3.214 \(\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=160 \[ \frac {25 a^{5/2} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}+\frac {13 a^3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{12 d \sqrt {a \cos (c+d x)+a}}+\frac {25 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{8 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d} \]
[Out]
25/8*a^(5/2)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+13/12*a^3*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*
cos(d*x+c))^(1/2)+25/8*a^3*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)+1/3*a^2*cos(d*x+c)^(3/2)*sin(d
*x+c)*(a+a*cos(d*x+c))^(1/2)/d
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Rubi [A] time = 0.30, antiderivative size = 160, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{2763, 2981, 2770, 2774, 216} \[ \frac {13 a^3 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{12 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {25 a^{5/2} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}+\frac {25 a^3 \sin (c+d x) \sqrt {\cos (c+d x)}}{8 d \sqrt {a \cos (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2),x]
[Out]
(25*a^(5/2)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (25*a^3*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(8*d*Sqrt[a + a*Cos[c + d*x]]) + (13*a^3*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x
]]) + (a^2*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*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 2763
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
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 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]
Rubi steps
\begin {align*} \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2} \, dx &=\frac {a^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \left (\frac {9 a^2}{2}+\frac {13}{2} a^2 \cos (c+d x)\right ) \, dx\\ &=\frac {13 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{8} \left (25 a^2\right ) \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {25 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {13 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{16} \left (25 a^2\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {25 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {13 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}-\frac {\left (25 a^2\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 )}{8 d}\\ &=\frac {25 a^{5/2} \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {25 a^3 \sqrt {\cos (c+d x)} \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {13 a^3 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}\\ \end {align*}
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Mathematica [C] time = 4.30, size = 182, normalized size = 1.14 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a (\cos (c+d x)+1))^{5/2} \left (-2 \sin ^4(c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (\frac {1}{2},\frac {3}{2},2;1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )-8 \sin ^2(c+d x) (\cos (c+d x)+3) \, _2F_1\left (\frac {1}{2},\frac {3}{2};\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )+7 (28 \cos (c+d x)+3 \cos (2 (c+d x))+89) \, _2F_1\left (-\frac {1}{2},\frac {1}{2};\frac {7}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )}{420 d} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(5/2),x]
[Out]
((a*(1 + Cos[c + d*x]))^(5/2)*Sec[(c + d*x)/2]^4*(7*(89 + 28*Cos[c + d*x] + 3*Cos[2*(c + d*x)])*Hypergeometric
2F1[-1/2, 1/2, 7/2, 2*Sin[(c + d*x)/2]^2] - 8*(3 + Cos[c + d*x])*Hypergeometric2F1[1/2, 3/2, 9/2, 2*Sin[(c + d
*x)/2]^2]*Sin[c + d*x]^2 - 2*Csc[(c + d*x)/2]^2*HypergeometricPFQ[{1/2, 3/2, 2}, {1, 9/2}, 2*Sin[(c + d*x)/2]^
2]*Sin[c + d*x]^4)*Tan[(c + d*x)/2])/(420*d)
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fricas [A] time = 1.77, size = 124, normalized size = 0.78 \[ \frac {{\left (8 \, a^{2} \cos \left (d x + c\right )^{2} + 34 \, a^{2} \cos \left (d x + c\right ) + 75 \, a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 75 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/24*((8*a^2*cos(d*x + c)^2 + 34*a^2*cos(d*x + c) + 75*a^2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*
x + c) - 75*(a^2*cos(d*x + 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 \[ \int {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cos \left (d x + c\right )}\,{d x} \]
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),x, algorithm="giac")
[Out]
integrate((a*cos(d*x + c) + a)^(5/2)*sqrt(cos(d*x + c)), x)
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maple [A] time = 0.20, size = 197, normalized size = 1.23 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (8 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+34 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+75 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+75 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) a^{2}}{24 d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{2}} \]
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),x)
[Out]
-1/24/d*(-1+cos(d*x+c))*(8*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*sin(d*x+c)+34*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)+75*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+75*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)/(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)/sin(d*x+c)^2*a^2
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maxima [B] time = 2.23, size = 1964, normalized size = 12.28 \[ \text {result too large to display} \]
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),x, algorithm="maxima")
[Out]
1/96*(4*(a^2*cos(3/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x +
3*c), cos(3*d*x + 3*c))) + 1))*sin(3*d*x + 3*c) - (a^2*cos(3*d*x + 3*c) - a^2)*sin(3/2*arctan2(sin(2/3*arctan
2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*(cos(2/3*a
rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos
(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*sqrt(a) + 30*(cos(2/3*arctan2(sin(3*d*x + 3*c), c
os(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3
*c), cos(3*d*x + 3*c))) + 1)^(1/4)*((a^2*sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 5*a^2*sin(1/3*
arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*
c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (a^2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(
3*d*x + 3*c))) + 3*a^2*cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) - 4*a^2)*sin(1/2*arctan2(sin(2/3*a
rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)))*sqrt(
a) + 75*(a^2*arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*
c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/2*arctan2(
sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1
))*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) - cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))
*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3
*d*x + 3*c))) + 1))), (cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*
c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/3*arctan2(
sin(3*d*x + 3*c), cos(3*d*x + 3*c)))*cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos
(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) + sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))
*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3
*d*x + 3*c))) + 1))) + 1) - a^2*arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arc
tan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4
)*(cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos
(3*d*x + 3*c))) + 1))*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) - cos(1/3*arctan2(sin(3*d*x + 3*c),
cos(3*d*x + 3*c)))*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(
3*d*x + 3*c), cos(3*d*x + 3*c))) + 1))), (cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arc
tan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4
)*(cos(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))*cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(
3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) + sin(1/3*arctan2(sin(3*d*x + 3*c),
cos(3*d*x + 3*c)))*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(
3*d*x + 3*c), cos(3*d*x + 3*c))) + 1))) - 1) - a^2*arctan2((cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)
))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x
+ 3*c))) + 1)^(1/4)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(
3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)), (cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arct
an2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)
*cos(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3
*d*x + 3*c))) + 1)) + 1) + a^2*arctan2((cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arcta
n2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*
sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*
d*x + 3*c))) + 1)), (cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c)
, cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*cos(1/2*arctan2(sin
(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1))
- 1))*sqrt(a))/d
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^(1/2)*(a + a*cos(c + d*x))^(5/2),x)
[Out]
int(cos(c + d*x)^(1/2)*(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)**(1/2)*(a+a*cos(d*x+c))**(5/2),x)
[Out]
Timed out
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