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

Optimal. Leaf size=176 \[ -\frac{(7 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(5 A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{2 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

-((7*A - 3*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)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[
c + d*x])^(3/2)) + ((5*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.519239, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2961, 2978, 2984, 12, 2782, 205} \[ -\frac{(7 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(5 A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{2 a d \sqrt{a \cos (c+d x)+a}}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A - 3*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)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[
c + d*x])^(3/2)) + ((5*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 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)) \sec ^{\frac{3}{2}}(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)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (5 A-B)-a (A-B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{a^2 (7 A-3 B)}{4 \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{a^3}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}-\frac{\left ((7 A-3 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) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{\left ((7 A-3 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{(7 A-3 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) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(5 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 4.45078, size = 443, normalized size = 2.52 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (\frac{(A+3 B) \csc ^3\left (\frac{1}{2} (c+d x)\right ) \left (5 (4 \cos (c+d x)+\cos (2 (c+d x))+1) \left (-\cos (c+d x)+\cos (c+d x) \sqrt{2-2 \sec (c+d x)} \tanh ^{-1}\left (\sqrt{\sin ^2\left (\frac{1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )+1\right )-2 \sin ^4\left (\frac{1}{2} (c+d x)\right ) \sin (c+d x) \tan (c+d x) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\sec (c+d x) \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 \cos ^{\frac{3}{2}}(c+d x)}+\frac{5 (A-B) \left (2 \sin \left (\frac{1}{2} (c+d x)\right )-1\right )}{\sqrt{\cos (c+d x)} \left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )^2}-\frac{5 (A-B) \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+1\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )-1\right ) \sqrt{\cos (c+d x)}}-\frac{20 (A-B) \sqrt{\cos (c+d x)}}{\sin \left (\frac{1}{2} (c+d x)\right )-1}-\frac{20 (A-B) \sqrt{\cos (c+d x)}}{\sin \left (\frac{1}{2} (c+d x)\right )+1}+30 (A-B) \tan ^{-1}\left (\frac{1-2 \sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )-30 (A-B) \tan ^{-1}\left (\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )+1}{\sqrt{\cos (c+d x)}}\right )\right )}{10 d (a (\cos (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(30*(A - B)*ArcTan[(1 - 2*Sin[(c + d*x)/2])/Sqrt[Cos
[c + d*x]]] - 30*(A - B)*ArcTan[(1 + 2*Sin[(c + d*x)/2])/Sqrt[Cos[c + d*x]]] - (20*(A - B)*Sqrt[Cos[c + d*x]])
/(-1 + Sin[(c + d*x)/2]) - (20*(A - B)*Sqrt[Cos[c + d*x]])/(1 + Sin[(c + d*x)/2]) + (5*(A - B)*(-1 + 2*Sin[(c
+ d*x)/2]))/(Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^2) - (5*(A - B)*(1 + 2*Sin[(c + d*x)/2])
)/(Sqrt[Cos[c + d*x]]*(-1 + Sin[(c + d*x)/2])) + ((A + 3*B)*Csc[(c + d*x)/2]^3*(5*(1 + 4*Cos[c + d*x] + Cos[2*
(c + d*x)])*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[
c + d*x]]) - 2*Hypergeometric2F1[2, 5/2, 7/2, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^4*Sin[c + d
*x]*Tan[c + d*x]))/(2*Cos[c + d*x]^(3/2))))/(10*d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 0.675, size = 312, normalized size = 1.8 \begin{align*} -{\frac{\sqrt{2}\cos \left ( dx+c \right ) }{4\,{a}^{2}d\sin \left ( dx+c \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( -7\,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 ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+3\,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 ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+5\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}-7\,A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}+3\,B\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-A\cos \left ( dx+c \right ) \sqrt{2}+B\cos \left ( dx+c \right ) \sqrt{2}-4\,A\sqrt{2} \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \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)^(3/2)/(a+cos(d*x+c)*a)^(3/2),x)

[Out]

-1/4/d*2^(1/2)/a^2*(-7*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)+3*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+5*A*cos(d
*x+c)^2*2^(1/2)-7*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-B*cos(d*x+
c)^2*2^(1/2)+3*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-A*cos(d*x+c)*
2^(1/2)+B*cos(d*x+c)*2^(1/2)-4*A*2^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)/
(1+cos(d*x+c))

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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*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.0023, size = 435, normalized size = 2.47 \begin{align*} \frac{\sqrt{2}{\left ({\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 3 \, 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 (5 \, A - B\right )} \cos \left (d x + c\right ) + 4 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{\cos \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/4*(sqrt(2)*((7*A - 3*B)*cos(d*x + c)^2 + 2*(7*A - 3*B)*cos(d*x + c) + 7*A - 3*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*((5*A - B)*cos(d*x + c) + 4*A)*sqrt(a*cos(
d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{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)^(3/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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