimal result

Integrand size = 25, antiderivative size = 196 \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {(2 a+b) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(2 a+b) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{3 (a+b) f}+\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3271, 423, 541, 538, 437, 435, 432, 430} \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {2 a \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 f (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {(2 a+b) \tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f (a+b)}+\frac {\tan (e+f x) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f} \]

[In]

Int[Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-1/3*((2*a + b)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f
*x]^2])/((a + b)*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (2*a*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]],
-(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(3*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((2*a + b)*Sqrt[a + b
*Sin[e + f*x]^2]*Tan[e + f*x])/(3*(a + b)*f) + (Sec[e + f*x]^2*Sqrt[a + b*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 3271

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

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

Mathematica [A] (verified)

Time = 2.19 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.95 \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {-2 a (2 a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+4 a (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\frac {\left (\left (8 a^2-4 b^2\right ) \cos (2 (e+f x))+(2 a+b) (8 a+5 b-b \cos (4 (e+f x)))\right ) \sec ^2(e+f x) \tan (e+f x)}{2 \sqrt {2}}}{6 (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

Integrate[Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(-2*a*(2*a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 4*a*(a + b)*Sqrt[(2*a + b
- b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + (((8*a^2 - 4*b^2)*Cos[2*(e + f*x)] + (2*a + b)*(8*a + 5*
b - b*Cos[4*(e + f*x)]))*Sec[e + f*x]^2*Tan[e + f*x])/(2*Sqrt[2]))/(6*(a + b)*f*Sqrt[2*a + b - b*Cos[2*(e + f*
x)]])

Maple [A] (verified)

Time = 3.69 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.88

method	result	size
default	\(\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (2 a +b \right ) \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-2 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, a \left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, a \left (2 F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +2 F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )}{3 \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (a +b \right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(368\)

[In]

int(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b*(2*a+b)*cos(f*x+e)^4*sin(f*x+e)-2*(-b*cos(f*x+e)^4+(a+b)*cos
(f*x+e)^2)^(1/2)*a*(a+b)*cos(f*x+e)^2*sin(f*x+e)-(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(
f*x+e)^2)^(1/2)*(cos(f*x+e)^2)^(1/2)*a*(2*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a+2*EllipticF(sin(f*x+e),(-1/a*
b)^(1/2))*b-2*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a-EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b)*cos(f*x+e)^2-(-b*
cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a^2+2*a*b+b^2)*sin(f*x+e))/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(
f*x+e)))^(1/2)/(a+b)/(sin(f*x+e)-1)/(1+sin(f*x+e))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 789, normalized size of antiderivative = 4.03 \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {{\left (2 \, {\left (-2 i \, a b - i \, b^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right )^{3} - {\left (4 i \, a^{2} + 4 i \, a b + i \, b^{2}\right )} \sqrt {-b} \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left (2 i \, a b + i \, b^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right )^{3} - {\left (-4 i \, a^{2} - 4 i \, a b - i \, b^{2}\right )} \sqrt {-b} \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - 2 \, {\left (2 \, {\left (-i \, a b - i \, b^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right )^{3} + {\left (-2 i \, a^{2} - i \, a b\right )} \sqrt {-b} \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - 2 \, {\left (2 \, {\left (i \, a b + i \, b^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right )^{3} + {\left (2 i \, a^{2} + i \, a b\right )} \sqrt {-b} \cos \left (f x + e\right )^{3}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + 2 \, {\left ({\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{6 \, {\left (a b + b^{2}\right )} f \cos \left (f x + e\right )^{3}} \]

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sec ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate(sec(f*x+e)**4*(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*sec(e + f*x)**4, x)

Maxima [F]

\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sec(f*x + e)^4, x)

Giac [F]

\[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{4} \,d x } \]

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sec(f*x + e)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \frac {\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}}{{\cos \left (e+f\,x\right )}^4} \,d x \]

[In]

int((a + b*sin(e + f*x)^2)^(1/2)/cos(e + f*x)^4,x)

[Out]

int((a + b*sin(e + f*x)^2)^(1/2)/cos(e + f*x)^4, x)