3.6.45 \(\int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx\) [545]

Optimal. Leaf size=380 \[ -\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {5 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}} \]

[Out]

1/3*b*(3*a^2-5*b^2)*sin(d*x+c)/a^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)+1/3*b*(3*a^4-26*a^2*b^2+15*b^4)*sin(d*x+
c)/a^3/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)-1/3*(3*a^4-26*a^2*b^2+15*b^4)*(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)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/a^3/(a^2-b^2)^2/d/((a
+b*cos(d*x+c))/(a+b))^(1/2)+1/3*(3*a^2-5*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)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/a^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-5*
b*(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)*(b/(a+b))^(1/2))*((a
+b*cos(d*x+c))/(a+b))^(1/2)/a^3/d/(a+b*cos(d*x+c))^(1/2)+tan(d*x+c)/a/d/(a+b*cos(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.74, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {2881, 3135, 3134, 3138, 2734, 2732, 3081, 2742, 2740, 2886, 2884} \begin {gather*} -\frac {5 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

-1/3*((3*a^4 - 26*a^2*b^2 + 15*b^4)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(a^3*(a^2
- b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((3*a^2 - 5*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF
[(c + d*x)/2, (2*b)/(a + b)])/(3*a^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (5*b*Sqrt[(a + b*Cos[c + d*x])/
(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(a^3*d*Sqrt[a + b*Cos[c + d*x]]) + (b*(3*a^2 - 5*b^2)*Sin[
c + d*x])/(3*a^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (b*(3*a^4 - 26*a^2*b^2 + 15*b^4)*Sin[c + d*x])/(3
*a^3*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) + Tan[c + d*x]/(a*d*(a + b*Cos[c + d*x])^(3/2))

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 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2
- b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])
^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m +
n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||
!(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3134

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3135

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c
+ d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C
)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n +
3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3138

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

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (-\frac {5 b}{2}+\frac {3}{2} b \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx}{a}\\ &=\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}+\frac {2 \int \frac {\left (-\frac {15}{4} b \left (a^2-b^2\right )+\frac {3}{2} a b^2 \cos (c+d x)+\frac {1}{4} b \left (3 a^2-5 b^2\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}+\frac {4 \int \frac {\left (-\frac {15}{8} b \left (a^2-b^2\right )^2+\frac {1}{4} a b^2 \left (9 a^2-5 b^2\right ) \cos (c+d x)-\frac {1}{8} b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {4 \int \frac {\left (\frac {15}{8} b^2 \left (a^2-b^2\right )^2-\frac {1}{8} a b \left (3 a^4-8 a^2 b^2+5 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 a^3 b \left (a^2-b^2\right )^2}-\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{6 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {(5 b) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{2 a^3}+\frac {\left (3 a^2-5 b^2\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}-\frac {\left (\left (3 a^4-26 a^2 b^2+15 b^4\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{6 a^3 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}\\ &=-\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}-\frac {\left (5 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{2 a^3 \sqrt {a+b \cos (c+d x)}}+\frac {\left (\left (3 a^2-5 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{6 a^2 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {\left (3 a^4-26 a^2 b^2+15 b^4\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (3 a^2-5 b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 a^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {5 b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a^3 d \sqrt {a+b \cos (c+d x)}}+\frac {b \left (3 a^2-5 b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac {b \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {\tan (c+d x)}{a d (a+b \cos (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 16.61, size = 638, normalized size = 1.68 \begin {gather*} -\frac {b \left (\frac {2 \left (-36 a^3 b+20 a b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (33 a^4-86 a^2 b^2+45 b^4\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i \left (3 a^4-26 a^2 b^2+15 b^4\right ) \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a F\left (i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )-b \Pi \left (\frac {a+b}{a};i \sinh ^{-1}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}\right )}{12 a^3 (-a+b)^2 (a+b)^2 d}+\frac {\sqrt {a+b \cos (c+d x)} \left (-\frac {2 b^3 \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {4 \left (5 a^2 b^3 \sin (c+d x)-3 b^5 \sin (c+d x)\right )}{3 a^3 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {\tan (c+d x)}{a^3}\right )}{d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^2/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

-1/12*(b*((2*(-36*a^3*b + 20*a*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/
Sqrt[a + b*Cos[c + d*x]] + (2*(33*a^4 - 86*a^2*b^2 + 45*b^4)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2,
(c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(3*a^4 - 26*a^2*b^2 + 15*b^4)*Sqrt[(b - b*Cos[c
+ d*x])/(a + b)]*Sqrt[-((b + b*Cos[c + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt
[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*S
qrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*
Cos[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[-((a^2 -
b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b*Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a
+ b*Cos[c + d*x])^2))))/(a^3*(-a + b)^2*(a + b)^2*d) + (Sqrt[a + b*Cos[c + d*x]]*((-2*b^3*Sin[c + d*x])/(3*a^
2*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) - (4*(5*a^2*b^3*Sin[c + d*x] - 3*b^5*Sin[c + d*x]))/(3*a^3*(a^2 - b^2)^2
*(a + b*Cos[c + d*x])) + Tan[c + d*x]/a^3))/d

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1323\) vs. \(2(441)=882\).
time = 0.91, size = 1324, normalized size = 3.48

method	result	size
default	\(\text {Expression too large to display}\)	\(1324\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2/(a+b*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*b^2/a^3/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x
+1/2*c)^2*b-a-b)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*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*b+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ellipt
icE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+
b)/(a-b))^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))+2*b^2/a^2*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1
/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+8/3*si
n(1/2*d*x+1/2*c)^2*b/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)
^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(
a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a
-b))^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*
sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-Ell
ipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))+2/a^2*(-cos(1/2*d*x+1/2*c)/a*(-2*sin(1/2*d*x+1/2*c)^4*b+(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)
^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1
/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/
2*d*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli
pticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2)))+4/a^3*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2
*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c
),2,(-2*b/(a-b))^(1/2)))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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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(sec(d*x+c)^2/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**2/(a + b*cos(c + d*x))**(5/2), x)

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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(sec(d*x+c)^2/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^2/(b*cos(d*x + c) + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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