3.754 \(\int \frac{A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=626 \[ \frac{2 \left (a^3 b^2 (13 A+5 C)+36 a^2 A b^3-3 a^4 b (15 A+8 C)+3 a^5 C-5 a A b^4-15 A b^5\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 a^3 b d \sqrt{a+b} \left (a^2-b^2\right )^2}-\frac{2 \left (-29 a^4 b^2 (2 A+C)+41 a^2 A b^4-3 a^6 C-15 A b^6\right ) \tan (c+d x)}{15 a^3 d \left (a^2-b^2\right )^3 \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-a^2 b^2 (13 A+5 C)-3 a^4 C+5 A b^4\right ) \tan (c+d x)}{15 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (-29 a^4 b^2 (2 A+C)+41 a^2 A b^4-3 a^6 C-15 A b^6\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 a^3 b^2 d \sqrt{a+b} \left (a^2-b^2\right )^2}-\frac{2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^4 d} \]

[Out]

(-2*(41*a^2*A*b^4 - 15*A*b^6 - 3*a^6*C - 29*a^4*b^2*(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*a^3*b^2*Sqrt[a + b]*(a^2 - b^2)^2*d) + (2*(36*a^2*A*b^3 - 5*a*A*b^4 - 15*A*b^5 + 3*a^5*C + a^3*b^2*
(13*A + 5*C) - 3*a^4*b*(15*A + 8*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*a^3*b*Sqrt[a +
b]*(a^2 - b^2)^2*d) - (2*A*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/(a^4*
d) + (2*(A*b^2 + a^2*C)*Tan[c + d*x])/(5*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(5/2)) - (2*(5*A*b^4 - 3*a^4*C -
a^2*b^2*(13*A + 5*C))*Tan[c + d*x])/(15*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^(3/2)) - (2*(41*a^2*A*b^4 -
15*A*b^6 - 3*a^6*C - 29*a^4*b^2*(2*A + C))*Tan[c + d*x])/(15*a^3*(a^2 - b^2)^3*d*Sqrt[a + b*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.2757, antiderivative size = 626, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {4061, 4060, 4058, 3921, 3784, 3832, 4004} \[ -\frac{2 \left (-29 a^4 b^2 (2 A+C)+41 a^2 A b^4-3 a^6 C-15 A b^6\right ) \tan (c+d x)}{15 a^3 d \left (a^2-b^2\right )^3 \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-a^2 b^2 (13 A+5 C)-3 a^4 C+5 A b^4\right ) \tan (c+d x)}{15 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}+\frac{2 \left (a^3 b^2 (13 A+5 C)+36 a^2 A b^3-3 a^4 b (15 A+8 C)+3 a^5 C-5 a A b^4-15 A b^5\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 a^3 b d \sqrt{a+b} \left (a^2-b^2\right )^2}-\frac{2 \left (-29 a^4 b^2 (2 A+C)+41 a^2 A b^4-3 a^6 C-15 A b^6\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 a^3 b^2 d \sqrt{a+b} \left (a^2-b^2\right )^2}-\frac{2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

(-2*(41*a^2*A*b^4 - 15*A*b^6 - 3*a^6*C - 29*a^4*b^2*(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*a^3*b^2*Sqrt[a + b]*(a^2 - b^2)^2*d) + (2*(36*a^2*A*b^3 - 5*a*A*b^4 - 15*A*b^5 + 3*a^5*C + a^3*b^2*
(13*A + 5*C) - 3*a^4*b*(15*A + 8*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*a^3*b*Sqrt[a +
b]*(a^2 - b^2)^2*d) - (2*A*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/(a^4*
d) + (2*(A*b^2 + a^2*C)*Tan[c + d*x])/(5*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(5/2)) - (2*(5*A*b^4 - 3*a^4*C -
a^2*b^2*(13*A + 5*C))*Tan[c + d*x])/(15*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^(3/2)) - (2*(41*a^2*A*b^4 -
15*A*b^6 - 3*a^6*C - 29*a^4*b^2*(2*A + C))*Tan[c + d*x])/(15*a^3*(a^2 - b^2)^3*d*Sqrt[a + b*Sec[c + d*x]])

Rule 4061

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

Rule 4060

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

Rule 4058

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

Rule 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && 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 \frac{A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx &=\frac{2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \int \frac{-\frac{5}{2} A \left (a^2-b^2\right )+\frac{5}{2} a b (A+C) \sec (c+d x)-\frac{3}{2} \left (A b^2+a^2 C\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{5 a \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac{4 \int \frac{\frac{15}{4} A \left (a^2-b^2\right )^2+\frac{3}{2} a b \left (A b^2-a^2 (5 A+4 C)\right ) \sec (c+d x)-\frac{1}{4} \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{15 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}-\frac{8 \int \frac{-\frac{15}{8} A \left (a^2-b^2\right )^3+\frac{1}{8} a b \left (10 A b^4-a^2 b^2 (23 A-5 C)+9 a^4 (5 A+3 C)\right ) \sec (c+d x)-\frac{1}{8} \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}\\ &=\frac{2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}-\frac{8 \int \frac{-\frac{15}{8} A \left (a^2-b^2\right )^3+\left (\frac{1}{8} \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right )+\frac{1}{8} a b \left (10 A b^4-a^2 b^2 (23 A-5 C)+9 a^4 (5 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}+\frac{\left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}\\ &=-\frac{2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (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 a^3 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac{2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}+\frac{A \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{a^3}+\frac{\left (36 a^2 A b^3-5 a A b^4-15 A b^5+3 a^5 C+a^3 b^2 (13 A+5 C)-3 a^4 b (15 A+8 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^3 (a-b)^2 (a+b)^3}\\ &=-\frac{2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (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 a^3 (a-b)^2 b^2 (a+b)^{5/2} d}+\frac{2 \left (36 a^2 A b^3-5 a A b^4-15 A b^5+3 a^5 C+a^3 b^2 (13 A+5 C)-3 a^4 b (15 A+8 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}}}{15 a^3 (a-b)^2 b (a+b)^{5/2} d}-\frac{2 A \sqrt{a+b} \cot (c+d x) \Pi \left (\frac{a+b}{a};\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}}}{a^4 d}+\frac{2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac{2 \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 22.0851, size = 2204, normalized size = 3.52 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(7/2),x]

[Out]

((b + a*Cos[c + d*x])^4*Sec[c + d*x]^2*(A + C*Sec[c + d*x]^2)*((4*(58*a^4*A*b^2 - 41*a^2*A*b^4 + 15*A*b^6 + 3*
a^6*C + 29*a^4*b^2*C)*Sin[c + d*x])/(15*a^3*b*(-a^2 + b^2)^3) + (4*(A*b^4*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x
]))/(5*a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])^3) + (4*(-19*a^2*A*b^3*Sin[c + d*x] + 11*A*b^5*Sin[c + d*x] - 9*a^
4*b*C*Sin[c + d*x] + a^2*b^3*C*Sin[c + d*x]))/(15*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (4*(74*a^4*A*b^2
*Sin[c + d*x] - 65*a^2*A*b^4*Sin[c + d*x] + 23*A*b^6*Sin[c + d*x] + 9*a^6*C*Sin[c + d*x] + 25*a^4*b^2*C*Sin[c
+ d*x] - 2*a^2*b^4*C*Sin[c + d*x]))/(15*a^3*(a^2 - b^2)^3*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*
d*x])*(a + b*Sec[c + d*x])^(7/2)) - (4*(b + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*Sq
rt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x
)/2]^2)]*(58*a^5*A*b^2*Tan[(c + d*x)/2] + 58*a^4*A*b^3*Tan[(c + d*x)/2] - 41*a^3*A*b^4*Tan[(c + d*x)/2] - 41*a
^2*A*b^5*Tan[(c + d*x)/2] + 15*a*A*b^6*Tan[(c + d*x)/2] + 15*A*b^7*Tan[(c + d*x)/2] + 3*a^7*C*Tan[(c + d*x)/2]
+ 3*a^6*b*C*Tan[(c + d*x)/2] + 29*a^5*b^2*C*Tan[(c + d*x)/2] + 29*a^4*b^3*C*Tan[(c + d*x)/2] - 116*a^5*A*b^2*
Tan[(c + d*x)/2]^3 + 82*a^3*A*b^4*Tan[(c + d*x)/2]^3 - 30*a*A*b^6*Tan[(c + d*x)/2]^3 - 6*a^7*C*Tan[(c + d*x)/2
]^3 - 58*a^5*b^2*C*Tan[(c + d*x)/2]^3 + 58*a^5*A*b^2*Tan[(c + d*x)/2]^5 - 58*a^4*A*b^3*Tan[(c + d*x)/2]^5 - 41
*a^3*A*b^4*Tan[(c + d*x)/2]^5 + 41*a^2*A*b^5*Tan[(c + d*x)/2]^5 + 15*a*A*b^6*Tan[(c + d*x)/2]^5 - 15*A*b^7*Tan
[(c + d*x)/2]^5 + 3*a^7*C*Tan[(c + d*x)/2]^5 - 3*a^6*b*C*Tan[(c + d*x)/2]^5 + 29*a^5*b^2*C*Tan[(c + d*x)/2]^5
- 29*a^4*b^3*C*Tan[(c + d*x)/2]^5 - 30*a^6*A*b*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt
[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 90*a^4*A*b^3*El
lipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c +
d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 90*a^2*A*b^5*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a - b)/(
a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*
A*b^7*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*
Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 30*a^6*A*b*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a
- b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c +
d*x)/2]^2)/(a + b)] + 90*a^4*A*b^3*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]
^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 90*a^2*A
*b^5*EllipticPi[-1, -ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2
]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 30*A*b^7*EllipticPi[-1, -ArcSin[Tan[(c
+ d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]
^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(-41*a^2*A*b^4 + 15*A*b^6 + 3*a^6*C + 29*a^4*b^2*(2*A + C))*Elli
pticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a
+ b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - a*b*(a + b)*(-6*a*A*b^3 + 10*A*b^4 + 3*a^4*(5*A
+ C) + 6*a^3*b*(5*A + 4*C) + a^2*b^2*(-17*A + 5*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt
[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(
a + b)]))/(15*a^3*b*(a^2 - b^2)^3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(7/2)*Sqrt[1 + Tan[(c
+ d*x)/2]^2]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))

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Maple [B]  time = 0.721, size = 11805, normalized size = 18.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x)

[Out]

result too large to display

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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+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{b^{4} \sec \left (d x + c\right )^{4} + 4 \, a b^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{3} b \sec \left (d x + c\right ) + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)*sqrt(b*sec(d*x + c) + a)/(b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x + c)^3 + 6*a^2*
b^2*sec(d*x + c)^2 + 4*a^3*b*sec(d*x + c) + a^4), 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+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c) + a)^(7/2), x)