\(\int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx\) [754]

Optimal result

Integrand size = 27, antiderivative size = 626 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=-\frac {2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 b^2 \sqrt {a+b} \left (a^2-b^2\right )^2 d}+\frac {2 \left (36 a^2 A b^3-5 a A b^4-15 A b^5+3 a^5 C+a^3 b^2 (13 A+5 C)-3 a^4 b (15 A+8 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 b \sqrt {a+b} \left (a^2-b^2\right )^2 d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (5 A b^4-3 a^4 C-a^2 b^2 (13 A+5 C)\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (41 a^2 A b^4-15 A b^6-3 a^6 C-29 a^4 b^2 (2 A+C)\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2/15*(41*a^2*A*b^4-15*A*b^6-3*a^6*C-29*a^4*b^2*(2*A+C))*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/
2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/b^2/(a^2-b^2)^2/d/(
a+b)^(1/2)+2/15*(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(d*x+c)*Ell
ipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c
))/(a-b))^(1/2)/a^3/b/(a^2-b^2)^2/d/(a+b)^(1/2)-2*A*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(
a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/d+2
/5*(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(5/2)-2/15*(5*A*b^4-3*a^4*C-a^2*b^2*(13*A+5*C))*tan
(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(3/2)-2/15*(41*a^2*A*b^4-15*A*b^6-3*a^6*C-29*a^4*b^2*(2*A+C))*tan(d
*x+c)/a^3/(a^2-b^2)^3/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 626, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4146, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=-\frac {2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^4 d}+\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (-3 a^4 C-a^2 b^2 (13 A+5 C)+5 A b^4\right ) \tan (c+d x)}{15 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (-3 a^6 C-29 a^4 b^2 (2 A+C)+41 a^2 A b^4-15 A b^6\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 a^3 b^2 d \sqrt {a+b} \left (a^2-b^2\right )^2}-\frac {2 \left (-3 a^6 C-29 a^4 b^2 (2 A+C)+41 a^2 A b^4-15 A b^6\right ) \tan (c+d x)}{15 a^3 d \left (a^2-b^2\right )^3 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^5 C-3 a^4 b (15 A+8 C)+a^3 b^2 (13 A+5 C)+36 a^2 A b^3-5 a A b^4-15 A b^5\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{15 a^3 b d \sqrt {a+b} \left (a^2-b^2\right )^2} \]

[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 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*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)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(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)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4006

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4143

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 4145

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 4146

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]

Rubi steps \begin{align*} \text {integral}& = \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 (\arcsin \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 (\arcsin \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) \operatorname {EllipticF}\left (\arcsin \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) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \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] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2188\) vs. \(2(626)=1252\).

Time = 24.28 (sec) , antiderivative size = 2188, normalized size of antiderivative = 3.50 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\text {Result too large to show} \]

[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*Ell
ipticPi[-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*Sqr
t[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*E
llipticPi[-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*T
an[(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))*EllipticE[Ar
cSin[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)))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(14041\) vs. \(2(583)=1166\).

Time = 15.93 (sec) , antiderivative size = 14042, normalized size of antiderivative = 22.43

method	result	size
parts	\(\text {Expression too large to display}\)	\(14042\)
default	\(\text {Expression too large to display}\)	\(14230\)

[In]

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

[Out]

result too large to display

Fricas [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{7/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \]

[In]

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

[Out]

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