3.398 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=215 \[ \frac {a^{5/2} (163 A+200 B+304 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {a^3 (299 A+392 B+432 C) \tan (c+d x)}{192 d \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (17 A+24 B+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{32 d}+\frac {a (5 A+8 B) \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{24 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d} \]

[Out]

1/64*a^(5/2)*(163*A+200*B+304*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d+1/24*a*(5*A+8*B)*(a+a*co
s(d*x+c))^(3/2)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^3*tan(d*x+c)/d+1/192*a^3*(29
9*A+392*B+432*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/32*a^2*(17*A+24*B+16*C)*sec(d*x+c)*(a+a*cos(d*x+c))^(1/
2)*tan(d*x+c)/d

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Rubi [A]  time = 0.79, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3043, 2975, 2980, 2773, 206} \[ \frac {a^3 (299 A+392 B+432 C) \tan (c+d x)}{192 d \sqrt {a \cos (c+d x)+a}}+\frac {a^{5/2} (163 A+200 B+304 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{64 d}+\frac {a^2 (17 A+24 B+16 C) \tan (c+d x) \sec (c+d x) \sqrt {a \cos (c+d x)+a}}{32 d}+\frac {a (5 A+8 B) \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^{3/2}}{24 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{5/2}}{4 d} \]

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]^5,x]

[Out]

(a^(5/2)*(163*A + 200*B + 304*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^3*(299*
A + 392*B + 432*C)*Tan[c + d*x])/(192*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(17*A + 24*B + 16*C)*Sqrt[a + a*Cos[c
+ d*x]]*Sec[c + d*x]*Tan[c + d*x])/(32*d) + (a*(5*A + 8*B)*(a + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^2*Tan[c +
d*x])/(24*d) + (A*(a + a*Cos[c + d*x])^(5/2)*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 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))^{5/2} \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))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^{5/2} \left (\frac {1}{2} a (5 A+8 B)+\frac {1}{2} a (A+8 C) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac {a (5 A+8 B) (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int (a+a \cos (c+d x))^{3/2} \left (\frac {3}{4} a^2 (17 A+24 B+16 C)+\frac {1}{4} a^2 (11 A+8 B+48 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac {a^2 (17 A+24 B+16 C) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 d}+\frac {a (5 A+8 B) (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {\int \sqrt {a+a \cos (c+d x)} \left (\frac {1}{8} a^3 (299 A+392 B+432 C)+\frac {1}{8} a^3 (95 A+104 B+240 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac {a^3 (299 A+392 B+432 C) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 d}+\frac {a (5 A+8 B) (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{128} \left (a^2 (163 A+200 B+304 C)\right ) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac {a^3 (299 A+392 B+432 C) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 d}+\frac {a (5 A+8 B) (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (a^3 (163 A+200 B+304 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^{5/2} (163 A+200 B+304 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{64 d}+\frac {a^3 (299 A+392 B+432 C) \tan (c+d x)}{192 d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 (17 A+24 B+16 C) \sqrt {a+a \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{32 d}+\frac {a (5 A+8 B) (a+a \cos (c+d x))^{3/2} \sec ^2(c+d x) \tan (c+d x)}{24 d}+\frac {A (a+a \cos (c+d x))^{5/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.99, size = 176, normalized size = 0.82 \[ \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt {a (\cos (c+d x)+1)} \left (\sin \left (\frac {1}{2} (c+d x)\right ) ((2203 A+2056 B+1584 C) \cos (c+d x)+4 (163 A+136 B+48 C) \cos (2 (c+d x))+489 A \cos (3 (c+d x))+844 A+600 B \cos (3 (c+d x))+544 B+528 C \cos (3 (c+d x))+192 C)+6 \sqrt {2} (163 A+200 B+304 C) \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^4*(6*Sqrt[2]*(163*A + 200*B + 304*C)*ArcTanh[Sqr
t[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^4 + (844*A + 544*B + 192*C + (2203*A + 2056*B + 1584*C)*Cos[c + d*x] + 4*(
163*A + 136*B + 48*C)*Cos[2*(c + d*x)] + 489*A*Cos[3*(c + d*x)] + 600*B*Cos[3*(c + d*x)] + 528*C*Cos[3*(c + d*
x)])*Sin[(c + d*x)/2]))/(768*d)

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fricas [A]  time = 0.69, size = 244, normalized size = 1.13 \[ \frac {3 \, {\left ({\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} + {\left (163 \, A + 200 \, B + 304 \, C\right )} a^{2} \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (163 \, A + 200 \, B + 176 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (163 \, A + 136 \, B + 48 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (23 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right ) + 48 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/768*(3*((163*A + 200*B + 304*C)*a^2*cos(d*x + c)^5 + (163*A + 200*B + 304*C)*a^2*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)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(163*A + 200*B + 176*C)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B
+ 48*C)*a^2*cos(d*x + c)^2 + 8*(23*A + 8*B)*a^2*cos(d*x + c) + 48*A*a^2)*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))^(5/2)*(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 = 3.00, size = 2369, 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))^(5/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

1/24*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*a*(163*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))+163*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))+200*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))+200*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)*c
os(1/2*d*x+1/2*c)+2*a))+304*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))+304*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*(163*A*2^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)*a^(1/2)+200*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)+326*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+326*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+400*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+400*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+608*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+
608*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*(1793*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2072*B*2^(1/
2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1680*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1467*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+1467*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+1800*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+1800*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+2736*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+2736*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+(-9212*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-3912*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-3912*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-9632
*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-4800*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-4800*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-7104*C*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-7296*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-7296*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+2094*A*2^(1/2)*(a*
sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+489*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+489*A*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+1872*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)+600*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+600*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+1248*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+912*
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+912*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))^(5/2)*(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 \[ \int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^5} \,d x \]

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)^5,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)^5, 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)**5,x)

[Out]

Timed out

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