3.6.80 \(\int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [580]

Optimal. Leaf size=131 \[ \frac {\sqrt {a} (8 A+4 B+3 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a (4 B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d} \]

[Out]

1/4*(8*A+4*B+3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*a^(1/2)/d+1/4*a*(4*B+C)*sec(d*x+c)^(3/2)*
sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*C*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.22, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {4173, 4101, 3886, 221} \begin {gather*} \frac {\sqrt {a} (8 A+4 B+3 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {a (4 B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Sqrt[a]*(8*A + 4*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a*(4*B + C)*Sec[
c + d*x]^(3/2)*Sin[c + d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) + (C*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*S
in[c + d*x])/(2*d)

Rule 221

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]

Rule 3886

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], 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 4101

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] &&  !LtQ[n, 0]

Rule 4173

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]

Rubi steps

\begin {align*} \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {\int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \left (\frac {1}{2} a (4 A+C)+\frac {1}{2} a (4 B+C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=\frac {a (4 B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}+\frac {1}{8} (8 A+4 B+3 C) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a (4 B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {(8 A+4 B+3 C) \text {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 )}{4 d}\\ &=\frac {\sqrt {a} (8 A+4 B+3 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a (4 B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.86, size = 109, normalized size = 0.83 \begin {gather*} \frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (\sqrt {2} (8 A+4 B+3 C) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sec (c+d x) (4 B+3 C+2 C \sec (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*(8*A + 4*B + 3*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + 2*
Sec[c + d*x]*(4*B + 3*C + 2*C*Sec[c + d*x])*Sin[(c + d*x)/2]))/(8*d*Sqrt[Sec[c + d*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(451\) vs. \(2(111)=222\).
time = 0.19, size = 452, normalized size = 3.45

method result size
default \(\frac {\left (8 A \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-1-\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+8 A \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+4 B \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-1-\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+4 B \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 C \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-1-\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+3 C \sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+8 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+4 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}{16 d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}}\) \(452\)

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,method=_RETURNVERBOSE)

[Out]

1/16/d*(8*A*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(-1-cos(d*x+c)+sin(d*x+c))*2^(1/2))*cos(d*x+c)^2+8*A*
2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(1+cos(d*x+c)+sin(d*x+c))*2^(1/2))*cos(d*x+c)^2+4*B*2^(1/2)*arcta
n(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(-1-cos(d*x+c)+sin(d*x+c))*2^(1/2))*cos(d*x+c)^2+4*B*2^(1/2)*arctan(1/4*(-2/(1
+cos(d*x+c)))^(1/2)*(1+cos(d*x+c)+sin(d*x+c))*2^(1/2))*cos(d*x+c)^2+3*C*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))
^(1/2)*(-1-cos(d*x+c)+sin(d*x+c))*2^(1/2))*cos(d*x+c)^2+3*C*2^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(1+co
s(d*x+c)+sin(d*x+c))*2^(1/2))*cos(d*x+c)^2+8*B*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+6*C*(-2/(1+cos(
d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+4*C*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c))*(1/cos(d*x+c))^(1/2)*(a*(1+cos(
d*x+c))/cos(d*x+c))^(1/2)*(-2/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)/sin(d*x+c)^2*(cos(d*x+c)^2-1)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 2167 vs. \(2 (111) = 222\).
time = 0.76, size = 2167, normalized size = 16.54 \begin {gather*} \text {Too large to display} \end {gather*}

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]

1/16*(8*A*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) +
2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(
1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)
^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*s
in(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)) - 4*(4*sqrt(2)*c
os(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(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*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) + (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*arctan
2(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*arctan
2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 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*arctan2(sin(d*x + c), cos(d*x + c))))*B*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
 + 2*c) + 1) - (12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(d*x + c), cos(d
*x + c))) + 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin(d*x + c), cos(d*x +
c))) - 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))
- 12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 3*
(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2
+ 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(s
in(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(s
in(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x
+ 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x +
4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*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) - 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(
4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x +
 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*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) + 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + c
os(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*
d*x + 2*c)^2 + 4*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*arctan
2(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) - 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2
))*sin(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c) + s
qrt(2))*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x + 2*c
) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(2*d*x
 + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*C*sqrt(a)/(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d
*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*
c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1))/d

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Fricas [A]
time = 3.15, size = 414, normalized size = 3.16 \begin {gather*} \left [\frac {{\left ({\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left ({\left (4 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (8 \, A + 4 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left ({\left (4 \, B + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end {gather*}

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]

[1/16*(((8*A + 4*B + 3*C)*cos(d*x + c)^2 + (8*A + 4*B + 3*C)*cos(d*x + 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)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*((4*B + 3*C)*cos(d*x + c) + 2*C)*sqrt((a
*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), 1/8*(((
8*A + 4*B + 3*C)*cos(d*x + c)^2 + (8*A + 4*B + 3*C)*cos(d*x + 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*((4*B +
3*C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x +
 c)^2 + d*cos(d*x + c))]

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

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]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(a*sec(d*x + c) + a)*sqrt(sec(d*x + c)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int((a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

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