3.6.17 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx\) [517]
Optimal. Leaf size=200 \[ \frac {a^{5/2} (38 A+25 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a^3 (54 A+49 B) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]
[Out]
1/3*a*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+1/24*a^3*(54*A+49*B)*sin(d*x+c)/d/(a+a*cos(d*x+c)
)^(1/2)/sec(d*x+c)^(1/2)+1/4*a^2*(2*A+3*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)+1/8*a^(5/2)*(3
8*A+25*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d
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Rubi [A]
time = 0.42, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3040, 3055,
3060, 2853, 222} \begin {gather*} \frac {a^{5/2} (38 A+25 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \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)}{24 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (2 A+3 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{4 d \sqrt {\sec (c+d x)}}+\frac {a B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sqrt[Sec[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]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/(8*d) + (a^3*(54*A + 49*B)*Sin[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (a^2*(2*
A + 3*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]) + (a*B*(a + a*Cos[c + d*x])^(3/2)*Sin
[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])
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 3040
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] && !IntegerQ[p] && !(IntegerQ[m] && IntegerQ[n])
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 (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sqrt {\sec (c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{6} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{16} \left (a^2 (38 A+25 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {a^3 (54 A+49 B) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {\left (a^2 (38 A+25 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a^3 (54 A+49 B) \sin (c+d x)}{24 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {a^2 (2 A+3 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{4 d \sqrt {\sec (c+d x)}}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 1.04, size = 141, normalized size = 0.70 \begin {gather*} \frac {a^2 \sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \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[Sec[c + d*x]],x]
[Out]
(a^2*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(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 [A]
time = 0.44, size = 305, normalized size = 1.52
| | |
| method |
result |
size |
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| default |
\(-\frac {\left (8 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+12 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+34 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+66 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+75 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+114 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+75 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right ) a^{2}}{24 d \sin \left (d x +c \right )^{2}}\) |
\(305\) |
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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))*sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
[Out]
-1/24/d*(8*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+12*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
cos(d*x+c)*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)+66*A*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*sin(d*x+c)+75*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(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))+75*B*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)))*(1/cos(
d*x+c))^(1/2)*(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)^2-1)*a^2
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 3071 vs.
\(2 (170) = 340\).
time = 0.87, size = 3071, normalized size = 15.36 \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))*sec(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 = 0.45, size = 163, normalized size = 0.82 \begin {gather*} -\frac {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 ) - \frac {{\left (8 \, B a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (6 \, A + 17 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (22 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\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))*sec(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
-1/24*(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))) - (8*B*a^2*cos(d*x + c)^3 + 2*(6*A + 17*B)*a^2*cos(d*x + c)^2 + 3*(22*A + 2
5*B)*a^2*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)
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Sympy [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))*sec(d*x+c)**(1/2),x)
[Out]
Timed out
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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))*sec(d*x+c)^(1/2),x, algorithm="giac")
[Out]
Timed out
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(5/2),x)
[Out]
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(5/2), x)
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