3.1320 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=97 \[ \frac {2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}-\frac {2 (A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

2*A*(cos(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))/a/d-2*(A*b-B*a)*(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))/a^2/d+2*(A*b^2-a*(B*b-C*a))*
(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^2/(a+b)/d

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Rubi [A]  time = 0.39, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4112, 3059, 2639, 3002, 2641, 2805} \[ \frac {2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}-\frac {2 (A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*EllipticE[(c + d*x)/2, 2])/(a*d) - (2*(A*b - a*B)*EllipticF[(c + d*x)/2, 2])/(a^2*d) + (2*(A*b^2 - a*(b*B
- a*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^2*(a + b)*d)

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 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx\\ &=-\frac {\int \frac {-a C+(A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a}+\frac {A \int \sqrt {\cos (c+d x)} \, dx}{a}\\ &=\frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {(A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{a^2}+\left (\frac {b (A b-a B)}{a^2}+C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx\\ &=\frac {2 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 (A b-a B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 \left (\frac {b (A b-a B)}{a^2}+C\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{(a+b) d}\\ \end {align*}

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Mathematica [F]  time = 64.19, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 151.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 5.13, size = 323, normalized size = 3.33 \[ \frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}-A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +A \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) b^{2}-B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -B \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) a b +C \EllipticPi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 a}{a -b}, \sqrt {2}\right ) a^{2}\right )}{a^{2} \left (a -b \right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^
2+1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+A*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b+A*EllipticPi(cos(1/2*d*x+1/2*c),
2*a/(a-b),2^(1/2))*b^2-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b
-B*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a*b+C*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a^2
)/a^2/(a-b)/(-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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)*cos(d*x+c)**(1/2)/(a+b*sec(d*x+c)),x)

[Out]

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

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