3.1017 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\)

Optimal. Leaf size=269 \[ -\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^2 b (A+3 C)+a^3 (-B)-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}+\frac{2 b^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a
^3*d) - (2*(3*A*b^3 - a^3*B - 3*a*b^2*B + a^2*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[S
ec[c + d*x]])/(3*a^4*d) + (2*b^2*(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^4*(a + b)*d) + (2*A*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (2*(A*b - a*B)*S
in[c + d*x])/(3*a^2*d*Sqrt[Sec[c + d*x]])

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Rubi [A]  time = 0.855461, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 b (A+3 C)+a^3 (-B)-3 a b^2 B+3 A b^3\right )}{3 a^4 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}+\frac{2 b^2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])),x]

[Out]

(2*(5*A*b^2 - 5*a*b*B + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*a
^3*d) - (2*(3*A*b^3 - a^3*B - 3*a*b^2*B + a^2*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[S
ec[c + d*x]])/(3*a^4*d) + (2*b^2*(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^4*(a + b)*d) + (2*A*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (2*(A*b - a*B)*S
in[c + d*x])/(3*a^2*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{5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx &=\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 \int \frac{\frac{5}{2} (A b-a B)-\frac{1}{2} a (3 A+5 C) \sec (c+d x)-\frac{3}{2} A b \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx}{5 a}\\ &=\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right )+\frac{1}{4} a (4 A b+5 a B) \sec (c+d x)-\frac{5}{4} b (A b-a B) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 a^2}\\ &=\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} a \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right )-\left (-\frac{1}{4} a^2 (4 A b+5 a B)+\frac{3}{4} b \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^4}+\frac{\left (b^2 \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^4}\\ &=\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{\left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) \int \sqrt{\sec (c+d x)} \, dx}{3 a^4}+\frac{\left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{5 a^3}+\frac{\left (b^2 \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^4}\\ &=\frac{2 b^2 \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^4 (a+b) d}+\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{\left (\left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (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^4}+\frac{\left (\left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 a^3 d}-\frac{2 \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (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^4 d}+\frac{2 b^2 \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^4 (a+b) d}+\frac{2 A \sin (c+d x)}{5 a d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [F]  time = 55.2983, size = 0, normalized size = 0. \[ \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx \]

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]^(5/2)*(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]^(5/2)*(a + b*Sec[c + d*x])), x]

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Maple [B]  time = 5.605, size = 801, normalized size = 3. \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)^(5/2)/(a+b*sec(d*x+c)),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5*A/a*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(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))*
(sin(1/2*d*x+1/2*c)^2)^(1/2)-6*sin(1/2*d*x+1/2*c)^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)-4/3/a^2*(3*A*a+A*b-B*a)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(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
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2/a^3*(3*A*a^2+2*A*a*b+A*b^2-2*B*a^2-B*a*b+C*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)*(Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(A*a^3+A*a^2*b+A*a*b^2+A*b^3-B*a^3
-B*a^2*b-B*a*b^2+C*a^3+C*a^2*b)/a^4*(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*b^2*(A*b^2-B*a*b+C*a^2)/a^3/
(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)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c)),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)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c)),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)/sec(d*x+c)**(5/2)/(a+b*sec(d*x+c)),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 )} \sec \left (d x + c\right )^{\frac{5}{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)^(5/2)/(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)^(5/2)), x)