3.6.19 \(\int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [519]

Optimal. Leaf size=294 \[ \frac {a^{5/2} (326 A+283 B) \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{128 d}+\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{128 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

1/5*a*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(5/2)+1/240*a^3*(170*A+157*B)*sin(d*x+c)/d/sec(d*x+c)^(
5/2)/(a+a*cos(d*x+c))^(1/2)+1/192*a^3*(326*A+283*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+1/40*
a^2*(10*A+13*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)+1/128*a^3*(326*A+283*B)*sin(d*x+c)/d/(a+a
*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+1/128*a^(5/2)*(326*A+283*B)*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/
2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.54, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3040, 3055, 3060, 2849, 2853, 222} \begin {gather*} \frac {a^{5/2} (326 A+283 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {ArcSin}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{128 d}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{192 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{128 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {a^2 (10 A+13 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(326*A + 283*B)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c
+ d*x]])/(128*d) + (a^3*(170*A + 157*B)*Sin[c + d*x])/(240*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(5/2)) + (
a^2*(10*A + 13*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(40*d*Sec[c + d*x]^(5/2)) + (a*B*(a + a*Cos[c + d*x])
^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(5/2)) + (a^3*(326*A + 283*B)*Sin[c + d*x])/(192*d*Sqrt[a + a*Cos[c + d
*x]]*Sec[c + d*x]^(3/2)) + (a^3*(326*A + 283*B)*Sin[c + d*x])/(128*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x
]])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2849

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[2*n*((b*c + a*d)
/(b*(2*n + 1))), 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] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 3040

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Sin[e + f*x])^m*((
c + d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3055

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)))
*Sin[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 3060

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)*Sqrt
[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]

Rubi steps

\begin {align*} \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\\ &=\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \left (\frac {5}{2} a (2 A+B)+\frac {1}{2} a (10 A+13 B) \cos (c+d x)\right ) \, dx\\ &=\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{20} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {5}{4} a^2 (26 A+21 B)+\frac {1}{4} a^2 (170 A+157 B) \cos (c+d x)\right ) \, dx\\ &=\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{96} \left (a^2 (326 A+283 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {1}{128} \left (a^2 (326 A+283 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{128 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {1}{256} \left (a^2 (326 A+283 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{128 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left (a^2 (326 A+283 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{128 d}\\ &=\frac {a^{5/2} (326 A+283 B) \sin ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{128 d}+\frac {a^3 (170 A+157 B) \sin (c+d x)}{240 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^2 (10 A+13 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{40 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{192 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {a^3 (326 A+283 B) \sin (c+d x)}{128 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.50, size = 181, normalized size = 0.62 \begin {gather*} \frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (15 \sqrt {2} (326 A+283 B) \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+(5810 A+5521 B+(3620 A+3874 B) \cos (c+d x)+4 (230 A+331 B) \cos (2 (c+d x))+120 A \cos (3 (c+d x))+348 B \cos (3 (c+d x))+48 B \cos (4 (c+d x))) \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{3840 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(15*Sqrt[2]*(326*A + 283*B)*ArcSin[Sqrt[2]
*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + (5810*A + 5521*B + (3620*A + 3874*B)*Cos[c + d*x] + 4*(230*A + 331*B)*
Cos[2*(c + d*x)] + 120*A*Cos[3*(c + d*x)] + 348*B*Cos[3*(c + d*x)] + 48*B*Cos[4*(c + d*x)])*(-Sin[(c + d*x)/2]
+ Sin[(3*(c + d*x))/2])))/(3840*d)

________________________________________________________________________________________

Maple [A]
time = 0.38, size = 455, normalized size = 1.55

method	result	size
default	\(-\frac {\left (-1+\cos \left (d x +c \right )\right )^{3} \left (384 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{4}\left (d x +c \right )\right )+480 A \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+1392 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )+1840 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+2264 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+3260 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+2830 B \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+4890 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+4245 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+4890 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )+4245 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{1920 d \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{6}}\)	\(455\)

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))/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/1920/d*(-1+cos(d*x+c))^3*(384*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4+480*A*cos(d*x+c)^
3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+1392*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^
3+1840*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+2264*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*cos(d*x+c)^2+3260*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)+2830*B*sin(d*x+c)*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+4890*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+4245*B*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*sin(d*x+c)+4890*A*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+4245*B*
arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)))*cos(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)/(cos(d*x+
c)/(1+cos(d*x+c)))^(5/2)/(1/cos(d*x+c))^(3/2)/sin(d*x+c)^6*a^2

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 10042 vs. \(2 (252) = 504\).
time = 1.15, size = 10042, normalized size = 34.16 \begin {gather*} \text {Too large to display} \end {gather*}

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)^(3/2),x, algorithm="maxima")

[Out]

1/7680*((10*(cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*
d*x + 5*c)))^2 + 2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)^(3/4)*((75*a^2*sin(4/5*arctan2(si
n(5*d*x + 5*c), cos(5*d*x + 5*c))) + 88*a^2*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 75*a^2*sin(
1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))))*cos(3/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x
+ 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)) - (75*a^2*cos(4/5*arctan2(sin(5*d*x + 5*c
), cos(5*d*x + 5*c))) + 88*a^2*cos(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) - 75*a^2*cos(1/5*arctan2(s
in(5*d*x + 5*c), cos(5*d*x + 5*c))) - 88*a^2)*sin(3/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*
c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)))*sqrt(a) + 6*(cos(2/5*arctan2(sin(5*d*x + 5*c
), cos(5*d*x + 5*c)))^2 + sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + 2*cos(2/5*arctan2(sin(5*d*x
+ 5*c), cos(5*d*x + 5*c))) + 1)^(1/4)*(8*(a^2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2*sin(5*d*
x + 5*c) + a^2*sin(5*d*x + 5*c)*sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + 2*a^2*cos(2/5*arctan2
(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))*sin(5*d*x + 5*c) + a^2*sin(5*d*x + 5*c))*cos(5/2*arctan2(sin(2/5*arctan2
(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)) - 5*(15*a^2*
sin(4/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 15*a^2*sin(3/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*
c))) - 60*a^2*sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) - 268*a^2*sin(1/5*arctan2(sin(5*d*x + 5*c),
cos(5*d*x + 5*c))))*cos(1/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin
(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)) - 8*(a^2*cos(5*d*x + 5*c) + (a^2*cos(5*d*x + 5*c) - a^2)*cos(2/5*arcta
n2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + (a^2*cos(5*d*x + 5*c) - a^2)*sin(2/5*arctan2(sin(5*d*x + 5*c), cos
(5*d*x + 5*c)))^2 - a^2 + 2*(a^2*cos(5*d*x + 5*c) - a^2)*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))))
*sin(5/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5
*d*x + 5*c))) + 1)) + 5*(15*a^2*cos(4/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) - 15*a^2*cos(3/5*arctan2(
sin(5*d*x + 5*c), cos(5*d*x + 5*c))) - 60*a^2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) - 148*a^2*c
os(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 208*a^2)*sin(1/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c
), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)))*sqrt(a) + 4245*(a^2*arctan2
(-(cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)
))^2 + 2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)^(1/4)*(cos(1/2*arctan2(sin(2/5*arctan2(sin(
5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1))*sin(1/5*arctan2(s
in(5*d*x + 5*c), cos(5*d*x + 5*c))) - cos(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))*sin(1/2*arctan2(sin
(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)))
, (cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)
))^2 + 2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)^(1/4)*(cos(1/5*arctan2(sin(5*d*x + 5*c), co
s(5*d*x + 5*c)))*cos(1/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d
*x + 5*c), cos(5*d*x + 5*c))) + 1)) + sin(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))*sin(1/2*arctan2(sin
(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)))
+ 1) - a^2*arctan2(-(cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + sin(2/5*arctan2(sin(5*d*x + 5*c
), cos(5*d*x + 5*c)))^2 + 2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)^(1/4)*(cos(1/2*arctan2(s
in(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)
)*sin(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) - cos(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))*
sin(1/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*
d*x + 5*c))) + 1))), (cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + sin(2/5*arctan2(sin(5*d*x + 5*c
), cos(5*d*x + 5*c)))^2 + 2*cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)^(1/4)*(cos(1/5*arctan2(s
in(5*d*x + 5*c), cos(5*d*x + 5*c)))*cos(1/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(
2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))) + 1)) + sin(1/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))*
sin(1/2*arctan2(sin(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c))), cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*
d*x + 5*c))) + 1))) - 1) - a^2*arctan2((cos(2/5*arctan2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + sin(2/5*arcta
n2(sin(5*d*x + 5*c), cos(5*d*x + 5*c)))^2 + 2*c...

________________________________________________________________________________________

Fricas [A]
time = 0.44, size = 203, normalized size = 0.69 \begin {gather*} -\frac {15 \, {\left ({\left (326 \, A + 283 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (326 \, A + 283 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (384 \, B a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (10 \, A + 29 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 8 \, {\left (230 \, A + 283 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 10 \, {\left (326 \, A + 283 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (326 \, A + 283 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{1920 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

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)^(3/2),x, algorithm="fricas")

[Out]

-1/1920*(15*((326*A + 283*B)*a^2*cos(d*x + c) + (326*A + 283*B)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*s
qrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (384*B*a^2*cos(d*x + c)^5 + 48*(10*A + 29*B)*a^2*cos(d*x + c)^4 +
8*(230*A + 283*B)*a^2*cos(d*x + c)^3 + 10*(326*A + 283*B)*a^2*cos(d*x + c)^2 + 15*(326*A + 283*B)*a^2*cos(d*x
+ c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(5/2)/sec(d*x + c)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(3/2), x)

________________________________________________________________________________________