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

Optimal. Leaf size=324 \[ \frac{2 (a-b) \sqrt{a+b} \cot (c+d x) (2 a C+15 A b-5 b B+9 b C) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b^2 d}-\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (5 b B-2 a C) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d} \]

[Out]

(-2*(a - b)*Sqrt[a + b]*(3*b^2*(5*A + 3*C) + a*(5*b*B - 2*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))])/(15*b^3*d) + (2*(a - b)*Sqrt[a + b]*(15*A*b - 5*b*B + 2*a*C + 9*b*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqr
t[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) + (2*(5*b*B - 2*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 = 0.592077, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4082, 4002, 4005, 3832, 4004} \[ \frac{2 (a-b) \sqrt{a+b} \cot (c+d x) (2 a C+15 A b-5 b B+9 b C) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d}-\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (5 b B-2 a C) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b d}+\frac{2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(3*b^2*(5*A + 3*C) + a*(5*b*B - 2*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))])/(15*b^3*d) + (2*(a - b)*Sqrt[a + b]*(15*A*b - 5*b*B + 2*a*C + 9*b*C)*Cot[c + d*x]*EllipticF[ArcSin[Sqr
t[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) + (2*(5*b*B - 2*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 4082

Int[csc[(e_.) + (f_.)*(x_)]*((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 + 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) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, 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+B \sec (c+d x)+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)+\frac{1}{2} (5 b B-2 a C) \sec (c+d x)\right ) \, dx}{5 b}\\ &=\frac{2 (5 b B-2 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} b (15 a A+5 b B+7 a C)+\frac{1}{4} \left (3 b^2 (5 A+3 C)+a (5 b B-2 a C)\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}\\ &=\frac{2 (5 b B-2 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-5 b 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 (3 b^2 (5 A+3 C)+a (5 b B-2 a 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 (3 b^2 (5 A+3 C)+a (5 b B-2 a 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-5 b 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{2 (5 b B-2 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 = 20.5021, size = 579, normalized size = 1.79 \[ \frac{4 \sqrt{2} \sqrt{\frac{\cos (c+d x)}{(\cos (c+d x)+1)^2}} \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \sqrt{a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((a+b) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2} \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \left (b (-2 a C+15 A b+5 b B+9 b C) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )-\left (-2 a^2 C+5 a b B+15 A b^2+9 b^2 C\right ) E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )-\cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (-2 a^2 C+5 a b B+15 A b^2+9 b^2 C\right ) (a \cos (c+d x)+b)\right )}{15 b^2 d \sqrt{\frac{1}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b) (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{\cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 \sin (c+d x) \left (-2 a^2 C+5 a b B+15 A b^2+9 b^2 C\right )}{15 b^2}+\frac{4 \sec (c+d x) (a C \sin (c+d x)+5 b B \sin (c+d x))}{15 b}+\frac{4}{5} C \tan (c+d x) \sec (c+d x)\right )}{d (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((a + b)*(-((15*A*b^2 + 5*
a*b*B - 2*a^2*C + 9*b^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]) + b*(15*A*b + 5*b*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 + 5*a*b*B - 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 + 2*B*Cos[c + d*x] + 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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(15*A*b^2 + 5*a*b*B
- 2*a^2*C + 9*b^2*C)*Sin[c + d*x])/(15*b^2) + (4*Sec[c + d*x]*(5*b*B*Sin[c + d*x] + a*C*Sin[c + d*x]))/(15*b)
+ (4*C*Sec[c + d*x]*Tan[c + d*x])/5))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))

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Maple [B]  time = 0.821, size = 3344, normalized size = 10.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+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*(2*C*cos(d*x+c)^3*a^3+15*A*
cos(d*x+c)^4*a*b^2-15*A*cos(d*x+c)^3*a*b^2-5*B*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^2*b+9*C
*sin(d*x+c)*cos(d*x+c)^3*(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)*Ell
ipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+2*C*sin(d*x+c)*cos(d*x+c)^2*(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*sin(d*x+c)*cos(d*x+c)^2*(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*sin(d*x+c)*cos(d*x+c)^2*(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+5*B*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+5*B*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^3-5*B*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+5*B*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)*Ellipti
cF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-5*B*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-5*B*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+5*B*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*sin(d*x+c)*cos(d*x+c)^2*(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*sin(d*x+c)*co
s(d*x+c)^2*(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*sin(d*x+c)*cos(d*x+c)^2*(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*sin(d*x+c)*cos(d*x+c)^2*(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*sin(d*x+c)*cos(d*x+c)^3*(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*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(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*sin(d*x+c)*cos(d*x+c)^3
*(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))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+7*C*sin(d*x+c)*cos(d*x+c)^3*(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*(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))*sin(d*x+c)*a*b^2-15*A*cos(d*x+c)^2*(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))*sin(d*
x+c)*a*b^2-15*A*cos(d*x+c)^3*(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))*sin(d*x+c)*a*b^2+C*cos(d*x+c)^4*a^2*b+9*C*cos(d*x+c
)^4*a*b^2-2*C*cos(d*x+c)^3*a^2*b-5*C*cos(d*x+c)^3*a*b^2-4*C*cos(d*x+c)*a*b^2+5*B*cos(d*x+c)^3*a*b^2-10*B*cos(d
*x+c)^2*a*b^2+5*B*cos(d*x+c)^4*a^2*b+5*B*cos(d*x+c)^4*a*b^2-5*B*cos(d*x+c)^3*a^2*b+C*cos(d*x+c)^2*a^2*b-2*C*co
s(d*x+c)^4*a^3+9*C*cos(d*x+c)^3*b^3-6*C*cos(d*x+c)^2*b^3+5*B*cos(d*x+c)^3*b^3-5*B*cos(d*x+c)*b^3+2*C*sin(d*x+c
)*cos(d*x+c)^3*(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*sin(d*x+c)*cos(d*x+c)^3*(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*(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))*sin(d*x+c)*a*b^2-15*A*cos(d*x+c)^2*b^3+15*A*cos(d*x+c)^3*b^3+
15*A*cos(d*x+c)^2*(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))*sin(d*x+c)*b^3-15*A*cos(d*x+c)^2*(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))
*sin(d*x+c)*b^3+15*A*cos(d*x+c)^3*(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))*sin(d*x+c)*b^3-15*A*cos(d*x+c)^3*(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))*sin(d*x+c)*b^3-3*C*b^3)/(b+a*cos(d*x+c))/cos(d*x+c)^2/sin(d*x+c)^5

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \sec \left (d x + c\right )^{3} + B \sec \left (d x + c\right )^{2} + 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+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)^3 + B*sec(d*x + c)^2 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)

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

[Out]

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

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