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

Optimal. Leaf size=541 \[ \frac{2 \cot (c+d x) \left (-a^2 b (6 A+B+3 C)+a^3 (3 B+C)+a A b^2+3 A b^3\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 a^2 b d \sqrt{a+b} \left (a^2-b^2\right )}-\frac{2 \tan (c+d x) \left (-a^2 b^2 (7 A+3 C)+4 a^3 b B+a^4 (-C)+3 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \cot (c+d x) \left (7 a^2 A b^2+3 a^2 b^2 C-4 a^3 b B+a^4 C-3 A b^4\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b^2 d (a-b) (a+b)^{3/2}}-\frac{2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^3 d} \]

[Out]

(2*(7*a^2*A*b^2 - 3*A*b^4 - 4*a^3*b*B + a^4*C + 3*a^2*b^2*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*a^2*(a - b)*b^2*(a + b)^(3/2)*d) + (2*(a*A*b^2 + 3*A*b^3 + a^3*(3*B + C) - a^2*b*(6*A + B + 3*C))*Cot[
c + d*x]*EllipticF[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*a^2*b*Sqrt[a + b]*(a^2 - b^2)*d) - (2*A*Sqrt[a + b]*Cot[c
+ d*x]*EllipticPi[(a + b)/a, 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))])/(a^3*d) + (2*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x]
)/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (2*(3*A*b^4 + 4*a^3*b*B - a^4*C - a^2*b^2*(7*A + 3*C))*Tan[
c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

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Rubi [A]  time = 0.910364, antiderivative size = 541, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4060, 4058, 3921, 3784, 3832, 4004} \[ -\frac{2 \tan (c+d x) \left (-a^2 b^2 (7 A+3 C)+4 a^3 b B+a^4 (-C)+3 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{2 \cot (c+d x) \left (-a^2 b (6 A+B+3 C)+a^3 (3 B+C)+a A b^2+3 A b^3\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b d \sqrt{a+b} \left (a^2-b^2\right )}+\frac{2 \cot (c+d x) \left (7 a^2 A b^2+3 a^2 b^2 C-4 a^3 b B+a^4 C-3 A b^4\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 a^2 b^2 d (a-b) (a+b)^{3/2}}-\frac{2 A \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a^2*A*b^2 - 3*A*b^4 - 4*a^3*b*B + a^4*C + 3*a^2*b^2*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*a^2*(a - b)*b^2*(a + b)^(3/2)*d) + (2*(a*A*b^2 + 3*A*b^3 + a^3*(3*B + C) - a^2*b*(6*A + B + 3*C))*Cot[
c + d*x]*EllipticF[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*a^2*b*Sqrt[a + b]*(a^2 - b^2)*d) - (2*A*Sqrt[a + b]*Cot[c
+ d*x]*EllipticPi[(a + b)/a, 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))])/(a^3*d) + (2*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x]
)/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (2*(3*A*b^4 + 4*a^3*b*B - a^4*C - a^2*b^2*(7*A + 3*C))*Tan[
c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])

Rule 4060

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

Rule 4058

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

Rule 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3784

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && 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{A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{-\frac{3}{2} A \left (a^2-b^2\right )+\frac{3}{2} a (A b-a B+b C) \sec (c+d x)-\frac{1}{2} \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} A \left (a^2-b^2\right )^2+\frac{1}{4} a \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sec (c+d x)+\frac{1}{4} \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} A \left (a^2-b^2\right )^2+\left (\frac{1}{4} a \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right )+\frac{1}{4} \left (-3 A b^4-4 a^3 b B+a^4 C+a^2 b^2 (7 A+3 C)\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}+\frac{\left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 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^2 (a-b) b^2 (a+b)^{3/2} d}+\frac{2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{A \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx}{a^2}+\frac{\left (a A b^2+3 A b^3+a^3 (3 B+C)-a^2 b (6 A+B+3 C)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 (a-b) (a+b)^2}\\ &=-\frac{2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 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^2 (a-b) b^2 (a+b)^{3/2} d}+\frac{2 \left (a A b^2+3 A b^3+a^3 (3 B+C)-a^2 b (6 A+B+3 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^2 (a-b) b (a+b)^{3/2} d}-\frac{2 A \sqrt{a+b} \cot (c+d x) \Pi \left (\frac{a+b}{a};\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}}}{a^3 d}+\frac{2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 28.012, size = 11444, normalized size = 21.15 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.465, size = 8177, normalized size = 15.1 \begin{align*} \text{output 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)/(a+b*sec(d*x+c))^(5/2),x)

[Out]

result too large to display

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

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*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)

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

[Out]

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

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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 )}^{\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)/(a+b*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)^(5/2), x)