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[(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))

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Rubi [A]  time = 0.481825, antiderivative size = 120, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.499836, 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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Maple [B]  time = 3.66, size = 322, normalized size = 2.7 \begin{align*}{\frac{1}{32}\sqrt{a \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{2}} \left ( 3\,A\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( x/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( x/2 \right ) \right ) ^{4}a-43\,B\sqrt{2}\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( x/2 \right ) }} \right ) a \left ( \cos \left ( x/2 \right ) \right ) ^{4}+32\,B\ln \left ( -4\,{\frac{a\sqrt{2}\cos \left ( x/2 \right ) -\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}-2\,a}{2\,\cos \left ( x/2 \right ) -\sqrt{2}}} \right ) a \left ( \cos \left ( x/2 \right ) \right ) ^{4}+32\,B\ln \left ( 4\,{\frac{a\sqrt{2}\cos \left ( x/2 \right ) +\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}+2\,a}{2\,\cos \left ( x/2 \right ) +\sqrt{2}}} \right ) a \left ( \cos \left ( x/2 \right ) \right ) ^{4}+3\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}} \left ( \cos \left ( x/2 \right ) \right ) ^{2}-11\,B\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}} \left ( \cos \left ( x/2 \right ) \right ) ^{2}+2\,A\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a}-2\,B\sqrt{2}\sqrt{a \left ( \sin \left ( x/2 \right ) \right ) ^{2}}\sqrt{a} \right ){a}^{-{\frac{7}{2}}} \left ( \cos \left ({\frac{x}{2}} \right ) \right ) ^{-3} \left ( \sin \left ({\frac{x}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{x}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

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*ln(2*(2*a^(1/2)*(a*sin(1/2*x)^2)^(1/2)+2*a)/cos(1/2*x))*
2^(1/2)*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*(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+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+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., 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+B*sec(x))/(a+a*cos(x))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.6152, size = 689, normalized size = 5.74 \begin{align*} -\frac{\sqrt{2}{\left ({\left (3 \, A - 43 \, B\right )} \cos \left (x\right )^{3} + 3 \,{\left (3 \, A - 43 \, B\right )} \cos \left (x\right )^{2} + 3 \,{\left (3 \, A - 43 \, B\right )} \cos \left (x\right ) + 3 \, A - 43 \, B\right )} \sqrt{a} \log \left (-\frac{a \cos \left (x\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (x\right ) + a} \sqrt{a} \sin \left (x\right ) - 2 \, a \cos \left (x\right ) - 3 \, a}{\cos \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1}\right ) - 32 \,{\left (B \cos \left (x\right )^{3} + 3 \, B \cos \left (x\right )^{2} + 3 \, B \cos \left (x\right ) + B\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 ({\left (3 \, A - 11 \, B\right )} \cos \left (x\right ) + 7 \, A - 15 \, B\right )} \sqrt{a \cos \left (x\right ) + a} \sin \left (x\right )}{64 \,{\left (a^{3} \cos \left (x\right )^{3} + 3 \, a^{3} \cos \left (x\right )^{2} + 3 \, a^{3} \cos \left (x\right ) + a^{3}\right )}} \end{align*}

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

[Out]

Timed out

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Giac [B]  time = 2.43033, size = 269, normalized size = 2.24 \begin{align*} \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}}} \end{align*}

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)