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

Optimal. Leaf size=206 \[ -\frac{5 (A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}-\frac{5 (A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

[Out]

((4*A - 7*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) - (5*(A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[S
ec[c + d*x]]) + ((4*A - 7*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) + ((A - B)*Sin[c +
d*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.433914, antiderivative size = 206, 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, 4020, 3787, 3769, 3771, 2641, 2639} \[ -\frac{5 (A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}-\frac{5 (A-2 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(4 A-7 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*A - 7*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) - (5*(A - 2*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[S
ec[c + d*x]]) + ((4*A - 7*B)*Sin[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) + ((A - B)*Sin[c +
d*x])/(3*d*Sqrt[Sec[c + d*x]]*(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 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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 3769

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

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 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)}{(a+a \cos (c+d x))^2 \sec ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{B+A \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{-\frac{3}{2} a (A-3 B)+\frac{5}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{-\frac{15}{2} a^2 (A-2 B)+\frac{3}{2} a^2 (4 A-7 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{(4 A-7 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}-\frac{(5 (A-2 B)) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac{5 (A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{(5 (A-2 B)) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}+\frac{\left ((4 A-7 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-7 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) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^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{(4 A-7 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{5 (A-2 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{(4 A-7 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}+\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 6.84035, size = 777, normalized size = 3.77 \[ \frac{\cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} \left (\frac{8 (A-2 B) \cos (c) \sin (d x)}{d}-\frac{2 \sec \left (\frac{c}{2}\right ) \sec ^3\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 )}{3 d}-\frac{2 (A-B) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{4 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (7 A \sin \left (\frac{d x}{2}\right )-10 B \sin \left (\frac{d x}{2}\right )\right )}{3 d}-\frac{2 \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos (d x) (A \cos (2 c)+3 A-2 B \cos (2 c)-5 B)}{d}+\frac{4 (7 A-10 B) \tan \left (\frac{c}{2}\right )}{3 d}+\frac{4 B \sin (2 c) \cos (2 d x)}{3 d}+\frac{4 B \cos (2 c) \sin (2 d x)}{3 d}\right )}{(a \cos (c+d x)+a)^2}-\frac{4 \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 ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a \cos (c+d x)+a)^2}-\frac{10 A \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^4\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)^2}+\frac{7 \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 ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (a \cos (c+d x)+a)^2}+\frac{20 B \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^4\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)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*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]
^4*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])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) + (7*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]^4*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])/(3*d*E^(I*d*x)*(a + a*Cos[c + d*x])^2) - (10*A*Cos[c/2 + (d*x)/2]^4*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])^2) + (20*B*Cos[c/2 + (d*x)/2]^4*
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])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*((-2*(3*A - 5*B + A*Cos[2*c] - 2*B*Cos[2*c])*Cos[d*x]*Csc[
c/2]*Sec[c/2])/d + (4*B*Cos[2*d*x]*Sin[2*c])/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(7*A*Sin[(d*x)/2] - 10*B*S
in[(d*x)/2]))/(3*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/(3*d) + (8*(A - 2*B)
*Cos[c]*Sin[d*x])/d + (4*B*Cos[2*c]*Sin[2*d*x])/(3*d) + (4*(7*A - 10*B)*Tan[c/2])/(3*d) - (2*(A - B)*Sec[c/2 +
(d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2

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Maple [A]  time = 3.851, size = 435, normalized size = 2.1 \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))/(a+cos(d*x+c)*a)^2/sec(d*x+c)^(5/2),x)

[Out]

1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-16*B*cos(1/2*d*x+1/2*c)^8+24*A*cos(1/2*d*x+1/2*c
)^6+10*A*cos(1/2*d*x+1/2*c)^3*(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))+24*A*cos(1/2*d*x+1/2*c)^3*(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)^6-20*B*cos(1/2*d*x+1/2*c)^3*(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))-42*B*cos(1/2*d*x+1/2*c)^3*(si
n(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))-38*A*cos(1/2
*d*x+1/2*c)^4+48*B*cos(1/2*d*x+1/2*c)^4+15*A*cos(1/2*d*x+1/2*c)^2-21*B*cos(1/2*d*x+1/2*c)^2-A+B)/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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

[Out]

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