3.98 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\)
Optimal. Leaf size=209 \[ \frac{a^3 (163 A+200 B) \tan (c+d x)}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{5/2} (163 A+200 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^2 (11 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt{a \cos (c+d x)+a}}{24 d}+\frac{a^3 (95 A+104 B) \tan (c+d x) \sec (c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d} \]
[Out]
(a^(5/2)*(163*A + 200*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^3*(163*A + 200*
B)*Tan[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(95*A + 104*B)*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a
+ a*Cos[c + d*x]]) + (a^2*(11*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(24*d) + (a*A*(a
+ a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
________________________________________________________________________________________
Rubi [A] time = 0.608621, 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} \[ \frac{a^3 (163 A+200 B) \tan (c+d x)}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{5/2} (163 A+200 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^2 (11 A+8 B) \tan (c+d x) \sec ^2(c+d x) \sqrt{a \cos (c+d x)+a}}{24 d}+\frac{a^3 (95 A+104 B) \tan (c+d x) \sec (c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]
[Out]
(a^(5/2)*(163*A + 200*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^3*(163*A + 200*
B)*Tan[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(95*A + 104*B)*Sec[c + d*x]*Tan[c + d*x])/(96*d*Sqrt[a
+ a*Cos[c + d*x]]) + (a^2*(11*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(24*d) + (a*A*(a
+ a*Cos[c + d*x])^(3/2)*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))^{5/2} (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx &=\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (11 A+8 B)+\frac{1}{2} a (3 A+8 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^2 (11 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+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{12} \int \sqrt{a+a \cos (c+d x)} \left (\frac{1}{4} a^2 (95 A+104 B)+\frac{3}{4} a^2 (17 A+24 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (95 A+104 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (11 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+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{64} \left (a^2 (163 A+200 B)\right ) \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (163 A+200 B) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (95 A+104 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (11 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+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{128} \left (a^2 (163 A+200 B)\right ) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{a^3 (163 A+200 B) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (95 A+104 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (11 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+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (a^3 (163 A+200 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^{5/2} (163 A+200 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{64 d}+\frac{a^3 (163 A+200 B) \tan (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (95 A+104 B) \sec (c+d x) \tan (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (11 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+a \cos (c+d x))^{3/2} \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.65608, size = 152, normalized size = 0.73 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) ((2203 A+2056 B) \cos (c+d x)+(652 A+544 B) \cos (2 (c+d x))+489 A \cos (3 (c+d x))+844 A+600 B \cos (3 (c+d x))+544 B)+6 \sqrt{2} (163 A+200 B) \cos ^4(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{768 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*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)*ArcTanh[Sqrt[2]*Sin
[(c + d*x)/2]]*Cos[c + d*x]^4 + (844*A + 544*B + (2203*A + 2056*B)*Cos[c + d*x] + (652*A + 544*B)*Cos[2*(c + d
*x)] + 489*A*Cos[3*(c + d*x)] + 600*B*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(768*d)
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Maple [B] time = 4.174, size = 1630, 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)^(5/2)*(A+B*cos(d*x+c))*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))*(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))+163*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))+200*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))+200*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)*c
os(1/2*d*x+1/2*c)+2*a)))*sin(1/2*d*x+1/2*c)^8-48*(163*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+200*B*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))*(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+326*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+400*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+400*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)^6+8*(1793*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2072*B*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))*(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+1467*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+1800*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+1800*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)*si
n(1/2*d*x+1/2*c)^4+(-9212*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-3912*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-3912*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-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))*(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-4800*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+2094*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+489*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+489*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+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))*(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+600*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
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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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="maxima")
[Out]
Timed out
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Fricas [A] time = 2.18219, size = 608, normalized size = 2.91 \begin{align*} \frac{3 \,{\left ({\left (163 \, A + 200 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} +{\left (163 \, A + 200 \, B\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\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \,{\left (163 \, A + 136 \, B\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 )}} \end{align*}
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)^5,x, algorithm="fricas")
[Out]
1/768*(3*((163*A + 200*B)*a^2*cos(d*x + c)^5 + (163*A + 200*B)*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)*a^2*cos(d*x + c)^3 + 2*(163*A + 136*B)*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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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))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)
[Out]
Timed out
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Giac [B] time = 3.40891, 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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")
[Out]
1/384*(3*(163*A*a^(5/2) + 200*B*a^(5/2))*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*(163*A*a^(5/2) + 200*B*a^(5/2))*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3))) + 4*sqrt(2)*(489*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt
(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(7/2) + 600*(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^(7/2) - 10269*(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^(9/2)
- 12600*(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^(9/2) + 69885*(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^(11/2) + 103992*(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^(11/2) - 259233*(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^(13/2) - 339864*(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^(13/2) + 209979*(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^(15/2) + 262920*
(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^(15/2) - 55511*(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^(17/2) - 73640*(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^(17/2) + 6687*(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^(19/2) + 8808*(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^(19/2) - 299
*A*a^(21/2) - 392*B*a^(21/2))/((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