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

Optimal. Leaf size=176 \[ \frac{(19 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[Out]

((19*A + 5*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)*Sq
rt[Sec[c + d*x]]) - ((9*A - B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.526173, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2961, 2978, 12, 2782, 205} \[ \frac{(19 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(9 A-B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((19*A + 5*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)*Sq
rt[Sec[c + d*x]]) - ((9*A - B)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

Rule 2961

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*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 205

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

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (7 A+B)-a (A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(9 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a^2 (19 A+5 B)}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(9 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}+\frac{\left ((19 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(9 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}-\frac{\left ((19 A+5 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac{(19 A+5 B) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt{\sec (c+d x)}}-\frac{(9 A-B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.73673, size = 216, normalized size = 1.23 \[ \frac{i \cos ^5\left (\frac{1}{2} (c+d x)\right ) \left ((19 A+5 B) e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{1-e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )-\frac{1}{4} i \left (\sin \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{3}{2} (c+d x)\right )\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} ((9 A-B) \cos (c+d x)+13 A-5 B)\right )}{4 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*Cos[(c + d*x)/2]^5*(((19*A + 5*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c
+ d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/E^((I/2)*(c + d*x)) - (I/4)*(
13*A - 5*B + (9*A - B)*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x
))/2])))/(d*(a*(1 + Cos[c + d*x]))^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.682, size = 376, normalized size = 2.1 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{32\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{-1}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 9\,A\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,A\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) -19\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -4\,B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) -5\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -13\,A\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-19\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) +5\,B\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-5\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/d*2^(1/2)/a^3*(1/cos(d*x+c))^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^2*(9*A*2^(1/2)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+4*A*2^(1/
2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-19*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)-
4*B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-5*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos
(d*x+c)-13*A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-19*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+5*B*
2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-5*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c))/sin(d*x+c)^5/(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)

________________________________________________________________________________________

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*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.84354, size = 551, normalized size = 3.13 \begin{align*} -\frac{\sqrt{2}{\left ({\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \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{2 \,{\left ({\left (9 \, A - B\right )} \cos \left (d x + c\right )^{2} +{\left (13 \, A - 5 \, B\right )} \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 )}}}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(sqrt(2)*((19*A + 5*B)*cos(d*x + c)^3 + 3*(19*A + 5*B)*cos(d*x + c)^2 + 3*(19*A + 5*B)*cos(d*x + c) + 19
*A + 5*B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((9*A
- B)*cos(d*x + c)^2 + (13*A - 5*B)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a
^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

________________________________________________________________________________________

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*cos(d*x+c))*sec(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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