\(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\) [516]

Optimal result

Integrand size = 35, antiderivative size = 198 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^{5/2} (20 A+19 B) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

[Out]

-1/4*a^3*(4*A-9*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)-1/2*a^2*(4*A-B)*sin(d*x+c)*(a+a*cos(d*
x+c))^(1/2)/d/sec(d*x+c)^(1/2)+2*a*A*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+1/4*a^(5/2)*(20*A+19
*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] (verified)

Time = 0.79 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3040, 3054, 3055, 3060, 2853, 222} \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^{5/2} (20 A+19 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 d}-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {a^2 (4 A-B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}{d} \]

[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)*(20*A + 19*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]])/(4*d) - (a^3*(4*A - 9*B)*Sin[c + d*x])/(4*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (a^2*(4*A -
B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(2*d*Sqrt[Sec[c + d*x]]) + (2*a*A*(a + a*Cos[c + d*x])^(3/2)*Sqrt[S
ec[c + d*x]]*Sin[c + d*x])/d

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 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 3054

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
*Sin[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 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*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+a \cos (c+d x))^{3/2} \left (\frac {1}{2} a (4 A+B)-\frac {1}{2} a (4 A-B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{4} a^2 (12 A+5 B)-\frac {1}{4} a^2 (4 A-9 B) \cos (c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{8} \left (a^2 (20 A+19 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 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d}-\frac {\left (a^2 (20 A+19 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 )}{4 d} \\ & = \frac {a^{5/2} (20 A+19 B) \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)}}{4 d}-\frac {a^3 (4 A-9 B) \sin (c+d x)}{4 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {a^2 (4 A-B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{2 d \sqrt {\sec (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.64 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {a^2 \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} (20 A+19 B) \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 (8 A+B+(4 A+11 B) \cos (c+d x)+B \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d} \]

[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]]*(Sqrt[2]*(20*A + 19*B)*ArcSin[Sqrt[2]*Sin[
(c + d*x)/2]]*Sqrt[Cos[c + d*x]] + 2*(8*A + B + (4*A + 11*B)*Cos[c + d*x] + B*Cos[2*(c + d*x)])*Sin[(c + d*x)/
2]))/(8*d)

Maple [A] (verified)

Time = 22.91 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.57

method	result	size
default	\(\frac {a^{2} \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (20 A \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+2 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+19 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )+4 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+20 A \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+11 B \sin \left (d x +c \right ) \cos \left (d x +c \right )+19 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+8 A \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{4 d \left (1+\cos \left (d x +c \right )\right )}\)	\(310\)
parts	\(\frac {A \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (5 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )+5 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+2 \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a^{2}}{d \left (1+\cos \left (d x +c \right )\right )}+\frac {B \left (\sec ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (19 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+2 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+19 \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+11 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right ) a^{2}}{4 d \left (1+\cos \left (d x +c \right )\right )}\)	\(352\)

[In]

int((a+cos(d*x+c)*a)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/4*a^2/d*sec(d*x+c)^(3/2)*(20*A*cos(d*x+c)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)+2*B*sin(d*x+c)*cos(d*x+c)^2+19*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(tan(d*x+c)*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)+4*A*sin(d*x+c)*cos(d*x+c)+20*A*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+11*B*sin(d*x+c)*cos(d*x+c)+19*B*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+8*A*sin(d*x+c))*cos(d*x+c)*(a*(1+cos(d*x+c)))^(1/2)
/(1+cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.74 \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (20 \, A + 19 \, B\right )} a^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, A a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[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]

-1/4*(((20*A + 19*B)*a^2*cos(d*x + c) + (20*A + 19*B)*a^2)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*
x + c))/(sqrt(a)*sin(d*x + c))) - (2*B*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 8*A*a^2)*sqrt(a*co
s(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]

[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

Maxima [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]

[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]

Timed out

Giac [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]

[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]

Timed out

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]

[In]

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

[Out]

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