3.198 \(\int \frac {A+B \sec (x)}{(a+a \cos (x))^{5/2}} \, dx\)

Optimal. Leaf size=120 \[ \frac {(3 A-43 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {2} \sqrt {a \cos (x)+a}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a \cos (x)+a}}\right )}{a^{5/2}}+\frac {(3 A-11 B) \sin (x)}{16 a (a \cos (x)+a)^{3/2}}+\frac {(A-B) \sin (x)}{4 (a \cos (x)+a)^{5/2}} \]

[Out]

2*B*arctanh(sin(x)*a^(1/2)/(a+a*cos(x))^(1/2))/a^(5/2)+1/4*(A-B)*sin(x)/(a+a*cos(x))^(5/2)+1/16*(3*A-11*B)*sin
(x)/a/(a+a*cos(x))^(3/2)+1/32*(3*A-43*B)*arctanh(1/2*sin(x)*a^(1/2)*2^(1/2)/(a+a*cos(x))^(1/2))/a^(5/2)*2^(1/2
)

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Rubi [A]  time = 0.48, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2828, 2978, 2985, 2649, 206, 2773} \[ \frac {(3 A-43 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {2} \sqrt {a \cos (x)+a}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a \cos (x)+a}}\right )}{a^{5/2}}+\frac {(3 A-11 B) \sin (x)}{16 a (a \cos (x)+a)^{3/2}}+\frac {(A-B) \sin (x)}{4 (a \cos (x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*ArcTanh[(Sqrt[a]*Sin[x])/Sqrt[a + a*Cos[x]]])/a^(5/2) + ((3*A - 43*B)*ArcTanh[(Sqrt[a]*Sin[x])/(Sqrt[2]*S
qrt[a + a*Cos[x]])])/(16*Sqrt[2]*a^(5/2)) + ((A - B)*Sin[x])/(4*(a + a*Cos[x])^(5/2)) + ((3*A - 11*B)*Sin[x])/
(16*a*(a + a*Cos[x])^(3/2))

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])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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 2828

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 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]

Rubi steps

\begin {align*} \int \frac {A+B \sec (x)}{(a+a \cos (x))^{5/2}} \, dx &=\int \frac {(B+A \cos (x)) \sec (x)}{(a+a \cos (x))^{5/2}} \, dx\\ &=\frac {(A-B) \sin (x)}{4 (a+a \cos (x))^{5/2}}+\frac {\int \frac {\left (4 a B+\frac {3}{2} a (A-B) \cos (x)\right ) \sec (x)}{(a+a \cos (x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \sin (x)}{4 (a+a \cos (x))^{5/2}}+\frac {(3 A-11 B) \sin (x)}{16 a (a+a \cos (x))^{3/2}}+\frac {\int \frac {\left (8 a^2 B+\frac {1}{4} a^2 (3 A-11 B) \cos (x)\right ) \sec (x)}{\sqrt {a+a \cos (x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \sin (x)}{4 (a+a \cos (x))^{5/2}}+\frac {(3 A-11 B) \sin (x)}{16 a (a+a \cos (x))^{3/2}}+\frac {(3 A-43 B) \int \frac {1}{\sqrt {a+a \cos (x)}} \, dx}{32 a^2}+\frac {B \int \sqrt {a+a \cos (x)} \sec (x) \, dx}{a^3}\\ &=\frac {(A-B) \sin (x)}{4 (a+a \cos (x))^{5/2}}+\frac {(3 A-11 B) \sin (x)}{16 a (a+a \cos (x))^{3/2}}-\frac {(3 A-43 B) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (x)}{\sqrt {a+a \cos (x)}}\right )}{16 a^2}-\frac {(2 B) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,-\frac {a \sin (x)}{\sqrt {a+a \cos (x)}}\right )}{a^2}\\ &=\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {a+a \cos (x)}}\right )}{a^{5/2}}+\frac {(3 A-43 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (x)}{\sqrt {2} \sqrt {a+a \cos (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(A-B) \sin (x)}{4 (a+a \cos (x))^{5/2}}+\frac {(3 A-11 B) \sin (x)}{16 a (a+a \cos (x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 95, normalized size = 0.79 \[ \frac {\tan \left (\frac {x}{2}\right ) (3 A \cos (x)+7 A-11 B \cos (x)-15 B)+2 (3 A-43 B) \cos ^3\left (\frac {x}{2}\right ) \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right )+64 \sqrt {2} B \cos ^3\left (\frac {x}{2}\right ) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {x}{2}\right )\right )}{16 a (a (\cos (x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*A - 43*B)*ArcTanh[Sin[x/2]]*Cos[x/2]^3 + 64*Sqrt[2]*B*ArcTanh[Sqrt[2]*Sin[x/2]]*Cos[x/2]^3 + (7*A - 15*B
+ 3*A*Cos[x] - 11*B*Cos[x])*Tan[x/2])/(16*a*(a*(1 + Cos[x]))^(3/2))

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fricas [B]  time = 0.73, size = 234, normalized size = 1.95 \[ -\frac {\sqrt {2} {\left ({\left (3 \, A - 43 \, B\right )} \cos \relax (x)^{3} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \relax (x)^{2} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \relax (x) + 3 \, A - 43 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \relax (x)^{2} + 2 \, \sqrt {2} \sqrt {a \cos \relax (x) + a} \sqrt {a} \sin \relax (x) - 2 \, a \cos \relax (x) - 3 \, a}{\cos \relax (x)^{2} + 2 \, \cos \relax (x) + 1}\right ) - 32 \, {\left (B \cos \relax (x)^{3} + 3 \, B \cos \relax (x)^{2} + 3 \, B \cos \relax (x) + B\right )} \sqrt {a} \log \left (\frac {a \cos \relax (x)^{3} - 7 \, a \cos \relax (x)^{2} - 4 \, \sqrt {a \cos \relax (x) + a} \sqrt {a} {\left (\cos \relax (x) - 2\right )} \sin \relax (x) + 8 \, a}{\cos \relax (x)^{3} + \cos \relax (x)^{2}}\right ) - 4 \, {\left ({\left (3 \, A - 11 \, B\right )} \cos \relax (x) + 7 \, A - 15 \, B\right )} \sqrt {a \cos \relax (x) + a} \sin \relax (x)}{64 \, {\left (a^{3} \cos \relax (x)^{3} + 3 \, a^{3} \cos \relax (x)^{2} + 3 \, a^{3} \cos \relax (x) + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(sqrt(2)*((3*A - 43*B)*cos(x)^3 + 3*(3*A - 43*B)*cos(x)^2 + 3*(3*A - 43*B)*cos(x) + 3*A - 43*B)*sqrt(a)*
log(-(a*cos(x)^2 + 2*sqrt(2)*sqrt(a*cos(x) + a)*sqrt(a)*sin(x) - 2*a*cos(x) - 3*a)/(cos(x)^2 + 2*cos(x) + 1))
- 32*(B*cos(x)^3 + 3*B*cos(x)^2 + 3*B*cos(x) + B)*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 - 11*B)*cos(x) + 7*A - 15*B)*sqrt(a*cos(
x) + a)*sin(x))/(a^3*cos(x)^3 + 3*a^3*cos(x)^2 + 3*a^3*cos(x) + a^3)

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giac [B]  time = 0.95, size = 199, normalized size = 1.66 \[ \frac {1}{32} \, \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5}\right )} \tan \left (\frac {1}{2} \, x\right )^{2}}{a^{8}} + \frac {\sqrt {2} {\left (5 \, A a^{5} - 13 \, B a^{5}\right )}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, x\right ) - \frac {\sqrt {2} {\left (3 \, A \sqrt {a} - 43 \, B \sqrt {a}\right )} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, x\right ) - \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a}\right )}^{2}\right )}{64 \, a^{3}} + \frac {B \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, x\right ) - \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{\frac {5}{2}}} - \frac {B \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, x\right ) - \sqrt {a \tan \left (\frac {1}{2} \, x\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(a*tan(1/2*x)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5)*tan(1/2*x)^2/a^8 + sqrt(2)*(5*A*a^5 - 13*B*a^5)/a^8)*
tan(1/2*x) - 1/64*sqrt(2)*(3*A*sqrt(a) - 43*B*sqrt(a))*log((sqrt(a)*tan(1/2*x) - sqrt(a*tan(1/2*x)^2 + a))^2)/
a^3 + B*log(abs((sqrt(a)*tan(1/2*x) - sqrt(a*tan(1/2*x)^2 + a))^2 - a*(2*sqrt(2) + 3)))/a^(5/2) - B*log(abs((s
qrt(a)*tan(1/2*x) - sqrt(a*tan(1/2*x)^2 + a))^2 + a*(2*sqrt(2) - 3)))/a^(5/2)

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maple [B]  time = 0.36, size = 322, normalized size = 2.68 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}\, \left (3 A \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}+4 a}{\cos \left (\frac {x}{2}\right )}\right ) \left (\cos ^{4}\left (\frac {x}{2}\right )\right ) a -43 B \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}+4 a}{\cos \left (\frac {x}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {x}{2}\right )\right )+32 B \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {x}{2}\right )+4 \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}+8 a}{2 \cos \left (\frac {x}{2}\right )+\sqrt {2}}\right ) a \left (\cos ^{4}\left (\frac {x}{2}\right )\right )+32 B \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {x}{2}\right )-\sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}-2 a \right )}{2 \cos \left (\frac {x}{2}\right )-\sqrt {2}}\right ) a \left (\cos ^{4}\left (\frac {x}{2}\right )\right )+3 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {x}{2}\right )\right )-11 B \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {x}{2}\right )\right )+2 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {a}-2 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {x}{2}\right )\right )}\, \sqrt {a}\right )}{32 a^{\frac {7}{2}} \cos \left (\frac {x}{2}\right )^{3} \sin \left (\frac {x}{2}\right ) \sqrt {\left (\cos ^{2}\left (\frac {x}{2}\right )\right ) a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/a^(7/2)/cos(1/2*x)^3*(a*sin(1/2*x)^2)^(1/2)*(3*A*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*x)^2)^(1/2)+2*a)/cos(
1/2*x))*cos(1/2*x)^4*a-43*B*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*x)^2)^(1/2)+2*a)/cos(1/2*x))*a*cos(1/2*x)^4+32*
B*ln(4/(2*cos(1/2*x)+2^(1/2))*(a*2^(1/2)*cos(1/2*x)+a^(1/2)*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)+2*a))*a*cos(1/2*x)^
4+32*B*ln(-4*(a*2^(1/2)*cos(1/2*x)-a^(1/2)*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)-2*a)/(2*cos(1/2*x)-2^(1/2)))*a*cos(1
/2*x)^4+3*A*a^(1/2)*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*cos(1/2*x)^2-11*B*a^(1/2)*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*co
s(1/2*x)^2+2*A*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*a^(1/2)-2*B*2^(1/2)*(a*sin(1/2*x)^2)^(1/2)*a^(1/2))/sin(1/2*x)/(
cos(1/2*x)^2*a)^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {B}{\cos \relax (x)}}{{\left (a+a\,\cos \relax (x)\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sec {\relax (x )}}{\left (a \left (\cos {\relax (x )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(x))/(a*(cos(x) + 1))**(5/2), x)

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