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

Optimal. Leaf size=161 \[ \frac{(2 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{A \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d (\sec (c+d x)+1)}+\frac{A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.388056, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{(2 A+B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{A \sin (c+d x) \sqrt{\sec (c+d x)}}{a^2 d (\sec (c+d x)+1)}+\frac{A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + ((2*A + B)*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^2*d) - (A*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a^2*d*(1 + Sec[c
+ d*x])) - ((A - B)*Sec[c + d*x]^(3/2)*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 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))^2} \, dx &=\int \frac{\sec ^{\frac{3}{2}}(c+d x) (B+A \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\\ &=-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{2} a (A-B)+\frac{1}{2} a (5 A+B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{A \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{3 a^2 A}{2}+\frac{1}{2} a^2 (2 A+B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{A \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{A \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}+\frac{(2 A+B) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}\\ &=-\frac{A \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\left (A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}+\frac{\left ((2 A+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{A \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{(2 A+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{A \sqrt{\sec (c+d x)} \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 2.00357, size = 256, normalized size = 1.59 \[ \frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)} \left (\cos \left (\frac{1}{2} (c+3 d x)\right )+i \sin \left (\frac{1}{2} (c+3 d x)\right )\right ) \left (2 i \cos (c+d x) (i (A-B) \sin (c+d x)+(5 A+B) \cos (c+d x)+7 A-B)+8 (2 A+B) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-i \sin \left (\frac{1}{2} (c+d x)\right )\right )-i A e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^3 \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )\right )}{6 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(((-I)*A*(1 + E^(I*(c + d*x)))^3*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeome
tric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(c + d*x)) + 8*(2*A + B)*Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*
x]]*EllipticF[(c + d*x)/2, 2]*(Cos[(c + d*x)/2] - I*Sin[(c + d*x)/2]) + (2*I)*Cos[c + d*x]*(7*A - B + (5*A + B
)*Cos[c + d*x] + I*(A - B)*Sin[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)

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Maple [A]  time = 3.796, size = 350, normalized size = 2.2 \begin{align*}{\frac{1}{6\,{a}^{2}d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 12\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-4\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +6\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -2\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -16\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-2\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+3\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+3\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+A-B \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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)^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)*(12*A*cos(1/2*d*x+1/2*c)^6-4*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))+6*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))-2*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))-16*A*cos(1/2*d*x+1/2*c)^4-2*B*cos(1/2*d*x+1/2*c)^4+3*A*cos(1/2*d*x+1/2*c)^2+3*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{{\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 )}^{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)^(1/2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{A \sqrt{\sec{\left (c + d x \right )}}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \sqrt{\sec{\left (c + d x \right )}}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

(Integral(A*sqrt(sec(c + d*x))/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sqrt(sec(c
+ d*x))/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x))/a**2

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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 )}^{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)^(1/2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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