3.11.49 \(\int \frac {(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [1049]

Optimal. Leaf size=441 \[ -\frac {2 \left (15 A b^4-56 a^3 b B+56 a b^3 B+10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{105 a d \sqrt {a+b \sec (c+d x)}}+\frac {2 b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{105 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/35*(5*A*b+7*B*a)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/7*A*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c
)/d/sec(d*x+c)^(5/2)-2/105*(15*A*b^4-56*a^3*b*B+56*a*b^3*B+10*a^2*b^2*(A-7*C)-5*a^4*(5*A+7*C))*(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/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b)
)^(1/2)*sec(d*x+c)^(1/2)/a/d/(a+b*sec(d*x+c))^(1/2)+2*b^3*C*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/d/(a+b*
sec(d*x+c))^(1/2)+2/105*(15*A*b^2+56*a*b*B+5*a^2*(5*A+7*C))*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/
2)+2/105*(15*A*b^3+63*a^3*B+161*a*b^2*B+5*a^2*b*(29*A+49*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a/d/((b+a*cos(d*x+c))/(a+b))^(1/2)
/sec(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.11, antiderivative size = 441, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4179, 4193, 3944, 2886, 2884, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {2 \sin (c+d x) \left (5 a^2 (5 A+7 C)+56 a b B+15 A b^2\right ) \sqrt {a+b \sec (c+d x)}}{105 d \sqrt {\sec (c+d x)}}+\frac {2 \left (63 a^3 B+5 a^2 b (29 A+49 C)+161 a b^2 B+15 A b^3\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \sqrt {\sec (c+d x)} \left (-5 a^4 (5 A+7 C)-56 a^3 b B+10 a^2 b^2 (A-7 C)+56 a b^3 B+15 A b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {a+b \sec (c+d x)}}+\frac {2 (7 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 b^3 C \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {a+b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

(-2*(15*A*b^4 - 56*a^3*b*B + 56*a*b^3*B + 10*a^2*b^2*(A - 7*C) - 5*a^4*(5*A + 7*C))*Sqrt[(b + a*Cos[c + d*x])/
(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(105*a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*b^3
*C*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(d*Sqrt[a
+ b*Sec[c + d*x]]) + (2*(15*A*b^3 + 63*a^3*B + 161*a*b^2*B + 5*a^2*b*(29*A + 49*C))*EllipticE[(c + d*x)/2, (2*
a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(105*a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*(15
*A*b^2 + 56*a*b*B + 5*a^2*(5*A + 7*C))*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]) + (2*
(5*A*b + 7*a*B)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d*Sec[c + d*x]^(3/2)) + (2*A*(a + b*Sec[c + d*x])
^(5/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))

Rule 2732

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

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

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 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*Sin[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] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 3941

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

Rule 3943

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

Rule 3944

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

Rule 4120

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

Rule 4179

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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*
Csc[e + f*x])^n/(f*n)), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4193

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

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+7 a B)+\frac {1}{2} (5 a A+7 b B+7 a C) \sec (c+d x)+\frac {7}{2} b C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {1}{4} \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right )+\frac {1}{4} \left (40 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \sec (c+d x)+\frac {35}{4} b^2 C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8}{105} \int \frac {\frac {1}{8} \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )+\frac {1}{8} \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right ) \sec (c+d x)+\frac {105}{8} b^3 C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {8}{105} \int \frac {\frac {1}{8} \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )+\frac {1}{8} \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx+\left (b^3 C\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {\left (15 A b^4-56 a^3 b B+56 a b^3 B+10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 a}+\frac {\left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{105 a}+\frac {\left (b^3 C \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{\sqrt {a+b \sec (c+d x)}}\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {\left (\left (15 A b^4-56 a^3 b B+56 a b^3 B+10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{105 a \sqrt {a+b \sec (c+d x)}}+\frac {\left (b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{\sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{105 a \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {\left (\left (15 A b^4-56 a^3 b B+56 a b^3 B+10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{105 a \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{105 a \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=-\frac {2 \left (15 A b^4-56 a^3 b B+56 a b^3 B+10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{105 a d \sqrt {a+b \sec (c+d x)}}+\frac {2 b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{105 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt {\sec (c+d x)}}+\frac {2 (5 A b+7 a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [F]
time = 84.89, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2),x]

[Out]

Integrate[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(7/2), x]

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Maple [C] Result contains complex when optimal does not.
time = 0.39, size = 5602, normalized size = 12.70

method	result	size
default	\(\text {Expression too large to display}\)	\(5602\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(7/2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(7/2),x)

[Out]

int(((a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(7/2), x)

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