3.1.97 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\) [97]
Optimal. Leaf size=164 \[ \frac {a^{5/2} (25 A+38 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (3 A+2 B) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
[Out]
1/8*a^(5/2)*(25*A+38*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/3*a*A*(a+a*cos(d*x+c))^(3/2)*se
c(d*x+c)^2*tan(d*x+c)/d+1/24*a^3*(49*A+54*B)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/4*a^2*(3*A+2*B)*sec(d*x+c)*
(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
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Rubi [A]
time = 0.33, antiderivative size = 164, 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 = {3054, 3059,
2852, 212} \begin {gather*} \frac {a^{5/2} (25 A+38 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \tan (c+d x)}{24 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (3 A+2 B) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{4 d}+\frac {a A \tan (c+d x) \sec ^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]^4,x]
[Out]
(a^(5/2)*(25*A + 38*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(8*d) + (a^3*(49*A + 54*B)*Ta
n[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(3*A + 2*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d
*x])/(4*d) + (a*A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)
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 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 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 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 (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx &=\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+a \cos (c+d x))^{3/2} \left (\frac {3}{2} a (3 A+2 B)+\frac {1}{2} a (A+6 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a^2 (3 A+2 B) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (49 A+54 B)+\frac {1}{4} a^2 (13 A+30 B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a^3 (49 A+54 B) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (3 A+2 B) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{16} \left (a^2 (25 A+38 B)\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac {a^3 (49 A+54 B) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (3 A+2 B) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (a^3 (25 A+38 B)\right ) \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 {a^{5/2} (25 A+38 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 d}+\frac {a^3 (49 A+54 B) \tan (c+d x)}{24 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (3 A+2 B) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a A (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A]
time = 1.13, size = 131, normalized size = 0.80 \begin {gather*} \frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (3 \sqrt {2} (25 A+38 B) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3(c+d x)+(91 A+66 B+4 (17 A+6 B) \cos (c+d x)+(75 A+66 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])*Sec[c + d*x]^4,x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^3*(3*Sqrt[2]*(25*A + 38*B)*ArcTanh[Sqrt[2]*Sin[(
c + d*x)/2]]*Cos[c + d*x]^3 + (91*A + 66*B + 4*(17*A + 6*B)*Cos[c + d*x] + (75*A + 66*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. \(1325\) vs.
\(2(144)=288\).
time = 0.42, size = 1326, normalized size = 8.09
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result |
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\(\text {Expression too large to display}\) |
\(1326\) |
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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)^4,x,method=_RETURNVERBOSE)
[Out]
1/6*a^(3/2)*cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*(-24*a*(25*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))+25*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))+38*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
))+38*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*(50*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+44*B*2^(1/2)*(
sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)+75*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+75*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+114*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+114*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+(-736*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)-450*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-450*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-57
6*B*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)*a^(1/2)-684*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-684*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+23
4*A*a^(1/2)*2^(1/2)*(sin(1/2*d*x+1/2*c)^2*a)^(1/2)+75*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+75*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+156*B*2^(1/2)*(sin(1/2*d*x
+1/2*c)^2*a)^(1/2)*a^(1/2)+114*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+114*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)/(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. 7994 vs.
\(2 (144) = 288\).
time = 3.87, size = 7994, normalized size = 48.74 \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)^4,x, algorithm="maxima")
[Out]
-1/96*((1530*a^2*cos(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 1530*a^2*cos(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) +
1530*a^2*sin(4*d*x + 4*c)^2*sin(3/2*d*x + 3/2*c) + 1530*a^2*sin(2*d*x + 2*c)^2*sin(3/2*d*x + 3/2*c) + 4176*a^2
*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) + 2430*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + 678*a^2*cos(3/2*d*x
+ 3/2*c)*sin(2*d*x + 2*c) + 342*a^2*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) + 10*(a^2*sin(9/2*d*x + 9/2*c) + 17*
a^2*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c)^2 + 10*(a^2*sin(9/2*d*x + 9/2*c) + 17*a^2*sin(3/2*d*x + 3/2*c))*sin
(6*d*x + 6*c)^2 - 56*a^2*sin(3/2*d*x + 3/2*c) + 10*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(
2*d*x + 2*c))*cos(21/2*d*x + 21/2*c) - 30*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2
*c))*cos(19/2*d*x + 19/2*c) - 48*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2*c))*cos(
17/2*d*x + 17/2*c) + 80*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2*c))*cos(15/2*d*x
+ 15/2*c) + 396*(a^2*sin(6*d*x + 6*c) + 3*a^2*sin(4*d*x + 4*c) + 3*a^2*sin(2*d*x + 2*c))*cos(13/2*d*x + 13/2*c
) + 6*(170*a^2*cos(4*d*x + 4*c)*sin(3/2*d*x + 3/2*c) + 170*a^2*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 170*a^2
*sin(11/2*d*x + 11/2*c) - 232*a^2*sin(7/2*d*x + 7/2*c) - 135*a^2*sin(5/2*d*x + 5/2*c) + 19*a^2*sin(3/2*d*x + 3
/2*c) + 10*(a^2*cos(4*d*x + 4*c) + a^2*cos(2*d*x + 2*c) - 25*a^2)*sin(9/2*d*x + 9/2*c))*cos(6*d*x + 6*c) + 306
0*(a^2*sin(4*d*x + 4*c) + a^2*sin(2*d*x + 2*c))*cos(11/2*d*x + 11/2*c) + 4560*(a^2*sin(4*d*x + 4*c) + a^2*sin(
2*d*x + 2*c))*cos(9/2*d*x + 9/2*c) + 18*(170*a^2*cos(2*d*x + 2*c)*sin(3/2*d*x + 3/2*c) - 232*a^2*sin(7/2*d*x +
7/2*c) - 135*a^2*sin(5/2*d*x + 5/2*c) + 19*a^2*sin(3/2*d*x + 3/2*c))*cos(4*d*x + 4*c) - 75*(sqrt(2)*a^2*cos(6
*d*x + 6*c)^2 + 9*sqrt(2)*a^2*cos(4*d*x + 4*c)^2 + 9*sqrt(2)*a^2*cos(2*d*x + 2*c)^2 + sqrt(2)*a^2*sin(6*d*x +
6*c)^2 + 9*sqrt(2)*a^2*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*a^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*a^2*s
in(2*d*x + 2*c)^2 + 6*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2 + 2*(3*sqrt(2)*a^2*cos(4*d*x + 4*c) + 3*sqrt(
2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*cos
(4*d*x + 4*c) + 6*(sqrt(2)*a^2*sin(4*d*x + 4*c) + sqrt(2)*a^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(2*cos(1/
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 75*(sqrt(2)*a^2*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*a^2*co
s(4*d*x + 4*c)^2 + 9*sqrt(2)*a^2*cos(2*d*x + 2*c)^2 + sqrt(2)*a^2*sin(6*d*x + 6*c)^2 + 9*sqrt(2)*a^2*sin(4*d*x
+ 4*c)^2 + 18*sqrt(2)*a^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*a^2*sin(2*d*x + 2*c)^2 + 6*sqrt(2)*a^
2*cos(2*d*x + 2*c) + sqrt(2)*a^2 + 2*(3*sqrt(2)*a^2*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2
)*a^2)*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*cos(4*d*x + 4*c) + 6*(sqrt(2)*a^2*s
in(4*d*x + 4*c) + sqrt(2)*a^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c))) + 2) - 75*(sqrt(2)*a^2*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*a^2*cos(4*d*x + 4*c)^2 + 9*sqrt(2)*a^2
*cos(2*d*x + 2*c)^2 + sqrt(2)*a^2*sin(6*d*x + 6*c)^2 + 9*sqrt(2)*a^2*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*a^2*sin(4
*d*x + 4*c)*sin(2*d*x + 2*c) + 9*sqrt(2)*a^2*sin(2*d*x + 2*c)^2 + 6*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2
+ 2*(3*sqrt(2)*a^2*cos(4*d*x + 4*c) + 3*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*cos(6*d*x + 6*c) + 6*(3*s
qrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*cos(4*d*x + 4*c) + 6*(sqrt(2)*a^2*sin(4*d*x + 4*c) + sqrt(2)*a^2*si
n(2*d*x + 2*c))*sin(6*d*x + 6*c))*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin
(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + 75*(sqr
t(2)*a^2*cos(6*d*x + 6*c)^2 + 9*sqrt(2)*a^2*cos(4*d*x + 4*c)^2 + 9*sqrt(2)*a^2*cos(2*d*x + 2*c)^2 + sqrt(2)*a^
2*sin(6*d*x + 6*c)^2 + 9*sqrt(2)*a^2*sin(4*d*x + 4*c)^2 + 18*sqrt(2)*a^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9
*sqrt(2)*a^2*sin(2*d*x + 2*c)^2 + 6*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2 + 2*(3*sqrt(2)*a^2*cos(4*d*x +
4*c) + 3*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*cos(6*d*x + 6*c) + 6*(3*sqrt(2)*a^2*cos(2*d*x + 2*c) + sq
rt(2)*a^2)*cos(4*d*x + 4*c) + 6*(sqrt(2)*a^2*sin(4*d*x + 4*c) + sqrt(2)*a^2*sin(2*d*x + 2*c))*sin(6*d*x + 6*c)
)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c
), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3...
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Fricas [A]
time = 0.38, size = 212, normalized size = 1.29 \begin {gather*} \frac {3 \, {\left ({\left (25 \, A + 38 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (25 \, A + 38 \, B\right )} a^{2} \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 (25 \, A + 22 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (17 \, A + 6 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")
[Out]
1/96*(3*((25*A + 38*B)*a^2*cos(d*x + c)^4 + (25*A + 38*B)*a^2*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*(25*A + 22*B)*a^2*cos(d*x + c)^2 + 2*(17*A + 6*B)*a^2*cos(d*x + c) + 8*A*a^2)*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))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)
[Out]
Timed out
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Giac [A]
time = 0.78, size = 268, normalized size = 1.63 \begin {gather*} -\frac {\sqrt {2} {\left (3 \, \sqrt {2} {\left (25 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 38 \, B a^{2} \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 (300 \, A a^{2} \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} + 264 \, B a^{2} \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} - 368 \, A a^{2} \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} - 288 \, B a^{2} \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} + 117 \, A a^{2} \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 ) + 78 \, B a^{2} \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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")
[Out]
-1/96*sqrt(2)*(3*sqrt(2)*(25*A*a^2*sgn(cos(1/2*d*x + 1/2*c)) + 38*B*a^2*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*(300*A*a^2*sgn(cos(1/2*d*x + 1/
2*c))*sin(1/2*d*x + 1/2*c)^5 + 264*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^5 - 368*A*a^2*sgn(cos(
1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 - 288*B*a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)^3 + 117*A*
a^2*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 78*B*a^2*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 )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\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))^(5/2))/cos(c + d*x)^4,x)
[Out]
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/cos(c + d*x)^4, x)
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