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

Optimal. Leaf size=162 \[ -\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac{b c d^2 \sqrt{c^2 x^2-1}}{\sqrt{c^2 x^2}}-\frac{b e x \left (12 c^2 d+e\right ) \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{6 c^2 \sqrt{c^2 x^2}}-\frac{b e^2 x^2 \sqrt{c^2 x^2-1}}{6 c \sqrt{c^2 x^2}} \]

[Out]

(b*c*d^2*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*x^2] - (b*e^2*x^2*Sqrt[-1 + c^2*x^2])/(6*c*Sqrt[c^2*x^2]) - (d^2*(a + b*
ArcSec[c*x]))/x + 2*d*e*x*(a + b*ArcSec[c*x]) + (e^2*x^3*(a + b*ArcSec[c*x]))/3 - (b*e*(12*c^2*d + e)*x*ArcTan
h[(c*x)/Sqrt[-1 + c^2*x^2]])/(6*c^2*Sqrt[c^2*x^2])

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Rubi [A]  time = 0.127189, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {270, 5238, 12, 1265, 388, 217, 206} \[ -\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac{b c d^2 \sqrt{c^2 x^2-1}}{\sqrt{c^2 x^2}}-\frac{b e x \left (12 c^2 d+e\right ) \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{6 c^2 \sqrt{c^2 x^2}}-\frac{b e^2 x^2 \sqrt{c^2 x^2-1}}{6 c \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*d^2*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*x^2] - (b*e^2*x^2*Sqrt[-1 + c^2*x^2])/(6*c*Sqrt[c^2*x^2]) - (d^2*(a + b*
ArcSec[c*x]))/x + 2*d*e*x*(a + b*ArcSec[c*x]) + (e^2*x^3*(a + b*ArcSec[c*x]))/3 - (b*e*(12*c^2*d + e)*x*ArcTan
h[(c*x)/Sqrt[-1 + c^2*x^2]])/(6*c^2*Sqrt[c^2*x^2])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 5238

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1265

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(d*f*(m + 1)), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[(d*f*(m + 1)*Qx)/x - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

Rule 388

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

Rule 217

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

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{\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \int \frac{-3 d^2+6 d e x^2+e^2 x^4}{3 x^2 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \int \frac{-3 d^2+6 d e x^2+e^2 x^4}{x^2 \sqrt{-1+c^2 x^2}} \, dx}{3 \sqrt{c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac{(b c x) \int \frac{6 d e+e^2 x^2}{\sqrt{-1+c^2 x^2}} \, dx}{3 \sqrt{c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{b e^2 x^2 \sqrt{-1+c^2 x^2}}{6 c \sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )--\frac{\left (b \left (-12 c^2 d e-e^2\right ) x\right ) \int \frac{1}{\sqrt{-1+c^2 x^2}} \, dx}{6 c \sqrt{c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{b e^2 x^2 \sqrt{-1+c^2 x^2}}{6 c \sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )--\frac{\left (b \left (-12 c^2 d e-e^2\right ) x\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\frac{x}{\sqrt{-1+c^2 x^2}}\right )}{6 c \sqrt{c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{b e^2 x^2 \sqrt{-1+c^2 x^2}}{6 c \sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e^2 x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac{b e \left (12 c^2 d+e\right ) x \tanh ^{-1}\left (\frac{c x}{\sqrt{-1+c^2 x^2}}\right )}{6 c^2 \sqrt{c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.194875, size = 136, normalized size = 0.84 \[ \frac{c^2 \left (2 a c \left (-3 d^2+6 d e x^2+e^2 x^4\right )+b x \sqrt{1-\frac{1}{c^2 x^2}} \left (6 c^2 d^2-e^2 x^2\right )\right )+2 b c^3 \sec ^{-1}(c x) \left (-3 d^2+6 d e x^2+e^2 x^4\right )-b e x \left (12 c^2 d+e\right ) \log \left (x \left (\sqrt{1-\frac{1}{c^2 x^2}}+1\right )\right )}{6 c^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(b*Sqrt[1 - 1/(c^2*x^2)]*x*(6*c^2*d^2 - e^2*x^2) + 2*a*c*(-3*d^2 + 6*d*e*x^2 + e^2*x^4)) + 2*b*c^3*(-3*d^
2 + 6*d*e*x^2 + e^2*x^4)*ArcSec[c*x] - b*e*(12*c^2*d + e)*x*Log[(1 + Sqrt[1 - 1/(c^2*x^2)])*x])/(6*c^3*x)

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Maple [A]  time = 0.174, size = 286, normalized size = 1.8 \begin{align*}{\frac{a{x}^{3}{e}^{2}}{3}}+2\,aedx-{\frac{a{d}^{2}}{x}}+{\frac{b{\rm arcsec} \left (cx\right ){x}^{3}{e}^{2}}{3}}+2\,b{\rm arcsec} \left (cx\right )edx-{\frac{b{\rm arcsec} \left (cx\right ){d}^{2}}{x}}+{cb{d}^{2}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{d}^{2}}{c{x}^{2}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-2\,{\frac{b\sqrt{{c}^{2}{x}^{2}-1}ed\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ) }{{c}^{2}x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{e}^{2}{x}^{2}}{6\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{b{e}^{2}}{6\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{e}^{2}}{6\,{c}^{4}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3*e^2+2*a*e*d*x-a*d^2/x+1/3*b*arcsec(c*x)*x^3*e^2+2*b*arcsec(c*x)*e*d*x-b*arcsec(c*x)*d^2/x+c*b/((c^2*
x^2-1)/c^2/x^2)^(1/2)*d^2-b/c/x^2/((c^2*x^2-1)/c^2/x^2)^(1/2)*d^2-2*b/c^2*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x
^2)^(1/2)/x*e*d*ln(c*x+(c^2*x^2-1)^(1/2))-1/6*b/c/((c^2*x^2-1)/c^2/x^2)^(1/2)*e^2*x^2+1/6*b/c^3/((c^2*x^2-1)/c
^2/x^2)^(1/2)*e^2-1/6*b/c^4*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*e^2*ln(c*x+(c^2*x^2-1)^(1/2))

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Maxima [A]  time = 0.993497, size = 267, normalized size = 1.65 \begin{align*} \frac{1}{3} \, a e^{2} x^{3} +{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} - \frac{\operatorname{arcsec}\left (c x\right )}{x}\right )} b d^{2} + \frac{1}{12} \,{\left (4 \, x^{3} \operatorname{arcsec}\left (c x\right ) - \frac{\frac{2 \, \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac{\log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} - \frac{\log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b e^{2} + 2 \, a d e x + \frac{{\left (2 \, c x \operatorname{arcsec}\left (c x\right ) - \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right ) + \log \left (-\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )\right )} b d e}{c} - \frac{a d^{2}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e^2*x^3 + (c*sqrt(-1/(c^2*x^2) + 1) - arcsec(c*x)/x)*b*d^2 + 1/12*(4*x^3*arcsec(c*x) - (2*sqrt(-1/(c^2*x
^2) + 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + log(sqrt(-1/(c^2*x^2) + 1) + 1)/c^2 - log(sqrt(-1/(c^2*x^2) + 1) - 1)
/c^2)/c)*b*e^2 + 2*a*d*e*x + (2*c*x*arcsec(c*x) - log(sqrt(-1/(c^2*x^2) + 1) + 1) + log(-sqrt(-1/(c^2*x^2) + 1
) + 1))*b*d*e/c - a*d^2/x

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Fricas [A]  time = 2.57997, size = 502, normalized size = 3.1 \begin{align*} \frac{2 \, a c^{3} e^{2} x^{4} + 6 \, b c^{4} d^{2} x + 12 \, a c^{3} d e x^{2} - 6 \, a c^{3} d^{2} - 4 \,{\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (12 \, b c^{2} d e + b e^{2}\right )} x \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 2 \,{\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2} +{\left (3 \, b c^{3} d^{2} - 6 \, b c^{3} d e - b c^{3} e^{2}\right )} x\right )} \operatorname{arcsec}\left (c x\right ) +{\left (6 \, b c^{3} d^{2} - b c e^{2} x^{2}\right )} \sqrt{c^{2} x^{2} - 1}}{6 \, c^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*e^2*x^4 + 6*b*c^4*d^2*x + 12*a*c^3*d*e*x^2 - 6*a*c^3*d^2 - 4*(3*b*c^3*d^2 - 6*b*c^3*d*e - b*c^3*e
^2)*x*arctan(-c*x + sqrt(c^2*x^2 - 1)) + (12*b*c^2*d*e + b*e^2)*x*log(-c*x + sqrt(c^2*x^2 - 1)) + 2*(b*c^3*e^2
*x^4 + 6*b*c^3*d*e*x^2 - 3*b*c^3*d^2 + (3*b*c^3*d^2 - 6*b*c^3*d*e - b*c^3*e^2)*x)*arcsec(c*x) + (6*b*c^3*d^2 -
b*c*e^2*x^2)*sqrt(c^2*x^2 - 1))/(c^3*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{2}{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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