3.214 \(\int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=270 \[ \frac{4 a^2 (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a^2 (21 A+19 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{105 d}+\frac{4 a^2 (7 A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{16 a^2 (3 A+2 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}-\frac{16 a^2 (3 A+2 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{8 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{63 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d} \]

[Out]

(-16*a^2*(3*A + 2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^2*(7*A + 5
*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (16*a^2*(3*A + 2*C)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/(15*d) + (4*a^2*(7*A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a^2*(21*A + 19*C
)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d) + (2*C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(9*d
) + (8*C*Sec[c + d*x]^(5/2)*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(63*d)

________________________________________________________________________________________

Rubi [A]  time = 0.438974, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4089, 4018, 3997, 3787, 3768, 3771, 2639, 2641} \[ \frac{2 a^2 (21 A+19 C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{105 d}+\frac{4 a^2 (7 A+5 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{16 a^2 (3 A+2 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{16 a^2 (3 A+2 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{8 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{63 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^2}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(-16*a^2*(3*A + 2*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(15*d) + (4*a^2*(7*A + 5
*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (16*a^2*(3*A + 2*C)*Sqrt[Sec[c +
d*x]]*Sin[c + d*x])/(15*d) + (4*a^2*(7*A + 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (2*a^2*(21*A + 19*C
)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(105*d) + (2*C*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(9*d
) + (8*C*Sec[c + d*x]^(5/2)*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(63*d)

Rule 4089

Int[((A_.) + 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*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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 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 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 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]

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]

Rubi steps

\begin{align*} \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{2 \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (\frac{3}{2} a (3 A+C)+2 a C \sec (c+d x)\right ) \, dx}{9 a}\\ &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{8 C \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{4 \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x)) \left (\frac{3}{4} a^2 (21 A+11 C)+\frac{3}{4} a^2 (21 A+19 C) \sec (c+d x)\right ) \, dx}{63 a}\\ &=\frac{2 a^2 (21 A+19 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{8 C \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{8 \int \sec ^{\frac{3}{2}}(c+d x) \left (21 a^3 (3 A+2 C)+\frac{45}{4} a^3 (7 A+5 C) \sec (c+d x)\right ) \, dx}{315 a}\\ &=\frac{2 a^2 (21 A+19 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{8 C \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d}+\frac{1}{15} \left (8 a^2 (3 A+2 C)\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{7} \left (2 a^2 (7 A+5 C)\right ) \int \sec ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{16 a^2 (3 A+2 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{4 a^2 (7 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a^2 (21 A+19 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{8 C \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d}-\frac{1}{15} \left (8 a^2 (3 A+2 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{21} \left (2 a^2 (7 A+5 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{16 a^2 (3 A+2 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{4 a^2 (7 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a^2 (21 A+19 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{8 C \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d}-\frac{1}{15} \left (8 a^2 (3 A+2 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (2 a^2 (7 A+5 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{16 a^2 (3 A+2 C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{15 d}+\frac{4 a^2 (7 A+5 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{16 a^2 (3 A+2 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 d}+\frac{4 a^2 (7 A+5 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{2 a^2 (21 A+19 C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{105 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{9 d}+\frac{8 C \sec ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{63 d}\\ \end{align*}

Mathematica [C]  time = 6.80225, size = 821, normalized size = 3.04 \[ \frac{4 \sqrt{2} A e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (\cos (2 c+2 d x) A+A+2 C)}+\frac{8 \sqrt{2} C e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \cos ^4(c+d x) \csc (c) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{45 d (\cos (2 c+2 d x) A+A+2 C)}+\frac{2 A \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (\cos (2 c+2 d x) A+A+2 C) \sec ^{\frac{7}{2}}(c+d x)}+\frac{10 C \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) (\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{21 d (\cos (2 c+2 d x) A+A+2 C) \sec ^{\frac{7}{2}}(c+d x)}+\frac{(\sec (c+d x) a+a)^2 \left (C \sec ^2(c+d x)+A\right ) \left (\frac{C \sec (c) \sin (d x) \sec ^4(c+d x)}{9 d}+\frac{\sec (c) (7 C \sin (c)+18 C \sin (d x)) \sec ^3(c+d x)}{63 d}+\frac{\sec (c) (90 C \sin (c)+63 A \sin (d x)+112 C \sin (d x)) \sec ^2(c+d x)}{315 d}+\frac{\sec (c) (63 A \sin (c)+112 C \sin (c)+210 A \sin (d x)+150 C \sin (d x)) \sec (c+d x)}{315 d}+\frac{8 (3 A+2 C) \cos (d x) \csc (c)}{15 d}+\frac{2 (7 A+5 C) \tan (c)}{21 d}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(\cos (2 c+2 d x) A+A+2 C) \sec ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(4*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 + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C
+ A*Cos[2*c + 2*d*x])) + (8*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))*Hypergeometr
ic2F1[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 + C*Sec[c + d*x]^2)
)/(45*d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])) + (2*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 + C*Sec[c + d*x]^2))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(
7/2)) + (10*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 + C*
Sec[c + d*x]^2))/(21*d*(A + 2*C + 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 + C*Sec[c + d*x]^2)*((8*(3*A + 2*C)*Cos[d*x]*Csc[c])/(15*d) + (C*Sec[c]*Sec[c + d*x]^4*Sin[d*x])
/(9*d) + (Sec[c]*Sec[c + d*x]^3*(7*C*Sin[c] + 18*C*Sin[d*x]))/(63*d) + (Sec[c]*Sec[c + d*x]^2*(90*C*Sin[c] + 6
3*A*Sin[d*x] + 112*C*Sin[d*x]))/(315*d) + (Sec[c]*Sec[c + d*x]*(63*A*Sin[c] + 112*C*Sin[c] + 210*A*Sin[d*x] +
150*C*Sin[d*x]))/(315*d) + (2*(7*A + 5*C)*Tan[c])/(21*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(7/2))

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Maple [B]  time = 9.129, size = 1168, normalized size = 4.3 \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)^(3/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^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)*(4*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))
)+2*C*(-1/144*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)^5-7/180*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)^
3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15
*(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))-7/15*(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))-EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))))+4*A*(-1/6*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)/(co
s(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)))-8/5*(1/4*A+1/4*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)+2*A*(-(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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^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 )^{5} + 2 \, C a^{2} \sec \left (d x + c\right )^{4} +{\left (A + C\right )} a^{2} \sec \left (d x + c\right )^{3} + 2 \, A a^{2} \sec \left (d x + c\right )^{2} + A a^{2} \sec \left (d x + c\right )\right )} \sqrt{\sec \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*a^2*sec(d*x + c)^5 + 2*C*a^2*sec(d*x + c)^4 + (A + C)*a^2*sec(d*x + c)^3 + 2*A*a^2*sec(d*x + c)^2
+ A*a^2*sec(d*x + c))*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(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**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} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2*sec(d*x + c)^(3/2), x)