3.62 \(\int \frac{a+b \sec ^{-1}(c x)}{(d+e x)^3} \, dx\)
Optimal. Leaf size=172 \[ -\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b c e \sqrt{1-\frac{1}{c^2 x^2}}}{2 d \left (c^2 d^2-e^2\right ) \left (\frac{d}{x}+e\right )}-\frac{b \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\frac{c^2 d+\frac{e}{x}}{c \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 d^2-e^2}}\right )}{2 d^2 \left (c^2 d^2-e^2\right )^{3/2}}-\frac{b \csc ^{-1}(c x)}{2 d^2 e} \]
[Out]
(b*c*e*Sqrt[1 - 1/(c^2*x^2)])/(2*d*(c^2*d^2 - e^2)*(e + d/x)) - (b*ArcCsc[c*x])/(2*d^2*e) - (a + b*ArcSec[c*x]
)/(2*e*(d + e*x)^2) - (b*(2*c^2*d^2 - e^2)*ArcTanh[(c^2*d + e/x)/(c*Sqrt[c^2*d^2 - e^2]*Sqrt[1 - 1/(c^2*x^2)])
])/(2*d^2*(c^2*d^2 - e^2)^(3/2))
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Rubi [A] time = 0.29226, antiderivative size = 172, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used =
{5226, 1568, 1475, 1651, 844, 216, 725, 206} \[ -\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b c e \sqrt{1-\frac{1}{c^2 x^2}}}{2 d \left (c^2 d^2-e^2\right ) \left (\frac{d}{x}+e\right )}-\frac{b \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\frac{c^2 d+\frac{e}{x}}{c \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 d^2-e^2}}\right )}{2 d^2 \left (c^2 d^2-e^2\right )^{3/2}}-\frac{b \csc ^{-1}(c x)}{2 d^2 e} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcSec[c*x])/(d + e*x)^3,x]
[Out]
(b*c*e*Sqrt[1 - 1/(c^2*x^2)])/(2*d*(c^2*d^2 - e^2)*(e + d/x)) - (b*ArcCsc[c*x])/(2*d^2*e) - (a + b*ArcSec[c*x]
)/(2*e*(d + e*x)^2) - (b*(2*c^2*d^2 - e^2)*ArcTanh[(c^2*d + e/x)/(c*Sqrt[c^2*d^2 - e^2]*Sqrt[1 - 1/(c^2*x^2)])
])/(2*d^2*(c^2*d^2 - e^2)^(3/2))
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 1651
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]
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)^3} \, dx &=-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} x^2 (d+e x)^2} \, dx}{2 c e}\\ &=-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} \left (e+\frac{d}{x}\right )^2 x^4} \, dx}{2 c e}\\ &=-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}-\frac{b \operatorname{Subst}\left (\int \frac{x^2}{(e+d x)^2 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 c e}\\ &=\frac{b c e \sqrt{1-\frac{1}{c^2 x^2}}}{2 d \left (c^2 d^2-e^2\right ) \left (e+\frac{d}{x}\right )}-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}+\frac{(b c) \operatorname{Subst}\left (\int \frac{e-\left (d-\frac{e^2}{c^2 d}\right ) x}{(e+d x) \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 e \left (c^2 d^2-e^2\right )}\\ &=\frac{b c e \sqrt{1-\frac{1}{c^2 x^2}}}{2 d \left (c^2 d^2-e^2\right ) \left (e+\frac{d}{x}\right )}-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 c d^2 e}+\frac{\left (b c \left (2-\frac{e^2}{c^2 d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(e+d x) \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 \left (c^2 d^2-e^2\right )}\\ &=\frac{b c e \sqrt{1-\frac{1}{c^2 x^2}}}{2 d \left (c^2 d^2-e^2\right ) \left (e+\frac{d}{x}\right )}-\frac{b \csc ^{-1}(c x)}{2 d^2 e}-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}-\frac{\left (b c \left (2-\frac{e^2}{c^2 d^2}\right )\right ) \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 )}{2 \left (c^2 d^2-e^2\right )}\\ &=\frac{b c e \sqrt{1-\frac{1}{c^2 x^2}}}{2 d \left (c^2 d^2-e^2\right ) \left (e+\frac{d}{x}\right )}-\frac{b \csc ^{-1}(c x)}{2 d^2 e}-\frac{a+b \sec ^{-1}(c x)}{2 e (d+e x)^2}-\frac{b \left (2 c^2 d^2-e^2\right ) \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 )}{2 d^2 \left (c^2 d^2-e^2\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.480236, size = 247, normalized size = 1.44 \[ \frac{1}{2} \left (-\frac{a}{e (d+e x)^2}+\frac{b c e x \sqrt{1-\frac{1}{c^2 x^2}}}{d \left (c^2 d^2-e^2\right ) (d+e x)}+\frac{b \left (2 c^2 d^2-e^2\right ) \log \left (c x \left (c d-\sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 d^2-e^2}\right )+e\right )}{d^2 (c d-e) (c d+e) \sqrt{c^2 d^2-e^2}}+\frac{b \left (e^2-2 c^2 d^2\right ) \log (d+e x)}{d^2 (c d-e) (c d+e) \sqrt{c^2 d^2-e^2}}-\frac{b \sin ^{-1}\left (\frac{1}{c x}\right )}{d^2 e}-\frac{b \sec ^{-1}(c x)}{e (d+e x)^2}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcSec[c*x])/(d + e*x)^3,x]
[Out]
(-(a/(e*(d + e*x)^2)) + (b*c*e*Sqrt[1 - 1/(c^2*x^2)]*x)/(d*(c^2*d^2 - e^2)*(d + e*x)) - (b*ArcSec[c*x])/(e*(d
+ e*x)^2) - (b*ArcSin[1/(c*x)])/(d^2*e) + (b*(-2*c^2*d^2 + e^2)*Log[d + e*x])/(d^2*(c*d - e)*(c*d + e)*Sqrt[c^
2*d^2 - e^2]) + (b*(2*c^2*d^2 - e^2)*Log[e + c*(c*d - Sqrt[c^2*d^2 - e^2]*Sqrt[1 - 1/(c^2*x^2)])*x])/(d^2*(c*d
- e)*(c*d + e)*Sqrt[c^2*d^2 - e^2]))/2
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Maple [B] time = 0.28, size = 1005, normalized size = 5.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arcsec(c*x))/(e*x+d)^3,x)
[Out]
-1/2*c^2*a/(c*e*x+c*d)^2/e-1/2*c^2*b/(c*e*x+c*d)^2/e*arcsec(c*x)-1/2*c^2*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/
x^2)^(1/2)/(c^2*d^2-e^2)/(c*e*x+c*d)*arctan(1/(c^2*x^2-1)^(1/2))-1/2*c^2*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)/(c*e*x+c*d)*arctan(1/(c^2*x^2-1)^(1/2))+c^2*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^
2/x^2)^(1/2)/((c^2*d^2-e^2)/e^2)^(1/2)/(c^2*d^2-e^2)/(c*e*x+c*d)*ln(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))+c^2*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)/(c^2*d^2-e^2)/(c*e*x+c*d)*ln(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))+1/2
*b*e^2*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/d^2/(c^2*d^2-e^2)/(c*e*x+c*d)*arctan(1/(c^2*x^2-1)^(1/2))
+1/2*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)/(c*e*x+c*d)*arctan(1/(c^2*x^2-1)^(1/2
))+1/2*c^2*b*e/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d/(c^2*d^2-e^2)/(c*e*x+c*d)-1/2*b*e/((c^2*x^2-1)/c^2/x^2)^(1/2)/x
/d/(c^2*d^2-e^2)/(c*e*x+c*d)-1/2*b*e^2*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/d^2/((c^2*d^2-e^2)/e^2)^(
1/2)/(c^2*d^2-e^2)/(c*e*x+c*d)*ln(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))-1/2
*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)/(c^2*d^2-e^2)/(c*e*x+c*d)*ln(
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] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (c^{2} e^{3} x^{2} + 2 \, c^{2} d e^{2} x + c^{2} d^{2} e\right )} \int \frac{x e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (c x - 1\right )\right )}}{c^{2} e^{3} x^{4} + 2 \, c^{2} d e^{2} x^{3} - 2 \, d e^{2} x +{\left (c^{2} e^{3} x^{4} + 2 \, c^{2} d e^{2} x^{3} - 2 \, d e^{2} x - d^{2} e +{\left (c^{2} d^{2} e - e^{3}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )} - d^{2} e +{\left (c^{2} d^{2} e - e^{3}\right )} x^{2}}\,{d x} - \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )\right )} b}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} - \frac{a}{2 \,{\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsec(c*x))/(e*x+d)^3,x, algorithm="maxima")
[Out]
1/2*(2*(c^2*e^3*x^2 + 2*c^2*d*e^2*x + c^2*d^2*e)*integrate(1/2*x*e^(1/2*log(c*x + 1) + 1/2*log(c*x - 1))/(c^2*
e^3*x^4 + 2*c^2*d*e^2*x^3 - 2*d*e^2*x - d^2*e + (c^2*d^2*e - e^3)*x^2 + (c^2*e^3*x^4 + 2*c^2*d*e^2*x^3 - 2*d*e
^2*x - d^2*e + (c^2*d^2*e - e^3)*x^2)*e^(log(c*x + 1) + log(c*x - 1))), x) - arctan(sqrt(c*x + 1)*sqrt(c*x - 1
)))*b/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 1/2*a/(e^3*x^2 + 2*d*e^2*x + d^2*e)
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Fricas [B] time = 7.15157, size = 2223, normalized size = 12.92 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsec(c*x))/(e*x+d)^3,x, algorithm="fricas")
[Out]
[-1/2*(a*c^4*d^6 - b*c^3*d^5*e - 2*a*c^2*d^4*e^2 + b*c*d^3*e^3 + a*d^2*e^4 - (b*c^3*d^3*e^3 - b*c*d*e^5)*x^2 -
(2*b*c^2*d^4*e - b*d^2*e^3 + (2*b*c^2*d^2*e^3 - b*e^5)*x^2 + 2*(2*b*c^2*d^3*e^2 - b*d*e^4)*x)*sqrt(c^2*d^2 -
e^2)*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)*sq
rt(c^2*x^2 - 1))/(e*x + d)) - 2*(b*c^3*d^4*e^2 - b*c*d^2*e^4)*x + (b*c^4*d^6 - 2*b*c^2*d^4*e^2 + b*d^2*e^4)*ar
csec(c*x) - 2*(b*c^4*d^6 - 2*b*c^2*d^4*e^2 + b*d^2*e^4 + (b*c^4*d^4*e^2 - 2*b*c^2*d^2*e^4 + b*e^6)*x^2 + 2*(b*
c^4*d^5*e - 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*arctan(-c*x + sqrt(c^2*x^2 - 1)) - (b*c^2*d^4*e^2 - b*d^2*e^4 + (b*c
^2*d^3*e^3 - b*d*e^5)*x)*sqrt(c^2*x^2 - 1))/(c^4*d^8*e - 2*c^2*d^6*e^3 + d^4*e^5 + (c^4*d^6*e^3 - 2*c^2*d^4*e^
5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 - 2*c^2*d^5*e^4 + d^3*e^6)*x), -1/2*(a*c^4*d^6 - b*c^3*d^5*e - 2*a*c^2*d^4*e
^2 + b*c*d^3*e^3 + a*d^2*e^4 - (b*c^3*d^3*e^3 - b*c*d*e^5)*x^2 - 2*(2*b*c^2*d^4*e - b*d^2*e^3 + (2*b*c^2*d^2*e
^3 - b*e^5)*x^2 + 2*(2*b*c^2*d^3*e^2 - b*d*e^4)*x)*sqrt(-c^2*d^2 + e^2)*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)) - 2*(b*c^3*d^4*e^2 - b*c*d^2*e^4)*x + (b*c^
4*d^6 - 2*b*c^2*d^4*e^2 + b*d^2*e^4)*arcsec(c*x) - 2*(b*c^4*d^6 - 2*b*c^2*d^4*e^2 + b*d^2*e^4 + (b*c^4*d^4*e^2
- 2*b*c^2*d^2*e^4 + b*e^6)*x^2 + 2*(b*c^4*d^5*e - 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*arctan(-c*x + sqrt(c^2*x^2 -
1)) - (b*c^2*d^4*e^2 - b*d^2*e^4 + (b*c^2*d^3*e^3 - b*d*e^5)*x)*sqrt(c^2*x^2 - 1))/(c^4*d^8*e - 2*c^2*d^6*e^3
+ d^4*e^5 + (c^4*d^6*e^3 - 2*c^2*d^4*e^5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 - 2*c^2*d^5*e^4 + d^3*e^6)*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 )^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*asec(c*x))/(e*x+d)**3,x)
[Out]
Integral((a + b*asec(c*x))/(d + e*x)**3, 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 )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arcsec(c*x))/(e*x+d)^3,x, algorithm="giac")
[Out]
integrate((b*arcsec(c*x) + a)/(e*x + d)^3, x)