\(\int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [751]

Optimal result

Integrand size = 27, antiderivative size = 517 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (6 a^2 A b-a A b^2-3 A b^3-a^3 C+3 a^2 b C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^3 d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2/3*(6*A*a^2*b-A*a*b^2-3*A*b^3-C*a^3+3*C*a^2*b)*cot(d*x+c)*EllipticF((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^2/(a-b)/b/(a+b)^(3/2)/d-2/3*(
3*A*b^4-a^4*C-a^2*b^2*(7*A+3*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^2/b^2/(a^2-b^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^3/d+2/3*(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2
)-2/3*(3*A*b^4-a^4*C-a^2*b^2*(7*A+3*C))*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {4146, 4145, 4143, 4006, 3869, 3917, 4089} \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a^3 d}+\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (a^4 (-C)-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 a^2 b^2 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {2 \left (a^4 (-C)-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \tan (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (a^3 (-C)+6 a^2 A b+3 a^2 b C-a A b^2-3 A b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 a^2 b d (a-b) (a+b)^{3/2}} \]

[In]

Int[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(-2*(3*A*b^4 - a^4*C - a^2*b^2*(7*A + 3*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))])/(3*a^2*b^2*S
qrt[a + b]*(a^2 - b^2)*d) - (2*(6*a^2*A*b - a*A*b^2 - 3*A*b^3 - a^3*C + 3*a^2*b*C)*Cot[c + d*x]*EllipticF[ArcS
in[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))])/(3*a^2*(a - b)*b*(a + b)^(3/2)*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^3*d) + (2*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c
+ d*x])^(3/2)) - (2*(3*A*b^4 - a^4*C - a^2*b^2*(7*A + 3*C))*Tan[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*S
ec[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)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {3}{2} A \left (a^2-b^2\right )+\frac {3}{2} a b (A+C) \sec (c+d x)-\frac {1}{2} \left (A b^2+a^2 C\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {3}{4} A \left (a^2-b^2\right )^2+\frac {1}{2} a b \left (A b^2-a^2 (3 A+2 C)\right ) \sec (c+d x)+\frac {1}{4} \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {3}{4} A \left (a^2-b^2\right )^2+\left (\frac {1}{2} a b \left (A b^2-a^2 (3 A+2 C)\right )+\frac {1}{4} \left (-3 A b^4+a^4 C+a^2 b^2 (7 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}+\frac {\left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 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}}}{3 a^2 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {A \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^2}+\frac {\left (a A b^2+3 A b^3+a^3 C-3 a^2 b (2 A+C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 (a-b) (a+b)^2} \\ & = -\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 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}}}{3 a^2 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 \left (a A b^2+3 A b^3+a^3 C-3 a^2 b (2 A+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}}}{3 a^2 (a-b) b (a+b)^{3/2} 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^3 d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 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. \(1715\) vs. \(2(517)=1034\).

Time = 21.38 (sec) , antiderivative size = 1715, normalized size of antiderivative = 3.32 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (-7 a^2 A b^2+3 A b^4-a^4 C-3 a^2 b^2 C\right ) \sin (c+d x)}{3 a^2 b \left (-a^2+b^2\right )^2}-\frac {4 \left (A b^3 \sin (c+d x)+a^2 b C \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {8 \left (4 a^2 A b^2 \sin (c+d x)-2 A b^4 \sin (c+d x)+a^4 C \sin (c+d x)+a^2 b^2 C \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2}}-\frac {4 (b+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (7 a^3 A b^2 \tan \left (\frac {1}{2} (c+d x)\right )+7 a^2 A b^3 \tan \left (\frac {1}{2} (c+d x)\right )-3 a A b^4 \tan \left (\frac {1}{2} (c+d x)\right )-3 A b^5 \tan \left (\frac {1}{2} (c+d x)\right )+a^5 C \tan \left (\frac {1}{2} (c+d x)\right )+a^4 b C \tan \left (\frac {1}{2} (c+d x)\right )+3 a^3 b^2 C \tan \left (\frac {1}{2} (c+d x)\right )+3 a^2 b^3 C \tan \left (\frac {1}{2} (c+d x)\right )-14 a^3 A b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )+6 a A b^4 \tan ^3\left (\frac {1}{2} (c+d x)\right )-2 a^5 C \tan ^3\left (\frac {1}{2} (c+d x)\right )-6 a^3 b^2 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+7 a^3 A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )-7 a^2 A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a A b^4 \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 A b^5 \tan ^5\left (\frac {1}{2} (c+d x)\right )+a^5 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-a^4 b C \tan ^5\left (\frac {1}{2} (c+d x)\right )+3 a^3 b^2 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 a^2 b^3 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+6 a^4 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-12 a^2 A b^3 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+6 A b^5 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+6 a^4 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-12 a^2 A b^3 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+6 A b^5 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+(a+b) \left (-3 A b^4+a^4 C+a^2 b^2 (7 A+3 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-a b (a+b) \left (-2 A b^2+3 a b (A+C)+a^2 (3 A+C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{3 a^2 b \left (a^2-b^2\right )^2 d (A+2 C+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{5/2} \sqrt {1+\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((4*(-7*a^2*A*b^2 + 3*A*b^4 - a^4*C - 3*a^2*b^2*C)
*Sin[c + d*x])/(3*a^2*b*(-a^2 + b^2)^2) - (4*(A*b^3*Sin[c + d*x] + a^2*b*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*(
b + a*Cos[c + d*x])^2) + (8*(4*a^2*A*b^2*Sin[c + d*x] - 2*A*b^4*Sin[c + d*x] + a^4*C*Sin[c + d*x] + a^2*b^2*C*
Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d
*x])^(5/2)) - (4*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sqrt[(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)]*(7*a^3*A*b^2*
Tan[(c + d*x)/2] + 7*a^2*A*b^3*Tan[(c + d*x)/2] - 3*a*A*b^4*Tan[(c + d*x)/2] - 3*A*b^5*Tan[(c + d*x)/2] + a^5*
C*Tan[(c + d*x)/2] + a^4*b*C*Tan[(c + d*x)/2] + 3*a^3*b^2*C*Tan[(c + d*x)/2] + 3*a^2*b^3*C*Tan[(c + d*x)/2] -
14*a^3*A*b^2*Tan[(c + d*x)/2]^3 + 6*a*A*b^4*Tan[(c + d*x)/2]^3 - 2*a^5*C*Tan[(c + d*x)/2]^3 - 6*a^3*b^2*C*Tan[
(c + d*x)/2]^3 + 7*a^3*A*b^2*Tan[(c + d*x)/2]^5 - 7*a^2*A*b^3*Tan[(c + d*x)/2]^5 - 3*a*A*b^4*Tan[(c + d*x)/2]^
5 + 3*A*b^5*Tan[(c + d*x)/2]^5 + a^5*C*Tan[(c + d*x)/2]^5 - a^4*b*C*Tan[(c + d*x)/2]^5 + 3*a^3*b^2*C*Tan[(c +
d*x)/2]^5 - 3*a^2*b^3*C*Tan[(c + d*x)/2]^5 + 6*a^4*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)] - 12*a^2*A
*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Ta
n[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 6*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)] + 6*a
^4*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)] - 12*a^2*A*b^3*EllipticPi[-1, ArcSin[Ta
n[(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)] + 6*A*b^5*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Ta
n[(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)*(-3*A*b^4 + a^4*C + a^2*b^2*(7*A + 3*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sq
rt[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)*(-2*A*b^2 + 3*a*b*(A + C) + a^2*(3*A + 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)]))/(3*a^2*b*(a^2 - b^2)^2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)*Sqr
t[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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(7915\) vs. \(2(478)=956\).

Time = 9.39 (sec) , antiderivative size = 7916, normalized size of antiderivative = 15.31

method	result	size
parts	\(\text {Expression too large to display}\)	\(7916\)
default	\(\text {Expression too large to display}\)	\(8030\)

[In]

int((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)/(a + b*sec(c + d*x))**(5/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((A + C/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(5/2),x)

[Out]

int((A + C/cos(c + d*x)^2)/(a + b/cos(c + d*x))^(5/2), x)