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

Optimal. Leaf size=131 \[ -\frac {a+b \sec ^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac {b c x \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{2 d e \sqrt {c^2 x^2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{\sqrt {c^2 d+e}}\right )}{2 d \sqrt {e} \sqrt {c^2 x^2} \sqrt {c^2 d+e}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5236, 446, 86, 63, 205} \[ -\frac {a+b \sec ^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac {b c x \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{2 d e \sqrt {c^2 x^2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{\sqrt {c^2 d+e}}\right )}{2 d \sqrt {e} \sqrt {c^2 x^2} \sqrt {c^2 d+e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*ArcSec[c*x])/(2*e*(d + e*x^2)) + (b*c*x*ArcTan[Sqrt[-1 + c^2*x^2]])/(2*d*e*Sqrt[c^2*x^2]) - (b*c*x*Arc
Tan[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*d + e]])/(2*d*Sqrt[e]*Sqrt[c^2*d + e]*Sqrt[c^2*x^2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 205

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5236

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.64, size = 286, normalized size = 2.18 \[ \frac {-\frac {2 a}{d+e x^2}+\frac {b \sqrt {e} \log \left (\frac {4 c d \sqrt {e} x \left (c \sqrt {d}-i \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {c^2 (-d)-e}\right )+4 i d e}{b \sqrt {c^2 (-d)-e} \left (\sqrt {d}+i \sqrt {e} x\right )}\right )}{d \sqrt {c^2 (-d)-e}}+\frac {b \sqrt {e} \log \left (\frac {4 c d \sqrt {e} x \left (c \sqrt {d}+i \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {c^2 (-d)-e}\right )-4 i d e}{b \sqrt {c^2 (-d)-e} \left (\sqrt {d}-i \sqrt {e} x\right )}\right )}{d \sqrt {c^2 (-d)-e}}-\frac {2 b \sec ^{-1}(c x)}{d+e x^2}-\frac {2 b \sin ^{-1}\left (\frac {1}{c x}\right )}{d}}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a)/(d + e*x^2) - (2*b*ArcSec[c*x])/(d + e*x^2) - (2*b*ArcSin[1/(c*x)])/d + (b*Sqrt[e]*Log[((4*I)*d*e + 4*
c*d*Sqrt[e]*(c*Sqrt[d] - I*Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x)/(b*Sqrt[-(c^2*d) - e]*(Sqrt[d] + I*Sqr
t[e]*x))])/(d*Sqrt[-(c^2*d) - e]) + (b*Sqrt[e]*Log[((-4*I)*d*e + 4*c*d*Sqrt[e]*(c*Sqrt[d] + I*Sqrt[-(c^2*d) -
e]*Sqrt[1 - 1/(c^2*x^2)])*x)/(b*Sqrt[-(c^2*d) - e]*(Sqrt[d] - I*Sqrt[e]*x))])/(d*Sqrt[-(c^2*d) - e]))/(4*e)

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 384, normalized size = 2.93 \[ \left [-\frac {2 \, a c^{2} d^{2} + 2 \, a d e + \sqrt {-c^{2} d e - e^{2}} {\left (b e x^{2} + b d\right )} \log \left (\frac {c^{2} e x^{2} - c^{2} d + 2 \, \sqrt {-c^{2} d e - e^{2}} \sqrt {c^{2} x^{2} - 1} - 2 \, e}{e x^{2} + d}\right ) + 2 \, {\left (b c^{2} d^{2} + b d e\right )} \operatorname {arcsec}\left (c x\right ) - 4 \, {\left (b c^{2} d^{2} + b d e + {\left (b c^{2} d e + b e^{2}\right )} x^{2}\right )} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right )}{4 \, {\left (c^{2} d^{3} e + d^{2} e^{2} + {\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}, -\frac {a c^{2} d^{2} + a d e + \sqrt {c^{2} d e + e^{2}} {\left (b e x^{2} + b d\right )} \arctan \left (\frac {\sqrt {c^{2} d e + e^{2}} \sqrt {c^{2} x^{2} - 1}}{c^{2} d + e}\right ) + {\left (b c^{2} d^{2} + b d e\right )} \operatorname {arcsec}\left (c x\right ) - 2 \, {\left (b c^{2} d^{2} + b d e + {\left (b c^{2} d e + b e^{2}\right )} x^{2}\right )} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right )}{2 \, {\left (c^{2} d^{3} e + d^{2} e^{2} + {\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*c^2*d^2 + 2*a*d*e + sqrt(-c^2*d*e - e^2)*(b*e*x^2 + b*d)*log((c^2*e*x^2 - c^2*d + 2*sqrt(-c^2*d*e -
e^2)*sqrt(c^2*x^2 - 1) - 2*e)/(e*x^2 + d)) + 2*(b*c^2*d^2 + b*d*e)*arcsec(c*x) - 4*(b*c^2*d^2 + b*d*e + (b*c^
2*d*e + b*e^2)*x^2)*arctan(-c*x + sqrt(c^2*x^2 - 1)))/(c^2*d^3*e + d^2*e^2 + (c^2*d^2*e^2 + d*e^3)*x^2), -1/2*
(a*c^2*d^2 + a*d*e + sqrt(c^2*d*e + e^2)*(b*e*x^2 + b*d)*arctan(sqrt(c^2*d*e + e^2)*sqrt(c^2*x^2 - 1)/(c^2*d +
e)) + (b*c^2*d^2 + b*d*e)*arcsec(c*x) - 2*(b*c^2*d^2 + b*d*e + (b*c^2*d*e + b*e^2)*x^2)*arctan(-c*x + sqrt(c^
2*x^2 - 1)))/(c^2*d^3*e + d^2*e^2 + (c^2*d^2*e^2 + d*e^3)*x^2)]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(x)]sym2poly
/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [B]  time = 0.08, size = 354, normalized size = 2.70 \[ -\frac {c^{2} a}{2 e \left (c^{2} e \,x^{2}+c^{2} d \right )}-\frac {c^{2} b \,\mathrm {arcsec}\left (c x \right )}{2 e \left (c^{2} e \,x^{2}+c^{2} d \right )}-\frac {b \sqrt {c^{2} x^{2}-1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{2 c e \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x d}+\frac {b \sqrt {c^{2} x^{2}-1}\, \ln \left (\frac {2 \sqrt {-\frac {c^{2} d +e}{e}}\, \sqrt {c^{2} x^{2}-1}\, e -2 \sqrt {-c^{2} e d}\, c x -2 e}{c e x +\sqrt {-c^{2} e d}}\right )}{4 c e \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x d \sqrt {-\frac {c^{2} d +e}{e}}}+\frac {b \sqrt {c^{2} x^{2}-1}\, \ln \left (-\frac {2 \left (\sqrt {-\frac {c^{2} d +e}{e}}\, \sqrt {c^{2} x^{2}-1}\, e +\sqrt {-c^{2} e d}\, c x -e \right )}{-c e x +\sqrt {-c^{2} e d}}\right )}{4 c e \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, x d \sqrt {-\frac {c^{2} d +e}{e}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a/e/(c^2*e*x^2+c^2*d)-1/2*c^2*b/e/(c^2*e*x^2+c^2*d)*arcsec(c*x)-1/2/c*b/e*(c^2*x^2-1)^(1/2)/((c^2*x^2
-1)/c^2/x^2)^(1/2)/x/d*arctan(1/(c^2*x^2-1)^(1/2))+1/4/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+e)/e)^(1/2)*ln(2*((-(c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e-(-c^2*e*d)^(1/2)*c*x-e)/(c*e*x+(-c^2*e*d)
^(1/2)))+1/4/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+e)/e)^(1/2)*ln(-2*((-(c^2*d+e)/e
)^(1/2)*(c^2*x^2-1)^(1/2)*e+(-c^2*e*d)^(1/2)*c*x-e)/(-c*e*x+(-c^2*e*d)^(1/2)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left ({\left (c^{2} e^{2} x^{2} + c^{2} d e\right )} \int \frac {x e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (c x - 1\right )\right )}}{c^{2} e^{2} x^{4} + {\left (c^{2} e^{2} x^{4} + {\left (c^{2} d e - e^{2}\right )} x^{2} - d e\right )} {\left (c x + 1\right )} {\left (c x - 1\right )} + {\left (c^{2} d e - e^{2}\right )} x^{2} - d e}\,{d x} - \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} b}{2 \, {\left (e^{2} x^{2} + d e\right )}} - \frac {a}{2 \, {\left (e^{2} x^{2} + d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(c^2*e^2*x^2 + c^2*d*e)*integrate(1/2*x*e^(1/2*log(c*x + 1) + 1/2*log(c*x - 1))/(c^2*e^2*x^4 + (c^2*d*e
- e^2)*x^2 - d*e + (c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)*e^(log(c*x + 1) + log(c*x - 1))), x) - arctan(sq
rt(c*x + 1)*sqrt(c*x - 1)))*b/(e^2*x^2 + d*e) - 1/2*a/(e^2*x^2 + d*e)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a + b \operatorname {asec}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________