3.65 \(\int \frac{a+b \sec ^{-1}(c x)}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=212 \[ \frac{4 b \sqrt{1-c^2 x^2} \sqrt{\frac{c (d+e x)}{c d+e}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right ),\frac{2 e}{c d+e}\right )}{c^2 x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{d+e x}}+\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{4 b d \sqrt{1-c^2 x^2} \sqrt{\frac{c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c e x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{d+e x}} \]

[Out]

(2*Sqrt[d + e*x]*(a + b*ArcSec[c*x]))/e + (4*b*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticF[ArcSi
n[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c^2*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*x]) + (4*b*d*Sqrt[(c*(d +
e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticPi[2, ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c*e*Sqrt[1
- 1/(c^2*x^2)]*x*Sqrt[d + e*x])

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Rubi [A]  time = 0.303554, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5226, 1574, 944, 719, 419, 933, 168, 538, 537} \[ \frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{4 b \sqrt{1-c^2 x^2} \sqrt{\frac{c (d+e x)}{c d+e}} F\left (\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c^2 x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{d+e x}}+\frac{4 b d \sqrt{1-c^2 x^2} \sqrt{\frac{c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c e x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/Sqrt[d + e*x],x]

[Out]

(2*Sqrt[d + e*x]*(a + b*ArcSec[c*x]))/e + (4*b*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticF[ArcSi
n[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c^2*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*x]) + (4*b*d*Sqrt[(c*(d +
e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticPi[2, ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c*e*Sqrt[1
- 1/(c^2*x^2)]*x*Sqrt[d + e*x])

Rule 5226

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + b
*ArcSec[c*x]))/(e*(m + 1)), x] - Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x],
x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rule 1574

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Dist[(x^(2*n*Fr
acPart[p])*(a + c/x^(2*n))^FracPart[p])/(c + a*x^(2*n))^FracPart[p], Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + a*x^
(2*n))^p, x], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[mn2, -2*n] &&  !IntegerQ[p] &&  !IntegerQ[q] &&
PosQ[n]

Rule 944

Int[Sqrt[(f_.) + (g_.)*(x_)]/(((d_.) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[g/e, Int[1/(S
qrt[f + g*x]*Sqrt[a + c*x^2]), x], x] + Dist[(e*f - d*g)/e, Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[a + c*x^2]), x
], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0]

Rule 719

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*a*Rt[-(c/a), 2]*(d + e*x)^m*Sqrt[
1 + (c*x^2)/a])/(c*Sqrt[a + c*x^2]*((c*(d + e*x))/(c*d - a*e*Rt[-(c/a), 2]))^m), Subst[Int[(1 + (2*a*e*Rt[-(c/
a), 2]*x^2)/(c*d - a*e*Rt[-(c/a), 2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-(c/a), 2]*x)/2]], x] /; FreeQ[{a,
c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 933

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[Sqrt[1 + (c*x^2)/a]/Sqrt[a + c*x^2], Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[1 - q*x]*Sqrt[1 + q*x]
), x], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] &&  !GtQ[a, 0]

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 538

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

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rubi steps

\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{\sqrt{d+e x}} \, dx &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}-\frac{(2 b) \int \frac{\sqrt{d+e x}}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{c e}\\ &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}-\frac{\left (2 b \sqrt{-\frac{1}{c^2}+x^2}\right ) \int \frac{\sqrt{d+e x}}{x \sqrt{-\frac{1}{c^2}+x^2}} \, dx}{c e \sqrt{1-\frac{1}{c^2 x^2}} x}\\ &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}-\frac{\left (2 b \sqrt{-\frac{1}{c^2}+x^2}\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{-\frac{1}{c^2}+x^2}} \, dx}{c \sqrt{1-\frac{1}{c^2 x^2}} x}-\frac{\left (2 b d \sqrt{-\frac{1}{c^2}+x^2}\right ) \int \frac{1}{x \sqrt{d+e x} \sqrt{-\frac{1}{c^2}+x^2}} \, dx}{c e \sqrt{1-\frac{1}{c^2 x^2}} x}\\ &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}-\frac{\left (2 b d \sqrt{1-c^2 x^2}\right ) \int \frac{1}{x \sqrt{1-c x} \sqrt{1+c x} \sqrt{d+e x}} \, dx}{c e \sqrt{1-\frac{1}{c^2 x^2}} x}+\frac{\left (4 b \sqrt{\frac{d+e x}{d+\frac{e}{c}}} \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1-\frac{2 e x^2}{c \left (d+\frac{e}{c}\right )}}} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{2}}\right )}{c^2 \sqrt{1-\frac{1}{c^2 x^2}} x \sqrt{d+e x}}\\ &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{4 b \sqrt{\frac{c (d+e x)}{c d+e}} \sqrt{1-c^2 x^2} F\left (\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c^2 \sqrt{1-\frac{1}{c^2 x^2}} x \sqrt{d+e x}}+\frac{\left (4 b d \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{2-x^2} \sqrt{d+\frac{e}{c}-\frac{e x^2}{c}}} \, dx,x,\sqrt{1-c x}\right )}{c e \sqrt{1-\frac{1}{c^2 x^2}} x}\\ &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{4 b \sqrt{\frac{c (d+e x)}{c d+e}} \sqrt{1-c^2 x^2} F\left (\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c^2 \sqrt{1-\frac{1}{c^2 x^2}} x \sqrt{d+e x}}+\frac{\left (4 b d \sqrt{\frac{c (d+e x)}{c d+e}} \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \sqrt{2-x^2} \sqrt{1-\frac{e x^2}{c \left (d+\frac{e}{c}\right )}}} \, dx,x,\sqrt{1-c x}\right )}{c e \sqrt{1-\frac{1}{c^2 x^2}} x \sqrt{d+e x}}\\ &=\frac{2 \sqrt{d+e x} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{4 b \sqrt{\frac{c (d+e x)}{c d+e}} \sqrt{1-c^2 x^2} F\left (\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c^2 \sqrt{1-\frac{1}{c^2 x^2}} x \sqrt{d+e x}}+\frac{4 b d \sqrt{\frac{c (d+e x)}{c d+e}} \sqrt{1-c^2 x^2} \Pi \left (2;\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{2}}\right )|\frac{2 e}{c d+e}\right )}{c e \sqrt{1-\frac{1}{c^2 x^2}} x \sqrt{d+e x}}\\ \end{align*}

Mathematica [C]  time = 2.65761, size = 212, normalized size = 1. \[ \frac{2 \left (\frac{2 i b \sqrt{\frac{e (c x+1)}{e-c d}} \sqrt{\frac{e-c e x}{c d+e}} \left (\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-\frac{c}{c d+e}} \sqrt{d+e x}\right ),\frac{c d+e}{c d-e}\right )-\Pi \left (\frac{e}{c d}+1;i \sinh ^{-1}\left (\sqrt{-\frac{c}{c d+e}} \sqrt{d+e x}\right )|\frac{c d+e}{c d-e}\right )\right )}{c x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{-\frac{c}{c d+e}}}+a \sqrt{d+e x}+b \sec ^{-1}(c x) \sqrt{d+e x}\right )}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/Sqrt[d + e*x],x]

[Out]

(2*(a*Sqrt[d + e*x] + b*Sqrt[d + e*x]*ArcSec[c*x] + ((2*I)*b*Sqrt[(e*(1 + c*x))/(-(c*d) + e)]*Sqrt[(e - c*e*x)
/(c*d + e)]*(EllipticF[I*ArcSinh[Sqrt[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c*d - e)] - EllipticPi[1 + e/
(c*d), I*ArcSinh[Sqrt[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c*d - e)]))/(c*Sqrt[-(c/(c*d + e))]*Sqrt[1 -
1/(c^2*x^2)]*x)))/e

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Maple [A]  time = 0.257, size = 254, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{e} \left ( a\sqrt{ex+d}+b \left ( \sqrt{ex+d}{\rm arcsec} \left (cx\right )-2\,{\frac{1}{cx}\sqrt{-{\frac{ \left ( ex+d \right ) c-dc+e}{dc-e}}}\sqrt{-{\frac{ \left ( ex+d \right ) c-dc-e}{dc+e}}} \left ({\it EllipticF} \left ( \sqrt{ex+d}\sqrt{{\frac{c}{dc-e}}},\sqrt{{\frac{dc-e}{dc+e}}} \right ) -{\it EllipticPi} \left ( \sqrt{ex+d}\sqrt{{\frac{c}{dc-e}}},{\frac{dc-e}{dc}},{\sqrt{{\frac{c}{dc+e}}}{\frac{1}{\sqrt{{\frac{c}{dc-e}}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{{c}^{2} \left ( ex+d \right ) ^{2}-2\,d{c}^{2} \left ( ex+d \right ) +{c}^{2}{d}^{2}-{e}^{2}}{{c}^{2}{e}^{2}{x}^{2}}}}}}{\frac{1}{\sqrt{{\frac{c}{dc-e}}}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/(e*x+d)^(1/2),x)

[Out]

2/e*(a*(e*x+d)^(1/2)+b*((e*x+d)^(1/2)*arcsec(c*x)-2/c*(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*c-d*c-e)/(
c*d+e))^(1/2)*(EllipticF((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),((c*d-e)/(c*d+e))^(1/2))-EllipticPi((e*x+d)^(1/2)*(c/
(c*d-e))^(1/2),1/c*(c*d-e)/d,(c/(c*d+e))^(1/2)/(c/(c*d-e))^(1/2)))/((c^2*(e*x+d)^2-2*d*c^2*(e*x+d)+c^2*d^2-e^2
)/c^2/e^2/x^2)^(1/2)/x/(c/(c*d-e))^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arcsec}\left (c x\right ) + a}{\sqrt{e x + d}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

integral((b*arcsec(c*x) + a)/sqrt(e*x + d), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asec}{\left (c x \right )}}{\sqrt{d + e x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/(e*x+d)**(1/2),x)

[Out]

Integral((a + b*asec(c*x))/sqrt(d + e*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcsec}\left (c x\right ) + a}{\sqrt{e x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)/sqrt(e*x + d), x)