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

Optimal. Leaf size=408 \[ -\frac{2 \left (6 a^2 b C+8 a^3 C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 b^3 d \sqrt{a+b} \left (a^2-b^2\right )}+\frac{2 \left (a^2 b^2 (A+9 C)-5 a^4 C+3 A b^4\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (a^2 b^2 (A+15 C)-8 a^4 C+3 b^4 (A-C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b} \left (a^2-b^2\right )} \]

[Out]

(2*(3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[
a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4
*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(3*b^3*(A - C) + 8*a^3*C + 6*a^2*b*C - a*b^2*(A + 9*C))*Cot[c + d*x]*Elliptic
F[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((
b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^3*Sqrt[a + b]*(a^2 - b^2)*d) + (2*a*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b^2
*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*(3*A*b^4 - 5*a^4*C + a^2*b^2*(A + 9*C))*Tan[c + d*x])/(3*b^2*(
a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.821386, antiderivative size = 408, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4091, 4080, 4005, 3832, 4004} \[ \frac{2 \left (a^2 b^2 (A+9 C)-5 a^4 C+3 A b^4\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 a \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (6 a^2 b C+8 a^3 C-a b^2 (A+9 C)+3 b^3 (A-C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d \sqrt{a+b} \left (a^2-b^2\right )}+\frac{2 \left (a^2 b^2 (A+15 C)-8 a^4 C+3 b^4 (A-C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b} \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[
a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4
*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(3*b^3*(A - C) + 8*a^3*C + 6*a^2*b*C - a*b^2*(A + 9*C))*Cot[c + d*x]*Elliptic
F[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((
b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^3*Sqrt[a + b]*(a^2 - b^2)*d) + (2*a*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*b^2
*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (2*(3*A*b^4 - 5*a^4*C + a^2*b^2*(A + 9*C))*Tan[c + d*x])/(3*b^2*(
a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

Rule 4091

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

Rule 4080

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

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \int \frac{\sec (c+d x) \left (-\frac{3}{2} b \left (A b^2+a^2 C\right )+\frac{1}{2} a \left (A b^2-2 a^2 C+3 b^2 C\right ) \sec (c+d x)+\frac{3}{2} b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{4 \int \frac{\sec (c+d x) \left (-\frac{1}{2} a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )+\frac{1}{4} b \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (3 b^3 (A-C)+8 a^3 C+6 a^2 b C-a b^2 (A+9 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 (a-b) b^2 (a+b)^2}-\frac{\left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac{2 \left (3 b^3 (A-C)+8 a^3 C+6 a^2 b C-a b^2 (A+9 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac{2 a \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 22.5669, size = 702, normalized size = 1.72 \[ \frac{4 \sqrt{2} \sqrt{\frac{\cos (c+d x)}{(\cos (c+d x)+1)^2}} \sqrt{\sec (c+d x)} \sqrt{\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b)^2 \left (A+C \sec ^2(c+d x)\right ) \left (\left (a^2 b^2 (A+15 C)-8 a^4 C+3 b^4 (A-C)\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)-(a+b) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2} \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \left (b \left (6 a^2 b C-8 a^3 C+a b^2 (A+9 C)+3 b^3 (A-C)\right ) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (-a^2 b^2 (A+15 C)+8 a^4 C+3 b^4 (C-A)\right ) E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{1}{\cos (c+d x)+1}} \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} (a+b \sec (c+d x))^{5/2} (A \cos (2 c+2 d x)+A+2 C)}+\frac{\sec (c+d x) (a \cos (c+d x)+b)^3 \left (A+C \sec ^2(c+d x)\right ) \left (-\frac{4 \left (a^2 A b^2+15 a^2 b^2 C-8 a^4 C+3 A b^4-3 b^4 C\right ) \sin (c+d x)}{3 b^3 \left (b^2-a^2\right )^2}+\frac{4 \left (a^2 C \sin (c+d x)+A b^2 \sin (c+d x)\right )}{3 b \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}+\frac{8 \left (a^2 A b^2 \sin (c+d x)+4 a^2 b^2 C \sin (c+d x)-2 a^4 C \sin (c+d x)+A b^4 \sin (c+d x)\right )}{3 b^2 \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}\right )}{d (a+b \sec (c+d x))^{5/2} (A \cos (2 c+2 d x)+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-4*(a^2*A*b^2 + 3*A*b^4 - 8*a^4*C + 15*a^2*b^2*C
- 3*b^4*C)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)^2) + (4*(A*b^2*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(3*b*(-a^2 +
b^2)*(b + a*Cos[c + d*x])^2) + (8*(a^2*A*b^2*Sin[c + d*x] + A*b^4*Sin[c + d*x] - 2*a^4*C*Sin[c + d*x] + 4*a^2*
b^2*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Se
c[c + d*x])^(5/2)) + (4*Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*(b + a*Cos[c + d*x])^2*Sqrt[Cos[c + d*
x]*Sec[(c + d*x)/2]^2]*Sqrt[Sec[c + d*x]]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*(-((a
+ b)*((8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] +
b*(3*b^3*(A - C) - 8*a^3*C + 6*a^2*b*C + a*b^2*(A + 9*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]
)*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x]
) + (3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c
+ d*x)/2]))/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*(A + 2*C + A*Cos[2*c + 2*d*x])*(Sec[(c + d*x
)/2]^2)^(3/2)*(a + b*Sec[c + d*x])^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.805, size = 6135, normalized size = 15. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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