3.90 \(\int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=209 \[ \frac{a^2 (75 A+88 B) \tan (c+d x)}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{3/2} (75 A+88 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^2 (9 A+8 B) \tan (c+d x) \sec ^2(c+d x)}{24 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (75 A+88 B) \tan (c+d x) \sec (c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{a A \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{4 d} \]

[Out]

(a^(3/2)*(75*A + 88*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^2*(75*A + 88*B)*T
an[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(75*A + 88*B)*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a + a*
Cos[c + d*x]]) + (a^2*(9*A + 8*B)*Sec[c + d*x]^2*Tan[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a*A*Sqrt[a +
a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.484684, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2975, 2980, 2772, 2773, 206} \[ \frac{a^2 (75 A+88 B) \tan (c+d x)}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{3/2} (75 A+88 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^2 (9 A+8 B) \tan (c+d x) \sec ^2(c+d x)}{24 d \sqrt{a \cos (c+d x)+a}}+\frac{a^2 (75 A+88 B) \tan (c+d x) \sec (c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{a A \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(a^(3/2)*(75*A + 88*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^2*(75*A + 88*B)*T
an[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(75*A + 88*B)*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a + a*
Cos[c + d*x]]) + (a^2*(9*A + 8*B)*Sec[c + d*x]^2*Tan[c + d*x])/(24*d*Sqrt[a + a*Cos[c + d*x]]) + (a*A*Sqrt[a +
a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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 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 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])

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx &=\frac{a A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int \sqrt{a+a \cos (c+d x)} \left (\frac{1}{2} a (9 A+8 B)+\frac{1}{2} a (5 A+8 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^2 (9 A+8 B) \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{48} (a (75 A+88 B)) \int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac{a^2 (75 A+88 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (9 A+8 B) \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{64} (a (75 A+88 B)) \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (75 A+88 B) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (75 A+88 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (9 A+8 B) \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{128} (a (75 A+88 B)) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{a^2 (75 A+88 B) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (75 A+88 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (9 A+8 B) \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (a^2 (75 A+88 B)\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) \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) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (75 A+88 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (9 A+8 B) \sec ^2(c+d x) \tan (c+d x)}{24 d \sqrt{a+a \cos (c+d x)}}+\frac{a A \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.47133, size = 151, normalized size = 0.72 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) ((1155 A+1048 B) \cos (c+d x)+4 (75 A+88 B) \cos (2 (c+d x))+225 A \cos (3 (c+d x))+492 A+264 B \cos (3 (c+d x))+352 B)+6 \sqrt{2} (75 A+88 B) \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x])*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)*ArcTanh[Sqrt[2]*Sin[(c
+ d*x)/2]]*Cos[c + d*x]^4 + (492*A + 352*B + (1155*A + 1048*B)*Cos[c + d*x] + 4*(75*A + 88*B)*Cos[2*(c + d*x)]
+ 225*A*Cos[3*(c + d*x)] + 264*B*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(768*d)

________________________________________________________________________________________

Maple [B]  time = 4.114, size = 1631, normalized size = 7.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(3/2)*(A+B*cos(d*x+c))*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))
*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+88*B*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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*
d*x+1/2*c)^2)^(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))*(a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+176*B*ln(-4/(-2*c
os(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+968*B*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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/
2*d*x+1/2*c)^2)^(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))*(a^(1/2)*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+1208*B*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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*co
s(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(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))*(a^(1/2)*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(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))*(a^(1
/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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+1086*A*a
^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+225*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*
sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+225*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^
(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(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))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(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))*(a^(1/2)*2^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(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)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 2.23668, size = 581, normalized size = 2.78 \begin{align*} \frac{3 \,{\left ({\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )^{5} +{\left (75 \, A + 88 \, B\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\right )} a \cos \left (d x + c\right )^{3} + 2 \,{\left (75 \, A + 88 \, B\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 )}} \end{align*}

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))*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/768*(3*((75*A + 88*B)*a*cos(d*x + c)^5 + (75*A + 88*B)*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)/(cos(d*x + c)^3 +
cos(d*x + c)^2)) + 4*(3*(75*A + 88*B)*a*cos(d*x + c)^3 + 2*(75*A + 88*B)*a*cos(d*x + c)^2 + 8*(15*A + 8*B)*a*c
os(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)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))*sec(d*x+c)**5,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 3.21408, size = 1083, normalized size = 5.18 \begin{align*} \text{result too large to display} \end{align*}

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))*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/384*(3*(75*A*a^(3/2) + 88*B*a^(3/2))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^2 - a*(2*sqrt(2) + 3))) - 3*(75*A*a^(3/2) + 88*B*a^(3/2))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*
tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(225*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(5/2) + 264*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^14*B*a^(5/2) - 6261*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(7/2) - 40
08*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*B*a^(7/2) + 35925*(sqrt(a)*tan(1/2*d
*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(9/2) + 33960*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*
tan(1/2*d*x + 1/2*c)^2 + a))^10*B*a^(9/2) - 127449*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a))^8*A*a^(11/2) - 131784*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*a^(11/2
) + 101667*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(13/2) + 108312*(sqrt(a)*
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(13/2) - 26079*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^(15/2) - 29432*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a))^4*B*a^(15/2) + 3303*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*a
^(17/2) + 3384*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(17/2) - 147*A*a^(19/
2) - 152*B*a^(19/2))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1
/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^4)/d