3.396 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)
Optimal. Leaf size=199 \[ \frac {a^{5/2} (19 A+20 B+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}-\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{12 d}+\frac {a (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d} \]
[Out]
1/4*a^(5/2)*(19*A+20*B+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d-1/12*a^3*(27*A-12*B-56*C)*sin
(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/12*a^2*(21*A+12*B-8*C)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+1/4*a*(5*A+4*B)*
(a+a*cos(d*x+c))^(3/2)*tan(d*x+c)/d+1/2*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)*tan(d*x+c)/d
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Rubi [A] time = 0.70, antiderivative size = 199, 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, 2976, 2981, 2773, 206} \[ -\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (21 A+12 B-8 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{12 d}+\frac {a^{5/2} (19 A+20 B+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}+\frac {a (5 A+4 B) \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^{5/2}}{2 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
[Out]
(a^(5/2)*(19*A + 20*B + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*d) - (a^3*(27*A - 12
*B - 56*C)*Sin[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) - (a^2*(21*A + 12*B - 8*C)*Sqrt[a + a*Cos[c + d*x]]*S
in[c + d*x])/(12*d) + (a*(5*A + 4*B)*(a + a*Cos[c + d*x])^(3/2)*Tan[c + d*x])/(4*d) + (A*(a + a*Cos[c + d*x])^
(5/2)*Sec[c + d*x]*Tan[c + d*x])/(2*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 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 2976
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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], 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] && GtQ[m, 1/2] && !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Rule 2981
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)*Sqr
t[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]
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))^{5/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+4 B)-\frac {1}{2} a (3 A-4 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac {a (5 A+4 B) (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^{3/2} \left (\frac {1}{4} a^2 (19 A+20 B+8 C)-\frac {1}{4} a^2 (21 A+12 B-8 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac {a^2 (21 A+12 B-8 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {a (5 A+4 B) (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \sqrt {a+a \cos (c+d x)} \left (\frac {3}{8} a^3 (19 A+20 B+8 C)-\frac {1}{8} a^3 (27 A-12 B-56 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{3 a}\\ &=-\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (21 A+12 B-8 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {a (5 A+4 B) (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{8} \left (a^2 (19 A+20 B+8 C)\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=-\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (21 A+12 B-8 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {a (5 A+4 B) (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}-\frac {\left (a^3 (19 A+20 B+8 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 )}{4 d}\\ &=\frac {a^{5/2} (19 A+20 B+8 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 d}-\frac {a^3 (27 A-12 B-56 C) \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}-\frac {a^2 (21 A+12 B-8 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{12 d}+\frac {a (5 A+4 B) (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{4 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 1.10, size = 153, normalized size = 0.77 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sqrt {a (\cos (c+d x)+1)} \left (4 \sin \left (\frac {1}{2} (c+d x)\right ) (3 (11 A+4 B+2 C) \cos (c+d x)+6 A+4 (3 B+8 C) \cos (2 (c+d x))+12 B+2 C \cos (3 (c+d x))+32 C)+6 \sqrt {2} (19 A+20 B+8 C) \cos ^2(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{48 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^2*(6*Sqrt[2]*(19*A + 20*B + 8*C)*ArcTanh[Sqrt[2]
*Sin[(c + d*x)/2]]*Cos[c + d*x]^2 + 4*(6*A + 12*B + 32*C + 3*(11*A + 4*B + 2*C)*Cos[c + d*x] + 4*(3*B + 8*C)*C
os[2*(c + d*x)] + 2*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)
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fricas [A] time = 0.92, size = 232, normalized size = 1.17 \[ \frac {3 \, {\left ({\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (19 \, A + 20 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (8 \, C a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, B + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (11 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]
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)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")
[Out]
1/48*(3*((19*A + 20*B + 8*C)*a^2*cos(d*x + c)^3 + (19*A + 20*B + 8*C)*a^2*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)/(co
s(d*x + c)^3 + cos(d*x + c)^2)) + 4*(8*C*a^2*cos(d*x + c)^3 + 8*(3*B + 8*C)*a^2*cos(d*x + c)^2 + 3*(11*A + 4*B
)*a^2*cos(d*x + c) + 6*A*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)
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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))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 2.69, size = 1512, normalized size = 7.60 \[ \text {result too large to display} \]
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)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)
[Out]
1/6*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-128*C*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)*sin(1/2*d*x+1/2*c)^6+4*(48*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+176*C*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)*a^(1/2)+57*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+57*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+60*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+60*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+24*C*ln(-4/(-2*c
os(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+24*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*(33*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+60*B*2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+152*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+57*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+57
*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+60*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+60*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+24*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+24*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+78*A*2
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+57*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+57*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+72*B*2^(1/2)*(a*sin(1/2*d*x+1/
2*c)^2)^(1/2)*a^(1/2)+60*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+60*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+144*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2
4*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+24*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))^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 [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))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^3,x)
[Out]
int(((a + a*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^3, x)
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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))**(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)
[Out]
Timed out
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