\(\int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx\) [193]

Optimal result

Integrand size = 17, antiderivative size = 98 \[ \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx=2 a^{5/2} B \text {arctanh}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a+a \cos (x)}}\right )+\frac {2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt {a+a \cos (x)}}+\frac {2}{15} a^2 (8 A+5 B) \sqrt {a+a \cos (x)} \sin (x)+\frac {2}{5} a A (a+a \cos (x))^{3/2} \sin (x) \]

[Out]

2*a^(5/2)*B*arctanh(sin(x)*a^(1/2)/(a+a*cos(x))^(1/2))+2/5*a*A*(a+a*cos(x))^(3/2)*sin(x)+2/15*a^3*(32*A+35*B)*
sin(x)/(a+a*cos(x))^(1/2)+2/15*a^2*(8*A+5*B)*sin(x)*(a+a*cos(x))^(1/2)

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2907, 3055, 3060, 2852, 212} \[ \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx=2 a^{5/2} B \text {arctanh}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a \cos (x)+a}}\right )+\frac {2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt {a \cos (x)+a}}+\frac {2}{15} a^2 (8 A+5 B) \sin (x) \sqrt {a \cos (x)+a}+\frac {2}{5} a A \sin (x) (a \cos (x)+a)^{3/2} \]

[In]

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

[Out]

2*a^(5/2)*B*ArcTanh[(Sqrt[a]*Sin[x])/Sqrt[a + a*Cos[x]]] + (2*a^3*(32*A + 35*B)*Sin[x])/(15*Sqrt[a + a*Cos[x]]
) + (2*a^2*(8*A + 5*B)*Sqrt[a + a*Cos[x]]*Sin[x])/15 + (2*a*A*(a + a*Cos[x])^(3/2)*Sin[x])/5

Rule 212

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

Rule 2852

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 2907

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[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*} \text {integral}& = \int (a+a \cos (x))^{5/2} (B+A \cos (x)) \sec (x) \, dx \\ & = \frac {2}{5} a A (a+a \cos (x))^{3/2} \sin (x)+\frac {2}{5} \int (a+a \cos (x))^{3/2} \left (\frac {5 a B}{2}+\frac {1}{2} a (8 A+5 B) \cos (x)\right ) \sec (x) \, dx \\ & = \frac {2}{15} a^2 (8 A+5 B) \sqrt {a+a \cos (x)} \sin (x)+\frac {2}{5} a A (a+a \cos (x))^{3/2} \sin (x)+\frac {4}{15} \int \sqrt {a+a \cos (x)} \left (\frac {15 a^2 B}{4}+\frac {1}{4} a^2 (32 A+35 B) \cos (x)\right ) \sec (x) \, dx \\ & = \frac {2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt {a+a \cos (x)}}+\frac {2}{15} a^2 (8 A+5 B) \sqrt {a+a \cos (x)} \sin (x)+\frac {2}{5} a A (a+a \cos (x))^{3/2} \sin (x)+\left (a^2 B\right ) \int \sqrt {a+a \cos (x)} \sec (x) \, dx \\ & = \frac {2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt {a+a \cos (x)}}+\frac {2}{15} a^2 (8 A+5 B) \sqrt {a+a \cos (x)} \sin (x)+\frac {2}{5} a A (a+a \cos (x))^{3/2} \sin (x)-\left (2 a^3 B\right ) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (x)}{\sqrt {a+a \cos (x)}}\right ) \\ & = 2 a^{5/2} B \text {arctanh}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a+a \cos (x)}}\right )+\frac {2 a^3 (32 A+35 B) \sin (x)}{15 \sqrt {a+a \cos (x)}}+\frac {2}{15} a^2 (8 A+5 B) \sqrt {a+a \cos (x)} \sin (x)+\frac {2}{5} a A (a+a \cos (x))^{3/2} \sin (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx=\frac {1}{30} a^2 \sqrt {a (1+\cos (x))} \sec \left (\frac {x}{2}\right ) \left (30 \sqrt {2} B \text {arctanh}\left (\sqrt {2} \sin \left (\frac {x}{2}\right )\right )+2 (89 A+80 B+2 (14 A+5 B) \cos (x)+3 A \cos (2 x)) \sin \left (\frac {x}{2}\right )\right ) \]

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[x])]*Sec[x/2]*(30*Sqrt[2]*B*ArcTanh[Sqrt[2]*Sin[x/2]] + 2*(89*A + 80*B + 2*(14*A + 5*B)*C
os[x] + 3*A*Cos[2*x])*Sin[x/2]))/30

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(224\) vs. \(2(80)=160\).

Time = 3.62 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.30

method	result	size
parts	\(\frac {8 A \,a^{3} \cos \left (\frac {x}{2}\right ) \sin \left (\frac {x}{2}\right ) \left (3 \cos \left (\frac {x}{2}\right )^{4}+4 \cos \left (\frac {x}{2}\right )^{2}+8\right ) \sqrt {2}}{15 \sqrt {a \cos \left (\frac {x}{2}\right )^{2}}}+\frac {B \,a^{\frac {3}{2}} \cos \left (\frac {x}{2}\right ) \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \left (-4 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \sin \left (\frac {x}{2}\right )^{2}+18 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}+3 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {x}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}+8 a}{2 \cos \left (\frac {x}{2}\right )+\sqrt {2}}\right ) a +3 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {x}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}-2 a \right )}{2 \cos \left (\frac {x}{2}\right )-\sqrt {2}}\right ) a \right )}{3 \sin \left (\frac {x}{2}\right ) \sqrt {a \cos \left (\frac {x}{2}\right )^{2}}}\)	\(225\)
default	\(\frac {a^{\frac {3}{2}} \cos \left (\frac {x}{2}\right ) \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \left (24 A \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \sqrt {a}\, \sin \left (\frac {x}{2}\right )^{4}-20 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \left (4 A +B \right ) \sin \left (\frac {x}{2}\right )^{2}+120 A \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \sqrt {a}+90 B \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}\, \sqrt {a}+15 B \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {x}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}+8 a}{2 \cos \left (\frac {x}{2}\right )+\sqrt {2}}\right ) a +15 B \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {x}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {\sin \left (\frac {x}{2}\right )^{2} a}-2 a \right )}{2 \cos \left (\frac {x}{2}\right )-\sqrt {2}}\right ) a \right )}{15 \sin \left (\frac {x}{2}\right ) \sqrt {a \cos \left (\frac {x}{2}\right )^{2}}}\)	\(230\)

[In]

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

[Out]

8/15*A*a^3*cos(1/2*x)*sin(1/2*x)*(3*cos(1/2*x)^4+4*cos(1/2*x)^2+8)*2^(1/2)/(a*cos(1/2*x)^2)^(1/2)+1/3*B*a^(3/2
)*cos(1/2*x)*(sin(1/2*x)^2*a)^(1/2)*(-4*a^(1/2)*2^(1/2)*(sin(1/2*x)^2*a)^(1/2)*sin(1/2*x)^2+18*a^(1/2)*2^(1/2)
*(sin(1/2*x)^2*a)^(1/2)+3*ln(4/(2*cos(1/2*x)+2^(1/2))*(a*2^(1/2)*cos(1/2*x)+a^(1/2)*2^(1/2)*(sin(1/2*x)^2*a)^(
1/2)+2*a))*a+3*ln(-4/(2*cos(1/2*x)-2^(1/2))*(a*2^(1/2)*cos(1/2*x)-a^(1/2)*2^(1/2)*(sin(1/2*x)^2*a)^(1/2)-2*a))
*a)/sin(1/2*x)/(a*cos(1/2*x)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.26 \[ \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx=\frac {15 \, {\left (B a^{2} \cos \left (x\right ) + B a^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (x\right )^{3} - 7 \, a \cos \left (x\right )^{2} - 4 \, \sqrt {a \cos \left (x\right ) + a} \sqrt {a} {\left (\cos \left (x\right ) - 2\right )} \sin \left (x\right ) + 8 \, a}{\cos \left (x\right )^{3} + \cos \left (x\right )^{2}}\right ) + 4 \, {\left (3 \, A a^{2} \cos \left (x\right )^{2} + {\left (14 \, A + 5 \, B\right )} a^{2} \cos \left (x\right ) + {\left (43 \, A + 40 \, B\right )} a^{2}\right )} \sqrt {a \cos \left (x\right ) + a} \sin \left (x\right )}{30 \, {\left (\cos \left (x\right ) + 1\right )}} \]

[In]

integrate((a+a*cos(x))^(5/2)*(A+B*sec(x)),x, algorithm="fricas")

[Out]

1/30*(15*(B*a^2*cos(x) + B*a^2)*sqrt(a)*log((a*cos(x)^3 - 7*a*cos(x)^2 - 4*sqrt(a*cos(x) + a)*sqrt(a)*(cos(x)
- 2)*sin(x) + 8*a)/(cos(x)^3 + cos(x)^2)) + 4*(3*A*a^2*cos(x)^2 + (14*A + 5*B)*a^2*cos(x) + (43*A + 40*B)*a^2)
*sqrt(a*cos(x) + a)*sin(x))/(cos(x) + 1)

Sympy [F(-1)]

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

[In]

integrate((a+a*cos(x))**(5/2)*(A+B*sec(x)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.44 \[ \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx=\frac {1}{30} \, {\left (3 \, \sqrt {2} a^{2} \sin \left (\frac {5}{2} \, x\right ) + 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{2} \, x\right ) + 150 \, \sqrt {2} a^{2} \sin \left (\frac {1}{2} \, x\right )\right )} A \sqrt {a} \]

[In]

integrate((a+a*cos(x))^(5/2)*(A+B*sec(x)),x, algorithm="maxima")

[Out]

1/30*(3*sqrt(2)*a^2*sin(5/2*x) + 25*sqrt(2)*a^2*sin(3/2*x) + 150*sqrt(2)*a^2*sin(1/2*x))*A*sqrt(a)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int (a+a \cos (x))^{5/2} (A+B \sec (x)) \, dx=\frac {1}{30} \, \sqrt {2} {\left (48 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) \sin \left (\frac {1}{2} \, x\right )^{5} - 160 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) \sin \left (\frac {1}{2} \, x\right )^{3} - 40 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) \sin \left (\frac {1}{2} \, x\right )^{3} - 15 \, \sqrt {2} B a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, x\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, x\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) + 240 \, A a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) \sin \left (\frac {1}{2} \, x\right ) + 180 \, B a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, x\right )\right ) \sin \left (\frac {1}{2} \, x\right )\right )} \sqrt {a} \]

[In]

integrate((a+a*cos(x))^(5/2)*(A+B*sec(x)),x, algorithm="giac")

[Out]

1/30*sqrt(2)*(48*A*a^2*sgn(cos(1/2*x))*sin(1/2*x)^5 - 160*A*a^2*sgn(cos(1/2*x))*sin(1/2*x)^3 - 40*B*a^2*sgn(co
s(1/2*x))*sin(1/2*x)^3 - 15*sqrt(2)*B*a^2*log(abs(-2*sqrt(2) + 4*sin(1/2*x))/abs(2*sqrt(2) + 4*sin(1/2*x)))*sg
n(cos(1/2*x)) + 240*A*a^2*sgn(cos(1/2*x))*sin(1/2*x) + 180*B*a^2*sgn(cos(1/2*x))*sin(1/2*x))*sqrt(a)

Mupad [F(-1)]

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

[In]

int((a + a*cos(x))^(5/2)*(A + B/cos(x)),x)

[Out]

int((a + a*cos(x))^(5/2)*(A + B/cos(x)), x)