3.2.6 \(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [106]
Optimal. Leaf size=165 \[ \frac {(7 A-4 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}} \]
[Out]
1/4*(7*A-4*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a^(1/2)-(A-B)*arctanh(1/2*sin(d*x+c)*a^(1/2
)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)-1/4*(A-4*B)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/2*A*sec(
d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
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Rubi [A]
time = 0.32, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3063, 3064,
2728, 212, 2852} \begin {gather*} -\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a \cos (c+d x)+a}}+\frac {(7 A-4 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {A \tan (c+d x) \sec (c+d x)}{2 d \sqrt {a \cos (c+d x)+a}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/Sqrt[a + a*Cos[c + d*x]],x]
[Out]
((7*A - 4*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*Sqrt[a]*d) - (Sqrt[2]*(A - B)*ArcTan
h[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) - ((A - 4*B)*Tan[c + d*x])/(4*d*Sqrt
[a + a*Cos[c + d*x]]) + (A*Sec[c + d*x]*Tan[c + d*x])/(2*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 2728
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
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 3063
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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Rule 3064
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]
Rubi steps
\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx &=\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-\frac {1}{2} a (A-4 B)+\frac {3}{2} a A \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a}\\ &=-\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{4} a^2 (7 A-4 B)-\frac {1}{4} a^2 (A-4 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}}+\frac {(7 A-4 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a}+(-A+B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx\\ &=-\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}}-\frac {(7 A-4 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 )}{4 d}+\frac {(2 (A-B)) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=\frac {(7 A-4 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {(A-4 B) \tan (c+d x)}{4 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec (c+d x) \tan (c+d x)}{2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.86, size = 114, normalized size = 0.69 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (-8 (A-B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\sqrt {2} (7 A-4 B) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sec (c+d x) (-A+4 B+2 A \sec (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 d \sqrt {a (1+\cos (c+d x))}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/Sqrt[a + a*Cos[c + d*x]],x]
[Out]
(Cos[(c + d*x)/2]*(-8*(A - B)*ArcTanh[Sin[(c + d*x)/2]] + Sqrt[2]*(7*A - 4*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]
] + 2*Sec[c + d*x]*(-A + 4*B + 2*A*Sec[c + d*x])*Sin[(c + d*x)/2]))/(4*d*Sqrt[a*(1 + Cos[c + d*x])])
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1251\) vs.
\(2(140)=280\).
time = 0.47, size = 1252, normalized size = 7.59
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| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(1252\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
[Out]
-1/2*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(4*a*(8*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1
/2*d*x+1/2*c)^2*a)^(1/2)+a))*A-8*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*B
-7*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))-7*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))+4*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))+4*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)^4-4*(A*a^(1/2)*2^(1/2)*(sin(1/2*d*x
+1/2*c)^2*a)^(1/2)+8*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*A-4*B*2^(1/
2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)-8*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^
(1/2)+a))*a*B-7*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-7*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+4*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+4*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+8*2^(1/2)*ln
(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*A-8*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/
2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+a))*a*B-2*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-7*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-7
*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-8*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+4*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+4*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^(3/2)/(2*co
s(1/2*d*x+1/2*c)-2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^2/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. 76209 vs.
\(2 (140) = 280\).
time = 3.88, size = 76209, normalized size = 461.87 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
-1/16*((4*sqrt(2)*cos(6*d*x + 6*c)^2*sin(3/2*d*x + 3/2*c) + 16*sqrt(2)*cos(5*d*x + 5*c)^2*sin(3/2*d*x + 3/2*c)
+ 36*sqrt(2)*cos(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 64*sqrt(2)*cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 3
6*sqrt(2)*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 4*sqrt(2)*sin(6*d*x + 6*c)^2*sin(3/2*d*x + 3/2*c) + 16*sqr
t(2)*sin(5*d*x + 5*c)^2*sin(3/2*d*x + 3/2*c) + 36*sqrt(2)*sin(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 64*sqrt(2)
*sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + 36*sqrt(2)*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) - 8*(3*sqrt(2)*c
os(3/2*d*x + 3/2*c)*sin(2*d*x + 2*c) - 3*sqrt(2)*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 2*sqrt(2)*cos(3/2*d*x
+ 3/2*c)*sin(d*x + c) + (sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*
x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c) + 2*(sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(
7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(5*d*x + 5*c) - (3*sqrt(2
)*sin(4*d*x + 4*c) + 4*sqrt(2)*sin(3*d*x + 3*c) + 3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*cos(9/2
*d*x + 9/2*c) + 3*(3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2
*c))*cos(4*d*x + 4*c) - 3*(4*sqrt(2)*sin(3*d*x + 3*c) + 3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*c
os(7/2*d*x + 7/2*c) - 4*(3*sqrt(2)*sin(5/2*d*x + 5/2*c) + sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(3*d*x + 3*c) + 3*(
3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*cos(5/2*d*x + 5/2*c) - (sqrt(2)*cos(9/2*d*x + 9/2*c) + 3*
sqrt(2)*cos(7/2*d*x + 7/2*c) - 3*sqrt(2)*cos(5/2*d*x + 5/2*c) - sqrt(2)*cos(3/2*d*x + 3/2*c))*sin(6*d*x + 6*c)
- 2*(sqrt(2)*cos(9/2*d*x + 9/2*c) + 3*sqrt(2)*cos(7/2*d*x + 7/2*c) - 3*sqrt(2)*cos(5/2*d*x + 5/2*c) - sqrt(2)
*cos(3/2*d*x + 3/2*c))*sin(5*d*x + 5*c) + (3*sqrt(2)*cos(4*d*x + 4*c) + 4*sqrt(2)*cos(3*d*x + 3*c) + 3*sqrt(2)
*cos(2*d*x + 2*c) + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*sin(9/2*d*x + 9/2*c) - 3*(3*sqrt(2)*cos(7/2*d*x + 7/2*c)
- 3*sqrt(2)*cos(5/2*d*x + 5/2*c) - sqrt(2)*cos(3/2*d*x + 3/2*c))*sin(4*d*x + 4*c) + 3*(4*sqrt(2)*cos(3*d*x +
3*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*sin(7/2*d*x + 7/2*c) + 4*(3*sqrt(2)*cos(
5/2*d*x + 5/2*c) + sqrt(2)*cos(3/2*d*x + 3/2*c))*sin(3*d*x + 3*c) - 3*(3*sqrt(2)*cos(2*d*x + 2*c) + 2*sqrt(2)*
cos(d*x + c) + sqrt(2))*sin(5/2*d*x + 5/2*c) - (2*sqrt(2)*cos(d*x + c) + sqrt(2))*sin(3/2*d*x + 3/2*c))*cos(4/
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 32*(3*sqrt(2)*cos(3/2*d*x + 3/2*c)*sin(2*d*x + 2*c)
- 3*sqrt(2)*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 2*sqrt(2)*cos(3/2*d*x + 3/2*c)*sin(d*x + c) + (sqrt(2)*si
n(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3
/2*c))*cos(6*d*x + 6*c) + 2*(sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2
*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(5*d*x + 5*c) - (3*sqrt(2)*sin(4*d*x + 4*c) + 4*sqrt(2)*sin(3
*d*x + 3*c) + 3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*cos(9/2*d*x + 9/2*c) + 3*(3*sqrt(2)*sin(7/2
*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(4*d*x + 4*c) - 3*(4*sqrt(2)
*sin(3*d*x + 3*c) + 3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(2)*sin(d*x + c))*cos(7/2*d*x + 7/2*c) - 4*(3*sqrt(2)*s
in(5/2*d*x + 5/2*c) + sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(3*d*x + 3*c) + 3*(3*sqrt(2)*sin(2*d*x + 2*c) + 2*sqrt(
2)*sin(d*x + c))*cos(5/2*d*x + 5/2*c) - (sqrt(2)*cos(9/2*d*x + 9/2*c) + 3*sqrt(2)*cos(7/2*d*x + 7/2*c) - 3*sqr
t(2)*cos(5/2*d*x + 5/2*c) - sqrt(2)*cos(3/2*d*x + 3/2*c))*sin(6*d*x + 6*c) - 2*(sqrt(2)*cos(9/2*d*x + 9/2*c) +
3*sqrt(2)*cos(7/2*d*x + 7/2*c) - 3*sqrt(2)*cos(5/2*d*x + 5/2*c) - sqrt(2)*cos(3/2*d*x + 3/2*c))*sin(5*d*x + 5
*c) + (3*sqrt(2)*cos(4*d*x + 4*c) + 4*sqrt(2)*cos(3*d*x + 3*c) + 3*sqrt(2)*cos(2*d*x + 2*c) + 2*sqrt(2)*cos(d*
x + c) + sqrt(2))*sin(9/2*d*x + 9/2*c) - 3*(3*sqrt(2)*cos(7/2*d*x + 7/2*c) - 3*sqrt(2)*cos(5/2*d*x + 5/2*c) -
sqrt(2)*cos(3/2*d*x + 3/2*c))*sin(4*d*x + 4*c) + 3*(4*sqrt(2)*cos(3*d*x + 3*c) + 3*sqrt(2)*cos(2*d*x + 2*c) +
2*sqrt(2)*cos(d*x + c) + sqrt(2))*sin(7/2*d*x + 7/2*c) + 4*(3*sqrt(2)*cos(5/2*d*x + 5/2*c) + sqrt(2)*cos(3/2*d
*x + 3/2*c))*sin(3*d*x + 3*c) - 3*(3*sqrt(2)*cos(2*d*x + 2*c) + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*sin(5/2*d*x
+ 5/2*c) - (2*sqrt(2)*cos(d*x + c) + sqrt(2))*sin(3/2*d*x + 3/2*c))*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c)))^2 + 48*(sqrt(2)*cos(d*x + c) + sqrt(2))*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 8*(3*sqrt(2
)*cos(3/2*d*x + 3/2*c)*sin(2*d*x + 2*c) - 3*sqrt(2)*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 2*sqrt(2)*cos(3/2*
d*x + 3/2*c)*sin(d*x + c) + (sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*sin(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2
*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c) + 2*(sqrt(2)*sin(9/2*d*x + 9/2*c) + 3*sqrt(2)*s
in(7/2*d*x + 7/2*c) - 3*sqrt(2)*sin(5/2*d*x + 5/2*c) - sqrt(2)*sin(3/2*d*x + 3/2*c))*cos(5*d*x + 5*c) - (3*sqr
t(2)*sin(4*d*x + 4*c) + 4*sqrt(2)*sin(3*d*x + 3...
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 284 vs.
\(2 (140) = 280\).
time = 0.41, size = 284, normalized size = 1.72 \begin {gather*} -\frac {{\left ({\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{2}\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 ({\left (A - 4 \, B\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {8 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{3} + {\left (A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-1/16*(((7*A - 4*B)*cos(d*x + c)^3 + (7*A - 4*B)*cos(d*x + c)^2)*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*((A - 4*B)*cos(d*x + c) - 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 8*sqrt(2)*((A - B)*a*cos(d*x
+ c)^3 + (A - B)*a*cos(d*x + c)^2)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqr
t(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x +
c)^2)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c))**(1/2),x)
[Out]
Integral((A + B*cos(c + d*x))*sec(c + d*x)**3/sqrt(a*(cos(c + d*x) + 1)), x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 299 vs.
\(2 (140) = 280\).
time = 0.49, size = 299, normalized size = 1.81 \begin {gather*} -\frac {\frac {4 \, \sqrt {2} {\left (A \sqrt {a} - B \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (A \sqrt {a} - B \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {{\left (7 \, A \sqrt {a} - 4 \, B \sqrt {a}\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {{\left (7 \, A \sqrt {a} - 4 \, B \sqrt {a}\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (2 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, \sqrt {2} B \sqrt {a} \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 )}^{2} a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
-1/8*(4*sqrt(2)*(A*sqrt(a) - B*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a*sgn(cos(1/2*d*x + 1/2*c))) - 4*sqrt(2
)*(A*sqrt(a) - B*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a*sgn(cos(1/2*d*x + 1/2*c))) - (7*A*sqrt(a) - 4*B*sq
rt(a))*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a*sgn(cos(1/2*d*x + 1/2*c))) + (7*A*sqrt(a) - 4*B*sqrt(a)
)*log(abs(-1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a*sgn(cos(1/2*d*x + 1/2*c))) - 2*(2*sqrt(2)*A*sqrt(a)*sin(1/2
*d*x + 1/2*c)^3 - 8*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 + sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 4*sqrt
(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)^2 - 1)^2*a*sgn(cos(1/2*d*x + 1/2*c))))/d
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^3\,\sqrt {a+a\,\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))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(1/2)),x)
[Out]
int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(1/2)), x)
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