3.711 \(\int \sec (c+d x) \sqrt{a+b \sec (c+d x)} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=308 \[ \frac{2 (a-b) \sqrt{a+b} (2 a C+15 A b+9 b C) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b^2 d}+\frac{2 (a-b) \sqrt{a+b} \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}-\frac{4 a C \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b d} \]

[Out]

(2*(a - b)*Sqrt[a + b]*(2*a^2*C - 3*b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq
rt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15
*b^3*d) + (2*(a - b)*Sqrt[a + b]*(15*A*b + 2*a*C + 9*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))]
)/(15*b^2*d) - (4*a*C*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b*d) + (2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c
+ d*x])/(5*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.505171, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4083, 4002, 4005, 3832, 4004} \[ \frac{2 (a-b) \sqrt{a+b} \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} (2 a C+15 A b+9 b C) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}-\frac{4 a C \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a - b)*Sqrt[a + b]*(2*a^2*C - 3*b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq
rt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15
*b^3*d) + (2*(a - b)*Sqrt[a + b]*(15*A*b + 2*a*C + 9*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))]
)/(15*b^2*d) - (4*a*C*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b*d) + (2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c
+ d*x])/(5*b*d)

Rule 4083

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + 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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C*Csc[e + f*x], x], x], x] /; FreeQ
[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rule 4002

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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, 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}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

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]

Rubi steps

\begin{align*} \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}+\frac{2 \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (\frac{1}{2} b (5 A+3 C)-a C \sec (c+d x)\right ) \, dx}{5 b}\\ &=-\frac{4 a C \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac{2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}+\frac{4 \int \frac{\sec (c+d x) \left (\frac{1}{4} a b (15 A+7 C)-\frac{1}{4} \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}\\ &=-\frac{4 a C \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac{2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}+\frac{((a-b) (15 A b+2 a C+9 b C)) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}+\frac{\left (-2 a^2 C+3 b^2 (5 A+3 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}\\ &=\frac{2 (a-b) \sqrt{a+b} \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \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}}}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} (15 A b+2 a C+9 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}}}{15 b^2 d}-\frac{4 a C \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac{2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 18.0112, size = 507, normalized size = 1.65 \[ \frac{4 \sqrt{2} \sqrt{\frac{\cos (c+d x)}{(\cos (c+d x)+1)^2}} \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left ((a+b) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2} \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \left (b (-2 a C+15 A b+9 b C) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (2 a^2 C-15 A b^2-9 b^2 C\right ) E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )-\left (-2 a^2 C+15 A b^2+9 b^2 C\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)\right )}{15 b^2 d \sqrt{\frac{1}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b) (A \cos (2 c+2 d x)+A+2 C)}+\frac{\cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac{4 \left (-2 a^2 C+15 A b^2+9 b^2 C\right ) \sin (c+d x)}{15 b^2}+\frac{4 a C \tan (c+d x)}{15 b}+\frac{4}{5} C \tan (c+d x) \sec (c+d x)\right )}{d (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*S
ec[c + d*x])^(3/2)*Sqrt[a + b*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((a + b)*((-15*A*b^2 + 2*a^2*C - 9*b^2*C)*E
llipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(15*A*b - 2*a*C + 9*b*C)*EllipticF[ArcSin[Tan[(c + d*x
)/2]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2
)/(a + b)]*Sec[c + d*x] - (15*A*b^2 - 2*a^2*C + 9*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*
Tan[(c + d*x)/2]))/(15*b^2*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*(b + a*Cos[c + d*x])*(A + 2*C + A*Cos[2*c + 2*d*x])
*(Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(A + C*Sec[c + d*x]
^2)*((4*(15*A*b^2 - 2*a^2*C + 9*b^2*C)*Sin[c + d*x])/(15*b^2) + (4*a*C*Tan[c + d*x])/(15*b) + (4*C*Sec[c + d*x
]*Tan[c + d*x])/5))/(d*(A + 2*C + A*Cos[2*c + 2*d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.754, size = 2453, normalized size = 8. \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)*(A+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)

[Out]

-2/15/d/b^2*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(15*A*cos(d*x+c)^3*b^3+2*C*
cos(d*x+c)^3*a^3-15*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos
(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-15*A*cos(d*x+c)^3*sin(d*x+c)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*b^3+15*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+2*C*cos(d*x+
c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-
1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3-9*C*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^
3+9*C*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2
)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-15*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(
a+b))^(1/2))*b^3+15*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos
(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+2*C*cos(d*x+c)^3*sin(d*x+c)*
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/si
n(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-9*C*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-2*C*cos(d*
x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+7*C*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*a*b^2-15*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+15*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*a*b^2+2*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*
x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-9*C*cos(d*x+c)^2*s
in(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(
d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-2*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+
7*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*
EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+15*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(
a+b))^(1/2))*b^3+2*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(
d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3-9*C*cos(d*x+c)^2*sin(d*x+c)*(co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d
*x+c),((a-b)/(a+b))^(1/2))*b^3+9*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos
(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+9*C*cos(d*x+c)^3*
b^3-15*A*cos(d*x+c)^2*b^3-6*C*cos(d*x+c)^2*b^3-2*C*cos(d*x+c)^4*a^3+15*A*cos(d*x+c)^4*a*b^2+C*cos(d*x+c)^4*a^2
*b+9*C*cos(d*x+c)^4*a*b^2-15*A*cos(d*x+c)^3*a*b^2-2*C*cos(d*x+c)^3*a^2*b-5*C*cos(d*x+c)^3*a*b^2+C*cos(d*x+c)^2
*a^2*b-4*C*cos(d*x+c)*a*b^2-3*C*b^3)/(b+a*cos(d*x+c))/cos(d*x+c)^2/sin(d*x+c)^5

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+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 + A)*sqrt(b*sec(d*x + c) + a)*sec(d*x + c), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \sec \left (d x + c\right )^{3} + A \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^3 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*sqrt(a + b*sec(c + d*x))*sec(c + d*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+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 + A)*sqrt(b*sec(d*x + c) + a)*sec(d*x + c), x)