3.1016 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\)
Optimal. Leaf size=207 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^2 (A+3 C)-3 a b B+3 A b^2\right )}{3 a^3 d}-\frac{2 b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}-\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}} \]
[Out]
(-2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + (2*(3*A*b^2 - 3*a*b
*B + a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^3*d) - (2*b*(A*b^2 -
a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a + b)*
d) + (2*A*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.530612, antiderivative size = 207, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used =
{4104, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (A+3 C)-3 a b B+3 A b^2\right )}{3 a^3 d}-\frac{2 b \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}-\frac{2 (A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])),x]
[Out]
(-2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*d) + (2*(3*A*b^2 - 3*a*b
*B + a^2*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*a^3*d) - (2*b*(A*b^2 -
a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^3*(a + b)*
d) + (2*A*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]])
Rule 4104
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Rule 4106
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2), Int[(d*Csc[
e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Dist[1/a^2, Int[(a*A - (A*b - a*B)*Csc[e + f*x])/Sqrt[d*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Rule 3849
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[d*Sqrt[d*S
in[e + f*x]]*Sqrt[d*Csc[e + f*x]], Int[1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d
, e, f}, x] && NeQ[a^2 - b^2, 0]
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 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 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 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]
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx &=\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{2 \int \frac{\frac{3}{2} (A b-a B)-\frac{1}{2} a (A+3 C) \sec (c+d x)-\frac{1}{2} A b \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{3 a}\\ &=\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{2 \int \frac{\frac{3}{2} a (A b-a B)-\left (\frac{3}{2} b (A b-a B)+\frac{1}{2} a^2 (A+3 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{3 a^3}-\frac{\left (b \left (A b^2-a (b B-a C)\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{a^3}\\ &=\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{(A b-a B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{a^2}+\frac{\left (3 A b^2-3 a b B+a^2 (A+3 C)\right ) \int \sqrt{\sec (c+d x)} \, dx}{3 a^3}-\frac{\left (b \left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a^3}\\ &=-\frac{2 b \left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^3 (a+b) d}+\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}-\frac{\left ((A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{a^2}+\frac{\left (\left (3 A b^2-3 a b B+a^2 (A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac{2 (A b-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^2 d}+\frac{2 \left (3 A b^2-3 a b B+a^2 (A+3 C)\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^3 d}-\frac{2 b \left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^3 (a+b) d}+\frac{2 A \sin (c+d x)}{3 a d \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [F] time = 59.8327, size = 0, normalized size = 0. \[ \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx \]
Verification is Not applicable to the result.
[In]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])),x]
[Out]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])), x]
________________________________________________________________________________________
Maple [B] time = 2.712, size = 945, normalized size = 4.6 \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)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x)
[Out]
-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((4*A*a^3-4*A*a^2*b)*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^4+(-2*A*a^3+2*A*a^2*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+A*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b+3*A*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(
cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b^3+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(co
s(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a*b^2+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^3-3*B*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a*b^2-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a*b^2+3*a^3*C*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*C*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b+3*C*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a^2*b)/a^3/(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
________________________________________________________________________________________
Maxima [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 )} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{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)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*sec(d*x + c)^(3/2)), 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)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
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)/sec(d*x+c)**(3/2)/(a+b*sec(d*x+c)),x)
[Out]
Timed out
________________________________________________________________________________________
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 )} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{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)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*sec(d*x + c)^(3/2)), x)