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 \[ -\frac{\left (4 a^2 A b-2 a^3 B+a b^2 B-3 A b^3\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{a^3 d \left (a^2-b^2\right )}+\frac{\left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{b \left (5 a^2 A b-3 a^3 B+a b^2 B-3 A b^3\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a-b) (a+b)^2}+\frac{b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]
[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]))
________________________________________________________________________________________
Rubi [A] time = 0.702195, antiderivative size = 223, normalized size of antiderivative = 1.,
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 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 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 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 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 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]
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*}
Mathematica [A] time = 2.81339, size = 283, normalized size = 1.27 \[ \frac{\frac{-\frac{2 \left (2 a^2 A+a b B-3 A b^2\right ) \sin (c+d x) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{8 (a B-A b) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{2 \left (2 a^2 A-a b B-A b^2\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(a-b) (a+b)}+\frac{4 b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a d} \]
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)*Ellipt
icPi[-(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)
________________________________________________________________________________________
Maple [B] time = 6.337, size = 843, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}
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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(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*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*cos(1/2*d*x+1/2*
c)^2*a-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)))-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)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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)
________________________________________________________________________________________
Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}
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)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}
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)