3.817 \(\int \sqrt {a+b \sec (c+d x)} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=256 \[ -\frac {2 (a-b) \sqrt {a+b} (a C+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 B-C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d} \]
[Out]
-2/3*(a-b)*(3*B*b+C*a)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((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)/b^2/d+2/3*(a-b)*(3*B-C)*cot(d*x+c)*EllipticF(
(a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((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)/b/d+2/3*C*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A] time = 0.29, antiderivative size = 256, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used =
{4056, 4058, 12, 3832, 4004} \[ -\frac {2 (a-b) \sqrt {a+b} (a C+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 B-C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d}+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(-2*(a - b)*Sqrt[a + b]*(3*b*B + a*C)*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))])/(3*b^2*d) + (2*(a
- b)*Sqrt[a + b]*(3*B - C)*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 + Sec[c + d*x]))/(a - b))])/(3*b*d) + (2*C*Sqrt[a + b*Sec
[c + d*x]]*Tan[c + d*x])/(3*d)
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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]
Rule 4056
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int
[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) + a
*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
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
]
Rubi steps
\begin {align*} \int \sqrt {a+b \sec (c+d x)} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {2}{3} \int \frac {\frac {1}{2} (3 a B+b C) \sec (c+d x)+\frac {1}{2} (3 b B+a C) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {2}{3} \int \frac {\left (\frac {1}{2} (-3 b B-a C)+\frac {1}{2} (3 a B+b C)\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{3} (3 b B+a C) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} (3 b B+a C) \cot (c+d x) E\left (\sin ^{-1}\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}}}{3 b^2 d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}+\frac {1}{3} ((a-b) (3 B-C)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} (3 b B+a C) \cot (c+d x) E\left (\sin ^{-1}\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}}}{3 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (3 B-C) \cot (c+d x) F\left (\sin ^{-1}\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}}}{3 b d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 20.51, size = 480, normalized size = 1.88 \[ \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {2 (a C+3 b B) \sin (c+d x)}{3 b}+\frac {2}{3} C \tan (c+d x)\right )}{d}+\frac {2 \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \sqrt {a+b \sec (c+d x)} \left ((a C+3 b B) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) \left (-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b\right )+b (a+b) (3 B+C) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-\left ((a+b) (a C+3 b B) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )\right )}{3 b d \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[a + b*Sec[c + d*x]]*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(2*Sqrt[a + b*Sec[c + d*x]]*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((3*b*B + a*C)*Tan[(c + d*x)/2]*(-1 + Tan[(c +
d*x)/2]^2)*(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2) - (a + b)*(3*b*B + a*C)*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)] + b*(a + b)*(3*B + C)*EllipticF[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)]))/(3*b*d*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(1 + Tan[(c + d*x)/2]^2)^(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)]) + (Sqrt[a + b*Sec[c + d*x]]*
((2*(3*b*B + a*C)*Sin[c + d*x])/(3*b) + (2*C*Tan[c + d*x])/3))/d
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fricas [F] time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
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maple [B] time = 2.27, size = 1754, normalized size = 6.85 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)
[Out]
2/3/d*(-1+cos(d*x+c))^2*(-3*B*cos(d*x+c)^2*b^2+b^2*C+C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b
-C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ell
ipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))
*a*b-C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*E
llipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+3*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(
d*x+c)*a*b-3*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b+3*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/
2))*sin(d*x+c)*a*b-3*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b-3*B*cos(d*x+c)^3*a*b+3*B*cos(d*x+
c)^2*a*b+3*b^2*B*cos(d*x+c)-b^2*C*cos(d*x+c)^2-C*cos(d*x+c)^3*a*b-C*cos(d*x+c)^2*a*b+2*C*cos(d*x+c)*a*b+C*sin(
d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2-C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2+
C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2-C*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^
2+3*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((
-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^2-3*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin
(d*x+c)*b^2+3*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ell
ipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^2-3*B*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*sin(d*x+c)*b^2-C*cos(d*x+c)^3*a^2+C*cos(d*x+c)^2*a^2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2/
(b+a*cos(d*x+c))/cos(d*x+c)/sin(d*x+c)^5/b
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/2),x)
[Out]
int((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \sqrt {a + b \sec {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral((B + C*sec(c + d*x))*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x)
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