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

Optimal. Leaf size=375 \[ -\frac{2 (a-b) \sqrt{a+b} \left (C \left (8 a^2+6 a b+25 b^2\right )+35 A b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{105 b^3 d}+\frac{2 \left (8 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{105 b^2 d}-\frac{2 a (a-b) \sqrt{a+b} \left (8 a^2 C+35 A b^2+19 b^2 C\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^4 d}-\frac{8 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 b^2 d}+\frac{2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d} \]

[Out]

(-2*a*(a - b)*Sqrt[a + b]*(35*A*b^2 + 8*a^2*C + 19*b^2*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))]
)/(105*b^4*d) - (2*(a - b)*Sqrt[a + b]*(35*A*b^2 + (8*a^2 + 6*a*b + 25*b^2)*C)*Cot[c + d*x]*EllipticF[ArcSin[S
qrt[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))])/(105*b^3*d) + (2*(8*a^2*C + 5*b^2*(7*A + 5*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(
105*b^2*d) - (8*a*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*b^2*d) + (2*C*Sec[c + d*x]*(a + b*Sec[c + d*x
])^(3/2)*Tan[c + d*x])/(7*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.772456, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4093, 4082, 4002, 4005, 3832, 4004} \[ \frac{2 \left (8 a^2 C+5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{105 b^2 d}-\frac{2 (a-b) \sqrt{a+b} \left (C \left (8 a^2+6 a b+25 b^2\right )+35 A b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^3 d}-\frac{2 a (a-b) \sqrt{a+b} \left (8 a^2 C+35 A b^2+19 b^2 C\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{105 b^4 d}-\frac{8 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{35 b^2 d}+\frac{2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(a - b)*Sqrt[a + b]*(35*A*b^2 + 8*a^2*C + 19*b^2*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))]
)/(105*b^4*d) - (2*(a - b)*Sqrt[a + b]*(35*A*b^2 + (8*a^2 + 6*a*b + 25*b^2)*C)*Cot[c + d*x]*EllipticF[ArcSin[S
qrt[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))])/(105*b^3*d) + (2*(8*a^2*C + 5*b^2*(7*A + 5*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(
105*b^2*d) - (8*a*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(35*b^2*d) + (2*C*Sec[c + d*x]*(a + b*Sec[c + d*x
])^(3/2)*Tan[c + d*x])/(7*b*d)

Rule 4093

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))
^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[
1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - 2*a
*C*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

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 ^2(c+d x) \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 b d}+\frac{2 \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (a C+\frac{1}{2} b (7 A+5 C) \sec (c+d x)-2 a C \sec ^2(c+d x)\right ) \, dx}{7 b}\\ &=-\frac{8 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 b d}+\frac{4 \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (-\frac{1}{2} a b C+\frac{1}{4} \left (8 a^2 C+5 b^2 (7 A+5 C)\right ) \sec (c+d x)\right ) \, dx}{35 b^2}\\ &=\frac{2 \left (8 a^2 C+5 b^2 (7 A+5 C)\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 b^2 d}-\frac{8 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 b d}+\frac{8 \int \frac{\sec (c+d x) \left (\frac{1}{8} b \left (35 A b^2+2 a^2 C+25 b^2 C\right )+\frac{1}{8} a \left (35 A b^2+8 a^2 C+19 b^2 C\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{105 b^2}\\ &=\frac{2 \left (8 a^2 C+5 b^2 (7 A+5 C)\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 b^2 d}-\frac{8 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 b d}+\frac{\left (a \left (35 A b^2+8 a^2 C+19 b^2 C\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{105 b^2}-\frac{\left ((a-b) \left (35 A b^2+\left (8 a^2+6 a b+25 b^2\right ) C\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{105 b^2}\\ &=-\frac{2 a (a-b) \sqrt{a+b} \left (35 A b^2+8 a^2 C+19 b^2 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}}}{105 b^4 d}-\frac{2 (a-b) \sqrt{a+b} \left (35 A b^2+\left (8 a^2+6 a b+25 b^2\right ) C\right ) \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}}}{105 b^3 d}+\frac{2 \left (8 a^2 C+5 b^2 (7 A+5 C)\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{105 b^2 d}-\frac{8 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 b d}\\ \end{align*}

Mathematica [A]  time = 19.9581, size = 560, normalized size = 1.49 \[ \frac{4 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (2 b (a+b) \left (C \left (8 a^2-6 a b+25 b^2\right )+35 A b^2\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )-a \left (8 a^2 C+35 A b^2+19 b^2 C\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)-2 a (a+b) \left (8 a^2 C+35 A b^2+19 b^2 C\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{105 b^3 d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b) (A \cos (2 c+2 d x)+A+2 C)}+\frac{\cos ^2(c+d x) \sqrt{a+b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (\frac{4 a \left (8 a^2 C+35 A b^2+19 b^2 C\right ) \sin (c+d x)}{105 b^3}+\frac{4 \sec (c+d x) \left (-4 a^2 C \sin (c+d x)+35 A b^2 \sin (c+d x)+25 b^2 C \sin (c+d x)\right )}{105 b^2}+\frac{4 a C \tan (c+d x) \sec (c+d x)}{35 b}+\frac{4}{7} C \tan (c+d x) \sec ^2(c+d x)\right )}{d (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*(-2*a*(a + b)*(35*A*b
^2 + 8*a^2*C + 19*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(35*A*b^2 + (8*a^2 - 6*a*b + 25*b^
2)*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[
ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - a*(35*A*b^2 + 8*a^2*C + 19*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x
])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(105*b^3*d*(b + a*Cos[c + d*x])*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[S
ec[(c + d*x)/2]^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*a
*(35*A*b^2 + 8*a^2*C + 19*b^2*C)*Sin[c + d*x])/(105*b^3) + (4*Sec[c + d*x]*(35*A*b^2*Sin[c + d*x] - 4*a^2*C*Si
n[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b^2) + (4*a*C*Sec[c + d*x]*Tan[c + d*x])/(35*b) + (4*C*Sec[c + d*x]^
2*Tan[c + d*x])/7))/(d*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [B]  time = 1., size = 2784, normalized size = 7.4 \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+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)

[Out]

2/105/d/b^3*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(35*A*cos(d*x+c)^2*b^4+35*A
*sin(d*x+c)*cos(d*x+c)^4*(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
ipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-8*C*cos(d*x+c)^5*a^4+8*C*sin(d*x+c)*cos(d*x+c)^
4*(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^4-25*C*sin(d*x+c)*cos(d*x+c)^4*(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^4-35*A*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))*b^4+8*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^4-25*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))*b^4+70*A*cos(d*x+c)^3*a*b^3+35*A*cos(d*x+c)^4*a
^2*b^2+35*A*sin(d*x+c)*cos(d*x+c)^4*(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^3-35*A*cos(d*x+c)^4*a*b^3-8*C*cos(d*x+c)
^4*a^3*b+20*C*cos(d*x+c)^4*a^2*b^2-19*C*cos(d*x+c)^4*a*b^3+4*C*cos(d*x+c)^3*a^3*b+26*C*cos(d*x+c)^3*a*b^3-C*co
s(d*x+c)^2*a^2*b^2+18*C*cos(d*x+c)*a*b^3-35*A*cos(d*x+c)^5*a^2*b^2-35*A*cos(d*x+c)^5*a*b^3+4*C*cos(d*x+c)^5*a^
3*b-19*C*cos(d*x+c)^5*a^2*b^2-25*C*cos(d*x+c)^5*a*b^3-35*A*sin(d*x+c)*cos(d*x+c)^4*(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^4-35*A*sin(d*x+c)*cos(d*x+c)^4*(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^3+8*C*sin(d*x+c)*cos(d*x+c)^4*(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*b+19*C*sin(d*x+c)*cos(d*x+c)^4*(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^2+19*C*sin(d*x+c)*c
os(d*x+c)^4*(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+co
s(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-8*C*sin(d*x+c)*cos(d*x+c)^4*(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^3*
b-2*C*sin(d*x+c)*cos(d*x+c)^4*(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^2-19*C*sin(d*x+c)*cos(d*x+c)^4*(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^3+35*A*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^2+35*A*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*b^3-35*A*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^3+
8*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*b+19*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^2+19*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*b^3-8*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^3*b-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))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-19*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))*a*b^3+8*C*cos(d*x+c)^4*a^4+10*C*cos(d*x+c)^2*b^4-35*A*c
os(d*x+c)^4*b^4-25*C*cos(d*x+c)^4*b^4+15*C*b^4)/(b+a*cos(d*x+c))/cos(d*x+c)^3/sin(d*x+c)^5

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

[Out]

Timed out

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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 )^{4} + A \sec \left (d x + c\right )^{2}\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)^2*(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)^4 + A*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)

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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 ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, 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} + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(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)^2, x)