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

Optimal. Leaf size=223 \[ \frac {\left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2} \]

[Out]

(2*A*a^2-3*A*b^2+B*a*b)*(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^2/(a^2-b^2)/d-(4*A*a^2*b-3*A*b^3-2*B*a^3+B*a*b^2)*(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^3/(a^2-b^2)/d+b*(5*A*a^2*b-3*A*b^3-3*B*a^3+B*a*b^2)*(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^3/(a-b)/(a+b)^2/d+b*(A*b-B*a)*sin(d*x+c
)*cos(d*x+c)^(1/2)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))

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Rubi [A]  time = 0.70, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2954, 2989, 3059, 2639, 3002, 2641, 2805} \[ -\frac {\left (4 a^2 A b-2 a^3 B+a b^2 B-3 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}+\frac {\left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac {b \left (5 a^2 A b-3 a^3 B+a b^2 B-3 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*A - 3*A*b^2 + a*b*B)*EllipticE[(c + d*x)/2, 2])/(a^2*(a^2 - b^2)*d) - ((4*a^2*A*b - 3*A*b^3 - 2*a^3*B
+ a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)*d) + (b*(5*a^2*A*b - 3*A*b^3 - 3*a^3*B + a*b^2*B)*Ellip
ticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^3*(a - b)*(a + b)^2*d) + (b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(a*(a^2 - b^2)*d*(b + a*Cos[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 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 2954

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[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 2989

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

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^{\frac {3}{2}}(c+d x) (B+A \cos (c+d x))}{(b+a \cos (c+d x))^2} \, dx\\ &=\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} b (A b-a B)-a (A b-a B) \cos (c+d x)+\frac {1}{2} \left (2 a^2 A-3 A b^2+a b B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac {\int \frac {-\frac {1}{2} a b (A b-a B)+\frac {1}{2} \left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^2 \left (a^2-b^2\right )}+\frac {\left (2 a^2 A-3 A b^2+a b B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 2.85, size = 281, normalized size = 1.26 \[ \frac {\frac {4 b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\frac {2 \left (2 a^2 A-a b B-A b^2\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) \sin (c+d x) \left (\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt {\sin ^2(c+d x)}}+\frac {8 (a B-A b) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}}{(a-b) (a+b)}}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) + ((2*(2*a^2*A - A*b^2 -
a*b*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(-(A*b) + a*B)*((a + b)*EllipticF[(c + d*x)/2,
2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + (2*(2*a^2*A - 3*A*b^2 + a*b*B)*(-2*a*b*EllipticE
[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*Ellip
ticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/
(4*a*d)

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fricas [F]  time = 123.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 13.83, size = 843, normalized size = 3.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/a^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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*b+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a)-2/a^2*b*(3*A*b-2*B
*a)/(a^2-a*b)*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*b^2*(A*b-B*a)/a^3*(a^2/b/(a^2-b^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*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b
*(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))+1/2*a/b/(a^2-b^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)/(-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))-1/2
*a/b/(a^2-b^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)/(-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))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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)*EllipticPi(cos(
1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),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 \[ \int \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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