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

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

[Out]

1/4*b*(24*a^4-65*a^2*b^2+35*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))/a^4/(a^2-b^2)^2/d+1/12*(8*a^4-61*a^2*b^2+35*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli
pticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/(a^2-b^2)^2/d+1/4*b^2*(63*a^4-86*a^2*b^2+35*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*b/(a+b),2^(1/2))/a^4/(a-b)^2/(a+b)^3/d+1/12*(8*a^4-61
*a^2*b^2+35*b^4)*sin(d*x+c)/a^3/(a^2-b^2)^2/d/cos(d*x+c)^(3/2)+1/2*b^2*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(3/
2)/(a+b*cos(d*x+c))^2+1/4*b^2*(13*a^2-7*b^2)*sin(d*x+c)/a^2/(a^2-b^2)^2/d/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))-1/
4*b*(24*a^4-65*a^2*b^2+35*b^4)*sin(d*x+c)/a^4/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 1.35, antiderivative size = 395, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 5.96, size = 349, normalized size = 0.88 \[ \frac {4 \sqrt {\cos (c+d x)} \left (\frac {3 b^4 \sin (c+d x) \left (19 a^3+b \left (17 a^2-11 b^2\right ) \cos (c+d x)-13 a b^2\right )}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+8 \tan (c+d x) (a \sec (c+d x)-9 b)\right )+\frac {\frac {16 \left (20 a^5-64 a^3 b^2+35 a b^4\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac {6 \left (24 a^4-65 a^2 b^2+35 b^4\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt {\sin ^2(c+d x)}}+\frac {2 \left (16 a^6+328 a^4 b^2-641 a^2 b^4+315 b^6\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b)^2 (a+b)^2}}{48 a^4 d} \]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [B]  time = 5.45, size = 2128, normalized size = 5.39 \[ \text {Expression too large to display} \]

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)*(2*b^2/a^2*(-1/2*b^2/a/(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*cos(1/2*d*x+1/2*c)^2*b+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*cos(1/2*d*x+1/2*c)^2*b
+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(co
s(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*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))+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)*Elliptic
E(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*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))+3/2/(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)))-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))-6/a^4*b*(-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+
1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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^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)/(-1/2+cos(1/2*d*x+1/2*c)^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)))+4*b^2/a^3*(-b^2/a
/(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*cos(1/2*d*x+1/2*c)^2*b+a
-b)-1/2/(a+b)/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/(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))+1/2*b/a/(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)*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))))/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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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