3.493 \(\int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=333 \[ \frac{a^3 (680 A+628 B+545 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{960 d \sqrt{a \cos (c+d x)+a}}+\frac{a^3 (1304 A+1132 B+1015 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{768 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (120 A+156 B+115 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{480 d}+\frac{a^{5/2} (1304 A+1132 B+1015 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{512 d}+\frac{a^3 (1304 A+1132 B+1015 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{512 d \sqrt{a \cos (c+d x)+a}}+\frac{a (12 B+5 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{60 d}+\frac{C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d} \]

[Out]

(a^(5/2)*(1304*A + 1132*B + 1015*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(512*d) + (a^3*(1
304*A + 1132*B + 1015*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(512*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1304*A + 11
32*B + 1015*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(768*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(680*A + 628*B + 545*C
)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(960*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(120*A + 156*B + 115*C)*Cos[c + d*x
]^(5/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(480*d) + (a*(12*B + 5*C)*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x
])^(3/2)*Sin[c + d*x])/(60*d) + (C*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.989966, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3045, 2976, 2981, 2770, 2774, 216} \[ \frac{a^3 (680 A+628 B+545 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{960 d \sqrt{a \cos (c+d x)+a}}+\frac{a^3 (1304 A+1132 B+1015 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{768 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (120 A+156 B+115 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{480 d}+\frac{a^{5/2} (1304 A+1132 B+1015 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{512 d}+\frac{a^3 (1304 A+1132 B+1015 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{512 d \sqrt{a \cos (c+d x)+a}}+\frac{a (12 B+5 C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{60 d}+\frac{C \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(1304*A + 1132*B + 1015*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(512*d) + (a^3*(1
304*A + 1132*B + 1015*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(512*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1304*A + 11
32*B + 1015*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(768*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(680*A + 628*B + 545*C
)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(960*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(120*A + 156*B + 115*C)*Cos[c + d*x
]^(5/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(480*d) + (a*(12*B + 5*C)*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x
])^(3/2)*Sin[c + d*x])/(60*d) + (C*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(6*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 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

Rule 2770

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

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac{\int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (12 A+5 C)+\frac{1}{2} a (12 B+5 C) \cos (c+d x)\right ) \, dx}{6 a}\\ &=\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac{\int \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \left (\frac{15}{4} a^2 (8 A+4 B+5 C)+\frac{1}{4} a^2 (120 A+156 B+115 C) \cos (c+d x)\right ) \, dx}{30 a}\\ &=\frac{a^2 (120 A+156 B+115 C) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{480 d}+\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac{\int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \left (\frac{5}{8} a^3 (312 A+252 B+235 C)+\frac{3}{8} a^3 (680 A+628 B+545 C) \cos (c+d x)\right ) \, dx}{120 a}\\ &=\frac{a^3 (680 A+628 B+545 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (120 A+156 B+115 C) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{480 d}+\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac{1}{384} \left (a^2 (1304 A+1132 B+1015 C)\right ) \int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a^3 (1304 A+1132 B+1015 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (680 A+628 B+545 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (120 A+156 B+115 C) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{480 d}+\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac{1}{512} \left (a^2 (1304 A+1132 B+1015 C)\right ) \int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a^3 (1304 A+1132 B+1015 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{512 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1304 A+1132 B+1015 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (680 A+628 B+545 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (120 A+156 B+115 C) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{480 d}+\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}+\frac{\left (a^2 (1304 A+1132 B+1015 C)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{1024}\\ &=\frac{a^3 (1304 A+1132 B+1015 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{512 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1304 A+1132 B+1015 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (680 A+628 B+545 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (120 A+156 B+115 C) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{480 d}+\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}-\frac{\left (a^2 (1304 A+1132 B+1015 C)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{512 d}\\ &=\frac{a^{5/2} (1304 A+1132 B+1015 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{512 d}+\frac{a^3 (1304 A+1132 B+1015 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{512 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1304 A+1132 B+1015 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{768 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (680 A+628 B+545 C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (120 A+156 B+115 C) \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{480 d}+\frac{a (12 B+5 C) \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{60 d}+\frac{C \cos ^{\frac{5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 2.16488, size = 205, normalized size = 0.62 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (15 \sqrt{2} (1304 A+1132 B+1015 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} (2 (7240 A+7748 B+8085 C) \cos (c+d x)+4 (920 A+1324 B+1575 C) \cos (2 (c+d x))+480 A \cos (3 (c+d x))+23240 A+1392 B \cos (3 (c+d x))+192 B \cos (4 (c+d x))+22084 B+2140 C \cos (3 (c+d x))+560 C \cos (4 (c+d x))+80 C \cos (5 (c+d x))+20965 C)\right )}{15360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(15*Sqrt[2]*(1304*A + 1132*B + 1015*C)*ArcSin[Sqrt[2]*Sin[(c
+ d*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(23240*A + 22084*B + 20965*C + 2*(7240*A + 7748*B + 8085*C)*Cos[c + d*x] + 4
*(920*A + 1324*B + 1575*C)*Cos[2*(c + d*x)] + 480*A*Cos[3*(c + d*x)] + 1392*B*Cos[3*(c + d*x)] + 2140*C*Cos[3*
(c + d*x)] + 192*B*Cos[4*(c + d*x)] + 560*C*Cos[4*(c + d*x)] + 80*C*Cos[5*(c + d*x)])*Sin[(c + d*x)/2]))/(1536
0*d)

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Maple [B]  time = 0.146, size = 841, normalized size = 2.5 \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))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

1/7680/d*a^2*(-1+cos(d*x+c))^4*(1920*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)*cos(d*x+c)^5+11200*A*(cos(
d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^4*sin(d*x+c)+1536*B*cos(d*x+c)^6*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(3/2)+29680*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+7104*B*cos(d*x+c)^5*sin(d*x+c)*(cos(
d*x+c)/(1+cos(d*x+c)))^(3/2)+1280*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^7+53000*A*sin(d*x+
c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+14624*B*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(3/2)+4480*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^6+52160*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+
c)/(1+cos(d*x+c)))^(5/2)+20376*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+6960*C*sin(d*x+c)*c
os(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+19560*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+28300*B*cos
(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+8120*C*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)+16980*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+10150*C*sin(d*x+c)*cos(d*x+c)^3*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)+15225*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+19560*A*cos(d*
x+c)^2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+16980*B*cos(d*x+c)^2*arctan(sin(d*x+c)*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+15225*C*cos(d*x+c)^2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)/cos(d*x+c)))*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(3/2)/sin(d*x+c)^8/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)

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Maxima [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))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 6.93179, size = 679, normalized size = 2.04 \begin{align*} \frac{{\left (1280 \, C a^{2} \cos \left (d x + c\right )^{5} + 128 \,{\left (12 \, B + 35 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 48 \,{\left (40 \, A + 116 \, B + 145 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (920 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \,{\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \,{\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \,{\left ({\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (1304 \, A + 1132 \, B + 1015 \, C\right )} a^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )}{7680 \,{\left (d \cos \left (d x + c\right ) + d\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))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/7680*((1280*C*a^2*cos(d*x + c)^5 + 128*(12*B + 35*C)*a^2*cos(d*x + c)^4 + 48*(40*A + 116*B + 145*C)*a^2*cos(
d*x + c)^3 + 8*(920*A + 1132*B + 1015*C)*a^2*cos(d*x + c)^2 + 10*(1304*A + 1132*B + 1015*C)*a^2*cos(d*x + c) +
15*(1304*A + 1132*B + 1015*C)*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*((1304*A + 1
132*B + 1015*C)*a^2*cos(d*x + c) + (1304*A + 1132*B + 1015*C)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqr
t(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c) + d)

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

[Out]

Timed out