3.99 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\)
Optimal. Leaf size=254 \[ \frac{a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{5/2} (283 A+326 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{128 d}+\frac{a^2 (13 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{40 d}+\frac{a^3 (157 A+170 B) \tan (c+d x) \sec ^2(c+d x)}{240 d \sqrt{a \cos (c+d x)+a}}+\frac{a^3 (283 A+326 B) \tan (c+d x) \sec (c+d x)}{192 d \sqrt{a \cos (c+d x)+a}}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]
[Out]
(a^(5/2)*(283*A + 326*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(128*d) + (a^3*(283*A + 326
*B)*Tan[c + d*x])/(128*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(283*A + 326*B)*Sec[c + d*x]*Tan[c + d*x])/(192*d*Sq
rt[a + a*Cos[c + d*x]]) + (a^3*(157*A + 170*B)*Sec[c + d*x]^2*Tan[c + d*x])/(240*d*Sqrt[a + a*Cos[c + d*x]]) +
(a^2*(13*A + 10*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(40*d) + (a*A*(a + a*Cos[c + d*x])^(
3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d)
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Rubi [A] time = 0.713067, antiderivative size = 254, normalized size of antiderivative = 1.,
number of steps used = 7, 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} \[ \frac{a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{5/2} (283 A+326 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{128 d}+\frac{a^2 (13 A+10 B) \tan (c+d x) \sec ^3(c+d x) \sqrt{a \cos (c+d x)+a}}{40 d}+\frac{a^3 (157 A+170 B) \tan (c+d x) \sec ^2(c+d x)}{240 d \sqrt{a \cos (c+d x)+a}}+\frac{a^3 (283 A+326 B) \tan (c+d x) \sec (c+d x)}{192 d \sqrt{a \cos (c+d x)+a}}+\frac{a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]
[Out]
(a^(5/2)*(283*A + 326*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(128*d) + (a^3*(283*A + 326
*B)*Tan[c + d*x])/(128*d*Sqrt[a + a*Cos[c + d*x]]) + (a^3*(283*A + 326*B)*Sec[c + d*x]*Tan[c + d*x])/(192*d*Sq
rt[a + a*Cos[c + d*x]]) + (a^3*(157*A + 170*B)*Sec[c + d*x]^2*Tan[c + d*x])/(240*d*Sqrt[a + a*Cos[c + d*x]]) +
(a^2*(13*A + 10*B)*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^3*Tan[c + d*x])/(40*d) + (a*A*(a + a*Cos[c + d*x])^(
3/2)*Sec[c + d*x]^4*Tan[c + d*x])/(5*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 ^6(c+d x) \, dx &=\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (13 A+10 B)+\frac{5}{2} a (A+2 B) \cos (c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a^2 (13 A+10 B) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{20} \int \sqrt{a+a \cos (c+d x)} \left (\frac{1}{4} a^2 (157 A+170 B)+\frac{5}{4} a^2 (21 A+26 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (13 A+10 B) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{96} \left (a^2 (283 A+326 B)\right ) \int \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (13 A+10 B) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{128} \left (a^2 (283 A+326 B)\right ) \int \sqrt{a+a \cos (c+d x)} \sec ^2(c+d x) \, dx\\ &=\frac{a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (13 A+10 B) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{256} \left (a^2 (283 A+326 B)\right ) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (13 A+10 B) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{\left (a^3 (283 A+326 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 )}{128 d}\\ &=\frac{a^{5/2} (283 A+326 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{128 d}+\frac{a^3 (283 A+326 B) \tan (c+d x)}{128 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (283 A+326 B) \sec (c+d x) \tan (c+d x)}{192 d \sqrt{a+a \cos (c+d x)}}+\frac{a^3 (157 A+170 B) \sec ^2(c+d x) \tan (c+d x)}{240 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (13 A+10 B) \sqrt{a+a \cos (c+d x)} \sec ^3(c+d x) \tan (c+d x)}{40 d}+\frac{a A (a+a \cos (c+d x))^{3/2} \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 2.02376, size = 176, normalized size = 0.69 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) (36 (781 A+650 B) \cos (c+d x)+4 (6509 A+6730 B) \cos (2 (c+d x))+5660 A \cos (3 (c+d x))+4245 A \cos (4 (c+d x))+24863 A+6520 B \cos (3 (c+d x))+4890 B \cos (4 (c+d x))+22030 B)+60 \sqrt{2} (283 A+326 B) \cos ^5(c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{15360 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]
[Out]
(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^5*(60*Sqrt[2]*(283*A + 326*B)*ArcTanh[Sqrt[2]*Si
n[(c + d*x)/2]]*Cos[c + d*x]^5 + (24863*A + 22030*B + 36*(781*A + 650*B)*Cos[c + d*x] + 4*(6509*A + 6730*B)*Co
s[2*(c + d*x)] + 5660*A*Cos[3*(c + d*x)] + 6520*B*Cos[3*(c + d*x)] + 4245*A*Cos[4*(c + d*x)] + 4890*B*Cos[4*(c
+ d*x)])*Sin[(c + d*x)/2]))/(15360*d)
________________________________________________________________________________________
Maple [B] time = 4.312, size = 1951, normalized size = 7.7 \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)^6,x)
[Out]
1/120*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-480*a*(283*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))+283*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))+326*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))+326*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)^10+240*(566*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+65
2*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+1415*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+1415*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+1630*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+1
630*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)^8-80*(3962*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+4564*B*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+4245*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+4245*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+4890*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+4890*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*(36224*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+40960*B*2^(1/2)*(a*sin(
1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+21225*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+21225*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+24450*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+24450*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)^4-10*(12556*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+13400*B*2^(1/2)*(a*sin(1/2*
d*x+1/2*c)^2)^(1/2)*a^(1/2)+4245*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+4245*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+4890*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+4890*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+22230*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+4245*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+4245*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+2
0940*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+4890*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+4890*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))^5/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^5/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))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 2.32016, size = 670, normalized size = 2.64 \begin{align*} \frac{15 \,{\left ({\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} +{\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{5}\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 (15 \,{\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \,{\left (283 \, A + 326 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (283 \, A + 230 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 48 \,{\left (29 \, A + 10 \, B\right )} a^{2} \cos \left (d x + c\right ) + 384 \, A a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{7680 \,{\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\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)^6,x, algorithm="fricas")
[Out]
1/7680*(15*((283*A + 326*B)*a^2*cos(d*x + c)^6 + (283*A + 326*B)*a^2*cos(d*x + c)^5)*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*(15*(283*A + 326*B)*a^2*cos(d*x + c)^4 + 10*(283*A + 326*B)*a^2*cos(d*x + c)^3
+ 8*(283*A + 230*B)*a^2*cos(d*x + c)^2 + 48*(29*A + 10*B)*a^2*cos(d*x + c) + 384*A*a^2)*sqrt(a*cos(d*x + c) +
a)*sin(d*x + c))/(d*cos(d*x + c)^6 + d*cos(d*x + c)^5)
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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)**6,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 3.56792, size = 1304, normalized size = 5.13 \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)^6,x, algorithm="giac")
[Out]
1/3840*(15*(283*A*a^(5/2) + 326*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))) - 15*(283*A*a^(5/2) + 326*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))) + 4*sqrt(2)*(4245*(sqrt(a)*tan(1/2*d*x + 1/2*c) -
sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^18*A*a^(7/2) + 4890*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1
/2*c)^2 + a))^18*B*a^(7/2) - 114615*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^16*A*a
^(9/2) - 132030*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^16*B*a^(9/2) + 1298820*(sq
rt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(11/2) + 1319880*(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^(11/2) - 6176700*(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^(13/2) - 6888120*(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^(13/2) + 16394598*(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
^(15/2) + 18352620*(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^(15/2) - 1404277
0*(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^(17/2) - 15746180*(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^(17/2) + 4791060*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(19/2) + 5497320*(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^(19/2) - 860300*(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^
(21/2) - 959320*(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^(21/2) + 75885*(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^(23/2) + 84810*(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^(23/2) - 2671*A*a^(25/2) - 2990*B*a^(25/2))/((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)^5)/d