3.1326 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=239 \[ -\frac{\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^2 (-(2 A+C))+a b B+A b^2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a b d \left (a^2-b^2\right )}+\frac{\left (-3 a^2 b^2 (A+C)+a^3 b B+a^4 C+a b^3 B+A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 b d (a-b) (a+b)^2}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]
[Out]
((A*b^2 - a*(b*B - a*C))*EllipticE[(c + d*x)/2, 2])/(a*b*(a^2 - b^2)*d) - ((A*b^2 + a*b*B - a^2*(2*A + C))*Ell
ipticF[(c + d*x)/2, 2])/(a^2*(a^2 - b^2)*d) + ((A*b^4 + a^3*b*B + a*b^3*B + a^4*C - 3*a^2*b^2*(A + C))*Ellipti
cPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^2*(a - b)*b*(a + b)^2*d) - ((A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*
Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))
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Rubi [A] time = 0.787957, antiderivative size = 239, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used =
{4112, 3055, 3059, 2639, 3002, 2641, 2805} \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (-(2 A+C))+a b B+A b^2\right )}{a^2 d \left (a^2-b^2\right )}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{a b d \left (a^2-b^2\right )}+\frac{\left (-3 a^2 b^2 (A+C)+a^3 b B+a^4 C+a b^3 B+A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 b d (a-b) (a+b)^2}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2),x]
[Out]
((A*b^2 - a*(b*B - a*C))*EllipticE[(c + d*x)/2, 2])/(a*b*(a^2 - b^2)*d) - ((A*b^2 + a*b*B - a^2*(2*A + C))*Ell
ipticF[(c + d*x)/2, 2])/(a^2*(a^2 - b^2)*d) + ((A*b^4 + a^3*b*B + a*b^3*B + a^4*C - 3*a^2*b^2*(A + C))*Ellipti
cPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^2*(a - b)*b*(a + b)^2*d) - ((A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*
Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))
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]
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 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 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{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^2} \, 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))^2} \, dx\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac{\int \frac{\frac{1}{2} \left (A b^2-a b B-a^2 C+2 b^2 C\right )+b (b B-a (A+C)) \cos (c+d x)-\frac{1}{2} \left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac{\int \frac{-\frac{1}{2} a \left (A b^2-a b B-a^2 C+2 b^2 C\right )-\frac{1}{2} b \left (A b^2+a b B-a^2 (2 A+C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a b \left (a^2-b^2\right )}-\frac{\left (-A b^2+a (b B-a C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac{\left (A b^4+a^3 b B+a b^3 B+a^4 C-3 a^2 b^2 (A+C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 b \left (a^2-b^2\right )}-\frac{\left (A b^2+a b B-a^2 (2 A+C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a b \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a b B-a^2 (2 A+C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^4+a^3 b B+a b^3 B+a^4 C-3 a^2 b^2 (A+C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 (a-b) b (a+b)^2 d}-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 4.20234, size = 301, normalized size = 1.26 \[ -\frac{\frac{\frac{2 \sin (c+d x) \left (a (a C-b B)+A b^2\right ) \left (2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )-\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{8 b (a (A+C)-b B) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a (a+b)}+\frac{2 \left (3 a^2 C+a b B-A b^2-4 b^2 C\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(b-a) (a+b)}+\frac{4 \sin (c+d x) \sqrt{\cos (c+d x)} \left (a (a C-b B)+A b^2\right )}{\left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 b d} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^2),x]
[Out]
-((4*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos[c + d*x])) + ((2*(-(A
*b^2) + a*b*B + 3*a^2*C - 4*b^2*C)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*b*(-(b*B) + a*(A +
C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a*(a + b)) + (2*(A*b^2
+ a*(-(b*B) + a*C))*(-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)*EllipticPi[-(a/b), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqr
t[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(4*b*d)
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Maple [B] time = 6.981, size = 809, normalized size = 3.4 \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)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),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/a^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)
)+2/a^2*(A*b^2-B*a*b+C*a^2)*(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*(-2*A*b+B*a)/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
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Maxima [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)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
Timed out
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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)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [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)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**2/cos(d*x+c)**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^2*sqrt(cos(d*x + c))), x)