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

Optimal. Leaf size=193 \[ -\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(5 A-3 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{3 (A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}+\frac{(5 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.304157, antiderivative size = 193, 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, 3768, 3771, 2639, 2641} \[ -\frac{(A-B) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac{(5 A-3 B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 a d}-\frac{3 (A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{a d}+\frac{(5 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{3 (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(A - B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + ((5*A - 3*B)*Sqrt[Cos[c +
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a*d) - (3*(A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*d
) + ((5*A - 3*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) - ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a +
a*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 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 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)) \sec ^{\frac{5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx &=\int \frac{\sec ^{\frac{5}{2}}(c+d x) (B+A \sec (c+d x))}{a+a \sec (c+d x)} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{\int \sec ^{\frac{3}{2}}(c+d x) \left (-\frac{3}{2} a (A-B)+\frac{1}{2} a (5 A-3 B) \sec (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{(5 A-3 B) \int \sec ^{\frac{5}{2}}(c+d x) \, dx}{2 a}-\frac{(3 (A-B)) \int \sec ^{\frac{3}{2}}(c+d x) \, dx}{2 a}\\ &=-\frac{3 (A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a d}+\frac{(5 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{(5 A-3 B) \int \sqrt{\sec (c+d x)} \, dx}{6 a}+\frac{(3 (A-B)) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a}\\ &=-\frac{3 (A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a d}+\frac{(5 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{\left ((5 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a}+\frac{\left (3 (A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a}\\ &=\frac{3 (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 d}+\frac{(5 A-3 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 d}-\frac{3 (A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{a d}+\frac{(5 A-3 B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{(A-B) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 7.18193, size = 650, normalized size = 3.37 \[ \frac{\cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} \left (\frac{2 \sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (A \sin \left (\frac{d x}{2}\right )-B \sin \left (\frac{d x}{2}\right )\right )}{d}-\frac{3 (A-B) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \cos (d x)}{d}+\frac{2 \tan \left (\frac{c}{2}\right ) \sec (c) (5 A \cos (c)+2 A-3 B \cos (c))}{3 d}+\frac{4 A \sec (c) \sin (d x) \sec (c+d x)}{3 d}\right )}{a \cos (c+d x)+a}-\frac{A \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{\sqrt{2} d (a \cos (c+d x)+a)}+\frac{5 A \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d (a \cos (c+d x)+a)}+\frac{B \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{\sqrt{2} d (a \cos (c+d x)+a)}-\frac{B \sin (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 9.714, size = 493, normalized size = 2.6 \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)^(5/2)/(a+cos(d*x+c)*a),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((A-B)*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)+(-2*A+2*B)*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)+2*A*(
-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticF(cos(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{{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{a \cos \left (d x + c\right ) + a}\,{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)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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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 )} \sec \left (d x + c\right )^{\frac{5}{2}}}{a \cos \left (d x + c\right ) + a}, 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)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*cos(d*x + c) + a), 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)**(5/2)/(a+a*cos(d*x+c)),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{5}{2}}}{a \cos \left (d x + c\right ) + a}\,{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)^(5/2)/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

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