3.255 \(\int \frac{1}{(a+b \sec (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=347 \[ -\frac{2 \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{a f \sqrt{a+b}}+\frac{2 b^2 \tan (e+f x)}{a f \left (a^2-b^2\right ) \sqrt{a+b \sec (e+f x)}}-\frac{2 \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^2 f}+\frac{2 \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a f \sqrt{a+b}} \]
[Out]
(2*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e +
f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(a*Sqrt[a + b]*f) - (2*Cot[e + f*x]*EllipticF[ArcSin[
Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Se
c[e + f*x]))/(a - b))])/(a*Sqrt[a + b]*f) - (2*Sqrt[a + b]*Cot[e + f*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a +
b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x
]))/(a - b))])/(a^2*f) + (2*b^2*Tan[e + f*x])/(a*(a^2 - b^2)*f*Sqrt[a + b*Sec[e + f*x]])
________________________________________________________________________________________
Rubi [A] time = 0.332869, antiderivative size = 347, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used =
{3785, 4058, 3921, 3784, 3832, 4004} \[ \frac{2 b^2 \tan (e+f x)}{a f \left (a^2-b^2\right ) \sqrt{a+b \sec (e+f x)}}-\frac{2 \sqrt{a+b} \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^2 f}-\frac{2 \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a f \sqrt{a+b}}+\frac{2 \cot (e+f x) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (\sec (e+f x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a f \sqrt{a+b}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sec[e + f*x])^(-3/2),x]
[Out]
(2*Cot[e + f*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e +
f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x]))/(a - b))])/(a*Sqrt[a + b]*f) - (2*Cot[e + f*x]*EllipticF[ArcSin[
Sqrt[a + b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Se
c[e + f*x]))/(a - b))])/(a*Sqrt[a + b]*f) - (2*Sqrt[a + b]*Cot[e + f*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a +
b*Sec[e + f*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[e + f*x
]))/(a - b))])/(a^2*f) + (2*b^2*Tan[e + f*x])/(a*(a^2 - b^2)*f*Sqrt[a + b*Sec[e + f*x]])
Rule 3785
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Rule 4058
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[(Csc[e + f*
x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]
Rule 3921
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Rule 3784
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4004
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rubi steps
\begin{align*} \int \frac{1}{(a+b \sec (e+f x))^{3/2}} \, dx &=\frac{2 b^2 \tan (e+f x)}{a \left (a^2-b^2\right ) f \sqrt{a+b \sec (e+f x)}}-\frac{2 \int \frac{\frac{1}{2} \left (-a^2+b^2\right )+\frac{1}{2} a b \sec (e+f x)+\frac{1}{2} b^2 \sec ^2(e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 b^2 \tan (e+f x)}{a \left (a^2-b^2\right ) f \sqrt{a+b \sec (e+f x)}}-\frac{2 \int \frac{\frac{1}{2} \left (-a^2+b^2\right )+\left (\frac{a b}{2}-\frac{b^2}{2}\right ) \sec (e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx}{a \left (a^2-b^2\right )}-\frac{b^2 \int \frac{\sec (e+f x) (1+\sec (e+f x))}{\sqrt{a+b \sec (e+f x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 \cot (e+f x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{a \sqrt{a+b} f}+\frac{2 b^2 \tan (e+f x)}{a \left (a^2-b^2\right ) f \sqrt{a+b \sec (e+f x)}}+\frac{\int \frac{1}{\sqrt{a+b \sec (e+f x)}} \, dx}{a}-\frac{b \int \frac{\sec (e+f x)}{\sqrt{a+b \sec (e+f x)}} \, dx}{a (a+b)}\\ &=\frac{2 \cot (e+f x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{a \sqrt{a+b} f}-\frac{2 \cot (e+f x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{a \sqrt{a+b} f}-\frac{2 \sqrt{a+b} \cot (e+f x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (e+f x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (e+f x))}{a+b}} \sqrt{-\frac{b (1+\sec (e+f x))}{a-b}}}{a^2 f}+\frac{2 b^2 \tan (e+f x)}{a \left (a^2-b^2\right ) f \sqrt{a+b \sec (e+f x)}}\\ \end{align*}
Mathematica [C] time = 6.14205, size = 1249, normalized size = 3.6 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Sec[e + f*x])^(-3/2),x]
[Out]
((b + a*Cos[e + f*x])^2*Sec[e + f*x]^2*((2*b*Sin[e + f*x])/(a*(-a^2 + b^2)) + (2*b^2*Sin[e + f*x])/(a*(a^2 - b
^2)*(b + a*Cos[e + f*x]))))/(f*(a + b*Sec[e + f*x])^(3/2)) + (2*(b + a*Cos[e + f*x])^(3/2)*Sec[e + f*x]^(3/2)*
Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(1 + Tan[(e + f*x)/2]^2)]*(a*b*Sqrt[(-a + b)/(a + b
)]*Tan[(e + f*x)/2] + b^2*Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2] - 2*a*b*Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)
/2]^3 + a*b*Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]^5 - b^2*Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]^5 - (2*I)*
a^2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1
- Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + (2*I)*b^2*Ellipti
cPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e +
f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - (2*I)*a^2*EllipticPi[-((a +
b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Tan[(e + f*x)/2]^2*Sqrt[1 -
Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + (2*I)*b^2*EllipticPi
[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Tan[(e + f*x)/2]^2*S
qrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] - I*(a - b)*b*
EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(1
+ Tan[(e + f*x)/2]^2)*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)] + I*(a^2 + a*b - 2*
b^2)*EllipticF[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(e + f*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(e + f*x)/2]^
2]*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(a + b - a*Tan[(e + f*x)/2]^2 + b*Tan[(e + f*x)/2]^2)/(a + b)]))/(a*Sqrt[(-a
+ b)/(a + b)]*(a^2 - b^2)*f*(a + b*Sec[e + f*x])^(3/2)*(-1 + Tan[(e + f*x)/2]^2)*Sqrt[(1 + Tan[(e + f*x)/2]^2)
/(1 - Tan[(e + f*x)/2]^2)]*(a*(-1 + Tan[(e + f*x)/2]^2) - b*(1 + Tan[(e + f*x)/2]^2)))
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Maple [B] time = 0.281, size = 1209, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*sec(f*x+e))^(3/2),x)
[Out]
1/f/a/(a+b)/(a-b)*4^(1/2)*(1/cos(f*x+e)*(a*cos(f*x+e)+b))^(1/2)*(cos(f*x+e)*EllipticF((-1+cos(f*x+e))/sin(f*x+
e),((a-b)/(a+b))^(1/2))*a^2*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*
sin(f*x+e)+cos(f*x+e)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*(cos(f*x+e)/(1+cos(f*x+e)))^(1
/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*a*b-cos(f*x+e)*EllipticE((-1+cos(f*x+e))/sin(f*
x+e),((a-b)/(a+b))^(1/2))*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*si
n(f*x+e)*a*b-cos(f*x+e)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*b^2*(cos(f*x+e)/(1+cos(f*x+e
)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-2*cos(f*x+e)*EllipticPi((-1+cos(f*x+e))/s
in(f*x+e),-1,((a-b)/(a+b))^(1/2))*a^2*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e
)))^(1/2)*sin(f*x+e)+2*sin(f*x+e)*cos(f*x+e)*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*(co
s(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*b^2+(cos(f*x+e)/(1+cos(f*x+e)))
^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2
))*a^2*sin(f*x+e)+EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*
(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)*a*b-(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*c
os(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a*b*sin(f*x+e)-(c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*EllipticE((-1+cos(f*x+e))/sin(
f*x+e),((a-b)/(a+b))^(1/2))*b^2*sin(f*x+e)-2*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f*x+e)+b)/(1+co
s(f*x+e)))^(1/2)*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*a^2*sin(f*x+e)+2*EllipticPi((-1
+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*sin(f*x+e)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(1/(a+b)*(a*cos(f
*x+e)+b)/(1+cos(f*x+e)))^(1/2)*b^2+cos(f*x+e)^2*a*b-cos(f*x+e)^2*b^2-a*b*cos(f*x+e)+b^2*cos(f*x+e))/(a*cos(f*x
+e)+b)/sin(f*x+e)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e) + a)^(-3/2), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (f x + e\right ) + a}}{b^{2} \sec \left (f x + e\right )^{2} + 2 \, a b \sec \left (f x + e\right ) + a^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(f*x + e) + a)/(b^2*sec(f*x + e)^2 + 2*a*b*sec(f*x + e) + a^2), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e))**(3/2),x)
[Out]
Integral((a + b*sec(e + f*x))**(-3/2), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sec(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e) + a)^(-3/2), x)