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

Optimal. Leaf size=312 \[ \frac {a^{5/2} (1304 A+1015 C) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{512 d}+\frac {a^3 (136 A+109 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{192 d \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (1304 A+1015 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{768 d \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (1304 A+1015 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{512 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (24 A+23 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{96 d}+\frac {a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d} \]

[Out]

1/512*a^(5/2)*(1304*A+1015*C)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/12*a*C*cos(d*x+c)^(5/2)*(a
+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+1/6*C*cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(5/2)*sin(d*x+c)/d+1/768*a^3*(1304*A
+1015*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/192*a^3*(136*A+109*C)*cos(d*x+c)^(5/2)*sin(d*x
+c)/d/(a+a*cos(d*x+c))^(1/2)+1/512*a^3*(1304*A+1015*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)+1/
96*a^2*(24*A+23*C)*cos(d*x+c)^(5/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.91, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3046, 2976, 2981, 2770, 2774, 216} \[ \frac {a^3 (136 A+109 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{192 d \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (1304 A+1015 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{768 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (24 A+23 C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{96 d}+\frac {a^{5/2} (1304 A+1015 C) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{512 d}+\frac {a^3 (1304 A+1015 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{512 d \sqrt {a \cos (c+d x)+a}}+\frac {a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{12 d}+\frac {C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(1304*A + 1015*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(512*d) + (a^3*(1304*A + 1
015*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(512*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(1304*A + 1015*C)*Cos[c + d*x]
^(3/2)*Sin[c + d*x])/(768*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(136*A + 109*C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/
(192*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(24*A + 23*C)*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x]
)/(96*d) + (a*C*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (C*Cos[c + d*x]^(5/2)*(a
+ a*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(6*d)

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]

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 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 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, 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]

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} \left (A+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 {5}{2} a C \cos (c+d x)\right ) \, dx}{6 a}\\ &=\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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+5 C)+\frac {5}{4} a^2 (24 A+23 C) \cos (c+d x)\right ) \, dx}{30 a}\\ &=\frac {a^2 (24 A+23 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{96 d}+\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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+235 C)+\frac {15}{8} a^3 (136 A+109 C) \cos (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^3 (136 A+109 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (24 A+23 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{96 d}+\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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+1015 C)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {a^3 (1304 A+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 (136 A+109 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (24 A+23 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{96 d}+\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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+1015 C)\right ) \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {a^3 (1304 A+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+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 (136 A+109 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (24 A+23 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{96 d}+\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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+1015 C)\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{1024}\\ &=\frac {a^3 (1304 A+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+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 (136 A+109 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (24 A+23 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{96 d}+\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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+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+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+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+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 (136 A+109 C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (24 A+23 C) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{96 d}+\frac {a C \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{12 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*}

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Mathematica [A]  time = 2.47, size = 170, normalized size = 0.54 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} \left (3 \sqrt {2} (1304 A+1015 C) \sin ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} ((2896 A+3234 C) \cos (c+d x)+4 (184 A+315 C) \cos (2 (c+d x))+96 A \cos (3 (c+d x))+4648 A+428 C \cos (3 (c+d x))+112 C \cos (4 (c+d x))+16 C \cos (5 (c+d x))+4193 C)\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(1304*A + 1015*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]
+ 2*Sqrt[Cos[c + d*x]]*(4648*A + 4193*C + (2896*A + 3234*C)*Cos[c + d*x] + 4*(184*A + 315*C)*Cos[2*(c + d*x)]
+ 96*A*Cos[3*(c + d*x)] + 428*C*Cos[3*(c + d*x)] + 112*C*Cos[4*(c + d*x)] + 16*C*Cos[5*(c + d*x)])*Sin[(c + d
*x)/2]))/(3072*d)

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fricas [A]  time = 0.59, size = 208, normalized size = 0.67 \[ \frac {{\left (256 \, C a^{2} \cos \left (d x + c\right )^{5} + 896 \, C a^{2} \cos \left (d x + c\right )^{4} + 48 \, {\left (8 \, A + 29 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (184 \, A + 203 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (1304 \, A + 1015 \, C\right )} a^{2} \cos \left (d x + c\right ) + 3 \, {\left (1304 \, A + 1015 \, C\right )} a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, {\left ({\left (1304 \, A + 1015 \, C\right )} a^{2} \cos \left (d x + c\right ) + {\left (1304 \, A + 1015 \, C\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{1536 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

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+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/1536*((256*C*a^2*cos(d*x + c)^5 + 896*C*a^2*cos(d*x + c)^4 + 48*(8*A + 29*C)*a^2*cos(d*x + c)^3 + 8*(184*A +
203*C)*a^2*cos(d*x + c)^2 + 2*(1304*A + 1015*C)*a^2*cos(d*x + c) + 3*(1304*A + 1015*C)*a^2)*sqrt(a*cos(d*x +
c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*((1304*A + 1015*C)*a^2*cos(d*x + c) + (1304*A + 1015*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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2), x)

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maple [B]  time = 0.41, size = 581, normalized size = 1.86 \[ \frac {a^{2} \left (-1+\cos \left (d x +c \right )\right )^{4} \left (384 A \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+2240 A \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+5936 A \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+256 C \left (\cos ^{7}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+10600 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+896 C \left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+10432 A \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+1392 C \sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3912 A \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+1624 C \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+2030 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3045 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3912 A \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+3045 C \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right )}{1536 d \sin \left (d x +c \right )^{8} \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {7}{2}}} \]

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+C*cos(d*x+c)^2),x)

[Out]

1/1536/d*a^2*(-1+cos(d*x+c))^4*(384*A*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+2240*A*cos(d*x
+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+5936*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^
(5/2)+256*C*cos(d*x+c)^7*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+10600*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x
+c)/(1+cos(d*x+c)))^(5/2)+896*C*cos(d*x+c)^6*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+10432*A*sin(d*x+c)*c
os(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+1392*C*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3
912*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+1624*C*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)+2030*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3045*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)+3912*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))+
3045*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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+C*cos(d*x+c)**2),x)

[Out]

Timed out

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