3.1012 \(\int \frac{\sec ^{\frac{5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=296 \[ \frac{2 (b B-a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 b^2 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 b^2 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 b d} \]

[Out]

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

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Rubi [A]  time = 1.11021, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 b^2 d}+\frac{2 (b B-a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^2 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4102

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[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*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] && GtQ[n, 0]

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{\sec ^{\frac{5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{2 \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3 a C}{2}+\frac{1}{2} b (5 A+3 C) \sec (c+d x)+\frac{5}{2} (b B-a C) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{5 b}\\ &=\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\sqrt{\sec (c+d x)} \left (\frac{5}{4} a (b B-a C)+\frac{1}{4} b (5 b B+4 a C) \sec (c+d x)+\frac{3}{4} \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{15 b^2}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{8 \int \frac{-\frac{3}{8} a \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )-\frac{1}{8} b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \sec (c+d x)+\frac{5}{8} \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 b^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{8 \int \frac{-\frac{3}{8} a^2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )-\left (-\frac{3}{8} a b \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )+\frac{1}{8} a b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^2 b^3}-\frac{\left (a \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}{b^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{(b B-a C) \int \sqrt{\sec (c+d x)} \, dx}{3 b^2}-\frac{\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{5 b^3}-\frac{\left (a \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}{b^3}\\ &=-\frac{2 a \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)}}{b^3 (a+b) d}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{\left ((b B-a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^2}-\frac{\left (\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3}\\ &=-\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 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 b^3 d}+\frac{2 (b B-a C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 b^2 d}-\frac{2 a \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)}}{b^3 (a+b) d}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}\\ \end{align*}

Mathematica [F]  time = 80.3292, size = 0, normalized size = 0. \[ \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 9.906, size = 800, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*(-2/5*C/b/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/
2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*(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)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)-12*(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)*sin(1
/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*(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)-8*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*b^2-B*a*b+C*a^2)*a^2/b^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))+2*(B*b-C*a)/b^2*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b^2-B*a*b+C*a^2)/b^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)*(-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))+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*
x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1))/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(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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(sec(d*x+c)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**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{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(b*sec(d*x + c) + a), x)