\(\int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx\) [165]

Optimal result

Integrand size = 33, antiderivative size = 254 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(33 A-13 B) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 (17 A-7 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

7/10*(17*A-7*B)*(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^3/d+1/
6*(33*A-13*B)*(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^3/d+1/6*
(33*A-13*B)*sin(d*x+c)/a^3/d/cos(d*x+c)^(3/2)-1/5*(A-B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^3-1/3*(
2*A-B)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-7/30*(17*A-7*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a^3+a
^3*cos(d*x+c))-7/10*(17*A-7*B)*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3057, 2827, 2716, 2720, 2719} \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(33 A-13 B) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 (17 A-7 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^3} \]

[In]

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

[Out]

(7*(17*A - 7*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((33*A - 13*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) +
((33*A - 13*B)*Sin[c + d*x])/(6*a^3*d*Cos[c + d*x]^(3/2)) - (7*(17*A - 7*B)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c
+ d*x]]) - ((A - B)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^3) - ((2*A - B)*Sin[c + d*x])/
(3*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^2) - (7*(17*A - 7*B)*Sin[c + d*x])/(30*d*Cos[c + d*x]^(3/2)*(a^
3 + a^3*Cos[c + d*x]))

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

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

Rule 2720

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3057

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (13 A-3 B)-\frac {7}{2} a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {3}{2} a^2 (23 A-8 B)-\frac {25}{2} a^2 (2 A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {\frac {15}{4} a^3 (33 A-13 B)-\frac {21}{4} a^3 (17 A-7 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{15 a^6} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(33 A-13 B) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{4 a^3}-\frac {(7 (17 A-7 B)) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{20 a^3} \\ & = \frac {(33 A-13 B) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 (17 A-7 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(33 A-13 B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}+\frac {(7 (17 A-7 B)) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3} \\ & = \frac {7 (17 A-7 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(33 A-13 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(33 A-13 B) \sin (c+d x)}{6 a^3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {7 (17 A-7 B) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B) \sin (c+d x)}{5 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \cos ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 8.36 (sec) , antiderivative size = 1110, normalized size of antiderivative = 4.37 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=-\frac {22 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}+\frac {26 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (a+a \cos (c+d x))^3 \sqrt {1+\cot ^2(c)}}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {2 (60 A-20 B+59 A \cos (c)-29 B \cos (c)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c)}{5 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (59 A \sin \left (\frac {d x}{2}\right )-29 B \sin \left (\frac {d x}{2}\right )\right )}{5 d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (16 A \sin \left (\frac {d x}{2}\right )-11 B \sin \left (\frac {d x}{2}\right )\right )}{15 d}-\frac {2 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )\right )}{5 d}+\frac {16 A \sec (c) \sec ^2(c+d x) \sin (d x)}{3 d}+\frac {16 \sec (c) \sec (c+d x) (A \sin (c)-9 A \sin (d x)+3 B \sin (d x))}{3 d}-\frac {4 (16 A-11 B) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}-\frac {2 (A-B) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{(a+a \cos (c+d x))^3}-\frac {119 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (a+a \cos (c+d x))^3}+\frac {49 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d (a+a \cos (c+d x))^3} \]

[In]

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

[Out]

(-22*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - Arc
Tan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) + (26*B*Cos[
c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x -
ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]
])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[c]^2]) + (Cos[c/2 + (d*x)/2]
^6*Sqrt[Cos[c + d*x]]*((-2*(60*A - 20*B + 59*A*Cos[c] - 29*B*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) - (4*Sec[
c/2]*Sec[c/2 + (d*x)/2]*(59*A*Sin[(d*x)/2] - 29*B*Sin[(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(16*
A*Sin[(d*x)/2] - 11*B*Sin[(d*x)/2]))/(15*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2
]))/(5*d) + (16*A*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (16*Sec[c]*Sec[c + d*x]*(A*Sin[c] - 9*A*Sin[d*x] + 3
*B*Sin[d*x]))/(3*d) - (4*(16*A - 11*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(A - B)*Sec[c/2 + (d*x)/2]^4
*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x])^3 - (119*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[
{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan
[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[
1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]
]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*
(a + a*Cos[c + d*x])^3) + (49*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}
, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 +
Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((
Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2
])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(5*d*(a + a*Cos[c + d*x]
)^3)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(875\) vs. \(2(282)=564\).

Time = 6.06 (sec) , antiderivative size = 876, normalized size of antiderivative = 3.45

method	result	size
default	\(\text {Expression too large to display}\)	\(876\)

[In]

int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/60*(4*(-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)*(165*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*B*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(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)^6-10*(-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)*(165*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*
B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^4*cos(
1/2*d*x+1/2*c)+8*(-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)*(165*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
-65*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(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*(-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)*(165*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-357*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))-65*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c
)-168*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(17*A-7*B)*sin(1/2*d*x+1/2*c)^10+8*(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(1167*A-482*B)*sin(1/2*d*x+1/2*c)^8-10*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)*(1111*A-461*B)*sin(1/2*d*x+1/2*c)^6+14*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*(404*A-169*B)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(1029*A-439*B)*sin(1/
2*d*x+1/2*c)^2)/a^3/cos(1/2*d*x+1/2*c)^5/(-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)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.13 (sec) , antiderivative size = 548, normalized size of antiderivative = 2.16 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=-\frac {2 \, {\left (21 \, {\left (17 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (453 \, A - 188 \, B\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (139 \, A - 59 \, B\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right ) - 20 \, A\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (33 i \, A - 13 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-33 i \, A + 13 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, {\left (\sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-17 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, {\left (\sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, \sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (17 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(2*(21*(17*A - 7*B)*cos(d*x + c)^4 + 2*(453*A - 188*B)*cos(d*x + c)^3 + 5*(139*A - 59*B)*cos(d*x + c)^2
+ 60*(2*A - B)*cos(d*x + c) - 20*A)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c
)^5 + 3*sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^3 + sqrt(2)*(33*I*
A - 13*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-33*I*A +
13*I*B)*cos(d*x + c)^5 + 3*sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x +
c)^3 + sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) +
21*(sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^5 + 3*sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(-17*I*A
+ 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInvers
e(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*(sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^5 + 3*sqrt(2)*(17*I*A - 7
*I*B)*cos(d*x + c)^4 + 3*sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^2)*we
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^5 + 3*a^3
*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

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

[Out]

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