3.1.81 \(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\) [81]
Optimal. Leaf size=160 \[ \frac {\sqrt {a} (5 A+6 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a (5 A+6 B) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {a (5 A+6 B) \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \]
[Out]
1/8*(5*A+6*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a^(1/2)/d+1/8*a*(5*A+6*B)*tan(d*x+c)/d/(a+a*c
os(d*x+c))^(1/2)+1/12*a*(5*A+6*B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/3*a*A*sec(d*x+c)^2*tan(d*x+
c)/d/(a+a*cos(d*x+c))^(1/2)
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Rubi [A]
time = 0.18, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3059, 2851,
2852, 212} \begin {gather*} \frac {a (5 A+6 B) \tan (c+d x)}{8 d \sqrt {a \cos (c+d x)+a}}+\frac {\sqrt {a} (5 A+6 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}+\frac {a (5 A+6 B) \tan (c+d x) \sec (c+d x)}{12 d \sqrt {a \cos (c+d x)+a}}+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
[Out]
(Sqrt[a]*(5*A + 6*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (a*(5*A + 6*B)*Tan[c +
d*x])/(8*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(5*A + 6*B)*Sec[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]
]) + (a*A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]])
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2851
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]
+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Rule 2852
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*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]
Rule 3059
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((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]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), 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,
A, B}, 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 {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{6} (5 A+6 B) \int \sqrt {a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac {a (5 A+6 B) \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{8} (5 A+6 B) \int \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac {a (5 A+6 B) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {a (5 A+6 B) \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{16} (5 A+6 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac {a (5 A+6 B) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {a (5 A+6 B) \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {(a (5 A+6 B)) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}\\ &=\frac {\sqrt {a} (5 A+6 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a (5 A+6 B) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}+\frac {a (5 A+6 B) \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 1.94, size = 129, normalized size = 0.81 \begin {gather*} \frac {\sqrt {a (1+\cos (c+d x))} \sec ^3(c+d x) \left (3 \sqrt {2} (5 A+6 B) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3(c+d x) \sec \left (\frac {1}{2} (c+d x)\right )+(31 A+18 B+4 (5 A+6 B) \cos (c+d x)+3 (5 A+6 B) \cos (2 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
[Out]
(Sqrt[a*(1 + Cos[c + d*x])]*Sec[c + d*x]^3*(3*Sqrt[2]*(5*A + 6*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*
x]^3*Sec[(c + d*x)/2] + (31*A + 18*B + 4*(5*A + 6*B)*Cos[c + d*x] + 3*(5*A + 6*B)*Cos[2*(c + d*x)])*Tan[(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. \(1326\) vs.
\(2(140)=280\).
time = 0.46, size = 1327, normalized size = 8.29
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| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(1327\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x,method=_RETURNVERBOSE)
[Out]
1/6*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-24*a*(5*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1
/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))+5*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(
1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))+6*B*ln(-4/(2*cos(1/2*
d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))+6*B*ln(
4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+
2*a)))*sin(1/2*d*x+1/2*c)^6+12*(10*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+12*B*2^(1/2)*(sin(1/2*d*x+
1/2*c)^2*a)^(1/2)*a^(1/2)+15*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1
/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+15*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2
*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a+18*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2
)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+18*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^
(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c
)^4-2*(80*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+96*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)
+45*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2
*a)^(1/2)-2*a))*a+45*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(
1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a+54*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1
/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+54*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2
*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+66*A*a^(1/2)*2^(1/2)*
(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+15*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*
2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+2*a))*a+15*A*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*
x+1/2*c)-a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a+60*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^
(1/2)+18*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*
c)^2*a)^(1/2)+2*a))*a+18*B*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-a^(1/2)*2^(1/2)*
(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-2*a))*a)/a^(1/2)/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^3/(2*cos(1/2*d*x+1/2*c)-2^(1/2)
)^3/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 5021 vs.
\(2 (140) = 280\).
time = 3.55, size = 5021, normalized size = 31.38 \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))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="maxima")
[Out]
-1/96*((120*(sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 3*sin(2*d*x + 2*c))*cos(13/2*d*x + 13/2*c) - 8*(15*sin(11
/2*d*x + 11/2*c) + 50*sin(9/2*d*x + 9/2*c) + 42*sin(7/2*d*x + 7/2*c) + 3*sin(5/2*d*x + 5/2*c) - 5*sin(3/2*d*x
+ 3/2*c))*cos(6*d*x + 6*c) + 360*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*cos(11/2*d*x + 11/2*c) + 1200*(sin(4*d*
x + 4*c) + sin(2*d*x + 2*c))*cos(9/2*d*x + 9/2*c) - 24*(42*sin(7/2*d*x + 7/2*c) + 3*sin(5/2*d*x + 5/2*c) - 5*s
in(3/2*d*x + 3/2*c))*cos(4*d*x + 4*c) - 15*(sqrt(2)*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*cos(4*d*x + 4*c)^2 + 9*sqrt
(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(6*d*x + 6*c)^2 + 9*sqrt(2)*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*sin(4*d*x + 4*
c)*sin(2*d*x + 2*c) + 9*sqrt(2)*sin(2*d*x + 2*c)^2 + 2*(3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c
) + sqrt(2))*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(4*d*x + 4*c) + 6*(sqrt(2)*sin(4*d
*x + 4*c) + sqrt(2)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 6*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/2*a
rctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*a
rctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 15*(sqrt(2
)*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*cos(4*d*x + 4*c)^2 + 9*sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(6*d*x + 6*c)^
2 + 9*sqrt(2)*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*sin(2*d*x + 2*c)^2
+ 2*(3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*cos(2
*d*x + 2*c) + sqrt(2))*cos(4*d*x + 4*c) + 6*(sqrt(2)*sin(4*d*x + 4*c) + sqrt(2)*sin(2*d*x + 2*c))*sin(6*d*x +
6*c) + 6*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*
arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*si
n(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 15*(sqrt(2)*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*cos(4*d*x + 4*c)^
2 + 9*sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(6*d*x + 6*c)^2 + 9*sqrt(2)*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*sin(
4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*sin(2*d*x + 2*c)^2 + 2*(3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2
*d*x + 2*c) + sqrt(2))*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(4*d*x + 4*c) + 6*(sqrt(
2)*sin(4*d*x + 4*c) + sqrt(2)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 6*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2
*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)
*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) +
15*(sqrt(2)*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*cos(4*d*x + 4*c)^2 + 9*sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(6*d
*x + 6*c)^2 + 9*sqrt(2)*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*sin(2*d*
x + 2*c)^2 + 2*(3*sqrt(2)*cos(4*d*x + 4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(6*d*x + 6*c) + 6*(3*sqr
t(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(4*d*x + 4*c) + 6*(sqrt(2)*sin(4*d*x + 4*c) + sqrt(2)*sin(2*d*x + 2*c))*si
n(6*d*x + 6*c) + 6*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 +
2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*
sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 120*(cos(6*d*x + 6*c) + 3*cos(4*d*x + 4*c) + 3*cos
(2*d*x + 2*c) + 1)*sin(13/2*d*x + 13/2*c) + 8*(15*cos(11/2*d*x + 11/2*c) + 50*cos(9/2*d*x + 9/2*c) + 42*cos(7/
2*d*x + 7/2*c) + 3*cos(5/2*d*x + 5/2*c) - 5*cos(3/2*d*x + 3/2*c))*sin(6*d*x + 6*c) - 120*(3*cos(4*d*x + 4*c) +
3*cos(2*d*x + 2*c) + 1)*sin(11/2*d*x + 11/2*c) - 400*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*sin(9/2*d*
x + 9/2*c) + 24*(42*cos(7/2*d*x + 7/2*c) + 3*cos(5/2*d*x + 5/2*c) - 5*cos(3/2*d*x + 3/2*c))*sin(4*d*x + 4*c) -
336*(3*cos(2*d*x + 2*c) + 1)*sin(7/2*d*x + 7/2*c) - 24*(3*cos(2*d*x + 2*c) + 1)*sin(5/2*d*x + 5/2*c) + 1008*c
os(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 72*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 120*cos(3/2*d*x + 3/2*c)*sin
(2*d*x + 2*c) + 120*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 120*(2*(3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) +
1)*cos(6*d*x + 6*c) + cos(6*d*x + 6*c)^2 + 6*(3*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 9*cos(4*d*x + 4*c)^2
+ 9*cos(2*d*x + 2*c)^2 + 6*(sin(4*d*x + 4*c) + sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + sin(6*d*x + 6*c)^2 + 9*sin
(4*d*x + 4*c)^2 + 18*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c) + 1)*sin(1/
2*arctan2(sin(d*x + c), cos(d*x + c))) + 40*sin(3/2*d*x + 3/2*c))*A*sqrt(a)/(sqrt(2)*cos(6*d*x + 6*c)^2 + 9*sq
rt(2)*cos(4*d*x + 4*c)^2 + 9*sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(6*d*x + 6*c)^2 + 9*sqrt(2)*sin(4*d*x + 4
*c)^2 + 18*sqrt(2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*sin(2*d*x + 2*c)^2 + 2*(3*sqrt(2)*cos(4*d*x +
4*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(
4*d*x + 4*c) + 6*(sqrt(2)*sin(4*d*x + 4*c) + sq...
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Fricas [A]
time = 0.39, size = 197, normalized size = 1.23 \begin {gather*} \frac {3 \, {\left ({\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 6 \, B\right )} \cos \left (d x + c\right ) + 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")
[Out]
1/96*(3*((5*A + 6*B)*cos(d*x + c)^4 + (5*A + 6*B)*cos(d*x + c)^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x
+ c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x +
c)^2)) + 4*(3*(5*A + 6*B)*cos(d*x + c)^2 + 2*(5*A + 6*B)*cos(d*x + c) + 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x
+ c))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
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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))**(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)
[Out]
Timed out
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Giac [A]
time = 0.51, size = 244, normalized size = 1.52 \begin {gather*} -\frac {\sqrt {2} {\left (3 \, \sqrt {2} {\left (5 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 6 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) + \frac {4 \, {\left (60 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 96 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 30 \, B \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )} \sqrt {a}}{96 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")
[Out]
-1/96*sqrt(2)*(3*sqrt(2)*(5*A*sgn(cos(1/2*d*x + 1/2*c)) + 6*B*sgn(cos(1/2*d*x + 1/2*c)))*log(abs(-2*sqrt(2) +
4*sin(1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))) + 4*(60*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d
*x + 1/2*c)^5 + 72*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 80*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2
*d*x + 1/2*c)^3 - 96*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 33*A*sgn(cos(1/2*d*x + 1/2*c))*sin(1
/2*d*x + 1/2*c) + 30*B*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))/(2*sin(1/2*d*x + 1/2*c)^2 - 1)^3)*sqrt(
a)/d
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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 )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4} \,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))^(1/2))/cos(c + d*x)^4,x)
[Out]
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(1/2))/cos(c + d*x)^4, x)
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