3.614 \(\int \frac {(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=513 \[ \frac {\left (59 a^2 B+104 a A b+36 b^2 B\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (15 a^3 B+264 a^2 A b+284 a b^2 B+128 A b^3\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{192 b d \sqrt {\cos (c+d x)}}+\frac {\left (133 a^3 B+472 a^2 A b+356 a b^2 B+128 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{192 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (15 a^3 B+264 a^2 A b+284 a b^2 B+128 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{192 b d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (-5 a^4 B+40 a^3 A b+120 a^2 b^2 B+160 a A b^3+48 b^4 B\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{64 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b (11 a B+8 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{4 d \cos ^{\frac {5}{2}}(c+d x)} \]
[Out]
1/4*b*B*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+1/192*(472*A*a^2*b+128*A*b^3+133*B*a^3+356*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/(a+b))^(1/2))*((b+a
*cos(d*x+c))/(a+b))^(1/2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+1/64*(40*A*a^3*b+160*A*a*b^3-5*B*a^4+120*B
*a^2*b^2+48*B*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),2,2^(1/2)*(a/
(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/b/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+1/24*b*(8*A*b+11*B*a)
*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d/cos(d*x+c)^(5/2)+1/96*(104*A*a*b+59*B*a^2+36*B*b^2)*sin(d*x+c)*(a+b*sec(d
*x+c))^(1/2)/d/cos(d*x+c)^(3/2)+1/192*(264*A*a^2*b+128*A*b^3+15*B*a^3+284*B*a*b^2)*sin(d*x+c)*(a+b*sec(d*x+c))
^(1/2)/b/d/cos(d*x+c)^(1/2)-1/192*(264*A*a^2*b+128*A*b^3+15*B*a^3+284*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co
s(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)
/b/d/((b+a*cos(d*x+c))/(a+b))^(1/2)
________________________________________________________________________________________
Rubi [A] time = 2.25, antiderivative size = 513, normalized size of antiderivative = 1.00,
number of steps used = 16, number of rules used = 15, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used
= {2955, 4026, 4096, 4102, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac {\left (59 a^2 B+104 a A b+36 b^2 B\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+15 a^3 B+284 a b^2 B+128 A b^3\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{192 b d \sqrt {\cos (c+d x)}}+\frac {\left (472 a^2 A b+133 a^3 B+356 a b^2 B+128 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{192 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (264 a^2 A b+15 a^3 B+284 a b^2 B+128 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{192 b d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (40 a^3 A b+120 a^2 b^2 B-5 a^4 B+160 a A b^3+48 b^4 B\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{64 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b (11 a B+8 A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{4 d \cos ^{\frac {5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
[Out]
((472*a^2*A*b + 128*A*b^3 + 133*a^3*B + 356*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2,
(2*a)/(a + b)])/(192*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((40*a^3*A*b + 160*a*A*b^3 - 5*a^4*B +
120*a^2*b^2*B + 48*b^4*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(64*b*
d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - ((264*a^2*A*b + 128*A*b^3 + 15*a^3*B + 284*a*b^2*B)*Sqrt[Cos[
c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(192*b*d*Sqrt[(b + a*Cos[c + d*x])/(
a + b)]) + (b*(8*A*b + 11*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(24*d*Cos[c + d*x]^(5/2)) + ((104*a*A*b
+ 59*a^2*B + 36*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(96*d*Cos[c + d*x]^(3/2)) + ((264*a^2*A*b + 128*
A*b^3 + 15*a^3*B + 284*a*b^2*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(192*b*d*Sqrt[Cos[c + d*x]]) + (b*B*(a
+ b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(5/2))
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 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 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]
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 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 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 2955
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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] && !IntegerQ[p] && !(IntegerQ[m] && IntegerQ[n])
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 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 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 4026
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(m + n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*Simp[a^2*A*(m + n) + a*b*B*n + (a
*(2*A*b + a*B)*(m + n) + b^2*B*(m + n - 1))*Csc[e + f*x] + b*(A*b*(m + n) + a*B*(2*m + n - 1))*Csc[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && !
(IGtQ[n, 1] && !IntegerQ[m])
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 4096
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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^
n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C
*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] &&
!LeQ[n, -1]
Rule 4102
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 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]
Rubi steps
\begin {align*} \int \frac {(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\\ &=\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{4} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {1}{2} a (8 a A+3 b B)+\left (8 a A b+4 a^2 B+3 b^2 B\right ) \sec (c+d x)+\frac {1}{2} b (8 A b+11 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{12} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{4} a \left (16 a^2 A+8 A b^2+17 a b B\right )+\frac {1}{2} \left (72 a^2 A b+16 A b^3+24 a^3 B+49 a b^2 B\right ) \sec (c+d x)+\frac {1}{4} b \left (104 a A b+59 a^2 B+36 b^2 B\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{8} a b \left (104 a A b+59 a^2 B+36 b^2 B\right )+\frac {1}{4} b \left (96 a^3 A+152 a A b^2+161 a^2 b B+36 b^3 B\right ) \sec (c+d x)+\frac {1}{8} b \left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{24 b}\\ &=\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{16} a b \left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right )+\frac {1}{8} a b^2 \left (104 a A b+59 a^2 B+36 b^2 B\right ) \sec (c+d x)+\frac {3}{16} b \left (40 a^3 A b+160 a A b^3-5 a^4 B+120 a^2 b^2 B+48 b^4 B\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{24 b^2}\\ &=\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{16} a b \left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right )+\frac {1}{8} a b^2 \left (104 a A b+59 a^2 B+36 b^2 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{24 b^2}+\frac {\left (\left (40 a^3 A b+160 a A b^3-5 a^4 B+120 a^2 b^2 B+48 b^4 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{128 b}\\ &=\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\left (\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{384 b}+\frac {1}{384} \left (\left (472 a^2 A b+128 A b^3+133 a^3 B+356 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {\left (\left (40 a^3 A b+160 a A b^3-5 a^4 B+120 a^2 b^2 B+48 b^4 B\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{128 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ &=\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (\left (472 a^2 A b+128 A b^3+133 a^3 B+356 a b^2 B\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{384 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (40 a^3 A b+160 a A b^3-5 a^4 B+120 a^2 b^2 B+48 b^4 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{128 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{384 b \sqrt {b+a \cos (c+d x)}}\\ &=\frac {\left (40 a^3 A b+160 a A b^3-5 a^4 B+120 a^2 b^2 B+48 b^4 B\right ) \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 )}{64 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (\left (472 a^2 A b+128 A b^3+133 a^3 B+356 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{384 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{384 b \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {\left (472 a^2 A b+128 A b^3+133 a^3 B+356 a b^2 B\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 )}{192 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (40 a^3 A b+160 a A b^3-5 a^4 B+120 a^2 b^2 B+48 b^4 B\right ) \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 )}{64 b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{192 b d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {b (8 A b+11 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{24 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {\left (104 a A b+59 a^2 B+36 b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{96 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {\left (264 a^2 A b+128 A b^3+15 a^3 B+284 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{192 b d \sqrt {\cos (c+d x)}}+\frac {b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x)}\\ \end {align*}
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Mathematica [C] time = 35.40, size = 131553, normalized size = 256.44 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
[Out]
Result too large to show
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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))^(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
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))/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/cos(d*x + c)^(3/2), x)
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maple [C] time = 2.33, size = 3175, normalized size = 6.19 \[ \text {output too large to display} \]
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))/cos(d*x+c)^(3/2),x)
[Out]
1/192/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(1+cos(d*x+c))*(-264*A*((a-b)/(a+b))^(1/2)*cos(d*x
+c)^4*a^3*b*(1/(1+cos(d*x+c)))^(1/2)+288*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elliptic
Pi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^4-48*B*((a-b
)/(a+b))^(1/2)*b^4*(1/(1+cos(d*x+c)))^(1/2)-472*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b^2*(1/(1+cos(d*x+c)))^
(1/2)-133*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^3*b*(1/(1+cos(d*x+c)))^(1/2)-272*A*((a-b)/(a+b))^(1/2)*cos(d*x+
c)^2*a*b^3*(1/(1+cos(d*x+c)))^(1/2)-254*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)-18
4*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)+264*A*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^3*b
*(1/(1+cos(d*x+c)))^(1/2)+208*A*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)+128*A*cos(d*
x+c)^5*((a-b)/(a+b))^(1/2)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)+118*B*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^3*b*(1/(1+c
os(d*x+c)))^(1/2)+284*B*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)+72*B*cos(d*x+c)^5*((
a-b)/(a+b))^(1/2)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)+128*A*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*b^4+264*A*cos(d*
x+c)^4*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)+144*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a*b^3*(1/(1
+cos(d*x+c)))^(1/2)+15*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b*(1/(1+cos(d*x+c)))^(1/2)-30*B*cos(d*x+c)^4*((a
-b)/(a+b))^(1/2)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)+284*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a*b^3*(1/(1+cos(d*x+c
)))^(1/2)-172*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)+30*B*cos(d*x+c)^4*EllipticF((-
1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/
2)*sin(d*x+c)*a^4-144*B*cos(d*x+c)^4*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(
1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*b^4-15*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(
d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)
*a^4-30*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^
(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4+15*B*cos(d*x+c)^5*((a-b)/(a+b))^(1/2)*a^4*(
1/(1+cos(d*x+c)))^(1/2)+128*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*b^4*(1/(1+cos(d*x+c)))^(1/2)-64*A*cos(d*x+c)^3*
((a-b)/(a+b))^(1/2)*b^4*(1/(1+cos(d*x+c)))^(1/2)+72*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*b^4*(1/(1+cos(d*x+c)))^
(1/2)-24*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^4*(1/(1+cos(d*x+c)))^(1/2)+960*A*cos(d*x+c)^4*((b+a*cos(d*x+c))/
(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+
b))^(1/2))*sin(d*x+c)*a*b^3+118*B*cos(d*x+c)^4*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b
)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a^3*b-76*B*cos(d*x+c)^4*EllipticF((-1
+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*sin(d*x+c)*a^2*b^2+72*B*cos(d*x+c)^4*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))
^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a*b^3+15*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+
cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*
x+c)*a^3*b-284*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(
a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^2*b^2+284*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a
*b^3+720*B*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))
^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^2+144*A*cos(d*x+c)^4*EllipticF((-1+cos(d
*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(
d*x+c)*a^3*b+208*A*cos(d*x+c)^4*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2-352*A*cos(d*x+c)^4*EllipticF((-1+cos(d*x+c))
*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)
*a*b^3-264*A*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b)
)^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^3*b+264*A*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^2*b^2
-128*A*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2
)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a*b^3+240*A*cos(d*x+c)^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*sin(d*x+c)
*a^3*b-15*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^4*a^4*(1/(1+cos(d*x+c)))^(1/2)-64*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*
b^4*(1/(1+cos(d*x+c)))^(1/2))/b/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3/cos
(d*x+c)^(7/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
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))/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/cos(d*x + c)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^(3/2),x)
[Out]
int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2))/cos(c + d*x)^(3/2), 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))**(5/2)*(A+B*sec(d*x+c))/cos(d*x+c)**(3/2),x)
[Out]
Timed out
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