3.630 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=368 \[ \frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (-5 a^3 B+8 a^2 A b+a b^2 B-4 A b^3\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (-3 a^3 B+9 a^2 A b+2 a b^2 B-8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4 A+6 a^3 b B-15 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
[Out]
2/3*b*(A*b-B*a)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2)+2/3*b*(8*A*a^2*b-4*A*b^3-5*B*
a^3+B*a*b^2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-2/3*(9*A*a^2*b-8*A*b^3-3*B*a
^3+2*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)/a^3/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/3*(3*A*a^4-15*A
*a^2*b^2+8*A*b^4+6*B*a^3*b-2*B*a*b^3)*(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)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^3/(a^2-b^2)^2/d/((b+a*cos(d*x+c))/(a+b
))^(1/2)
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Rubi [A] time = 1.11, antiderivative size = 368, normalized size of antiderivative =
1.00, number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\)
= 0.286, Rules used = {2955, 4030, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac {2 b \left (8 a^2 A b-5 a^3 B+a b^2 B-4 A b^3\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (9 a^2 A b-3 a^3 B+2 a b^2 B-8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-15 a^2 A b^2+3 a^4 A+6 a^3 b B-2 a b^3 B+8 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(-2*(9*a^2*A*b - 8*A*b^3 - 3*a^3*B + 2*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)])/(3*a^3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*a^4*A - 15*a^2*A*b^2 + 8
*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*
x]])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*b*(A*b - a*B)*Sin[c + d*x])/(3*a*(a^2 - b
^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (2*b*(8*a^2*A*b - 4*A*b^3 - 5*a^3*B + a*b^2*B)*Sin[c +
d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]])
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 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 4030
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(b*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/
(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*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] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] && ILtQ[n, 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 4100
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*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*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] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])
Rubi steps
\begin {align*} \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (-3 a^2 A+4 A b^2-a b B\right )+\frac {3}{2} a (A b-a B) \sec (c+d x)-b (A b-a B) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right )-\frac {1}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 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}{3 a^3 \left (a^2-b^2\right )}+\frac {\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 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}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 a b^2 B\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)}}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (9 a^2 A b-8 A b^3-3 a^3 B+2 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}{3 a^3 \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 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}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=-\frac {2 \left (9 a^2 A b-8 A b^3-3 a^3 B+2 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 )}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4 A-15 a^2 A b^2+8 A b^4+6 a^3 b B-2 a b^3 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)}}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 b (A b-a B) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {2 b \left (8 a^2 A b-4 A b^3-5 a^3 B+a b^2 B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [C] time = 19.78, size = 621, normalized size = 1.69 \[ \frac {(a \cos (c+d x)+b)^3 (A+B \sec (c+d x)) \left (-\frac {2 \left (A b^3 \sin (c+d x)-a b^2 B \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}-\frac {2 \left (6 a^3 b B \sin (c+d x)-9 a^2 A b^2 \sin (c+d x)-2 a b^3 B \sin (c+d x)+5 A b^4 \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}\right )}{d \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A \cos (c+d x)+B)}-\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sec ^{\frac {3}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b)^2 (A+B \sec (c+d x)) \left (i a (a+b) \left (3 a^3 (A+B)+3 a^2 b (B-3 A)-2 a b^2 (3 A+B)+8 A b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-\left (3 a^4 A+6 a^3 b B-15 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i (a+b) \left (3 a^4 A+6 a^3 b B-15 a^2 A b^2-2 a b^3 B+8 A b^4\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{3 a d \left (a^3-a b^2\right )^2 (a+b \sec (c+d x))^{5/2} (A \cos (c+d x)+B)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
((b + a*Cos[c + d*x])^3*(A + B*Sec[c + d*x])*((-2*(A*b^3*Sin[c + d*x] - a*b^2*B*Sin[c + d*x]))/(3*a^2*(a^2 - b
^2)*(b + a*Cos[c + d*x])^2) - (2*(-9*a^2*A*b^2*Sin[c + d*x] + 5*A*b^4*Sin[c + d*x] + 6*a^3*b*B*Sin[c + d*x] -
2*a*b^3*B*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*Cos[c + d*x]^(3/2)*(B + A*Cos[c + d*x
])*(a + b*Sec[c + d*x])^(5/2)) - (2*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(3/2)*(Cos[(c + d*x
)/2]^2*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x])*((-I)*(a + b)*(3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B -
2*a*b^3*B)*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*
x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(8*A*b^3 + 3*a^2*b*(-3*A + B) + 3*a^3*(A + B) - 2*a*b^2*(3*A +
B))*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec
[(c + d*x)/2]^2)/(a + b)] - (3*a^4*A - 15*a^2*A*b^2 + 8*A*b^4 + 6*a^3*b*B - 2*a*b^3*B)*(b + a*Cos[c + d*x])*(S
ec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(3*a*(a^3 - a*b^2)^2*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^(
5/2))
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fricas [F] time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*
x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(b*sec(d*x + c) + a)^(5/2), x)
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maple [B] time = 2.69, size = 3337, normalized size = 9.07 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x)
[Out]
2/3/d*(-1+cos(d*x+c))*(1+cos(d*x+c))^2*(6*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^4*b*(1/(1+cos(d*x+c)))^(1/2)-3*A*
(1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^3*b^2-6*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*b^2*(
1/(1+cos(d*x+c)))^(1/2)+8*A*((a-b)/(a+b))^(1/2)*b^5*(1/(1+cos(d*x+c)))^(1/2)+3*A*(1/(1+cos(d*x+c)))^(1/2)*cos(
d*x+c)^2*((a-b)/(a+b))^(1/2)*a^4*b-4*A*(1/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^3-3*A*(
(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-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)*cos(d*x+c)*a^5+3*A*((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)*cos(d*x+c)*a^5+B*(1/(1+cos(d*x+c)))^(1/2
)*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b^2+3*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-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)*cos(d*x+c)*a^5+18*A*((a-b)/(a+b))^(1/
2)*cos(d*x+c)^2*a^3*b^2*(1/(1+cos(d*x+c)))^(1/2)-12*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b^4*(1/(1+cos(d*x+c))
)^(1/2)-6*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^4*b*(1/(1+cos(d*x+c)))^(1/2)+3*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)
^2*a^2*b^3*(1/(1+cos(d*x+c)))^(1/2)-6*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^4*b*(1/(1+cos(d*x+c)))^(1/2)-12*A*((a
-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*b^2*(1/(1+cos(d*x+c)))^(1/2)+18*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b^3*(1/(1
+cos(d*x+c)))^(1/2)+8*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^4*(1/(1+cos(d*x+c)))^(1/2)+3*A*((a-b)/(a+b))^(1/2)*
cos(d*x+c)^3*a^4*b*(1/(1+cos(d*x+c)))^(1/2)-3*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b^3*(1/(1+cos(d*x+c)))^(1
/2)+3*A*sin(d*x+c)*EllipticE((-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)*a^4*b-15*A*sin(d*x+c)*EllipticE((-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)*a^2*b^3+3*B*sin(d*x+c)*((b+a*cos(d*x+c)
)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a
^4*b-3*B*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2
)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b^2-2*B*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^3+6*B*sin(d*x+c)*EllipticE((-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)
*a^3*b^2-2*B*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*c
os(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a*b^4-3*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elli
pticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*b-9*A*sin(d*x+c)*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*a^3*b^2+6*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^
(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^3+8*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*El
lipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^4-8*A*((a-b)/(a+b))^(1/2)*cos
(d*x+c)*b^5*(1/(1+cos(d*x+c)))^(1/2)-3*A*((a-b)/(a+b))^(1/2)*a^3*b^2*(1/(1+cos(d*x+c)))^(1/2)-11*A*((a-b)/(a+b
))^(1/2)*a^2*b^3*(1/(1+cos(d*x+c)))^(1/2)+4*A*((a-b)/(a+b))^(1/2)*a*b^4*(1/(1+cos(d*x+c)))^(1/2)-9*A*((b+a*cos
(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-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)*cos(d*x+c)*a^4*b+6*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-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)*cos(d*x+c)*a^3*b^2+8*A*((b+a*cos(d*x+c))/(1+co
s(d*x+c))/(a+b))^(1/2)*EllipticF((-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)*cos(d*x+c)*a^2*b^3-15*A*((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)*cos(d*x+c)*a^3*b^2+8*A*((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)*cos(d*x+
c)*a*b^4-3*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-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)*cos(d*x+c)*a^4*b-2*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elli
pticF((-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)*cos(d*x+c)*a^3*b^2+6*B*(
(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)*cos(d*x+c)*a^4*b-2*B*((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)*cos(d*x+c)*a^2*b^3+3*A*(1/(1+cos(d*x+c
)))^(1/2)*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^5+5*B*((a-b)/(a+b))^(1/2)*a^3*b^2*(1/(1+cos(d*x+c)))^(1/2)-B*((a-
b)/(a+b))^(1/2)*a^2*b^3*(1/(1+cos(d*x+c)))^(1/2)-2*B*((a-b)/(a+b))^(1/2)*a*b^4*(1/(1+cos(d*x+c)))^(1/2)-3*A*((
a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^5*(1/(1+cos(d*x+c)))^(1/2)+8*A*sin(d*x+c)*EllipticE((-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)*b^5-2*B*((a-b)/(a+b)
)^(1/2)*cos(d*x+c)*a^2*b^3*(1/(1+cos(d*x+c)))^(1/2)+2*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^4*(1/(1+cos(d*x+c))
)^(1/2))*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)/a^3
/(a+b)/(a-b)^2/(b+a*cos(d*x+c))^2/sin(d*x+c)^3
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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)^(1/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^(5/2),x)
[Out]
int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^(5/2), 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))*cos(d*x+c)**(1/2)/(a+b*sec(d*x+c))**(5/2),x)
[Out]
Timed out
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