3.5.80 \(\int \frac {A+B \cos (c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [480]

Optimal. Leaf size=163 \[ \frac {3 (A-B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {(3 A-5 B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(3 A-5 B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \]

[Out]

-1/3*(3*A-5*B)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)+(A-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))/sec(d*x+c)^(1/2)+3*(A-B)*(c
os(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c
)^(1/2)/a/d-1/3*(3*A-5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)
)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d

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Rubi [A]
time = 0.18, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3039, 4105, 3872, 3854, 3856, 2720, 2719} \begin {gather*} -\frac {(3 A-5 B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}+\frac {(A-B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}-\frac {(3 A-5 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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2719

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

Rule 2720

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

Rule 3039

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 3854

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 3856

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 3872

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 4105

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]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 5.35, size = 444, normalized size = 2.72 \begin {gather*} \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-6 \sqrt {2} A 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)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )+6 \sqrt {2} B 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)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )-12 A \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}+20 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}-\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left ((12 A-13 B) \cos \left (\frac {1}{2} (c-d x)\right )+(6 A-5 B) \cos \left (\frac {1}{2} (3 c+d x)\right )-2 B \sin (c) \sin \left (\frac {3}{2} (c+d x)\right )\right )}{\sqrt {\sec (c+d x)}}\right )}{6 a d (1+\cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^2*((-6*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)
)]*Csc[c]*(-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))]))/E^(I*d*x) + (6*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^
((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2
F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - 12*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[S
ec[c + d*x]] + 20*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - (Csc[c/2]*Sec[c/2]*Sec[(
c + d*x)/2]*((12*A - 13*B)*Cos[(c - d*x)/2] + (6*A - 5*B)*Cos[(3*c + d*x)/2] - 2*B*Sin[c]*Sin[(3*(c + d*x))/2]
))/Sqrt[Sec[c + d*x]]))/(6*a*d*(1 + Cos[c + d*x]))

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Maple [A]
time = 0.38, size = 262, normalized size = 1.61

method	result	size
default	\(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (3 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+8 B \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A -18 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A +7 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(262\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))/(a+a*cos(d*x+c))/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*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)^(1/2)*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))-5*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+8*B*sin(1/2*d*x+1/
2*c)^6+(6*A-18*B)*sin(1/2*d*x+1/2*c)^4+(-3*A+7*B)*sin(1/2*d*x+1/2*c)^2)/a/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)/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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 261, normalized size = 1.60 \begin {gather*} \frac {{\left (\sqrt {2} {\left (3 i \, A - 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 5 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-3 i \, A + 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 5 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (2 \, B \cos \left (d x + c\right )^{2} - {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((sqrt(2)*(3*I*A - 5*I*B)*cos(d*x + c) + sqrt(2)*(3*I*A - 5*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c)
+ I*sin(d*x + c)) + (sqrt(2)*(-3*I*A + 5*I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A + 5*I*B))*weierstrassPInverse(-4,
0, cos(d*x + c) - I*sin(d*x + c)) - 9*(sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*weierstrassZ
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*(sqrt(2)*(I*A - I*B)*cos(d*x + c) +
sqrt(2)*(I*A - I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(2*
B*cos(d*x + c)^2 - (3*A - 5*B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {A}{\cos {\left (c + d x \right )} \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\cos {\left (c + d x \right )} \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (a+a\,\cos \left (c+d\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))),x)

[Out]

int((A + B*cos(c + d*x))/((1/cos(c + d*x))^(3/2)*(a + a*cos(c + d*x))), x)

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