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

Optimal. Leaf size=261 \[ -\frac{2 \sqrt{a+b} (3 A b-B (2 a+b)) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 b^2 d}-\frac{2 (a-b) \sqrt{a+b} (3 A b-2 a B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d}+\frac{2 B \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{3 b d} \]

[Out]

(-2*(a - b)*Sqrt[a + b]*(3*A*b - 2*a*B)*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^3*d) - (2*S
qrt[a + b]*(3*A*b - (2*a + b)*B)*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*b^2*d) + (2*B*Sqrt[a
+ b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d)

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Rubi [A]  time = 0.39711, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {4010, 4005, 3832, 4004} \[ -\frac{2 \sqrt{a+b} (3 A b-B (2 a+b)) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^2 d}-\frac{2 (a-b) \sqrt{a+b} (3 A b-2 a B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d}+\frac{2 B \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - b)*Sqrt[a + b]*(3*A*b - 2*a*B)*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^3*d) - (2*S
qrt[a + b]*(3*A*b - (2*a + b)*B)*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*b^2*d) + (2*B*Sqrt[a
+ b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d)

Rule 4010

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

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) (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx &=\frac{2 B \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b d}+\frac{2 \int \frac{\sec (c+d x) \left (\frac{b B}{2}+\frac{1}{2} (3 A b-2 a B) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b}\\ &=\frac{2 B \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b d}+\frac{(3 A b-2 a B) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b}+\frac{(-3 A b+(2 a+b) B) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b}\\ &=-\frac{2 (a-b) \sqrt{a+b} (3 A b-2 a B) \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 b^3 d}-\frac{2 \sqrt{a+b} (3 A b-(2 a+b) B) \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 b^2 d}+\frac{2 B \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b d}\\ \end{align*}

Mathematica [A]  time = 16.407, size = 372, normalized size = 1.43 \[ \frac{2 \sqrt{\sec (c+d x)} \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 b (B (b-2 a)+3 A b) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )-(3 A b-2 a B) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 (a+b) (2 a B-3 A b) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{3 b^2 d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{a+b \sec (c+d x)}}+\frac{\sec (c+d x) (a \cos (c+d x)+b) \left (\frac{2 (3 A b-2 a B) \sin (c+d x)}{3 b^2}+\frac{2 B \tan (c+d x)}{3 b}\right )}{d \sqrt{a+b \sec (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-3*A*b + 2*a*B)*Sqrt[Cos[c + d*x]/(1 +
Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a
- b)/(a + b)] + 2*b*(3*A*b + (-2*a + b)*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((
a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - (3*A*b - 2*a*B)*Cos[c + d*x
]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^2*d*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[a + b*Sec[
c + d*x]]) + ((b + a*Cos[c + d*x])*Sec[c + d*x]*((2*(3*A*b - 2*a*B)*Sin[c + d*x])/(3*b^2) + (2*B*Tan[c + d*x])
/(3*b)))/(d*Sqrt[a + b*Sec[c + d*x]])

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Maple [B]  time = 0.495, size = 1567, normalized size = 6. \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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x)

[Out]

2/3/d/b^2*(-1+cos(d*x+c))^2*(B*b^2+2*B*cos(d*x+c)^3*a^2-2*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/
2))*a^2-3*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^
(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2+3*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/
(a+b))^(1/2))*b^2-B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2-2*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*a^2-3*A*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2+3*A*cos(d*x+c)^2*sin(d
*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+
c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2-B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2-3*A*cos(d*
x+c)^3*a*b-B*cos(d*x+c)^3*a*b+3*A*cos(d*x+c)^2*a*b+2*B*cos(d*x+c)^2*a*b-B*cos(d*x+c)*a*b+3*A*cos(d*x+c)^2*sin(
d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x
+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+2*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b
)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b-2*B*cos
(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipti
cE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b+3*A*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))
*a*b+2*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/
2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b-2*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+
b))^(1/2))*a*b-3*A*cos(d*x+c)^2*b^2-2*B*cos(d*x+c)^2*a^2+3*A*cos(d*x+c)*b^2-B*cos(d*x+c)^2*b^2)*((b+a*cos(d*x+
c))/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)^2/(b+a*cos(d*x+c))/cos(d*x+c)/sin(d*x+c)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt{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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((B*sec(d*x + c)^3 + A*sec(d*x + c)^2)/sqrt(b*sec(d*x + c) + a), 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 ) \sec ^{2}{\left (c + d x \right )}}{\sqrt{a + b \sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt{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)^2*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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