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

Optimal. Leaf size=303 \[ \frac{4 a^3 (121 A+105 B+95 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (221 A+195 B+175 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{20 a^3 (286 A+273 B+236 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{9009 d}+\frac{4 a^3 (221 A+195 B+175 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{585 d}+\frac{2 (143 A+195 B+145 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{1287 d}+\frac{4 a^3 (121 A+105 B+95 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d}+\frac{2 (13 B+6 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{143 a d}+\frac{2 C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^3}{13 d} \]

[Out]

(4*a^3*(221*A + 195*B + 175*C)*EllipticE[(c + d*x)/2, 2])/(195*d) + (4*a^3*(121*A + 105*B + 95*C)*EllipticF[(c
+ d*x)/2, 2])/(231*d) + (4*a^3*(121*A + 105*B + 95*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^3*(221*
A + 195*B + 175*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(585*d) + (20*a^3*(286*A + 273*B + 236*C)*Cos[c + d*x]^(5/
2)*Sin[c + d*x])/(9009*d) + (2*C*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(13*d) + (2*(13*B + 6
*C)*Cos[c + d*x]^(5/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(143*a*d) + (2*(143*A + 195*B + 145*C)*Cos[c +
d*x]^(5/2)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(1287*d)

________________________________________________________________________________________

Rubi [A]  time = 0.734794, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {3045, 2976, 2968, 3023, 2748, 2635, 2641, 2639} \[ \frac{4 a^3 (121 A+105 B+95 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (221 A+195 B+175 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{20 a^3 (286 A+273 B+236 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{9009 d}+\frac{4 a^3 (221 A+195 B+175 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{585 d}+\frac{2 (143 A+195 B+145 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{1287 d}+\frac{4 a^3 (121 A+105 B+95 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{231 d}+\frac{2 (13 B+6 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{143 a d}+\frac{2 C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^3}{13 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(221*A + 195*B + 175*C)*EllipticE[(c + d*x)/2, 2])/(195*d) + (4*a^3*(121*A + 105*B + 95*C)*EllipticF[(c
+ d*x)/2, 2])/(231*d) + (4*a^3*(121*A + 105*B + 95*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^3*(221*
A + 195*B + 175*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(585*d) + (20*a^3*(286*A + 273*B + 236*C)*Cos[c + d*x]^(5/
2)*Sin[c + d*x])/(9009*d) + (2*C*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(13*d) + (2*(13*B + 6
*C)*Cos[c + d*x]^(5/2)*(a^2 + a^2*Cos[c + d*x])^2*Sin[c + d*x])/(143*a*d) + (2*(143*A + 195*B + 145*C)*Cos[c +
d*x]^(5/2)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(1287*d)

Rule 3045

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

Rule 2976

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)))*S
in[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 2968

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 3023

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]

Rule 2748

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 2635

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] && Integer
Q[2*n]

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 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]

Rubi steps

\begin{align*} \int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 \int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^3 \left (\frac{1}{2} a (13 A+5 C)+\frac{1}{2} a (13 B+6 C) \cos (c+d x)\right ) \, dx}{13 a}\\ &=\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{4 \int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^2 \left (\frac{1}{4} a^2 (143 A+65 B+85 C)+\frac{1}{4} a^2 (143 A+195 B+145 C) \cos (c+d x)\right ) \, dx}{143 a}\\ &=\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{2 (143 A+195 B+145 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{1287 d}+\frac{8 \int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x)) \left (\frac{1}{4} a^3 (1001 A+780 B+745 C)+\frac{5}{4} a^3 (286 A+273 B+236 C) \cos (c+d x)\right ) \, dx}{1287 a}\\ &=\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{2 (143 A+195 B+145 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{1287 d}+\frac{8 \int \cos ^{\frac{3}{2}}(c+d x) \left (\frac{1}{4} a^4 (1001 A+780 B+745 C)+\left (\frac{5}{4} a^4 (286 A+273 B+236 C)+\frac{1}{4} a^4 (1001 A+780 B+745 C)\right ) \cos (c+d x)+\frac{5}{4} a^4 (286 A+273 B+236 C) \cos ^2(c+d x)\right ) \, dx}{1287 a}\\ &=\frac{20 a^3 (286 A+273 B+236 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d}+\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{2 (143 A+195 B+145 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{1287 d}+\frac{16 \int \cos ^{\frac{3}{2}}(c+d x) \left (\frac{117}{8} a^4 (121 A+105 B+95 C)+\frac{77}{8} a^4 (221 A+195 B+175 C) \cos (c+d x)\right ) \, dx}{9009 a}\\ &=\frac{20 a^3 (286 A+273 B+236 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d}+\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{2 (143 A+195 B+145 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{1287 d}+\frac{1}{77} \left (2 a^3 (121 A+105 B+95 C)\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{117} \left (2 a^3 (221 A+195 B+175 C)\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{4 a^3 (121 A+105 B+95 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{4 a^3 (221 A+195 B+175 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{585 d}+\frac{20 a^3 (286 A+273 B+236 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d}+\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{2 (143 A+195 B+145 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{1287 d}+\frac{1}{231} \left (2 a^3 (121 A+105 B+95 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{195} \left (2 a^3 (221 A+195 B+175 C)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 (221 A+195 B+175 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{195 d}+\frac{4 a^3 (121 A+105 B+95 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{231 d}+\frac{4 a^3 (121 A+105 B+95 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{231 d}+\frac{4 a^3 (221 A+195 B+175 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{585 d}+\frac{20 a^3 (286 A+273 B+236 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{9009 d}+\frac{2 C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^3 \sin (c+d x)}{13 d}+\frac{2 (13 B+6 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{143 a d}+\frac{2 (143 A+195 B+145 C) \cos ^{\frac{5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{1287 d}\\ \end{align*}

Mathematica [C]  time = 6.4697, size = 1426, normalized size = 4.71 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-((221*A + 195*B + 175*C)*Cot[c])/(390*d) + ((
2134*A + 1953*B + 1811*C)*Cos[d*x]*Sin[c])/(7392*d) + ((7592*A + 7800*B + 7825*C)*Cos[2*d*x]*Sin[2*c])/(74880*
d) + ((132*A + 189*B + 215*C)*Cos[3*d*x]*Sin[3*c])/(4928*d) + ((13*A + 39*B + 59*C)*Cos[4*d*x]*Sin[4*c])/(3744
*d) + ((B + 3*C)*Cos[5*d*x]*Sin[5*c])/(704*d) + (C*Cos[6*d*x]*Sin[6*c])/(1664*d) + ((2134*A + 1953*B + 1811*C)
*Cos[c]*Sin[d*x])/(7392*d) + ((7592*A + 7800*B + 7825*C)*Cos[2*c]*Sin[2*d*x])/(74880*d) + ((132*A + 189*B + 21
5*C)*Cos[3*c]*Sin[3*d*x])/(4928*d) + ((13*A + 39*B + 59*C)*Cos[4*c]*Sin[4*d*x])/(3744*d) + ((B + 3*C)*Cos[5*c]
*Sin[5*d*x])/(704*d) + (C*Cos[6*c]*Sin[6*d*x])/(1664*d)) - (11*A*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricP
FQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (5*B*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}
, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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]]]])/(22*d*Sq
rt[1 + Cot[c]^2]) - (95*C*(a + a*Cos[c + d*x])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[
Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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]]]])/(462*d*Sqrt[1 + Cot[c]^2])
- (17*A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((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 + A
rcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + Ar
cTan[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]]))/(60*d) - (B*(a + a*Cos[c + d*x])^3*C
sc[c]*Sec[c/2 + (d*x)/2]^6*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + Arc
Tan[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 + T
an[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]]))/(4*d) - (35*C*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((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]]))/(156*d)

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Maple [A]  time = 0.275, size = 576, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/45045*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-221760*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^14+(131040*B+1058400*C)*sin(1/2*d*x+1/2*c)^12*cos(1/2*d*x+1/2*c)+(-80080*A-567840*B-2122400*C)*sin(1/2
*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(314600*A+1004640*B+2331040*C)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-487
916*A-939120*B-1535860*C)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(386386*A+510510*B+633710*C)*sin(1/2*d*x+1/2
*c)^4*cos(1/2*d*x+1/2*c)+(-105534*A-114660*B-121230*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-51051*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))+23595*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))-45045*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))+20475*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))-40425*C*(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))+18525*C*(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+s
in(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{3} \cos \left (d x + c\right )^{6} +{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} +{\left (A + 3 \, B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} +{\left (3 \, A + 3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{3} +{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + A a^{3} \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*a^3*cos(d*x + c)^6 + (B + 3*C)*a^3*cos(d*x + c)^5 + (A + 3*B + 3*C)*a^3*cos(d*x + c)^4 + (3*A + 3*
B + C)*a^3*cos(d*x + c)^3 + (3*A + B)*a^3*cos(d*x + c)^2 + A*a^3*cos(d*x + c))*sqrt(cos(d*x + c)), x)

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

[Out]

Timed out

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

[Out]

Timed out