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

Optimal. Leaf size=255 \[ \frac{4 a^2 (14 A+7 B+6 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a^2 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{4 a^2 (5 A+4 B+3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{4 a^2 (5 A+4 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (7 B+4 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{35 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d} \]

[Out]

(-4*a^2*(5*A + 4*B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(14*
A + 7*B + 6*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^2*(5*A + 4*B + 3
*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a^2*(35*A + 49*B + 33*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105
*d) + (2*C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d) + (2*(7*B + 4*C)*Sec[c + d*x]^(3/2)*(
a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(35*d)

________________________________________________________________________________________

Rubi [A]  time = 0.505565, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4088, 4018, 3997, 3787, 3771, 2641, 3768, 2639} \[ \frac{2 a^2 (35 A+49 B+33 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{105 d}+\frac{4 a^2 (5 A+4 B+3 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a^2 (14 A+7 B+6 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{4 a^2 (5 A+4 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 (7 B+4 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{35 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^2*(5*A + 4*B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^2*(14*
A + 7*B + 6*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (4*a^2*(5*A + 4*B + 3
*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*a^2*(35*A + 49*B + 33*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(105
*d) + (2*C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(7*d) + (2*(7*B + 4*C)*Sec[c + d*x]^(3/2)*(
a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(35*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 4018

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

Rule 3997

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 \int \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2 \left (\frac{1}{2} a (7 A+C)+\frac{1}{2} a (7 B+4 C) \sec (c+d x)\right ) \, dx}{7 a}\\ &=\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (7 B+4 C) \sec ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{4 \int \sqrt{\sec (c+d x)} (a+a \sec (c+d x)) \left (\frac{1}{4} a^2 (35 A+7 B+9 C)+\frac{1}{4} a^2 (35 A+49 B+33 C) \sec (c+d x)\right ) \, dx}{35 a}\\ &=\frac{2 a^2 (35 A+49 B+33 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (7 B+4 C) \sec ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{8 \int \sqrt{\sec (c+d x)} \left (\frac{5}{4} a^3 (14 A+7 B+6 C)+\frac{21}{4} a^3 (5 A+4 B+3 C) \sec (c+d x)\right ) \, dx}{105 a}\\ &=\frac{2 a^2 (35 A+49 B+33 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (7 B+4 C) \sec ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{1}{5} \left (2 a^2 (5 A+4 B+3 C)\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{21} \left (2 a^2 (14 A+7 B+6 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{4 a^2 (5 A+4 B+3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 (35 A+49 B+33 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (7 B+4 C) \sec ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d}-\frac{1}{5} \left (2 a^2 (5 A+4 B+3 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (2 a^2 (14 A+7 B+6 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^2 (14 A+7 B+6 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{4 a^2 (5 A+4 B+3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 (35 A+49 B+33 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (7 B+4 C) \sec ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d}-\frac{1}{5} \left (2 a^2 (5 A+4 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{4 a^2 (5 A+4 B+3 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{4 a^2 (14 A+7 B+6 C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{4 a^2 (5 A+4 B+3 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 a^2 (35 A+49 B+33 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 (7 B+4 C) \sec ^{\frac{3}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{35 d}\\ \end{align*}

Mathematica [C]  time = 7.00799, size = 1216, normalized size = 4.77 \[ \frac{\sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{4 \sqrt{2} B e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{\sqrt{2} C e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}+\frac{4 A \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac{7}{2}}(c+d x)}+\frac{2 B \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac{7}{2}}(c+d x)}+\frac{4 C \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{7 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac{7}{2}}(c+d x)}+\frac{(\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac{C \sec (c) \sin (d x) \sec ^3(c+d x)}{7 d}+\frac{\sec (c) (5 C \sin (c)+7 B \sin (d x)+14 C \sin (d x)) \sec ^2(c+d x)}{35 d}+\frac{\sec (c) (21 B \sin (c)+42 C \sin (c)+35 A \sin (d x)+70 B \sin (d x)+60 C \sin (d x)) \sec (c+d x)}{105 d}+\frac{2 (5 A+4 B+3 C) \cos (d x) \csc (c)}{5 d}+\frac{(7 A+14 B+12 C) \tan (c)}{21 d}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^4*Csc[c]
*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*
I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*E^(I
*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c
+ d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^4*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x
)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c
+ d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c +
2*d*x])) + (Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*
x]^4*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(c + d*x))])*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)
)/(5*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d
*x)/2, 2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*(A + 2*C +
2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (2*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2
]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(3*d*(A + 2*C + 2*B*Cos
[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (4*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sec[c/
2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(7*d*(A + 2*C + 2*B*Cos[c + d*x
] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2)) + (Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]
+ C*Sec[c + d*x]^2)*((2*(5*A + 4*B + 3*C)*Cos[d*x]*Csc[c])/(5*d) + (C*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(7*d) +
(Sec[c]*Sec[c + d*x]^2*(5*C*Sin[c] + 7*B*Sin[d*x] + 14*C*Sin[d*x]))/(35*d) + (Sec[c]*Sec[c + d*x]*(21*B*Sin[c
] + 42*C*Sin[c] + 35*A*Sin[d*x] + 70*B*Sin[d*x] + 60*C*Sin[d*x]))/(105*d) + ((7*A + 14*B + 12*C)*Tan[c])/(21*d
)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2))

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Maple [B]  time = 9., size = 934, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)
)+2*C*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2
)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+
5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-8/5*(1/4*B+1/2*C)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2
*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)
-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/
2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+8*(1/4*A+1/2*B+1/4*C)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)
^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+8*(1/2
*A+1/4*B)*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)
)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(
2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*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 a^{2} \sec \left (d x + c\right )^{4} +{\left (B + 2 \, C\right )} a^{2} \sec \left (d x + c\right )^{3} +{\left (A + 2 \, B + C\right )} a^{2} \sec \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \sec \left (d x + c\right ) + A a^{2}\right )} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^2*sec(d*x + c)^4 + (B + 2*C)*a^2*sec(d*x + c)^3 + (A + 2*B + C)*a^2*sec(d*x + c)^2 + (2*A + B)*a
^2*sec(d*x + c) + A*a^2)*sqrt(sec(d*x + c)), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)