3.206 \(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=164 \[ \frac {(5 A-3 B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}+\frac {(5 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

1/3*(5*A-3*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)*(co
s(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*(5*A-3*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 = 0.19, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4020, 3787, 3769, 3771, 2641, 2639} \[ \frac {(5 A-3 B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}+\frac {(5 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a d}-\frac {3 (A-B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[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) + ((5*A - 3*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 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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {A+B \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 (5 A-3 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 {(5 A-3 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 {(5 A-3 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 {(5 A-3 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 {(5 A-3 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 ((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 {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 {(5 A-3 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]  time = 2.42, size = 232, normalized size = 1.41 \[ \frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (3 i (A-B) e^{\frac {1}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )+2 \cos (c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right ) (2 A \cos (c+d x)+5 A-3 B)-9 i (A-B) \cos \left (\frac {1}{2} (c+d x)\right )\right )+2 (5 A-3 B) \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \sec \left (d x + c\right )^{3} + a \sec \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 5.00, size = 262, normalized size = 1.60 \[ -\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (5 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 A \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (18 A -6 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-7 A +3 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*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))-3*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*A*sin(1/2*d*x+1
/2*c)^6+(18*A-6*B)*sin(1/2*d*x+1/2*c)^4+(-7*A+3*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 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(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 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A/(sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)**(5/2) + s
ec(c + d*x)**(3/2)), x))/a

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