3.1049 \(\int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 1.67083, antiderivative size = 441, normalized size of antiderivative = 1., 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 = {4094, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \sin (c+d x) \left (5 a^2 (5 A+7 C)+56 a b B+15 A b^2\right ) \sqrt{a+b \sec (c+d x)}}{105 d \sqrt{\sec (c+d x)}}-\frac{2 \sqrt{\sec (c+d x)} \left (10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)-56 a^3 b B+56 a b^3 B+15 A b^4\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{105 a d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (5 a^2 b (29 A+49 C)+63 a^3 B+161 a b^2 B+15 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{105 a d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 (7 a B+5 A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{2 b^3 C \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}} \]

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 4094

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 4108

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]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[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 2807

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*
Sin[e + f*x])/(c + d)]), 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 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 4035

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 3856

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 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

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

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[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 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Mathematica [F]  time = 51.1526, size = 0, normalized size = 0. \[ \int \frac{(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{7}{2}}(c+d x)} \, dx \]

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]  time = 0.731, size = 5602, normalized size = 12.7 \begin{align*} \text{output too large to display} \end{align*}

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)

[Out]

result too large to display

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

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)