\(\int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\) [134]

Optimal result

Integrand size = 33, antiderivative size = 120 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {4 a^2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {4 a^2 (2 A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-4*a^2*A*(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))/d+4/3*a^2*(2*A+
3*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))/d+2/3*A*(a^2+a^2*co
s(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/3*a^2*(5*A+3*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3054, 3047, 3100, 2827, 2720, 2719} \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 a^2 (2 A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}-\frac {4 a^2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 A \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d \cos ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

(-4*a^2*A*EllipticE[(c + d*x)/2, 2])/d + (4*a^2*(2*A + 3*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(5*A + 3
*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) + (2*A*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/
2))

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 3047

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

Rule 3054

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*
n] || EqQ[c, 0])

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a (5 A+3 B)-\frac {1}{2} a (A-3 B) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {1}{2} a^2 (5 A+3 B)+\left (-\frac {1}{2} a^2 (A-3 B)+\frac {1}{2} a^2 (5 A+3 B)\right ) \cos (c+d x)-\frac {1}{2} a^2 (A-3 B) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a^2 (5 A+3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4}{3} \int \frac {\frac {1}{2} a^2 (2 A+3 B)-\frac {3}{2} a^2 A \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a^2 (5 A+3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\left (2 a^2 A\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (2 A+3 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {4 a^2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {4 a^2 (2 A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (5 A+3 B) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 A \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 6.31 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.59 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (12 A \cos (c+d x) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))+\left (\csc (c) \left (-6 A \cos (c+d x) (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sec (c)+(12 A \cos (c)+2 A \cos (d x)-2 A \cos (2 c+d x)+12 A \cos (c+2 d x)+3 B \cos (c+2 d x)-3 B \cos (3 c+2 d x)) \sqrt {\sec ^2(c)}\right )-8 (2 A+3 B) \cos ^2(c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\arctan (\cot (c))) \sin (c)\right ) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{24 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}} \]

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(12*A*Cos[c + d*x]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x
+ ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x + ArcTan[Tan[c]]] + (Csc[c]*(-6*A*Cos[c + d*x]*(3*Cos[c - d*x - ArcTan[Ta
n[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sec[c] + (12*A*Cos[c] + 2*A*Cos[d*x] - 2*A*Cos[2*c + d*x] + 12*A*Cos[c
+ 2*d*x] + 3*B*Cos[c + 2*d*x] - 3*B*Cos[3*c + 2*d*x])*Sqrt[Sec[c]^2]) - 8*(2*A + 3*B)*Cos[c + d*x]^2*Sqrt[Cos
[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sqr
t[Sec[c]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(24*d*Cos[c + d*x]^(3/2)*Sqr
t[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(512\) vs. \(2(162)=324\).

Time = 7.33 (sec) , antiderivative size = 513, normalized size of antiderivative = 4.28

method	result	size
default	\(-\frac {4 \left (6 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 A +B \right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (7 A +3 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (2 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+3 A \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a^{2}}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\)	\(513\)
parts	\(\frac {2 \left (A \,a^{2}+2 B \,a^{2}\right ) \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d}-\frac {2 \left (2 A \,a^{2}+B \,a^{2}\right ) \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 A \,a^{2} \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}+\frac {2 B \,a^{2} \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(576\)

[In]

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

[Out]

-4/3*(6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A+B)*cos(1/2*d*x+1/2*c)*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)*(7*A+3*B)*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)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(
2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+3*A*(-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))+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)*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))*a^2/(-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 = 0.12 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.75 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (i \, \sqrt {2} {\left (2 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} {\left (2 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} A a^{2} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-2/3*(I*sqrt(2)*(2*A + 3*B)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - I*s
qrt(2)*(2*A + 3*B)*a^2*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*
A*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*s
qrt(2)*A*a^2*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)))
- (3*(2*A + B)*a^2*cos(d*x + c) + A*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.63 \[ \int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2\,A\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*A*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (4*B*a^2*ellipticF(c/2 + (
d*x)/2, 2))/d + (4*A*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(
c + d*x)^2)^(1/2)) + (2*A*a^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2
)*(sin(c + d*x)^2)^(1/2)) + (2*B*a^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)
^(1/2)*(sin(c + d*x)^2)^(1/2))