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

Optimal. Leaf size=281 \[ \frac{a^3 (1040 A+950 B+787 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{960 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (80 A+110 B+79 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{240 d}+\frac{a^{5/2} (400 A+326 B+283 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{128 d}+\frac{a^3 (400 A+326 B+283 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{128 d \sqrt{a \cos (c+d x)+a}}+\frac{a (2 B+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{8 d}+\frac{C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d} \]

[Out]

(a^(5/2)*(400*A + 326*B + 283*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(128*d) + (a^3*(400*
A + 326*B + 283*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(128*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1040*A + 950*B +
787*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(960*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(80*A + 110*B + 79*C)*Cos[c +
d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(240*d) + (a*(2*B + C)*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x
])^(3/2)*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.881471, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{a^3 (1040 A+950 B+787 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{960 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (80 A+110 B+79 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{240 d}+\frac{a^{5/2} (400 A+326 B+283 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{128 d}+\frac{a^3 (400 A+326 B+283 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{128 d \sqrt{a \cos (c+d x)+a}}+\frac{a (2 B+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{8 d}+\frac{C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(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)*(400*A + 326*B + 283*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(128*d) + (a^3*(400*
A + 326*B + 283*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(128*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1040*A + 950*B +
787*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(960*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(80*A + 110*B + 79*C)*Cos[c +
d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(240*d) + (a*(2*B + C)*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x
])^(3/2)*Sin[c + d*x])/(8*d) + (C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(5*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 \sqrt{\cos (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{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{\int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2} \left (\frac{1}{2} a (10 A+3 C)+\frac{5}{2} a (2 B+C) \cos (c+d x)\right ) \, dx}{5 a}\\ &=\frac{a (2 B+C) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{\int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (\frac{1}{4} a^2 (80 A+30 B+39 C)+\frac{1}{4} a^2 (80 A+110 B+79 C) \cos (c+d x)\right ) \, dx}{20 a}\\ &=\frac{a^2 (80 A+110 B+79 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 d}+\frac{a (2 B+C) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{\int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \left (\frac{3}{8} a^3 (240 A+170 B+157 C)+\frac{1}{8} a^3 (1040 A+950 B+787 C) \cos (c+d x)\right ) \, dx}{60 a}\\ &=\frac{a^3 (1040 A+950 B+787 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (80 A+110 B+79 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 d}+\frac{a (2 B+C) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{128} \left (a^2 (400 A+326 B+283 C)\right ) \int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a^3 (400 A+326 B+283 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{128 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1040 A+950 B+787 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (80 A+110 B+79 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 d}+\frac{a (2 B+C) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{256} \left (a^2 (400 A+326 B+283 C)\right ) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{a^3 (400 A+326 B+283 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{128 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1040 A+950 B+787 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (80 A+110 B+79 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 d}+\frac{a (2 B+C) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac{\left (a^2 (400 A+326 B+283 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 )}{128 d}\\ &=\frac{a^{5/2} (400 A+326 B+283 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{128 d}+\frac{a^3 (400 A+326 B+283 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{128 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (1040 A+950 B+787 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{960 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (80 A+110 B+79 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{240 d}+\frac{a (2 B+C) \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{8 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.70187, size = 171, normalized size = 0.61 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (15 \sqrt{2} (400 A+326 B+283 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} ((2720 A+3620 B+3874 C) \cos (c+d x)+4 (80 A+230 B+331 C) \cos (2 (c+d x))+6320 A+120 B \cos (3 (c+d x))+5810 B+348 C \cos (3 (c+d x))+48 C \cos (4 (c+d x))+5521 C)\right )}{3840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(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]*(400*A + 326*B + 283*C)*ArcSin[Sqrt[2]*Sin[(c + d
*x)/2]] + 2*Sqrt[Cos[c + d*x]]*(6320*A + 5810*B + 5521*C + (2720*A + 3620*B + 3874*C)*Cos[c + d*x] + 4*(80*A +
230*B + 331*C)*Cos[2*(c + d*x)] + 120*B*Cos[3*(c + d*x)] + 348*C*Cos[3*(c + d*x)] + 48*C*Cos[4*(c + d*x)])*Si
n[(c + d*x)/2]))/(3840*d)

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Maple [B]  time = 0.128, size = 733, normalized size = 2.6 \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)^(1/2)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

-1/1920/d*a^2*(-1+cos(d*x+c))^3*(640*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^4*sin(d*x+c)+4000*A*sin(d*
x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+480*B*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(3/2)+12080*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+2320*B*cos(d*x+c)^4*sin(d*x+c)*(cos(d*
x+c)/(1+cos(d*x+c)))^(3/2)+384*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^6+14720*A*sin(d*x+c)*
cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+5100*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+
1392*C*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+6000*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(5/2)+8150*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+2264*C*sin(d*x+c)*cos(d*x+c)^4*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)+4890*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+2830*C*sin(d*x+c)*co
s(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4245*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+
6000*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))+4890*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))+4245*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)^(1/2)/sin(d*x+c)^6/(cos(d*x+c)/(1+cos(d*x
+c)))^(5/2)

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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)^(1/2)*(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 = 5.84095, size = 593, normalized size = 2.11 \begin{align*} \frac{{\left (384 \, C a^{2} \cos \left (d x + c\right )^{4} + 48 \,{\left (10 \, B + 29 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (80 \, A + 230 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 10 \,{\left (272 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \,{\left (400 \, A + 326 \, B + 283 \, 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 (400 \, A + 326 \, B + 283 \, C\right )} a^{2} \cos \left (d x + c\right ) +{\left (400 \, A + 326 \, B + 283 \, 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 )}{1920 \,{\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)^(1/2)*(a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/1920*((384*C*a^2*cos(d*x + c)^4 + 48*(10*B + 29*C)*a^2*cos(d*x + c)^3 + 8*(80*A + 230*B + 283*C)*a^2*cos(d*x
+ c)^2 + 10*(272*A + 326*B + 283*C)*a^2*cos(d*x + c) + 15*(400*A + 326*B + 283*C)*a^2)*sqrt(a*cos(d*x + c) +
a)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*((400*A + 326*B + 283*C)*a^2*cos(d*x + c) + (400*A + 326*B + 283*C)*a^
2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c) + d)

________________________________________________________________________________________

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)**(1/2)*(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)^(1/2)*(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