3.6.15 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [515]
Optimal. Leaf size=193 \[ \frac {a^{5/2} (2 A+5 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {a^3 (14 A+3 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (2 A+B) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]
[Out]
2/3*a*A*(a+a*cos(d*x+c))^(3/2)*sec(d*x+c)^(3/2)*sin(d*x+c)/d-1/3*a^3*(14*A+3*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^
(1/2)/sec(d*x+c)^(1/2)+a^(5/2)*(2*A+5*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*se
c(d*x+c)^(1/2)/d+2*a^2*(2*A+B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)/d
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Rubi [A]
time = 0.43, antiderivative size = 193, 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, 3054,
3060, 2853, 222} \begin {gather*} \frac {a^{5/2} (2 A+5 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 )}{d}-\frac {a^3 (14 A+3 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (2 A+B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}{d}+\frac {2 a A \sin (c+d x) \sec ^{\frac {3}{2}}(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])*Sec[c + d*x]^(5/2),x]
[Out]
(a^(5/2)*(2*A + 5*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]])/d - (a^3*(14*A + 3*B)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^2*(2*A + B)
*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[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 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 3054
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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d
*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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)) \sec ^{\frac {5}{2}}(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))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {3}{2} a (2 A+B)-\frac {1}{2} a (2 A-3 B) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 (2 A+B) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (4 \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 (10 A+9 B)-\frac {1}{4} a^2 (14 A+3 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {a^3 (14 A+3 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (2 A+B) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{2} \left (a^2 (2 A+5 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 (14 A+3 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (2 A+B) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac {\left (a^2 (2 A+5 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 )}{d}\\ &=\frac {a^{5/2} (2 A+5 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)}}{d}-\frac {a^3 (14 A+3 B) \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {2 a^2 (2 A+B) \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A]
time = 0.78, size = 130, 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 ) \sec ^{\frac {3}{2}}(c+d x) \left (3 \sqrt {2} (2 A+5 B) \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+(4 A+3 B+4 (8 A+3 B) \cos (c+d x)+3 B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^(5/2),x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(3*Sqrt[2]*(2*A + 5*B)*ArcSin[Sqrt[2]*Sin[
(c + d*x)/2]]*Cos[c + d*x]^(3/2) + (4*A + 3*B + 4*(8*A + 3*B)*Cos[c + d*x] + 3*B*Cos[2*(c + d*x)])*Sin[(c + d*
x)/2]))/(6*d)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs.
\(2(167)=334\).
time = 0.45, size = 492, normalized size = 2.55
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| method |
result |
size |
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| default |
\(-\frac {\left (6 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \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 ) \left (\cos ^{2}\left (d x +c \right )\right )+15 B \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \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 ) \left (\cos ^{2}\left (d x +c \right )\right )+12 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \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 )+30 B \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \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 )+6 A \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \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 )+15 B \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \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 )+3 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+16 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+2 A \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}}{3 d \left (-1+\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2}}\) |
\(492\) |
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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)^(5/2),x,method=_RETURNVERBOSE)
[Out]
-1/3/d*(6*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))*
cos(d*x+c)^2+15*B*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*
x+c))*cos(d*x+c)^2+12*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/
cos(d*x+c))*cos(d*x+c)+30*B*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)/cos(d*x+c))*cos(d*x+c)+6*A*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)/cos(d*x+c))+15*B*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/
cos(d*x+c))+3*B*sin(d*x+c)*cos(d*x+c)^2+16*A*sin(d*x+c)*cos(d*x+c)+6*B*sin(d*x+c)*cos(d*x+c)+2*A*sin(d*x+c))*c
os(d*x+c)*(1/cos(d*x+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)^2/(-1+cos(d*x+c))/(1+cos(d*x+c))^2*a^2
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 2780 vs.
\(2 (167) = 334\).
time = 0.75, size = 2780, normalized size = 14.40 \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)^(5/2),x, algorithm="maxima")
[Out]
1/12*(2*(30*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(s
in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 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)*((12*a^2*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 3*a^2*sin(2*d*x + 2*c)
- 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (12*a^2*sin(2*d*x + 2*c)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c))) + 3*a^2*cos(2*d*x + 2*c) - a^2 + 4*(3*a^2*cos(2*d*x + 2*c) + 4*a^2)*cos(3/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c))))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*((a^2*cos(2*d*x + 2
*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + 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*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + 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(s
in(2*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x
+ 2*c) + 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*arctan
2(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*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x
+ 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + 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))*sqrt(a))*A/(c
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + 3*(18*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*a^(5/2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 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)*((4*a^2*sin(3*d*x + 3*c) + 5*a^2*sin(2*d*x
+ 2*c) + 4*a^2*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (a^2*cos(2*d*x + 2*c)
^2*sin(d*x + c) + a^2*sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*a^2*cos(2*d*x + 2*c)*sin(d*x + c) + a^2*sin(d*x + c)
)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (4*a^2*cos(3*d*x + 3*c) + 5*a^2*cos(2*d*x + 2*c)
+ 4*a^2*cos(d*x + c) + 5*a^2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((a^2*cos(d*x + c) -
a^2)*cos(2*d*x + 2*c)^2 + a^2*cos(d*x + c) + (a^2*cos(d*x + c) - a^2)*sin(2*d*x + 2*c)^2 - a^2 + 2*(a^2*cos(d*
x + c) - a^2)*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) + 5*((a^2*co
s(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + 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) + 1))*sin(d*
x + c) - cos(d*x + c)*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(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - (a^2*cos(2*d*x + 2*c)^2 +
a^2*sin(2*d*x + 2*c)^2 + 2*a^2*cos(2*d*x + 2*c) + 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) + 1))*sin(d*x + c) - cos(d*x +
c)*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*co
s(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c
)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - (a^2*cos(2*d*x + 2*c)^2 + a^2*sin(2*d*x + 2
*c)^2 + 2*a^2*cos(2*d*x + 2*c) + a^2)*arctan2((...
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Fricas [A]
time = 0.47, size = 166, normalized size = 0.86 \begin {gather*} -\frac {3 \, {\left ({\left (2 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (2 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )\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 (3 \, B a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, A + 3 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\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)^(5/2),x, algorithm="fricas")
[Out]
-1/3*(3*((2*A + 5*B)*a^2*cos(d*x + c)^2 + (2*A + 5*B)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a
)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (3*B*a^2*cos(d*x + c)^2 + 2*(8*A + 3*B)*a^2*cos(d*x + c) + 2*A*
a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
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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)**(5/2),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)^(5/2),x, algorithm="giac")
[Out]
Timed out
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\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))^(5/2)*(a + a*cos(c + d*x))^(5/2),x)
[Out]
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(5/2), x)
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