3.3.41 \(\int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [241]

Optimal. Leaf size=124 \[ -\frac {\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) f}+\frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {d} \tan (e+f x)}{\sqrt {c+d} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {d} \sqrt {c+d} f} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4061, 3880, 209, 4052, 211} \begin {gather*} \frac {2 c \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} \sqrt {d} 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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^2/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

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

Rule 4061

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=-\frac {\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx}{c-d}+\frac {c \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{a (c-d)}\\ &=\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 c) \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-d) 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 c \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-d) \sqrt {d} \sqrt {c+d} f}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 141, normalized size = 1.14 \begin {gather*} -\frac {2 \left (\sqrt {d} \sqrt {c+d} \text {ArcTan}\left (\frac {\sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\cos (e+f x)}}\right )-\sqrt {2} c \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d} \sqrt {\cos (e+f x)}}\right )\right ) \cos \left (\frac {1}{2} (e+f x)\right )}{(c-d) \sqrt {d} \sqrt {c+d} f \sqrt {\cos (e+f x)} \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^2/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

(-2*(Sqrt[d]*Sqrt[c + d]*ArcTan[Sin[(e + f*x)/2]/Sqrt[Cos[e + f*x]]] - Sqrt[2]*c*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(
e + f*x)/2])/(Sqrt[c + d]*Sqrt[Cos[e + f*x]])])*Cos[(e + f*x)/2])/((c - d)*Sqrt[d]*Sqrt[c + d]*f*Sqrt[Cos[e +
f*x]]*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(517\) vs. \(2(101)=202\).
time = 9.42, size = 518, normalized size = 4.18

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (2 \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}\, \sqrt {\frac {d}{c -d}}+c \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )-2 \sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )-2 c \sin \left (f x +e \right )+2 d \sin \left (f x +e \right )+2 \sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )-2 \sqrt {\left (c +d \right ) \left (c -d \right )}}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )-c \cos \left (f x +e \right )+d \cos \left (f x +e \right )+c -d}\right )-c \sqrt {2}\, \ln \left (-\frac {2 \left (\sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, c \sin \left (f x +e \right )-\sqrt {2}\, \sqrt {\frac {d}{c -d}}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, d \sin \left (f x +e \right )-\sqrt {\left (c +d \right ) \left (c -d \right )}\, \cos \left (f x +e \right )-c \sin \left (f x +e \right )+d \sin \left (f x +e \right )+\sqrt {\left (c +d \right ) \left (c -d \right )}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}\, \sin \left (f x +e \right )+c \cos \left (f x +e \right )-d \cos \left (f x +e \right )-c +d}\right )\right )}{2 f a \sqrt {\left (c +d \right ) \left (c -d \right )}\, \left (c -d \right ) \sqrt {\frac {d}{c -d}}}\)	\(518\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(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+d)*(c-d))^(1/2)*(d/(c-d))^(1/2)+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*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))-c*2^(1/2)*ln(-2*(2^(1/2)*(d/(c-d))^(1/2)*(-2*co
s(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)))/a/((c+d)*(c-d))^(1/2)/(c-d)/(d/(c-d))^(1/2)

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

[Out]

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

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Fricas [A]
time = 3.29, size = 1099, normalized size = 8.86 \begin {gather*} \left [-\frac {\sqrt {2} {\left (a c d + a d^{2}\right )} \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 ) - \sqrt {-a c d - a d^{2}} c \log \left (-\frac {{\left (a c^{2} + 8 \, a c d + 8 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} + a d^{2} + {\left (a c^{2} + 2 \, a c d\right )} \cos \left (f x + e\right )^{2} - 4 \, \sqrt {-a c d - a d^{2}} {\left ({\left (c + 2 \, d\right )} \cos \left (f x + e\right )^{2} - d \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )}{c^{2} \cos \left (f x + e\right )^{3} + {\left (c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + d^{2} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )}\right )}{2 \, {\left (a c^{2} d - a d^{3}\right )} f}, -\frac {\sqrt {2} {\left (a c d + a d^{2}\right )} \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 \, \sqrt {a c d + a d^{2}} c \arctan \left (\frac {2 \, \sqrt {a c d + a d^{2}} \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 (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}\right )}{2 \, {\left (a c^{2} d - a d^{3}\right )} f}, \frac {\sqrt {-a c d - a d^{2}} c \log \left (-\frac {{\left (a c^{2} + 8 \, a c d + 8 \, a d^{2}\right )} \cos \left (f x + e\right )^{3} + a d^{2} + {\left (a c^{2} + 2 \, a c d\right )} \cos \left (f x + e\right )^{2} - 4 \, \sqrt {-a c d - a d^{2}} {\left ({\left (c + 2 \, d\right )} \cos \left (f x + e\right )^{2} - d \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )}{c^{2} \cos \left (f x + e\right )^{3} + {\left (c^{2} + 2 \, c d\right )} \cos \left (f x + e\right )^{2} + d^{2} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )}\right ) + \frac {2 \, \sqrt {2} {\left (a c d + a d^{2}\right )} \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 )}{\sqrt {a}}}{2 \, {\left (a c^{2} d - a d^{3}\right )} f}, \frac {\sqrt {a c d + a d^{2}} c \arctan \left (\frac {2 \, \sqrt {a c d + a d^{2}} \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 (a c + 2 \, a d\right )} \cos \left (f x + e\right )^{2} - a d + {\left (a c + a d\right )} \cos \left (f x + e\right )}\right ) + \frac {\sqrt {2} {\left (a c d + a d^{2}\right )} \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 )}{\sqrt {a}}}{{\left (a c^{2} d - a d^{3}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(2)*(a*c*d + a*d^2)*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*c
os(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)) - sqr
t(-a*c*d - a*d^2)*c*log(-((a*c^2 + 8*a*c*d + 8*a*d^2)*cos(f*x + e)^3 + a*d^2 + (a*c^2 + 2*a*c*d)*cos(f*x + e)^
2 - 4*sqrt(-a*c*d - a*d^2)*((c + 2*d)*cos(f*x + e)^2 - d*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))
*sin(f*x + e) - (6*a*c*d + 7*a*d^2)*cos(f*x + e))/(c^2*cos(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (
2*c*d + d^2)*cos(f*x + e))))/((a*c^2*d - a*d^3)*f), -1/2*(sqrt(2)*(a*c*d + a*d^2)*sqrt(-1/a)*log(-(2*sqrt(2)*s
qrt((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*sqrt(a*c*d + a*d^2)*c*arctan(2*sqrt(a*c*d + a*d^2)*sqrt((a*c
os(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/((a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos
(f*x + e))))/((a*c^2*d - a*d^3)*f), 1/2*(sqrt(-a*c*d - a*d^2)*c*log(-((a*c^2 + 8*a*c*d + 8*a*d^2)*cos(f*x + e)
^3 + a*d^2 + (a*c^2 + 2*a*c*d)*cos(f*x + e)^2 - 4*sqrt(-a*c*d - a*d^2)*((c + 2*d)*cos(f*x + e)^2 - d*cos(f*x +
e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - (6*a*c*d + 7*a*d^2)*cos(f*x + e))/(c^2*cos(f*x + e
)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))) + 2*sqrt(2)*(a*c*d + a*d^2)*arctan(sqr
t(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))/sqrt(a))/((a*c^2*d - a*d^3)*
f), (sqrt(a*c*d + a*d^2)*c*arctan(2*sqrt(a*c*d + a*d^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*s
in(f*x + e)/((a*c + 2*a*d)*cos(f*x + e)^2 - a*d + (a*c + a*d)*cos(f*x + e))) + sqrt(2)*(a*c*d + a*d^2)*arctan(
sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))/sqrt(a))/((a*c^2*d - a*d^
3)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (e + f x \right )}}{\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(sec(f*x+e)**2/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral(sec(e + f*x)**2/(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(sec(f*x+e)^2/(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}{{\cos \left (e+f\,x\right )}^2\,\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/(cos(e + f*x)^2*(a + a/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x))),x)

[Out]

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

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