\(\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx\) [567]
Optimal result
Integrand size = 14, antiderivative size = 347 \[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {2 \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 b^2 \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}
\]
[Out]
2*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*
(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/d/(a+b)^(1/2)-2*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+
b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/d/(a+b)^(1/2)-2*cot(d*x+c)*E
llipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))
^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d+2*b^2*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.40 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3870, 4143, 4006, 3869, 3917,
4089} \[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 b^2 \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}+\frac {2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}
\]
[In]
Int[(a + b*Sec[c + d*x])^(-3/2),x]
[Out]
(2*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c +
d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*Sqrt[a + b]*d) - (2*Cot[c + d*x]*EllipticF[ArcSin[
Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Se
c[c + d*x]))/(a - b))])/(a*Sqrt[a + b]*d) - (2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a +
b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x
]))/(a - b))])/(a^2*d) + (2*b^2*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*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*}
\text {integral}& = \frac {2 b^2 \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-a^2+b^2\right )+\frac {1}{2} a b \sec (c+d x)+\frac {1}{2} b^2 \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {2 b^2 \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d 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 (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}-\frac {b^2 \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )} \\ & = \frac {2 \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}+\frac {2 b^2 \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{a}-\frac {b \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{a (a+b)} \\ & = \frac {2 \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {2 \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 b^2 \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \\
\end{align*}
Mathematica [B] (verified)
Leaf count is larger than twice the leaf count of optimal. \(972\) vs. \(2(347)=694\).
Time = 16.92 (sec) , antiderivative size = 972, normalized size of antiderivative = 2.80
\[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \left (\frac {2 b \sin (c+d x)}{a \left (-a^2+b^2\right )}+\frac {2 b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{3/2}}+\frac {2 (b+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a b \tan \left (\frac {1}{2} (c+d x)\right )+b^2 \tan \left (\frac {1}{2} (c+d x)\right )-2 a b \tan ^3\left (\frac {1}{2} (c+d x)\right )+a b \tan ^5\left (\frac {1}{2} (c+d x)\right )-b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+2 a^2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 b^2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+2 a^2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 b^2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2} \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )}
\]
[In]
Integrate[(a + b*Sec[c + d*x])^(-3/2),x]
[Out]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*b*Sin[c + d*x])/(a*(-a^2 + b^2)) + (2*b^2*Sin[c + d*x])/(a*(a^2 - b
^2)*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(3/2)) + (2*(b + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*
Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(a*b*Tan[(c + d*x)/2] + b
^2*Tan[(c + d*x)/2] - 2*a*b*Tan[(c + d*x)/2]^3 + a*b*Tan[(c + d*x)/2]^5 - b^2*Tan[(c + d*x)/2]^5 + 2*a^2*Ellip
ticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x
)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt
[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a^2*EllipticP
i[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b -
a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b
)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x
)/2]^2)/(a + b)] + b*(a + b)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]
*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - a*(a + b)*Elli
pticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a
+ b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)*(-1
+ Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2]^2)]*(a*(-1 + Tan[(c + d*x)/2]^2) -
b*(1 + Tan[(c + d*x)/2]^2)))
Maple [B] (warning: unable to verify)
Leaf count of result is larger than twice the leaf count of optimal. \(1798\) vs. \(2(318)=636\).
Time = 6.36 (sec) , antiderivative size = 1799, normalized size of antiderivative =
5.18
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| method | result | size |
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\(\text {Expression too large to display}\) |
\(1799\) |
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[In]
int(1/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
[Out]
2/d/a/(a+b)/(a-b)*(-((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*c
sc(d*x+c)^2-1))^(1/2)*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x
+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2-((a*(1-cos(d*x+
c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos(d*x+
c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/
2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b+((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))
^2*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1
-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),
((a-b)/(a+b))^(1/2))*a*b+((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)*((1-cos(d*x+c)
)^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-co
s(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^2+2*((a*(1-c
os(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-c
os(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+
b))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^2-2*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(
1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^
(1/2)*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*EllipticPi(cot(d*x+
c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*b^2-((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c))^4*b*csc(d*x+c)^4-2*a
*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)*a*b*(1-cos(d*x+c))^3*csc(d*x+c)^3+((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(
1-cos(d*x+c))^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)*b^2*(1-cos(d*x+c))^3*csc(d*x+c)^3+
((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c))^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)*
a*b*(-cot(d*x+c)+csc(d*x+c))-((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c))^4*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c
))^2*csc(d*x+c)^2+a+b)^(1/2)*b^2*(-cot(d*x+c)+csc(d*x+c)))*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^
2*csc(d*x+c)^2-a-b)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)/((1-cos(d*x+c))^4*a*csc(d*x+c)^4-(1-cos(d*x+c))^4
*b*csc(d*x+c)^4-2*a*(1-cos(d*x+c))^2*csc(d*x+c)^2+a+b)^(1/2)/(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))
^2*csc(d*x+c)^2-a-b)
Fricas [F]
\[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(1/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(d*x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
Sympy [F]
\[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx
\]
[In]
integrate(1/(a+b*sec(d*x+c))**(3/2),x)
[Out]
Integral((a + b*sec(c + d*x))**(-3/2), x)
Maxima [F]
\[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(1/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sec(d*x + c) + a)^(-3/2), x)
Giac [F]
\[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate(1/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((b*sec(d*x + c) + a)^(-3/2), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {1}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x
\]
[In]
int(1/(a + b/cos(c + d*x))^(3/2),x)
[Out]
int(1/(a + b/cos(c + d*x))^(3/2), x)