3.718 \(\int \frac{\sec ^{\frac{5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=341 \[ \frac{b^2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac{\left (2 a^2-5 b^2\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac{b \left (4 a^2-5 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{b^2 \left (7 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \]

[Out]

(b*(4*a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a^2 - b^2)*d) + ((2*
a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*(a^2 - b^2)*d) + (b^2*(7*
a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a - b)*(a
+ b)^2*d) - (b*(4*a^2 - 5*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((2*a^2 - 5*b^2)*Sec[c +
d*x]^(3/2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d) + (b^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(b +
a*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.965983, antiderivative size = 341, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {3238, 3845, 4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{b^2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{a d \left (a^2-b^2\right ) (a \sec (c+d x)+b)}+\frac{\left (2 a^2-5 b^2\right ) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a^2 d \left (a^2-b^2\right )}-\frac{b \left (4 a^2-5 b^2\right ) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d \left (a^2-b^2\right )}+\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{b^2 \left (7 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(4*a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a^2 - b^2)*d) + ((2*
a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*(a^2 - b^2)*d) + (b^2*(7*
a^2 - 5*b^2)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a - b)*(a
+ b)^2*d) - (b*(4*a^2 - 5*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((2*a^2 - 5*b^2)*Sec[c +
d*x]^(3/2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d) + (b^2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(b +
a*Sec[c + d*x]))

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3845

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

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 4106

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

Rule 3849

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sec ^{\frac{5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\int \frac{\sec ^{\frac{9}{2}}(c+d x)}{(b+a \sec (c+d x))^2} \, dx\\ &=\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3 b^2}{2}-a b \sec (c+d x)+\frac{1}{2} \left (2 a^2-5 b^2\right ) \sec ^2(c+d x)\right )}{b+a \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (2 a^2-5 b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{2 \int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{4} b \left (2 a^2-5 b^2\right )+\frac{1}{2} a \left (a^2+2 b^2\right ) \sec (c+d x)-\frac{3}{4} b \left (4 a^2-5 b^2\right ) \sec ^2(c+d x)\right )}{b+a \sec (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=-\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2-5 b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{4 \int \frac{\frac{3}{8} b^2 \left (4 a^2-5 b^2\right )+\frac{1}{4} a b \left (7 a^2-10 b^2\right ) \sec (c+d x)+\frac{1}{8} \left (2 a^4+16 a^2 b^2-15 b^4\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (b+a \sec (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}\\ &=-\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2-5 b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{4 \int \frac{\frac{3}{8} b^3 \left (4 a^2-5 b^2\right )-\left (-\frac{1}{4} a b^2 \left (7 a^2-10 b^2\right )+\frac{3}{8} a b^2 \left (4 a^2-5 b^2\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 a^3 b^2 \left (a^2-b^2\right )}+\frac{\left (b^2 \left (7 a^2-5 b^2\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{b+a \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2-5 b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{\left (2 a^2-5 b^2\right ) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2 \left (a^2-b^2\right )}+\frac{\left (b \left (4 a^2-5 b^2\right )\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}+\frac{\left (b^2 \left (7 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{b^2 \left (7 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}-\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2-5 b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}+\frac{\left (\left (2 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2 \left (a^2-b^2\right )}+\frac{\left (b \left (4 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \left (7 a^2-5 b^2\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^3 (a-b) (a+b)^2 d}-\frac{b \left (4 a^2-5 b^2\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac{\left (2 a^2-5 b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{b^2 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.71211, size = 661, normalized size = 1.94 \[ \frac{\sqrt{\sec (c+d x)} \left (-\frac{b \left (4 a^2-5 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right )}-\frac{b^3 \sin (c+d x)}{a^2 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{2 \tan (c+d x)}{3 a^2}\right )}{d}+\frac{-\frac{2 \left (40 a b^3-28 a^3 b\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{2 \left (-44 a^2 b^2-4 a^4+45 b^4\right ) \sin (c+d x) \cos ^2(c+d x) \sqrt{1-\sec ^2(c+d x)} (a \sec (c+d x)+b) \left (\Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac{\left (15 b^4-12 a^2 b^2\right ) \sin (c+d x) \cos (2 (c+d x)) (a \sec (c+d x)+b) \left (4 a^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt{\sec (c+d x)} \sqrt{1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}}{12 a^3 d (b-a) (a+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*(-28*a^3*b + 40*a*b^3)*Cos[c + d*x]^2*EllipticPi[-(a/b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c +
d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(-4*a^4 - 44*a
^2*b^2 + 45*b^4)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + EllipticPi[-(a/b), -ArcSin[Sqrt[S
ec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - C
os[c + d*x]^2)) + ((-12*a^2*b^2 + 15*b^4)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2
- 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b
*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 4*a^2*EllipticPi[-(a/
b), -ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 2*b^2*EllipticPi[-(a/b), -A
rcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a*b^2*(a + b*Cos[c
+ d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(12*a^3*(-a + b)*(a + b)*d) + (Sqrt[Sec
[c + d*x]]*(-((b*(4*a^2 - 5*b^2)*Sin[c + d*x])/(a^3*(a^2 - b^2))) - (b^3*Sin[c + d*x])/(a^2*(a^2 - b^2)*(a + b
*Cos[c + d*x])) + (2*Tan[c + d*x])/(3*a^2)))/d

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Maple [B]  time = 14.516, size = 1008, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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