3.389 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)
Optimal. Leaf size=215 \[ \frac {a^{3/2} (75 A+88 B+112 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{96 d \sqrt {a \cos (c+d x)+a}}+\frac {a (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{24 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d} \]
[Out]
1/64*a^(3/2)*(75*A+88*B+112*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/4*A*(a+a*cos(d*x+c))^(3/
2)*sec(d*x+c)^3*tan(d*x+c)/d+1/64*a^2*(75*A+88*B+112*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/96*a^2*(39*A+56*
B+48*C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/24*a*(3*A+8*B)*sec(d*x+c)^2*(a+a*cos(d*x+c))^(1/2)*ta
n(d*x+c)/d
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Rubi [A] time = 0.64, antiderivative size = 215, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used =
{3043, 2975, 2980, 2772, 2773, 206} \[ \frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a \cos (c+d x)+a}}+\frac {a^{3/2} (75 A+88 B+112 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {a^2 (39 A+56 B+48 C) \tan (c+d x) \sec (c+d x)}{96 d \sqrt {a \cos (c+d x)+a}}+\frac {a (3 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt {a \cos (c+d x)+a}}{24 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]
[Out]
(a^(3/2)*(75*A + 88*B + 112*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^2*(75*A +
88*B + 112*C)*Tan[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(39*A + 56*B + 48*C)*Sec[c + d*x]*Tan[c +
d*x])/(96*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(3*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(
24*d) + (A*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 2772
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 2773
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 2975
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*S
in[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 2980
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]
Rule 3043
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Rubi steps
\begin {align*} \int (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (3 A+8 B)+\frac {1}{2} a (3 A+8 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (39 A+56 B+48 C)+\frac {3}{4} a^2 (9 A+8 B+16 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {a^2 (39 A+56 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{64} (a (75 A+88 B+112 C)) \int \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{128} (a (75 A+88 B+112 C)) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (a^2 (75 A+88 B+112 C)\right ) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}\\ &=\frac {a^{3/2} (75 A+88 B+112 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \tan (c+d x)}{64 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \sec (c+d x) \tan (c+d x)}{96 d \sqrt {a+a \cos (c+d x)}}+\frac {a (3 A+8 B) \sqrt {a+a \cos (c+d x)} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}
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Mathematica [A] time = 2.09, size = 174, normalized size = 0.81 \[ \frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt {a (\cos (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) ((1155 A+1048 B+1008 C) \cos (c+d x)+4 (75 A+88 B+48 C) \cos (2 (c+d x))+225 A \cos (3 (c+d x))+492 A+264 B \cos (3 (c+d x))+352 B+336 C \cos (3 (c+d x))+192 C)+6 \sqrt {2} (75 A+88 B+112 C) \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{768 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]
[Out]
(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2]*(75*A + 88*B + 112*C)*ArcTanh[Sqrt[2]
*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (492*A + 352*B + 192*C + (1155*A + 1048*B + 1008*C)*Cos[c + d*x] + 4*(75*A
+ 88*B + 48*C)*Cos[2*(c + d*x)] + 225*A*Cos[3*(c + d*x)] + 264*B*Cos[3*(c + d*x)] + 336*C*Cos[3*(c + d*x)])*S
in[(c + d*x)/2]))/(768*d)
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fricas [A] time = 0.72, size = 232, normalized size = 1.08 \[ \frac {3 \, {\left ({\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{5} + {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B + 48 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right ) + 48 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")
[Out]
1/768*(3*((75*A + 88*B + 112*C)*a*cos(d*x + c)^5 + (75*A + 88*B + 112*C)*a*cos(d*x + c)^4)*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)/(c
os(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(75*A + 88*B + 112*C)*a*cos(d*x + c)^3 + 2*(75*A + 88*B + 48*C)*a*cos(
d*x + c)^2 + 8*(15*A + 8*B)*a*cos(d*x + c) + 48*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^5
+ d*cos(d*x + c)^4)
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 2.88, size = 2370, normalized size = 11.02 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)
[Out]
1/24*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*a*(75*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2
))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+75*A*ln(4/(2*cos(1/2*d*x
+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+88*B*ln(-4
/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+
2*a))+88*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1
/2*d*x+1/2*c)+2*a))+112*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2
)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))+112*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^
2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a)))*sin(1/2*d*x+1/2*c)^8-48*(75*A*2^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)*a^(1/2)+88*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+112*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*a^(1/2)+150*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2
^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+150*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^
(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+176*B*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+176*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2)
)*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+224*C*ln(-4/(-2*cos(1/2
*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+224*
C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/
2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^6+8*(825*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+968*B*2^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1104*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+675*A*ln(-4/(-2*cos(1/
2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+675
*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1
/2*c)+2*a))*a+792*B*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^
(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+792*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(
1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+1008*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+1008*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2
))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^4-
4*(1095*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1208*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)
+1200*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+450*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+450*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(
1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+528*B*ln(-4/(-2*cos
(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+
528*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*
x+1/2*c)+2*a))*a+672*C*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a
*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+672*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a)*sin(1/2*d*x+1/2*c)^2+225*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^
(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+1086*A*2^(1/2)*(a*
sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+225*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)
^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+264*B*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+1008*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)+264*B*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*
2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+336*C*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*a^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+864*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+336*C
*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-a*2^(1/2)*cos(1/2*d*x+1
/2*c)+2*a))*a)/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^4/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 [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^5} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^5,x)
[Out]
int(((a + a*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^5, x)
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)
[Out]
Timed out
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