3.592 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=420 \[ \frac {a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)^2}+\frac {\left (-5 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac {a \left (-5 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}+\frac {\left (-15 a^4 B+3 a^3 A b+29 a^2 b^2 B-9 a A b^3-8 b^4 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}-\frac {\left (-15 a^4 B+3 a^3 A b+29 a^2 b^2 B-9 a A b^3-8 b^4 B\right ) \sin (c+d x)}{4 b^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}}+\frac {\left (-15 a^5 B+3 a^4 A b+38 a^3 b^2 B-6 a^2 A b^3-35 a b^4 B+15 A b^5\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3} \]

[Out]

1/4*(3*A*a^3*b-9*A*a*b^3-15*B*a^4+29*B*a^2*b^2-8*B*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icE(sin(1/2*d*x+1/2*c),2^(1/2))/b^3/(a^2-b^2)^2/d+1/4*(A*a^2*b-7*A*b^3-5*B*a^3+11*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))/b^2/(a^2-b^2)^2/d+1/4*(3*A*a^4*b-6*A*a^2*b^
3+15*A*b^5-15*B*a^5+38*B*a^3*b^2-35*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-b)^2/b^3/(a+b)^3/d-1/4*(3*A*a^3*b-9*A*a*b^3-15*B*a^4+29*B*a^2*b^2-8*B*b^4)*
sin(d*x+c)/b^3/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)+1/2*a*(A*b-B*a)*sin(d*x+c)/b/(a^2-b^2)/d/(b+a*cos(d*x+c))^2/cos(
d*x+c)^(1/2)+1/4*a*(A*a^2*b-7*A*b^3-5*B*a^3+11*B*a*b^2)*sin(d*x+c)/b^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))/cos(d*x+
c)^(1/2)

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Rubi [A]  time = 1.48, antiderivative size = 420, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2954, 3000, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {\left (a^2 A b-5 a^3 B+11 a b^2 B-7 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac {\left (3 a^3 A b+29 a^2 b^2 B-15 a^4 B-9 a A b^3-8 b^4 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d \left (a^2-b^2\right )^2}+\frac {\left (-6 a^2 A b^3+3 a^4 A b+38 a^3 b^2 B-15 a^5 B-35 a b^4 B+15 A b^5\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 d (a-b)^2 (a+b)^3}-\frac {\left (3 a^3 A b+29 a^2 b^2 B-15 a^4 B-9 a A b^3-8 b^4 B\right ) \sin (c+d x)}{4 b^3 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)}}+\frac {a \left (a^2 A b-5 a^3 B+11 a b^2 B-7 A b^3\right ) \sin (c+d x)}{4 b^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}+\frac {a (A b-a B) \sin (c+d x)}{2 b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

((3*a^3*A*b - 9*a*A*b^3 - 15*a^4*B + 29*a^2*b^2*B - 8*b^4*B)*EllipticE[(c + d*x)/2, 2])/(4*b^3*(a^2 - b^2)^2*d
) + ((a^2*A*b - 7*A*b^3 - 5*a^3*B + 11*a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) + ((3*a^4*A
*b - 6*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 38*a^3*b^2*B - 35*a*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])
/(4*(a - b)^2*b^3*(a + b)^3*d) - ((3*a^3*A*b - 9*a*A*b^3 - 15*a^4*B + 29*a^2*b^2*B - 8*b^4*B)*Sin[c + d*x])/(4
*b^3*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]) + (a*(A*b - a*B)*Sin[c + d*x])/(2*b*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*
(b + a*Cos[c + d*x])^2) + (a*(a^2*A*b - 7*A*b^3 - 5*a^3*B + 11*a*b^2*B)*Sin[c + d*x])/(4*b^2*(a^2 - b^2)^2*d*S
qrt[Cos[c + d*x]]*(b + a*Cos[c + d*x]))

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]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 2954

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 3000

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)), Int
[(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 3002

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 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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*} \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\int \frac {B+A \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^3} \, dx\\ &=\frac {a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}-\frac {\int \frac {\frac {1}{2} \left (a A b-5 a^2 B+4 b^2 B\right )+2 b (A b-a B) \cos (c+d x)-\frac {3}{2} a (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(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) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} \left (-3 a^3 A b+9 a A b^3+15 a^4 B-29 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-7 A b^3-5 a^3 B+11 a b^2 B\right ) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{8} \left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right )+\frac {1}{2} b \left (a^3 A b-4 a A b^3-5 a^4 B+10 a^2 b^2 B-2 b^4 B\right ) \cos (c+d x)+\frac {1}{8} a \left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}-\frac {\int \frac {-\frac {1}{8} a \left (3 a^4 A b-5 a^2 A b^3+8 A b^5-15 a^5 B+33 a^3 b^2 B-24 a b^4 B\right )-\frac {1}{8} a^2 b \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b^3 \left (a^2-b^2\right )^2}+\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}+\frac {\left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}+\frac {\left (3 a^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^3 \left (a^2-b^2\right )^2 d}+\frac {\left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) F\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^4 A b-6 a^2 A b^3+15 A b^5-15 a^5 B+38 a^3 b^2 B-35 a b^4 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 (a-b)^2 b^3 (a+b)^3 d}-\frac {\left (3 a^3 A b-9 a A b^3-15 a^4 B+29 a^2 b^2 B-8 b^4 B\right ) \sin (c+d x)}{4 b^3 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)}}+\frac {a (A b-a B) \sin (c+d x)}{2 b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2}+\frac {a \left (a^2 A b-7 A b^3-5 a^3 B+11 a b^2 B\right ) \sin (c+d x)}{4 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.18, size = 458, normalized size = 1.09 \[ \frac {\frac {\sqrt {\cos (c+d x)} \left (16 B \left (b^3-a^2 b\right )^2 \tan (c+d x)+a^2 \left (15 a^4 B-3 a^3 A b-29 a^2 b^2 B+9 a A b^3+8 b^4 B\right ) \sin (2 (c+d x))+2 a b \left (25 a^4 B-5 a^3 A b-47 a^2 b^2 B+11 a A b^3+16 b^4 B\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a \cos (c+d x)+b)^2}-\frac {\frac {8 b \left (5 a^4 B-a^3 A b-10 a^2 b^2 B+4 a A b^3+2 b^4 B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a (a+b)}+\frac {\left (15 a^4 B-3 a^3 A b-29 a^2 b^2 B+9 a A b^3+8 b^4 B\right ) \sin (c+d x) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}+\frac {\left (45 a^5 B-9 a^4 A b-95 a^3 b^2 B+19 a^2 A b^3+56 a b^4 B-16 A b^5\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}}{8 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^3),x]

[Out]

(-((((-9*a^4*A*b + 19*a^2*A*b^3 - 16*A*b^5 + 45*a^5*B - 95*a^3*b^2*B + 56*a*b^4*B)*EllipticPi[(2*a)/(a + b), (
c + d*x)/2, 2])/(a + b) + (8*b*(-(a^3*A*b) + 4*a*A*b^3 + 5*a^4*B - 10*a^2*b^2*B + 2*b^4*B)*((a + b)*EllipticF[
(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a*(a + b)) + ((-3*a^3*A*b + 9*a*A*b^3 + 15*a^
4*B - 29*a^2*b^2*B + 8*b^4*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)) + (Sqrt[Cos[c + d*x]]*(2*a*b*(-5*a^3*A*b + 11*a*A*b^3 + 25*a^
4*B - 47*a^2*b^2*B + 16*b^4*B)*Sin[c + d*x] + a^2*(-3*a^3*A*b + 9*a*A*b^3 + 15*a^4*B - 29*a^2*b^2*B + 8*b^4*B)
*Sin[2*(c + d*x)] + 16*(-(a^2*b) + b^3)^2*B*Tan[c + d*x]))/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2))/(8*b^3*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(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)^(7/2)), x)

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maple [B]  time = 23.81, size = 2024, normalized size = 4.82 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x)

[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^2*B/b^3/(a^2-a*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)*EllipticPi(cos(1/2*d*x
+1/2*c),2*a/(a-b),2^(1/2))+2*(A*b-B*a)/b*(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)*EllipticPi(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)))+2/b^3*B*(-(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*(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*(-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)-2*a*B/b^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)-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)*Elliptic
E(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))))/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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^3),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^3), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(7/2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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