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

Optimal. Leaf size=208 \[ -\frac{(5 A-2 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(4 A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(5 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-(((4*A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) - ((5*A - 2*B)*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + ((4*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x
])/(a^2*d) - ((5*A - 2*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.430758, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2960, 4019, 3787, 3771, 2641, 3768, 2639} \[ -\frac{(5 A-2 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{(4 A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(5 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((4*A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d)) - ((5*A - 2*B)*Sqrt[Cos
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) + ((4*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x
])/(a^2*d) - ((5*A - 2*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Sec[c + d*x
]^(5/2)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 2960

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^(m + n), Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(
d + c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

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))^2} \, dx &=\int \frac{\sec ^{\frac{5}{2}}(c+d x) (B+A \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-\frac{3}{2} a (A-B)+\frac{1}{2} a (7 A-B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(5 A-2 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \sqrt{\sec (c+d x)} \left (-\frac{1}{2} a^2 (5 A-2 B)+\frac{3}{2} a^2 (4 A-B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(5 A-2 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(5 A-2 B) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}+\frac{(4 A-B) \int \sec ^{\frac{3}{2}}(c+d x) \, dx}{2 a^2}\\ &=\frac{(4 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac{(5 A-2 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(4 A-B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}-\frac{\left ((5 A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac{(5 A-2 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{(4 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac{(5 A-2 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\left ((4 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{(4 A-B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}-\frac{(5 A-2 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}+\frac{(4 A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d}-\frac{(5 A-2 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 3.14467, size = 303, normalized size = 1.46 \[ -\frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (\cos \left (\frac{1}{2} (c+3 d x)\right )+i \sin \left (\frac{1}{2} (c+3 d x)\right )\right ) \left (-i (4 A-B) e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )+2 i (25 A-7 B) \cos (c+d x)+8 (5 A-2 B) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-i \sin \left (\frac{1}{2} (c+d x)\right )\right )-12 A \sin (c+d x)-7 A \sin (2 (c+d x))+17 i A \cos (2 (c+d x))+29 i A+B \sin (2 (c+d x))-5 i B \cos (2 (c+d x))-5 i B\right )}{6 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*((29*I)*A - (5*I)*B + (2*I)*(25*A - 7*B)*Cos[c + d*x] + (17*I)*A*Cos[2*(
c + d*x)] - (5*I)*B*Cos[2*(c + d*x)] - (I*(4*A - B)*(1 + E^(I*(c + d*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hype
rgeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 8*(5*A - 2*B)*Cos[(c + d*x)/2]^3*Sqrt[Co
s[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x)/2]) - 12*A*Sin[c + d*x] - 7*A*Sin[2*
(c + d*x)] + B*Sin[2*(c + d*x)])*(Cos[(c + 3*d*x)/2] + I*Sin[(c + 3*d*x)/2]))/(6*a^2*d*E^(I*d*x)*(1 + Cos[c +
d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 4.286, size = 494, normalized size = 2.4 \begin{align*} \text{result too large to display} \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)^2,x)

[Out]

-1/6*(2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*B*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^2-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*B*EllipticF
(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*(-2*sin(1/2*d*x+
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A-B)*sin(1/2*d*x+1/2*c)^6+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*(43*A-10*B)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(37*A-7*B)*si
n(1/2*d*x+1/2*c)^2)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+
1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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))**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 )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{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))^2,x, algorithm="giac")

[Out]

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