3.512 \(\int \frac{\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (15 A-10 B+14 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (5 B-C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 a d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (2*(1
5*A - 10*B + 14*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt
[a + a*Sec[c + d*x]]) + (2*(5*B - C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.451997, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {4088, 4010, 4001, 3795, 203} \[ -\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (15 A-10 B+14 C) \tan (c+d x)}{15 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (5 B-C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{15 a d}+\frac{2 C \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (2*(1
5*A - 10*B + 14*C)*Tan[c + d*x])/(15*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt
[a + a*Sec[c + d*x]]) + (2*(5*B - C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(15*a*d)

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 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

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

Rule 3795

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 203

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

Rubi steps

\begin{align*} \int \frac{\sec ^2(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{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^2(c+d x) \left (\frac{1}{2} a (5 A+4 C)+\frac{1}{2} a (5 B-C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac{4 \int \frac{\sec (c+d x) \left (\frac{1}{4} a^2 (5 B-C)+\frac{1}{4} a^2 (15 A-10 B+14 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 (15 A-10 B+14 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+(-A+B-C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{2 (15 A-10 B+14 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}+\frac{(2 (A-B+C)) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{2 (15 A-10 B+14 C) \tan (c+d x)}{15 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^2(c+d x) \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{15 a d}\\ \end{align*}

Mathematica [C]  time = 10.6567, size = 1666, normalized size = 10.16 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2*S
in[(c + d*x)/2]^2]*(-(A*Sin[(c + d*x)/2])/(2*(1 - 2*Sin[(c + d*x)/2]^2)^(5/2)) + (2*B*Sin[(c + d*x)/2])/(5*(1
- 2*Sin[(c + d*x)/2]^2)^(5/2)) + (8*B*(Sin[(c + d*x)/2]/(1 - 2*Sin[(c + d*x)/2]^2)^(3/2) + (2*Sin[(c + d*x)/2]
)/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]))/15 - ((A - B + C)*Csc[(c + d*x)/2]^7*(4725*Sin[(c + d*x)/2]^2 - 48825*Sin[(
c + d*x)/2]^4 + 210105*Sin[(c + d*x)/2]^6 - 486630*Sin[(c + d*x)/2]^8 + 655812*Sin[(c + d*x)/2]^10 - 710*Hyper
geometric2F1[2, 9/2, 11/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^10 - 40*Cos[(c + d
*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(
c + d*x)/2]^10 - 518760*Sin[(c + d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, Sin[(c + d*x)/2]^2/(-1 + 2*
Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12 + 226656*Sin[(c + d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/2, S
in[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^14 - 42048*Sin[(c + d*x)/2]^16 + 440*Hypergeom
etric2F1[2, 9/2, 11/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^16 + 4725*ArcTanh[Sqrt
[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 56700
*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^2*Sqrt[Sin[(c + d*x)/2]^2/(-1
+ 2*Sin[(c + d*x)/2]^2)] + 291060*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/
2]^4*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 833760*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin
[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^6*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 1458000*ArcTanh[S
qrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^8*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c
+ d*x)/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^10*Sqr
t[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d
*x)/2]^2)]]*Sin[(c + d*x)/2]^12*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 414720*ArcTanh[Sqrt[Sin
[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^14*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x
)/2]^2)] + 69120*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^16*Sqrt[Sin[(c
+ d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 9/2}, {1, 11/2}, Si
n[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^10*(-5 + 4*Sin[(c + d*x)/2]^2)))/(675*(1 - 2*Si
n[(c + d*x)/2]^2)^(7/2)*(-1 + 2*Sin[(c + d*x)/2]^2)) + (A*((3*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^(5/
2) + 4*(Sin[(c + d*x)/2]/(1 - 2*Sin[(c + d*x)/2]^2)^(3/2) + (2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2
])))/30))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2)*Sqrt[a*(1 + Sec[c + d*x])])

________________________________________________________________________________________

Maple [B]  time = 0.353, size = 859, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60/d/a*(15*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*sin(d*x+c)-15*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(
d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*sin(d*x+c)+15*C*ln(-(-(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*si
n(d*x+c)+30*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(5/2)*cos(d*x+c)*sin(d*x+c)-30*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+
c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*sin(d*x+c)+30*C*ln(-(-(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*sin(d*x+c)
+15*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(5/2)*sin(d*x+c)-15*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+15*C*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(
d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+120*A*cos(d*x+c)^3-40*B*cos(d*x+c)^3+104
*C*cos(d*x+c)^3-120*A*cos(d*x+c)^2+80*B*cos(d*x+c)^2-112*C*cos(d*x+c)^2-40*B*cos(d*x+c)+32*C*cos(d*x+c)-24*C)*
(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)

________________________________________________________________________________________

Maxima [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)^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]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 0.618778, size = 1057, normalized size = 6.45 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{3} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt{-\frac{1}{a}} \log \left (\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left ({\left (15 \, A - 5 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (5 \, B - C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{30 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \,{\left ({\left (15 \, A - 5 \, B + 13 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (5 \, B - C\right )} \cos \left (d x + c\right ) + 3 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{3} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^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]

[1/30*(15*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a*cos(d*x + c)^2)*sqrt(-1/a)*log((2*sqrt(2)*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) -
1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((15*A - 5*B + 13*C)*cos(d*x + c)^2 + (5*B - C)*cos(d*x + c) +
3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2), 1/15*(2*
((15*A - 5*B + 13*C)*cos(d*x + c)^2 + (5*B - C)*cos(d*x + c) + 3*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si
n(d*x + c) + 15*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a*cos(d*x + c)^2)*arctan(sqrt(2)*sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x
+ c)^2)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/sqrt(a*(sec(c + d*x) + 1)), x)

________________________________________________________________________________________

Giac [B]  time = 9.15661, size = 464, normalized size = 2.83 \begin{align*} -\frac{\frac{15 \,{\left (\sqrt{2} A - \sqrt{2} B + \sqrt{2} C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{2 \,{\left (15 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 15 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) -{\left (30 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 10 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 20 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) -{\left (15 \, \sqrt{2} A a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 10 \, \sqrt{2} B a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) + 17 \, \sqrt{2} C a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^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]

-1/15*(15*(sqrt(2)*A - sqrt(2)*B + sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a)))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 2*(15*sqrt(2)*A*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
) + 15*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (30*sqrt(2)*A*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 10*
sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 20*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (15*sqrt(2)
*A*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 10*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 17*sqrt(2)*C*a^2*s
gn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1
/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d