3.602 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=395 \[ \frac{\left (-61 a^2 b^2+8 a^4+35 b^4\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{12 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (-65 a^2 b^2+24 a^4+35 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{b^2 \left (-86 a^2 b^2+63 a^4+35 b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{\left (-61 a^2 b^2+8 a^4+35 b^4\right ) \sin (c+d x)}{12 a^3 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x)}-\frac{b \left (-65 a^2 b^2+24 a^4+35 b^4\right ) \sin (c+d x)}{4 a^4 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}} \]

[Out]

(b*(24*a^4 - 65*a^2*b^2 + 35*b^4)*EllipticE[(c + d*x)/2, 2])/(4*a^4*(a^2 - b^2)^2*d) + ((8*a^4 - 61*a^2*b^2 +
35*b^4)*EllipticF[(c + d*x)/2, 2])/(12*a^3*(a^2 - b^2)^2*d) + (b^2*(63*a^4 - 86*a^2*b^2 + 35*b^4)*EllipticPi[(
2*b)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + ((8*a^4 - 61*a^2*b^2 + 35*b^4)*Sin[c + d*x])/(1
2*a^3*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)) - (b*(24*a^4 - 65*a^2*b^2 + 35*b^4)*Sin[c + d*x])/(4*a^4*(a^2 - b^2)
^2*d*Sqrt[Cos[c + d*x]]) + (b^2*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^2) +
(b^2*(13*a^2 - 7*b^2)*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.35036, antiderivative size = 395, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2802, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{\left (-61 a^2 b^2+8 a^4+35 b^4\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{12 a^3 d \left (a^2-b^2\right )^2}+\frac{b \left (-65 a^2 b^2+24 a^4+35 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac{b^2 \left (-86 a^2 b^2+63 a^4+35 b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{b^2 \sin (c+d x)}{2 a d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{\left (-61 a^2 b^2+8 a^4+35 b^4\right ) \sin (c+d x)}{12 a^3 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x)}-\frac{b \left (-65 a^2 b^2+24 a^4+35 b^4\right ) \sin (c+d x)}{4 a^4 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^3),x]

[Out]

(b*(24*a^4 - 65*a^2*b^2 + 35*b^4)*EllipticE[(c + d*x)/2, 2])/(4*a^4*(a^2 - b^2)^2*d) + ((8*a^4 - 61*a^2*b^2 +
35*b^4)*EllipticF[(c + d*x)/2, 2])/(12*a^3*(a^2 - b^2)^2*d) + (b^2*(63*a^4 - 86*a^2*b^2 + 35*b^4)*EllipticPi[(
2*b)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + ((8*a^4 - 61*a^2*b^2 + 35*b^4)*Sin[c + d*x])/(1
2*a^3*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)) - (b*(24*a^4 - 65*a^2*b^2 + 35*b^4)*Sin[c + d*x])/(4*a^4*(a^2 - b^2)
^2*d*Sqrt[Cos[c + d*x]]) + (b^2*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])^2) +
(b^2*(13*a^2 - 7*b^2)*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x]))

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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 3055

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] + Dis
t[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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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]

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 3002

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 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]

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]

Rubi steps

\begin{align*} \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^3} \, dx &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} \left (4 a^2-7 b^2\right )-2 a b \cos (c+d x)+\frac{5}{2} b^2 \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{4} \left (8 a^4-61 a^2 b^2+35 b^4\right )-a b \left (4 a^2-b^2\right ) \cos (c+d x)+\frac{3}{4} b^2 \left (13 a^2-7 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\int \frac{-\frac{3}{8} b \left (24 a^4-65 a^2 b^2+35 b^4\right )+\frac{1}{2} a \left (2 a^4+14 a^2 b^2-7 b^4\right ) \cos (c+d x)+\frac{1}{8} b \left (8 a^4-61 a^2 b^2+35 b^4\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{b \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{2 \int \frac{\frac{1}{16} \left (8 a^6+128 a^4 b^2-223 a^2 b^4+105 b^6\right )+\frac{1}{4} a b \left (20 a^4-64 a^2 b^2+35 b^4\right ) \cos (c+d x)+\frac{3}{16} b^2 \left (24 a^4-65 a^2 b^2+35 b^4\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{b \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{2 \int \frac{-\frac{1}{16} b \left (8 a^6+128 a^4 b^2-223 a^2 b^4+105 b^6\right )-\frac{1}{16} a b^2 \left (8 a^4-61 a^2 b^2+35 b^4\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^4 b \left (a^2-b^2\right )^2}+\frac{\left (b \left (24 a^4-65 a^2 b^2+35 b^4\right )\right ) \int \sqrt{\cos (c+d x)} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{b \left (24 a^4-65 a^2 b^2+35 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{b \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (b^2 \left (63 a^4-86 a^2 b^2+35 b^4\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{24 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{b \left (24 a^4-65 a^2 b^2+35 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{12 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \left (63 a^4-86 a^2 b^2+35 b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac{\left (8 a^4-61 a^2 b^2+35 b^4\right ) \sin (c+d x)}{12 a^3 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{b \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sin (c+d x)}{4 a^4 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}+\frac{b^2 \sin (c+d x)}{2 a \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))^2}+\frac{b^2 \left (13 a^2-7 b^2\right ) \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.33007, size = 353, normalized size = 0.89 \[ \frac{4 \sqrt{\cos (c+d x)} \left (\frac{3 b^4 \sin (c+d x) \left (b \left (17 a^2-11 b^2\right ) \cos (c+d x)+19 a^3-13 a b^2\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+8 \tan (c+d x) (a \sec (c+d x)-9 b)\right )+\frac{\frac{2 \left (328 a^4 b^2-641 a^2 b^4+16 a^6+315 b^6\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{16 \left (-64 a^3 b^2+20 a^5+35 a b^4\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{6 \left (-65 a^2 b^2+24 a^4+35 b^4\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{48 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^3),x]

[Out]

(((2*(16*a^6 + 328*a^4*b^2 - 641*a^2*b^4 + 315*b^6)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*(
20*a^5 - 64*a^3*b^2 + 35*a*b^4)*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2]))/(a + b) + (6*(24*a^4 - 65*a^2*b^2 + 35*b^4)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a +
b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -
1])*Sin[c + d*x])/(a*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2) + 4*Sqrt[Cos[c + d*x]]*((3*b^4*(19*a^3 - 13*
a*b^2 + b*(17*a^2 - 11*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + 8*(-9*b + a*S
ec[c + d*x])*Tan[c + d*x]))/(48*a^4*d)

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Maple [B]  time = 18.621, size = 2128, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-12*b^3/a^4/(-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))+2/a^2*b^2*(-1/2/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+s
in(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^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)-7/8/(a+b)/(a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+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))*b
+3/8/(a+b)/(a^2-b^2)/a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(a^2-b^2)^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)*EllipticF(cos
(1/2*d*x+1/2*c),2^(1/2))+3/8*b^3/a^2/(a^2-b^2)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9/8*b/(a^2-b^2)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2
))-15/4*a^2/(a^2-b^2)^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*si
n(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))+3/2/(a^2-b^2)
^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))-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^
2)*b^5*(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)))-6/a^4*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*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))+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)+4/a^3*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*c
os(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^3*(-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)*Ellipt
icF(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(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,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(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,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(1/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((b*cos(d*x + c) + a)^3*cos(d*x + c)^(5/2)), x)