3.2.84 \(\int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [184]
Optimal. Leaf size=180 \[ \frac {a^{5/2} (38 A+25 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a^3 (54 A+49 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d} \]
[Out]
1/8*a^(5/2)*(38*A+25*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/3*a*B*(a+a*cos(d*x+c))^(3/2)*sin
(d*x+c)*cos(d*x+c)^(1/2)/d+1/24*a^3*(54*A+49*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)+1/4*a^2*(
2*A+3*B)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(1/2)/d
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Rubi [A]
time = 0.34, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3055, 3060,
2853, 222} \begin {gather*} \frac {a^{5/2} (38 A+25 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}+\frac {a^3 (54 A+49 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{24 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (2 A+3 B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {a B \sin (c+d x) \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^{3/2}}{3 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Sqrt[Cos[c + d*x]],x]
[Out]
(a^(5/2)*(38*A + 25*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (a^3*(54*A + 49*B)*Sqr
t[Cos[c + d*x]]*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(2*A + 3*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + a
*Cos[c + d*x]]*Sin[c + d*x])/(4*d) + (a*B*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)
Rule 222
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 2853
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 3055
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)))
*Sin[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 3060
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)*Sqrt
[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 \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sqrt {\cos (c+d x)}} \, dx &=\frac {a B \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (6 A+B)+\frac {3}{2} a (2 A+3 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{6} \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (30 A+13 B)+\frac {1}{4} a^2 (54 A+49 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {a^3 (54 A+49 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{16} \left (a^2 (38 A+25 B)\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {a^3 (54 A+49 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac {\left (a^2 (38 A+25 B)\right ) \text {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 {a^{5/2} (38 A+25 B) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a^3 (54 A+49 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A]
time = 0.82, size = 121, normalized size = 0.67 \begin {gather*} \frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sqrt {2} (38 A+25 B) \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} (66 A+79 B+2 (6 A+17 B) \cos (c+d x)+4 B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]))/Sqrt[Cos[c + d*x]],x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(38*A + 25*B)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2
*Sqrt[Cos[c + d*x]]*(66*A + 79*B + 2*(6*A + 17*B)*Cos[c + d*x] + 4*B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(48*
d)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs.
\(2(154)=308\).
time = 0.38, size = 357, normalized size = 1.98
| | |
| method |
result |
size |
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| default |
\(-\frac {a^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (12 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+78 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \sin \left (d x +c \right )+8 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )+66 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+34 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+75 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+114 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \cos \left (d x +c \right )+75 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right ) \cos \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{24 d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{2} \sqrt {\cos \left (d x +c \right )}}\) |
\(357\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-1/24/d*a^2*(-1+cos(d*x+c))*(12*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*sin(d*x+c)+78*A*(cos(d*x+c)/(
1+cos(d*x+c)))^(3/2)*cos(d*x+c)*sin(d*x+c)+8*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+66*A*
(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)+34*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+75
*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+114*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)/cos(d*x+c))*cos(d*x+c)+75*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*cos(d*x+c))
*(a*(1+cos(d*x+c)))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^2/cos(d*x+c)^(1/2)
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 3071 vs.
\(2 (154) = 308\).
time = 0.85, size = 3071, normalized size = 17.06 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
1/96*(6*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((a^2*cos(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) + a^2*sin(2*d*x + 2*c) - (a^2*cos(2*d*x + 2*c) - 10*a^2)*sin
(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) +
(a^2*sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - a^2*cos(2*d*x + 2*c) + 10*a^2 + (
a^2*cos(2*d*x + 2*c) - 10*a^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sin(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 19*(a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*
d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) -
a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (cos(2*d*x + 2*c)^2
+ sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)
)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x +
2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) + 1)) + 1) + a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1))*sqrt(a))*A + (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*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)^(3/4)*sqrt(a) + 30*(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)*((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), co
s(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*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)))*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*arc
tan2(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*arc
tan2(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)*(cos(1/2*arctan2(sin(2/3*a
rctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), co...
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Fricas [A]
time = 0.43, size = 154, normalized size = 0.86 \begin {gather*} \frac {{\left (8 \, B a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 17 \, B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, {\left (22 \, A + 25 \, B\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 (38 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (38 \, A + 25 \, B\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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
1/24*((8*B*a^2*cos(d*x + c)^2 + 2*(6*A + 17*B)*a^2*cos(d*x + c) + 3*(22*A + 25*B)*a^2)*sqrt(a*cos(d*x + c) + a
)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*((38*A + 25*B)*a^2*cos(d*x + c) + (38*A + 25*B)*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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(1/2),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 4369 deep
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
Timed out
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^(1/2),x)
[Out]
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^(1/2), x)
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