3.3.55 \(\int \frac {1}{(a+b \sec (e+f x))^{3/2}} \, dx\) [255]

Optimal. Leaf size=347 \[ \frac {2 \cot (e+f x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{a \sqrt {a+b} f}-\frac {2 \cot (e+f x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{a \sqrt {a+b} f}-\frac {2 \sqrt {a+b} \cot (e+f x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{a^2 f}+\frac {2 b^2 \tan (e+f x)}{a \left (a^2-b^2\right ) f \sqrt {a+b \sec (e+f x)}} \]

[Out]

2*cot(f*x+e)*EllipticE((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(f*x+e))/(a+b))^(1/2)*
(-b*(1+sec(f*x+e))/(a-b))^(1/2)/a/f/(a+b)^(1/2)-2*cot(f*x+e)*EllipticF((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),((a+
b)/(a-b))^(1/2))*(b*(1-sec(f*x+e))/(a+b))^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/a/f/(a+b)^(1/2)-2*cot(f*x+e)*E
llipticPi((a+b*sec(f*x+e))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(f*x+e))/(a+b))
^(1/2)*(-b*(1+sec(f*x+e))/(a-b))^(1/2)/a^2/f+2*b^2*tan(f*x+e)/a/(a^2-b^2)/f/(a+b*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 347, normalized size of antiderivative = 1.00, 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 = {3870, 4143, 4006, 3869, 3917, 4089} \begin {gather*} -\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};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 f}+\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 \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 (\text {ArcSin}\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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a f \sqrt {a+b}} \end {gather*}

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 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*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)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3870

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 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(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)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4006

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4143

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
]

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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 6.11, size = 1249, normalized size = 3.60 \begin {gather*} \frac {(b+a \cos (e+f x))^2 \sec ^2(e+f x) \left (\frac {2 b \sin (e+f x)}{a \left (-a^2+b^2\right )}+\frac {2 b^2 \sin (e+f x)}{a \left (a^2-b^2\right ) (b+a \cos (e+f x))}\right )}{f (a+b \sec (e+f x))^{3/2}}+\frac {2 (b+a \cos (e+f x))^{3/2} \sec ^{\frac {3}{2}}(e+f x) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}} \left (a b \sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )+b^2 \sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )-2 a b \sqrt {\frac {-a+b}{a+b}} \tan ^3\left (\frac {1}{2} (e+f x)\right )+a b \sqrt {\frac {-a+b}{a+b}} \tan ^5\left (\frac {1}{2} (e+f x)\right )-b^2 \sqrt {\frac {-a+b}{a+b}} \tan ^5\left (\frac {1}{2} (e+f x)\right )-2 i a^2 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}+2 i b^2 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}-2 i a^2 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}+2 i b^2 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}-i (a-b) b E\left (i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}+i \left (a^2+a b-2 b^2\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (e+f x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+b \tan ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}}\right )}{a \sqrt {\frac {-a+b}{a+b}} \left (a^2-b^2\right ) f (a+b \sec (e+f x))^{3/2} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}{1-\tan ^2\left (\frac {1}{2} (e+f x)\right )}} \left (a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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] Leaf count of result is larger than twice the leaf count of optimal. \(1208\) vs. \(2(318)=636\).
time = 0.31, size = 1209, normalized size = 3.48

method	result	size
default	\(\text {Expression too large to display}\)	\(1209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*4^(1/2)*((a*cos(f*x+e)+b)/cos(f*x+e))^(1/2)*(EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*cos
(f*x+e)*a^2*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)+Ellipti
cF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*cos(f*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)
+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)*a*b-EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*cos(f
*x+e)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)*a*b-EllipticE
((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*cos(f*x+e)*b^2*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+
e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)-2*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*c
os(f*x+e)*a^2*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)+2*(co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*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))*b^2+(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(co
s(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*a^2*sin(f*x+e)+(cos(f*x+e)/
(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))/sin(f*x+e),((a-b
)/(a+b))^(1/2))*a*b*sin(f*x+e)-EllipticE((-1+cos(f*x+e))/sin(f*x+e),((a-b)/(a+b))^(1/2))*(cos(f*x+e)/(cos(f*x+
e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)*a*b-EllipticE((-1+cos(f*x+e))/sin(f*x+e)
,((a-b)/(a+b))^(1/2))*b^2*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(
f*x+e)-2*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*a^2*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(
(a*cos(f*x+e)+b)/(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)+2*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*((a*cos(f*x+e)+b)/
(cos(f*x+e)+1)/(a+b))^(1/2)*sin(f*x+e)*EllipticPi((-1+cos(f*x+e))/sin(f*x+e),-1,((a-b)/(a+b))^(1/2))*b^2+cos(f
*x+e)^2*a*b-cos(f*x+e)^2*b^2-cos(f*x+e)*a*b+cos(f*x+e)*b^2)/(a*cos(f*x+e)+b)/sin(f*x+e)/a/(a+b)/(a-b)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cos(e + f*x))^(3/2),x)

[Out]

int(1/(a + b/cos(e + f*x))^(3/2), x)

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