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

Optimal. Leaf size=225 \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{5 d}+\frac{2 a (5 a B+4 A b) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b^2 (A-5 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d} \]

[Out]

(2*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]
])/(5*d) + (2*(a^2*B + 3*b^2*B + 2*a*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*
x]])/(3*d) + (2*a*(4*A*b + 5*a*B)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) - (2*b^2*(A - 5*C)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(5*d) + (2*A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

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Rubi [A]  time = 0.517784, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.163, Rules used = {4094, 4074, 4047, 3771, 2641, 4046, 2639} \[ \frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )}{3 d}+\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )}{5 d}+\frac{2 a (5 a B+4 A b) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^2}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2 b^2 (A-5 C) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]
])/(5*d) + (2*(a^2*B + 3*b^2*B + 2*a*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*
x]])/(3*d) + (2*a*(4*A*b + 5*a*B)*Sin[c + d*x])/(15*d*Sqrt[Sec[c + d*x]]) - (2*b^2*(A - 5*C)*Sqrt[Sec[c + d*x]
]*Sin[c + d*x])/(5*d) + (2*A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))

Rule 4094

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*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4074

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_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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]

Rubi steps

\begin{align*} \int \frac{(a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \sec (c+d x)) \left (\frac{1}{2} (4 A b+5 a B)+\frac{1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)-\frac{1}{2} b (A-5 C) \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a (4 A b+5 a B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right )-\frac{5}{4} \left (a^2 B+3 b^2 B+2 a b (A+3 C)\right ) \sec (c+d x)+\frac{3}{4} b^2 (A-5 C) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a (4 A b+5 a B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} \left (4 A b^2+10 a b B+a^2 (3 A+5 C)\right )+\frac{3}{4} b^2 (A-5 C) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{3} \left (-a^2 B-3 b^2 B-2 a b (A+3 C)\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a (4 A b+5 a B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{1}{5} \left (-10 a b B-5 b^2 (A-C)-a^2 (3 A+5 C)\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-\frac{1}{3} \left (\left (-a^2 B-3 b^2 B-2 a 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\\ &=\frac{2 \left (a^2 B+3 b^2 B+2 a 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 d}+\frac{2 a (4 A b+5 a B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{1}{5} \left (\left (-10 a b B-5 b^2 (A-C)-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\\ &=\frac{2 \left (10 a b B+5 b^2 (A-C)+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 d}+\frac{2 \left (a^2 B+3 b^2 B+2 a 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 d}+\frac{2 a (4 A b+5 a B) \sin (c+d x)}{15 d \sqrt{\sec (c+d x)}}-\frac{2 b^2 (A-5 C) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 A (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 4.71443, size = 234, normalized size = 1.04 \[ \frac{2 (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (10 \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (a^2 B+2 a b (A+3 C)+3 b^2 B\right )+\sin (c+d x) \left (3 \left (a^2 A \cos (2 (c+d x))+a^2 A+10 b^2 C\right )+10 a (a B+2 A b) \cos (c+d x)\right )+6 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)+10 a b B+5 b^2 (A-C)\right )\right )}{15 d \sec ^{\frac{7}{2}}(c+d x) (a \cos (c+d x)+b)^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(6*(10*a*b*B + 5*b^2*(A - C) + a^2*(3*A + 5*
C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 10*(a^2*B + 3*b^2*B + 2*a*b*(A + 3*C))*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2] + (10*a*(2*A*b + a*B)*Cos[c + d*x] + 3*(a^2*A + 10*b^2*C + a^2*A*Cos[2*(c + d*x)]))*S
in[c + d*x]))/(15*d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(7/2
))

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Maple [B]  time = 2.896, size = 932, normalized size = 4.1 \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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)

[Out]

-2/15*(-24*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(6*A*a+10*A*b+5*B*a)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*a^2+10*A*a*b+5*B*a^2+15*C*b^2)*sin(1/2*d*x
+1/2*c)^2*cos(1/2*d*x+1/2*c)+10*A*a*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-9*A*(-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)^(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^2-15*A*(-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)^(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))*b^2+5*B*a^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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)+15*B*b^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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-30*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)^(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*b+30*a*b*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)-15*C*(-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)^(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+15*C*(-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)^(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))*b^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/s
in(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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{C b^{2} \sec \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )}{\sec \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*sec(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))/sec(d*x + c)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}{\left (b \sec \left (d x + c\right ) + a\right )}^{2}}{\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))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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