\(\int \frac {\sec ^2(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [362]

Optimal result

Integrand size = 25, antiderivative size = 240 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a-b) b \cos (e+f x) \sin (e+f x)}{a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a-b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a (a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3271, 425, 541, 538, 437, 435, 432, 430} \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{a f (a+b)^2 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan (e+f x)}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}}-\frac {b (a-b) \sin (e+f x) \cos (e+f x)}{a f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

-(((a - b)*b*Cos[e + f*x]*Sin[e + f*x])/(a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^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*(a + b)^2*f*Sqrt[1 +
(b*Sin[e + f*x]^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])/((a + b)*f*Sqrt[a + b*Sin[e + f*x]^2]) + Tan[e + f*x]/((a + b)*f*Sqrt[a + b*Sin[e + f*x
]^2])

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[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 {1}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {b+b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f} \\ & = -\frac {(a-b) b \cos (e+f x) \sin (e+f x)}{a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-2 a b+(a-b) b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a (a+b)^2 f} \\ & = -\frac {(a-b) b \cos (e+f x) \sin (e+f x)}{a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left ((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 )}{a (a+b)^2 f}+\frac {\left (\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 )}{(a+b) f} \\ & = -\frac {(a-b) b \cos (e+f x) \sin (e+f x)}{a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left ((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 )}{a (a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (\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 )}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \\ & = -\frac {(a-b) b \cos (e+f x) \sin (e+f x)}{a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(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)}}{a (a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\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}}}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.70 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\sqrt {2} a (a-b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} a (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\left (2 a^2+a b+b^2+b (-a+b) \cos (2 (e+f x))\right ) \tan (e+f x)}{\sqrt {2} a (a+b)^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

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

[Out]

(-(Sqrt[2]*a*(a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)]) + Sqrt[2]*a*(a + b)*Sq
rt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + (2*a^2 + a*b + b^2 + b*(-a + b)*Cos[2*(e + f
*x)])*Tan[e + f*x])/(Sqrt[2]*a*(a + b)^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

Maple [A] (verified)

Time = 2.98 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+a b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b -b \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a +b^{2} \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\sin \left (f x +e \right ) a^{2}+a b \sin \left (f x +e \right )\right )}{\left (a +b \right )^{2} \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, a \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(360\)

[In]

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

[Out]

(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*((cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(s
in(f*x+e),(-1/a*b)^(1/2))*a^2+a*b*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),
(-1/a*b)^(1/2))-(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^
2+(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b-b*cos(f*x+e)
^2*sin(f*x+e)*a+b^2*cos(f*x+e)^2*sin(f*x+e)+sin(f*x+e)*a^2+a*b*sin(f*x+e))/(a+b)^2/(-(a+b*sin(f*x+e)^2)*(sin(f
*x+e)-1)*(1+sin(f*x+e)))^(1/2)/a/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 = 0.17 (sec) , antiderivative size = 1079, normalized size of antiderivative = 4.50 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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