\(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\) [591]
Optimal result
Integrand size = 33, antiderivative size = 342 \[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}+\frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}
\]
[Out]
1/4*(A*a^2*b+5*A*b^3+3*B*a^3-9*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+
1/2*c),2^(1/2))/b^2/(a^2-b^2)^2/d+1/4*(3*A*a^2*b+3*A*b^3+B*a^3-7*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2
*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/b/(a^2-b^2)^2/d+1/4*(A*a^4*b-10*A*a^2*b^3-3*A*b^5+3*B*a^5-
6*B*a^3*b^2+15*B*a*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b
),2^(1/2))/a/(a-b)^2/b^2/(a+b)^3/d+1/2*a*(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+a*cos(d*x+c))^
2-1/4*a*(A*a^2*b+5*A*b^3+3*B*a^3-9*B*a*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/b^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
Rubi [A] (verified)
Time = 1.19 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.00, number of
steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3033, 3079, 3134, 3138, 2719,
3081, 2720, 2884} \[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\frac {a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {\left (a^3 B+3 a^2 A b-7 a b^2 B+3 A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a b d \left (a^2-b^2\right )^2}+\frac {\left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac {a \left (3 a^3 B+a^2 A b-9 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 b^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {\left (3 a^5 B+a^4 A b-6 a^3 b^2 B-10 a^2 A b^3+15 a b^4 B-3 A b^5\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a b^2 d (a-b)^2 (a+b)^3}
\]
[In]
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^3),x]
[Out]
((a^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) + ((3*a^2*A*b +
3*A*b^3 + a^3*B - 7*a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(4*a*b*(a^2 - b^2)^2*d) + ((a^4*A*b - 10*a^2*A*b^3 - 3
*A*b^5 + 3*a^5*B - 6*a^3*b^2*B + 15*a*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a*(a - b)^2*b^2*(a
+ b)^3*d) + (a*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) - (a*(a
^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*(b + a*Cos[c +
d*x]))
Rule 2719
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]
Rule 2720
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]
Rule 2884
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 3033
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 3079
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c +
d*Sin[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*
(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n
, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Rule 3081
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3134
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&& !IntegerQ[m]) || EqQ[a, 0])))
Rule 3138
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {B+A \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))^3} \, dx \\ & = \frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {\int \frac {\frac {1}{2} \left (-a A b-3 a^2 B+4 b^2 B\right )+2 b (A b-a B) \cos (c+d x)-\frac {1}{2} a (A b-a B) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )} \\ & = \frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right )+b \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \cos (c+d x)+\frac {1}{4} a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2} \\ & = \frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\int \frac {-\frac {1}{4} a \left (a^3 A b-7 a A b^3+3 a^4 B-5 a^2 b^2 B+8 b^4 B\right )-\frac {1}{4} a b \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a b^2 \left (a^2-b^2\right )^2}+\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a b \left (a^2-b^2\right )^2}+\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a b^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (a^4 A b-10 a^2 A b^3-3 A b^5+3 a^5 B-6 a^3 b^2 B+15 a b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}+\frac {a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 b \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}-\frac {a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \\
\end{align*}
Mathematica [A] (verified)
Time = 5.38 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.12
\[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 a \sqrt {\cos (c+d x)} \left (b \left (-a^2 A b+7 A b^3+5 a^3 B-11 a b^2 B\right )+a \left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (3 a^3 A b-9 a A b^3+9 a^4 B-19 a^2 b^2 B+16 b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 b \left (a^2 A b+2 A b^3+a^3 B-4 a b^2 B\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a (a+b)}+\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 b^2 d}
\]
[In]
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^3),x]
[Out]
((-2*a*Sqrt[Cos[c + d*x]]*(b*(-(a^2*A*b) + 7*A*b^3 + 5*a^3*B - 11*a*b^2*B) + a*(a^2*A*b + 5*A*b^3 + 3*a^3*B -
9*a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (((3*a^3*A*b - 9*a*A*b^3 + 9*a
^4*B - 19*a^2*b^2*B + 16*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*b*(a^2*A*b + 2*A*b^3 +
a^3*B - 4*a*b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a*(a +
b)) + ((a^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b^2*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a +
b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -
1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*b^2*d)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(1767\) vs. \(2(406)=812\).
Time = 29.88 (sec) , antiderivative size = 1768, normalized size of antiderivative =
5.17
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method | result | size |
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default |
\(\text {Expression too large to display}\) |
\(1768\) |
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[In]
int((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a*(a^2/b/(a^2-b^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*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b*(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))+1/2*a/b/(a^2-b^2)*(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))-1/2*a/b/(a^2-b^2)*(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)*El
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b
),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+2*(-A*b+B*a)/a*
(1/2*a^2/b/(a^2-b^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*cos(1/2*d*x+
1/2*c)^2-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^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*cos(1/2*d*x+1/2*c)^2-a+b)-3/8/(a+b)/(a^2-b^2)/b^2*(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
))*a^2-1/4/(a+b)/(a^2-b^2)/b*(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))*a+7/8/(a+b)/(a^2-b^2)*(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))+3/8*a^3/b^2/(a^2-b^2)^2*(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))-9/8*a/(a^2-b^
2)^2*(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))-3/8*a^3/b^2/(a^2-b^2)^2*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))+9/8*a/(a^2-b^2)^2*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8/(a-b)/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a
^5*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/4/(a-b)/(a+b)/(a^2-b^2)/(a^2-a*b)*a^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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/8/(a-b)/(a+b)/(a^2-b^2)*b^2/(a^2-a*b)*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)*EllipticPi(cos(1/2
*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Fricas [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
Sympy [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)**(5/2)/(a+b*sec(d*x+c))**3,x)
[Out]
Timed out
Maxima [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
Timed out
Giac [F]
\[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x }
\]
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^3*cos(d*x + c)^(5/2)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x
\]
[In]
int((A + B/cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^3),x)
[Out]
int((A + B/cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^3), x)