3.2.69 \(\int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [169]

Optimal. Leaf size=166 \[ \frac {2 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} c f}-\frac {\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) f}+\frac {2 d^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} c (c-d) \sqrt {c+d} f} \]

[Out]

2*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c/f/a^(1/2)-arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)/(a+a*sec
(f*x+e))^(1/2))*2^(1/2)/(c-d)/f/a^(1/2)+2*d^(3/2)*arctan(a^(1/2)*d^(1/2)*tan(f*x+e)/(c+d)^(1/2)/(a+a*sec(f*x+e
))^(1/2))/c/(c-d)/f/a^(1/2)/(c+d)^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4014, 4005, 3859, 209, 3880, 4052, 211} \begin {gather*} \frac {2 d^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}-\frac {\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}+\frac {2 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a} c f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(Sqrt[a]*c*f) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[e + f
*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*(c - d)*f) + (2*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Tan[e + f*x
])/(Sqrt[c + d]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*c*(c - d)*Sqrt[c + d]*f)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, 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] && EqQ[a^2 - b^2, 0]

Rule 4014

Int[1/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Dist[1
/(c*(b*c - a*d)), Int[(b*c - a*d - b*d*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d^2/(c*(b*c - a*d
)), Int[Csc[e + f*x]*(Sqrt[a + b*Csc[e + f*x]]/(c + d*Csc[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &
& NeQ[b*c - a*d, 0] && (EqQ[a^2 - b^2, 0] || EqQ[c^2 - d^2, 0])

Rule 4052

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)
), x_Symbol] :> Dist[-2*(b/f), Subst[Int[1/(b*c + a*d + d*x^2), x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]
])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=\frac {\int \frac {a c-a d-a d \sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx}{a c (c-d)}+\frac {d^2 \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{a c (c-d)}\\ &=\frac {\int \sqrt {a+a \sec (e+f x)} \, dx}{a c}-\frac {\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx}{c-d}-\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {1}{a c+a d+d x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c (c-d) f}\\ &=\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} c (c-d) \sqrt {c+d} f}-\frac {2 \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c f}+\frac {2 \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{(c-d) f}\\ &=\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} c f}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) f}+\frac {2 d^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} c (c-d) \sqrt {c+d} f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 39.28, size = 431980, normalized size = 2602.29 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

Result too large to show

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(662\) vs. \(2(137)=274\).
time = 4.65, size = 663, normalized size = 3.99

method	result	size
default	\(-\frac {\left (2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) c -2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) d +2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \sqrt {\frac {d}{c -d}}\, \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) c -\sqrt {2}\, \ln \left (\frac {-2 \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )+2 \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )+2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )+2 c \sin \left (f x +e \right )-2 d \sin \left (f x +e \right )-2 \sqrt {\left (c +d \right ) \left (c -d \right )}}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )+c \cos \left (f x +e \right )-d \cos \left (f x +e \right )-c +d}\right ) d^{2}+\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )-2 \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )-2 c \sin \left (f x +e \right )+2 d \sin \left (f x +e \right )+2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )-2 \sqrt {\left (c +d \right ) \left (c -d \right )}}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d}\right ) d^{2}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{2 f \sqrt {\frac {d}{c -d}}\, \left (c -d \right ) c \sqrt {\left (c +d \right ) \left (c -d \right )}\, a}\)	\(663\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(2*((c+d)*(c-d))^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x
+e)/cos(f*x+e)*2^(1/2))*c-2*((c+d)*(c-d))^(1/2)*2^(1/2)*(d/(c-d))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)
+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*d+2*((c+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)*ln((sin(f*x+e)*(-2*cos(f*x+e
)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*c-2^(1/2)*ln(2*(-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos
(f*x+e)+1))^(1/2)*c*sin(f*x+e)+2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)+((c+d
)*(c-d))^(1/2)*cos(f*x+e)+c*sin(f*x+e)-d*sin(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)+c*cos
(f*x+e)-d*cos(f*x+e)-c+d))*d^2+2^(1/2)*ln(2*(2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*c*si
n(f*x+e)-2^(1/2)*(d/(c-d))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*d*sin(f*x+e)-c*sin(f*x+e)+d*sin(f*x+e)+(
(c+d)*(c-d))^(1/2)*cos(f*x+e)-((c+d)*(c-d))^(1/2))/(((c+d)*(c-d))^(1/2)*sin(f*x+e)-c*cos(f*x+e)+d*cos(f*x+e)+c
-d))*d^2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)/(d/(c-d))^(1/2)/(c-d)/c/((c
+d)*(c-d))^(1/2)/a

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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/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(a*sec(f*x + e) + a)*(d*sec(f*x + e) + c)), x)

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Fricas [A]
time = 26.87, size = 1120, normalized size = 6.75 \begin {gather*} \left [-\frac {\sqrt {2} a c \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 2 \, a d \sqrt {-\frac {d}{a c + a d}} \log \left (\frac {2 \, {\left (c + d\right )} \sqrt {-\frac {d}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c + 2 \, d\right )} \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) - d}{c \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) + d}\right ) + 2 \, \sqrt {-a} {\left (c - d\right )} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{2 \, {\left (a c^{2} - a c d\right )} f}, -\frac {\sqrt {2} a c \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 4 \, a d \sqrt {\frac {d}{a c + a d}} \arctan \left (\frac {{\left (c + d\right )} \sqrt {\frac {d}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{d \sin \left (f x + e\right )}\right ) + 2 \, \sqrt {-a} {\left (c - d\right )} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right )}{2 \, {\left (a c^{2} - a c d\right )} f}, -\frac {a d \sqrt {-\frac {d}{a c + a d}} \log \left (\frac {2 \, {\left (c + d\right )} \sqrt {-\frac {d}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c + 2 \, d\right )} \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) - d}{c \cos \left (f x + e\right )^{2} + {\left (c + d\right )} \cos \left (f x + e\right ) + d}\right ) - \sqrt {2} \sqrt {a} c \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + 2 \, \sqrt {a} {\left (c - d\right )} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right )}{{\left (a c^{2} - a c d\right )} f}, -\frac {2 \, a d \sqrt {\frac {d}{a c + a d}} \arctan \left (\frac {{\left (c + d\right )} \sqrt {\frac {d}{a c + a d}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{d \sin \left (f x + e\right )}\right ) - \sqrt {2} \sqrt {a} c \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + 2 \, \sqrt {a} {\left (c - d\right )} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right )}{{\left (a c^{2} - a c d\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*a*c*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*
sin(f*x + e) - 3*cos(f*x + e)^2 - 2*cos(f*x + e) + 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 2*a*d*sqrt(-d/(
a*c + a*d))*log((2*(c + d)*sqrt(-d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x +
e) + (c + 2*d)*cos(f*x + e)^2 + (c + d)*cos(f*x + e) - d)/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) + 2*
sqrt(-a)*(c - d)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin
(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)))/((a*c^2 - a*c*d)*f), -1/2*(sqrt(2)*a*c*sqrt(-1/a)*log(-(2
*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*x + e) - 3*cos(f*x + e)^2 - 2*c
os(f*x + e) + 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 4*a*d*sqrt(d/(a*c + a*d))*arctan((c + d)*sqrt(d/(a*c
+ a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(d*sin(f*x + e))) + 2*sqrt(-a)*(c - d)*log((2*a*
cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e)
- a)/(cos(f*x + e) + 1)))/((a*c^2 - a*c*d)*f), -(a*d*sqrt(-d/(a*c + a*d))*log((2*(c + d)*sqrt(-d/(a*c + a*d))
*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + (c + 2*d)*cos(f*x + e)^2 + (c + d)*cos(f*
x + e) - d)/(c*cos(f*x + e)^2 + (c + d)*cos(f*x + e) + d)) - sqrt(2)*sqrt(a)*c*arctan(sqrt(2)*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + 2*sqrt(a)*(c - d)*arctan(sqrt((a*cos(f*x + e) +
a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))))/((a*c^2 - a*c*d)*f), -(2*a*d*sqrt(d/(a*c + a*d))*arcta
n((c + d)*sqrt(d/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(d*sin(f*x + e))) - sqrt(2)
*sqrt(a)*c*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + 2*sqr
t(a)*(c - d)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))))/((a*c^2 - a*
c*d)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral(1/(sqrt(a*(sec(e + f*x) + 1))*(c + d*sec(e + f*x))), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))),x)

[Out]

int(1/((a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))), x)

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