3.533 \(\int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=127 \[ \frac {(3 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

[Out]

-1/2*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(1/2)+1/4*(3*A+B)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1
/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {(3 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

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

Rule 12

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

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]

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 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])

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, 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))^{3/2}} \, dx\\ &=-\frac {(A-B) \sin (c+d x)}{2 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 (3 A+B)}{2 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left ((3 A+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}{4 a}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left ((3 A+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 )}{2 d}\\ &=\frac {(3 A+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)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.64, size = 196, normalized size = 1.54 \[ \frac {i \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left ((3 A+B) e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-\frac {1}{2} i (A-B) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}\right )}{d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x])*Sqrt[Sec[c + d*x]])/(a + a*Cos[c + d*x])^(3/2),x]

[Out]

(I*Cos[(c + d*x)/2]^3*(((3*A + 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/2)*(A - B)*
Sec[(c + d*x)/2]^2*Sqrt[Sec[c + d*x]]*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2])))/(d*(a*(1 + Cos[c + d*x]))^(3
/2))

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 144, normalized size = 1.13 \[ -\frac {\sqrt {2} {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right ) + 3 \, A + B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*((3*A + B)*cos(d*x + c)^2 + 2*(3*A + B)*cos(d*x + c) + 3*A + B)*sqrt(a)*arctan(sqrt(2)*sqrt(a*co
s(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*sqrt(a*cos(d*x + c) + a)*(A - B)*sqrt(cos(d*x +
c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(a*cos(d*x + c) + a)^(3/2), x)

________________________________________________________________________________________

maple [B]  time = 0.37, size = 235, normalized size = 1.85 \[ \frac {\sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (-A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+3 A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )-1\right ) \sqrt {2}}{4 d \sin \left (d x +c \right )^{3} a^{2}} \]

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+a*cos(d*x+c))^(3/2),x)

[Out]

1/4/d*(1/cos(d*x+c))^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*(-A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+B
*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+3*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+A*2^(1
/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)-B*2^(1/2)*(cos(d*x+c)/(1
+cos(d*x+c)))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3*(cos(d*x+c)^2-1)*2^(1/2)/a^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(a*cos(d*x + c) + a)^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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))**(3/2),x)

[Out]

Integral((A + B*cos(c + d*x))*sqrt(sec(c + d*x))/(a*(cos(c + d*x) + 1))**(3/2), x)

________________________________________________________________________________________