3.607 \(\int \frac{\sqrt{\sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)
Optimal. Leaf size=141 \[ \frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{(2 B-C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]
[Out]
((2*B - C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) + (Sqrt[2]*(A - B + C)*ArcTan
h[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + (C*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.417567, antiderivative size = 141, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{4088, 4023, 3808, 206, 3801, 215} \[ \frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{(2 B-C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
((2*B - C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) + (Sqrt[2]*(A - B + C)*ArcTan
h[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + (C*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])
Rule 4088
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Rule 4023
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Dist[B
/b, Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A
*b - a*B, 0] && EqQ[a^2 - b^2, 0]
Rule 3808
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3801
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]
Rule 215
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]
Rubi steps
\begin{align*} \int \frac{\sqrt{\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{C \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} a (2 A+C)+\frac{1}{2} a (2 B-C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{a}\\ &=\frac{C \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}+\frac{(2 B-C) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{2 a}+(A-B+C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{C \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}-\frac{(2 B-C) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a d}-\frac{(2 (A-B+C)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=\frac{(2 B-C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{C \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.623653, size = 107, normalized size = 0.76 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (2 (A-B+C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\sqrt{2} (2 B-C) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 C \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )}{d \sqrt{a (\sec (c+d x)+1)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
[Out]
(Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(2*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] + Sqrt[2]*(2*B - C)*ArcTanh[Sqrt
[2]*Sin[(c + d*x)/2]] + 2*C*Sec[c + d*x]*Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Sec[c + d*x])])
________________________________________________________________________________________
Maple [B] time = 0.409, size = 378, normalized size = 2.7 \begin{align*}{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1}{4\,ad \left ( \sin \left ( dx+c \right ) \right ) ^{2}}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 2\,B\sqrt{2}\cos \left ( dx+c \right ) \arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) -2\,B\sqrt{2}\cos \left ( dx+c \right ) \arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) -C\sqrt{2}\cos \left ( dx+c \right ) \arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) +C\sqrt{2}\cos \left ( dx+c \right ) \arctan \left ({\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) }{4}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) +4\,A\cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) -4\,B\cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +4\,C\cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +2\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)
[Out]
1/4/d/a*(1/cos(d*x+c))^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(2*B*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-
2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-2*B*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1
))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))-C*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x
+c)+1+sin(d*x+c)))+C*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))
)+4*A*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-4*B*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(co
s(d*x+c)+1))^(1/2))+4*C*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))+2*C*(-2/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c))*(-2/(cos(d*x+c)+1))^(1/2)/sin(d*x+c)^2*(cos(d*x+c)^2-1)
________________________________________________________________________________________
Maxima [B] time = 2.57597, size = 1947, normalized size = 13.81 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
1/4*(2*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*lo
g(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*A/sqrt(a) - 2*(sqrt(2)*log(co
s(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arct
an2(sin(d*x + c), cos(d*x + c))) + 1) - sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*a
rctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - log(2*cos(1/2*ar
ctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*ar
ctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + log(2*cos(1
/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1
/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - log(2*
cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*
cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + l
og(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqr
t(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2
))*B/sqrt(a) - (4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*ar
ctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)
))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*
x + c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d
*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d
*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 +
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*
arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*si
n(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c
) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2
- 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x +
c))) + 2) - 2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))
*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1
/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) + 2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*
sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d
*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - 4*(sqrt(2)*cos(2*d*x + 2*c)
+ sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arcta
n2(sin(d*x + c), cos(d*x + c))))*C/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sqrt(a)
))/d
________________________________________________________________________________________
Fricas [A] time = 0.890274, size = 1424, normalized size = 10.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[-1/4*(((2*B - C)*cos(d*x + c) + 2*B - C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)
^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(c
os(d*x + c)^3 + cos(d*x + c)^2)) - 2*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*log(-(cos(d*x + c)^2
- 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c)
- 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a) - 4*C*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)
/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d), -1/2*(2*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*sq
rt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) -
((2*B - C)*cos(d*x + c) + 2*B - C)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos
(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) - 2*C*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{\sqrt{a \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/sqrt(a*sec(d*x + c) + a), x)