\(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) [130]

Optimal result

Integrand size = 33, antiderivative size = 194 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {4 a^2 (9 A+8 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (6 A+5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^2 (6 A+5 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {4 a^2 (9 A+8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a^2 (9 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d} \]

[Out]

4/15*a^2*(9*A+8*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))/d+4/2
1*a^2*(6*A+5*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+4/45*a
^2*(9*A+8*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/63*a^2*(9*A+11*B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*B*cos(d*x+c)^
(5/2)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d+4/21*a^2*(6*A+5*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3055, 3047, 3102, 2827, 2715, 2720, 2719} \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {4 a^2 (6 A+5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^2 (9 A+8 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 a^2 (9 A+11 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d}+\frac {4 a^2 (9 A+8 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {4 a^2 (6 A+5 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{9 d} \]

[In]

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

[Out]

(4*a^2*(9*A + 8*B)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^2*(6*A + 5*B)*EllipticF[(c + d*x)/2, 2])/(21*d) +
(4*a^2*(6*A + 5*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (4*a^2*(9*A + 8*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x
])/(45*d) + (2*a^2*(9*A + 11*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*B*Cos[c + d*x]^(5/2)*(a^2 + a^2*C
os[c + d*x])*Sin[c + d*x])/(9*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 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 3055

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

Rule 3102

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac {2}{9} \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x)) \left (\frac {1}{2} a (9 A+5 B)+\frac {1}{2} a (9 A+11 B) \cos (c+d x)\right ) \, dx \\ & = \frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac {2}{9} \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a^2 (9 A+5 B)+\left (\frac {1}{2} a^2 (9 A+5 B)+\frac {1}{2} a^2 (9 A+11 B)\right ) \cos (c+d x)+\frac {1}{2} a^2 (9 A+11 B) \cos ^2(c+d x)\right ) \, dx \\ & = \frac {2 a^2 (9 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac {4}{63} \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {9}{2} a^2 (6 A+5 B)+\frac {7}{2} a^2 (9 A+8 B) \cos (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (9 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac {1}{7} \left (2 a^2 (6 A+5 B)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{9} \left (2 a^2 (9 A+8 B)\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx \\ & = \frac {4 a^2 (6 A+5 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {4 a^2 (9 A+8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a^2 (9 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d}+\frac {1}{21} \left (2 a^2 (6 A+5 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\frac {1}{15} \left (2 a^2 (9 A+8 B)\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {4 a^2 (9 A+8 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^2 (6 A+5 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^2 (6 A+5 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {4 a^2 (9 A+8 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 a^2 (9 A+11 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 B \cos ^{\frac {5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{9 d} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 6.56 (sec) , antiderivative size = 944, normalized size of antiderivative = 4.87 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {(9 A+8 B) \cot (c)}{15 d}+\frac {(51 A+46 B) \cos (d x) \sin (c)}{168 d}+\frac {(36 A+37 B) \cos (2 d x) \sin (2 c)}{360 d}+\frac {(A+2 B) \cos (3 d x) \sin (3 c)}{56 d}+\frac {B \cos (4 d x) \sin (4 c)}{144 d}+\frac {(51 A+46 B) \cos (c) \sin (d x)}{168 d}+\frac {(36 A+37 B) \cos (2 c) \sin (2 d x)}{360 d}+\frac {(A+2 B) \cos (3 c) \sin (3 d x)}{56 d}+\frac {B \cos (4 c) \sin (4 d x)}{144 d}\right )-\frac {2 A (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{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)))}}{7 d \sqrt {1+\cot ^2(c)}}-\frac {5 B (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{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)))}}{21 d \sqrt {1+\cot ^2(c)}}-\frac {3 A (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{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 )}{10 d}-\frac {4 B (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{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 )}{15 d} \]

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/15*((9*A + 8*B)*Cot[c])/d + ((51*A + 46*B)*
Cos[d*x]*Sin[c])/(168*d) + ((36*A + 37*B)*Cos[2*d*x]*Sin[2*c])/(360*d) + ((A + 2*B)*Cos[3*d*x]*Sin[3*c])/(56*d
) + (B*Cos[4*d*x]*Sin[4*c])/(144*d) + ((51*A + 46*B)*Cos[c]*Sin[d*x])/(168*d) + ((36*A + 37*B)*Cos[2*c]*Sin[2*
d*x])/(360*d) + ((A + 2*B)*Cos[3*c]*Sin[3*d*x])/(56*d) + (B*Cos[4*c]*Sin[4*d*x])/(144*d)) - (2*A*(a + a*Cos[c
+ d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*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]]]])/(7*d*Sqrt[1 + Cot[c]^2]) - (5*B*(a + a*Cos[c + d*x])^2*Csc[c]*Hype
rgeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*S
qrt[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]]]])/(21*d*Sqrt[1 + Cot[c]^2]) - (3*A*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((H
ypergeometricPFQ[{-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 + T
an[c]^2]]))/(10*d) - (4*B*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((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]]]]*Sq
rt[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 + Ta
n[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(15*d)

Maple [A] (verified)

Time = 13.62 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.13

method	result	size
default	\(-\frac {4 \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 )}\, a^{2} \left (-560 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (360 A +1840 B \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-1044 A -2368 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (1134 A +1568 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-351 A -387 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 )+90 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 )-189 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+75 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 )-168 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{315 \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}\)	\(413\)
parts	\(\text {Expression too large to display}\)	\(822\)

[In]

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

[Out]

-4/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(-560*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^10+(360*A+1840*B)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1044*A-2368*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)+(1134*A+1568*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-351*A-387*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)+90*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))-189*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+
75*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))-168*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)*EllipticE(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)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/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 = 223, normalized size of antiderivative = 1.15 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=-\frac {2 \, {\left (15 i \, \sqrt {2} {\left (6 \, A + 5 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (6 \, A + 5 \, B\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (9 \, A + 8 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (9 \, A + 8 \, B\right )} a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (35 \, B a^{2} \cos \left (d x + c\right )^{3} + 45 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 14 \, {\left (9 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 30 \, {\left (6 \, A + 5 \, B\right )} a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{315 \, d} \]

[In]

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

[Out]

-2/315*(15*I*sqrt(2)*(6*A + 5*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqrt(2)*
(6*A + 5*B)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(2)*(9*A + 8*B)*a^2*weier
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*I*sqrt(2)*(9*A + 8*B)*a^2*we
ierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (35*B*a^2*cos(d*x + c)^3 + 4
5*(A + 2*B)*a^2*cos(d*x + c)^2 + 14*(9*A + 8*B)*a^2*cos(d*x + c) + 30*(6*A + 5*B)*a^2)*sqrt(cos(d*x + c))*sin(
d*x + c))/d

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.37 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {2\,A\,a^2\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}-\frac {4\,A\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {4\,B\,a^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*A*a^2*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) - (4*A*a^2*cos(c + d*x)^(7/2)*
sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*A*a^2*cos(c + d*x)
^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (2*B*a^2*cos(c
+ d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (4*B*a^
2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) -
(2*B*a^2*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)
^(1/2))