3.61 \(\int \frac{a+b \sec ^{-1}(c x)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=104 \[ -\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}-\frac{b \tanh ^{-1}\left (\frac{c^2 d+\frac{e}{x}}{c \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 d^2-e^2}}\right )}{d \sqrt{c^2 d^2-e^2}}-\frac{b \csc ^{-1}(c x)}{d e} \]

[Out]

-((b*ArcCsc[c*x])/(d*e)) - (a + b*ArcSec[c*x])/(e*(d + e*x)) - (b*ArcTanh[(c^2*d + e/x)/(c*Sqrt[c^2*d^2 - e^2]
*Sqrt[1 - 1/(c^2*x^2)])])/(d*Sqrt[c^2*d^2 - e^2])

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Rubi [A]  time = 0.156008, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {5226, 1568, 1475, 844, 216, 725, 206} \[ -\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}-\frac{b \tanh ^{-1}\left (\frac{c^2 d+\frac{e}{x}}{c \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 d^2-e^2}}\right )}{d \sqrt{c^2 d^2-e^2}}-\frac{b \csc ^{-1}(c x)}{d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/(d + e*x)^2,x]

[Out]

-((b*ArcCsc[c*x])/(d*e)) - (a + b*ArcSec[c*x])/(e*(d + e*x)) - (b*ArcTanh[(c^2*d + e/x)/(c*Sqrt[c^2*d^2 - e^2]
*Sqrt[1 - 1/(c^2*x^2)])])/(d*Sqrt[c^2*d^2 - e^2])

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 1568

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(m + mn*q
)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (P
osQ[n2] ||  !IntegerQ[p])

Rule 1475

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x
] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 844

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

Rule 216

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{(d+e x)^2} \, dx &=-\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} x^2 (d+e x)} \, dx}{c e}\\ &=-\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} \left (e+\frac{d}{x}\right ) x^3} \, dx}{c e}\\ &=-\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}-\frac{b \operatorname{Subst}\left (\int \frac{x}{(e+d x) \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{c e}\\ &=-\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}+\frac{b \operatorname{Subst}\left (\int \frac{1}{(e+d x) \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{c d}-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{c d e}\\ &=-\frac{b \csc ^{-1}(c x)}{d e}-\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}-\frac{b \operatorname{Subst}\left (\int \frac{1}{d^2-\frac{e^2}{c^2}-x^2} \, dx,x,\frac{d+\frac{e}{c^2 x}}{\sqrt{1-\frac{1}{c^2 x^2}}}\right )}{c d}\\ &=-\frac{b \csc ^{-1}(c x)}{d e}-\frac{a+b \sec ^{-1}(c x)}{e (d+e x)}-\frac{b \tanh ^{-1}\left (\frac{c^2 d+\frac{e}{x}}{c \sqrt{c^2 d^2-e^2} \sqrt{1-\frac{1}{c^2 x^2}}}\right )}{d \sqrt{c^2 d^2-e^2}}\\ \end{align*}

Mathematica [A]  time = 0.222711, size = 142, normalized size = 1.37 \[ -\frac{a}{e (d+e x)}+\frac{b \log \left (c x \left (c d-\sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 d^2-e^2}\right )+e\right )}{d \sqrt{c^2 d^2-e^2}}-\frac{b \log (d+e x)}{d \sqrt{c^2 d^2-e^2}}-\frac{b \sin ^{-1}\left (\frac{1}{c x}\right )}{d e}-\frac{b \sec ^{-1}(c x)}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/(d + e*x)^2,x]

[Out]

-(a/(e*(d + e*x))) - (b*ArcSec[c*x])/(e*(d + e*x)) - (b*ArcSin[1/(c*x)])/(d*e) - (b*Log[d + e*x])/(d*Sqrt[c^2*
d^2 - e^2]) + (b*Log[e + c*(c*d - Sqrt[c^2*d^2 - e^2]*Sqrt[1 - 1/(c^2*x^2)])*x])/(d*Sqrt[c^2*d^2 - e^2])

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Maple [B]  time = 0.25, size = 214, normalized size = 2.1 \begin{align*} -{\frac{ac}{ \left ( cex+dc \right ) e}}-{\frac{cb{\rm arcsec} \left (cx\right )}{ \left ( cex+dc \right ) e}}-{\frac{b}{cexd}\sqrt{{c}^{2}{x}^{2}-1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{b}{cexd}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( 2\,{\frac{1}{cex+dc} \left ( \sqrt{{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}\sqrt{{c}^{2}{x}^{2}-1}e-d{c}^{2}x-e \right ) } \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}{\frac{1}{\sqrt{{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*a/(c*e*x+c*d)/e-c*b/(c*e*x+c*d)/e*arcsec(c*x)-1/c*b/e*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x/d*arc
tan(1/(c^2*x^2-1)^(1/2))+1/c*b/e*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x/d/((c^2*d^2-e^2)/e^2)^(1/2)*l
n(2*(((c^2*d^2-e^2)/e^2)^(1/2)*(c^2*x^2-1)^(1/2)*e-d*c^2*x-e)/(c*e*x+c*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*arcsec(c*x))/(e*x+d)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 3.28274, size = 960, normalized size = 9.23 \begin{align*} \left [-\frac{a c^{2} d^{3} - a d e^{2} - \sqrt{c^{2} d^{2} - e^{2}}{\left (b e^{2} x + b d e\right )} \log \left (\frac{c^{3} d^{2} x + c d e - \sqrt{c^{2} d^{2} - e^{2}}{\left (c^{2} d x + e\right )} +{\left (c^{2} d^{2} - \sqrt{c^{2} d^{2} - e^{2}} c d - e^{2}\right )} \sqrt{c^{2} x^{2} - 1}}{e x + d}\right ) +{\left (b c^{2} d^{3} - b d e^{2}\right )} \operatorname{arcsec}\left (c x\right ) - 2 \,{\left (b c^{2} d^{3} - b d e^{2} +{\left (b c^{2} d^{2} e - b e^{3}\right )} x\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right )}{c^{2} d^{4} e - d^{2} e^{3} +{\left (c^{2} d^{3} e^{2} - d e^{4}\right )} x}, -\frac{a c^{2} d^{3} - a d e^{2} - 2 \, \sqrt{-c^{2} d^{2} + e^{2}}{\left (b e^{2} x + b d e\right )} \arctan \left (-\frac{\sqrt{-c^{2} d^{2} + e^{2}} \sqrt{c^{2} x^{2} - 1} e - \sqrt{-c^{2} d^{2} + e^{2}}{\left (c e x + c d\right )}}{c^{2} d^{2} - e^{2}}\right ) +{\left (b c^{2} d^{3} - b d e^{2}\right )} \operatorname{arcsec}\left (c x\right ) - 2 \,{\left (b c^{2} d^{3} - b d e^{2} +{\left (b c^{2} d^{2} e - b e^{3}\right )} x\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right )}{c^{2} d^{4} e - d^{2} e^{3} +{\left (c^{2} d^{3} e^{2} - d e^{4}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(a*c^2*d^3 - a*d*e^2 - sqrt(c^2*d^2 - e^2)*(b*e^2*x + b*d*e)*log((c^3*d^2*x + c*d*e - sqrt(c^2*d^2 - e^2)*(c
^2*d*x + e) + (c^2*d^2 - sqrt(c^2*d^2 - e^2)*c*d - e^2)*sqrt(c^2*x^2 - 1))/(e*x + d)) + (b*c^2*d^3 - b*d*e^2)*
arcsec(c*x) - 2*(b*c^2*d^3 - b*d*e^2 + (b*c^2*d^2*e - b*e^3)*x)*arctan(-c*x + sqrt(c^2*x^2 - 1)))/(c^2*d^4*e -
d^2*e^3 + (c^2*d^3*e^2 - d*e^4)*x), -(a*c^2*d^3 - a*d*e^2 - 2*sqrt(-c^2*d^2 + e^2)*(b*e^2*x + b*d*e)*arctan(-
(sqrt(-c^2*d^2 + e^2)*sqrt(c^2*x^2 - 1)*e - sqrt(-c^2*d^2 + e^2)*(c*e*x + c*d))/(c^2*d^2 - e^2)) + (b*c^2*d^3
- b*d*e^2)*arcsec(c*x) - 2*(b*c^2*d^3 - b*d*e^2 + (b*c^2*d^2*e - b*e^3)*x)*arctan(-c*x + sqrt(c^2*x^2 - 1)))/(
c^2*d^4*e - d^2*e^3 + (c^2*d^3*e^2 - d*e^4)*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 )}}{\left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asec(c*x))/(d + e*x)**2, 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}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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