\(\int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) [415]

Optimal result

Integrand size = 33, antiderivative size = 277 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^3 d}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^2 d}+\frac {2 a^2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}-\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d} \]

[Out]

2/3*(A*b-B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/b^2/d+2/5*B*sec(d*x+c)^(5/2)*sin(d*x+c)/b/d-2/5*(5*A*a*b-5*B*a^2-3*B
*b^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/b^3/d+2/5*(5*A*a*b-5*B*a^2-3*B*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/d+2/3*(A*b-B*a)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/
b^2/d+2*a^2*(A*b-B*a)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a+b)/d

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4118, 4187, 4191, 3934, 2884, 3872, 3856, 2719, 2720} \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {2 a^2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 d (a+b)}-\frac {2 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 b^3 d}+\frac {2 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 b^2 d}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d}+\frac {2 B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 b d} \]

[In]

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

[Out]

(2*(5*a*A*b - 5*a^2*B - 3*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*b^3*d) +
(2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*b^2*d) + (2*a^2*(A*b - a*B)
*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^3*(a + b)*d) - (2*(5*a*A*
b - 5*a^2*B - 3*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*b^3*d) + (2*(A*b - a*B)*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(3*b^2*d) + (2*B*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*b*d)

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 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3934

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4118

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

Rule 4187

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

Rule 4191

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {2 \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3 a B}{2}+\frac {3}{2} b B \sec (c+d x)+\frac {5}{2} (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{5 b} \\ & = \frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {4 \int \frac {\sqrt {\sec (c+d x)} \left (\frac {5}{4} a (A b-a B)+\frac {1}{4} b (5 A b+4 a B) \sec (c+d x)-\frac {3}{4} \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{15 b^2} \\ & = -\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {8 \int \frac {\frac {3}{8} a \left (5 a A b-5 a^2 B-3 b^2 B\right )+\frac {1}{8} b \left (20 a A b-20 a^2 B-9 b^2 B\right ) \sec (c+d x)+\frac {5}{8} \left (3 a^2+b^2\right ) (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 b^3} \\ & = -\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {8 \int \frac {\frac {3}{8} a^2 \left (5 a A b-5 a^2 B-3 b^2 B\right )-\left (-\frac {1}{8} a b \left (20 a A b-20 a^2 B-9 b^2 B\right )+\frac {3}{8} a b \left (5 a A b-5 a^2 B-3 b^2 B\right )\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx}{15 a^2 b^3}+\frac {\left (a^2 (A b-a B)\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{b^3} \\ & = -\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {(A b-a B) \int \sqrt {\sec (c+d x)} \, dx}{3 b^2}+\frac {\left (5 a A b-5 a^2 B-3 b^2 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{5 b^3}+\frac {\left (a^2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3} \\ & = \frac {2 a^2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}-\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {\left ((A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^2}+\frac {\left (\left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^3} \\ & = \frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^3 d}+\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^2 d}+\frac {2 a^2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}-\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac {2 B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 b d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(664\) vs. \(2(277)=554\).

Time = 10.73 (sec) , antiderivative size = 664, normalized size of antiderivative = 2.40 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {\frac {2 \left (-45 a^2 A b-10 A b^3+45 a^3 B+19 a b^2 B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (-40 a A b^2+40 a^2 b B+18 b^3 B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (-15 a^2 A b+15 a^3 B+9 a b^2 B\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{30 b^3 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {2 \left (-5 a A b+5 a^2 B+3 b^2 B\right ) \sin (c+d x)}{5 b^3}+\frac {2 \sec (c+d x) (A b \sin (c+d x)-a B \sin (c+d x))}{3 b^2}+\frac {2 B \sec (c+d x) \tan (c+d x)}{5 b}\right )}{d} \]

[In]

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

[Out]

-1/30*((2*(-45*a^2*A*b - 10*A*b^3 + 45*a^3*B + 19*a*b^2*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]
], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin
[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(-40*a*A*b^2 + 40*a^2*b*B + 18*b^3*B)*Cos[c + d*
x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c +
d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((-15*a^2*A*b + 15*a^3*B + 9*a*b^2*B)*Cos[2*(c + d*x)]*(
a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[
c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*a*(a - 2*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]
]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[
1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec
[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d
*x]^2)))/(b^3*d) + (Sqrt[Sec[c + d*x]]*((2*(-5*a*A*b + 5*a^2*B + 3*b^2*B)*Sin[c + d*x])/(5*b^3) + (2*Sec[c + d
*x]*(A*b*Sin[c + d*x] - a*B*Sin[c + d*x]))/(3*b^2) + (2*B*Sec[c + d*x]*Tan[c + d*x])/(5*b)))/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(757\) vs. \(2(327)=654\).

Time = 37.93 (sec) , antiderivative size = 758, normalized size of antiderivative = 2.74

method	result	size
default	\(\text {Expression too large to display}\)	\(758\)

[In]

int(sec(d*x+c)^(7/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),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)*(2/5*B/b/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2
*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4
-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(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)+2*(A*b-B*a)/b^2*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(A*b-B*a)/
b^3*a/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2))-2*(A*b-B*a)*a^3/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)))/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. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]

[In]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + b/cos(c + d*x)),x)

[Out]

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + b/cos(c + d*x)), x)