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

Optimal. Leaf size=222 \[ -\frac{(9 A+B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(3 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(9 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(6 A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

[Out]

((9*A + B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((3*A + B)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(
5*d*(a + a*Sec[c + d*x])^3) - ((6*A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((
9*A + B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.581253, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2960, 4019, 3787, 3771, 2639, 2641} \[ -\frac{(9 A+B) \sin (c+d x) \sqrt{\sec (c+d x)}}{10 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(3 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(9 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{(6 A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((9*A + B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) + ((3*A + B)*Sqrt[Cos[c
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) - ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(
5*d*(a + a*Sec[c + d*x])^3) - ((6*A - B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((
9*A + B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Sec[c + d*x]))

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 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]

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]

Rubi steps

\begin{align*} \int \frac{(A+B \cos (c+d x)) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^3} \, dx &=\int \frac{\sec ^{\frac{5}{2}}(c+d x) (B+A \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-\frac{3}{2} a (A-B)+\frac{1}{2} a (9 A+B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(6 A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{2} a^2 (6 A-B)+\frac{1}{2} a^2 (21 A+4 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(6 A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(9 A+B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{\frac{3}{4} a^3 (9 A+B)+\frac{5}{4} a^3 (3 A+B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(6 A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(9 A+B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(3 A+B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}+\frac{(9 A+B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(6 A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(9 A+B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\left ((3 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}+\frac{\left ((9 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=\frac{(9 A+B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}+\frac{(3 A+B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(6 A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(9 A+B) \sqrt{\sec (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 6.88651, size = 793, normalized size = 3.57 \[ \frac{\cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} \left (\frac{2 \sec \left (\frac{c}{2}\right ) \sec ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )-B \sin \left (\frac{d x}{2}\right )\right )}{5 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (3 A \sin \left (\frac{d x}{2}\right )+2 B \sin \left (\frac{d x}{2}\right )\right )}{15 d}+\frac{2 (A-B) \tan \left (\frac{c}{2}\right ) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d}+\frac{4 (3 A+2 B) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (3 A \sin \left (\frac{d x}{2}\right )+B \sin \left (\frac{d x}{2}\right )\right )}{3 d}-\frac{2 (9 A+B) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos (d x)}{5 d}+\frac{4 (3 A+B) \tan \left (\frac{c}{2}\right )}{3 d}\right )}{(a \cos (c+d x)+a)^3}-\frac{3 \sqrt{2} A \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{5 d (a \cos (c+d x)+a)^3}+\frac{2 A \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d (a \cos (c+d x)+a)^3}-\frac{\sqrt{2} B \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right )}{15 d (a \cos (c+d x)+a)^3}+\frac{2 B \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^6\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]
^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7
/4, -E^((2*I)*(c + d*x))])*Sec[c/2])/(5*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) - (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/
(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(
c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]
)/(15*d*E^(I*d*x)*(a + a*Cos[c + d*x])^3) + (2*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c
+ d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(d*(a + a*Cos[c + d*x])^3) + (2*B*Cos[c/2 + (d*x)/2]^6*Sqrt[
Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*Sin[c])/(3*d*(a + a*Cos[c + d*x])
^3) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Sec[c + d*x]]*((-2*(9*A + B)*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) + (2*Sec[c/2]*
Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/
2] + B*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(3*A*Sin[(d*x)/2] + 2*B*Sin[(d*x)/2]))/(15*d) +
(4*(3*A + B)*Tan[c/2])/(3*d) + (4*(3*A + 2*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (2*(A - B)*Sec[c/2 + (d
*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x])^3

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Maple [A]  time = 3.971, size = 451, normalized size = 2. \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)^(1/2)/(a+cos(d*x+c)*a)^3,x)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(108*A*cos(1/2*d*x+1/2*c)^8-30*A*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+54*A
*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2
*c),2^(1/2))+12*B*cos(1/2*d*x+1/2*c)^8-10*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+6*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-138*A*cos(1/2*d*x+1/2*c)^6-22*B*cos(1/
2*d*x+1/2*c)^6+24*A*cos(1/2*d*x+1/2*c)^4+6*B*cos(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d*x+1/2*c)^2+7*B*cos(1/2*d*x+1/2
*c)^2+3*A-3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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

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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)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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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 )} \sqrt{\sec \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, 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)^(1/2)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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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)**(1/2)/(a+a*cos(d*x+c))**3,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 )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

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