3.73 \(\int \frac{(d+e x^2) (a+b \sec ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=105 \[ -\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \sec ^{-1}(c x)\right )}{x}+\frac{b c \sqrt{c^2 x^2-1} \left (2 c^2 d+9 e\right )}{9 \sqrt{c^2 x^2}}+\frac{b c d \sqrt{c^2 x^2-1}}{9 x^2 \sqrt{c^2 x^2}} \]

[Out]

(b*c*(2*c^2*d + 9*e)*Sqrt[-1 + c^2*x^2])/(9*Sqrt[c^2*x^2]) + (b*c*d*Sqrt[-1 + c^2*x^2])/(9*x^2*Sqrt[c^2*x^2])
- (d*(a + b*ArcSec[c*x]))/(3*x^3) - (e*(a + b*ArcSec[c*x]))/x

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Rubi [A]  time = 0.0748212, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 5238, 12, 453, 264} \[ -\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \sec ^{-1}(c x)\right )}{x}+\frac{b c \sqrt{c^2 x^2-1} \left (2 c^2 d+9 e\right )}{9 \sqrt{c^2 x^2}}+\frac{b c d \sqrt{c^2 x^2-1}}{9 x^2 \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcSec[c*x]))/x^4,x]

[Out]

(b*c*(2*c^2*d + 9*e)*Sqrt[-1 + c^2*x^2])/(9*Sqrt[c^2*x^2]) + (b*c*d*Sqrt[-1 + c^2*x^2])/(9*x^2*Sqrt[c^2*x^2])
- (d*(a + b*ArcSec[c*x]))/(3*x^3) - (e*(a + b*ArcSec[c*x]))/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 264

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{(b c x) \int \frac{-d-3 e x^2}{3 x^4 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{(b c x) \int \frac{-d-3 e x^2}{x^4 \sqrt{-1+c^2 x^2}} \, dx}{3 \sqrt{c^2 x^2}}\\ &=\frac{b c d \sqrt{-1+c^2 x^2}}{9 x^2 \sqrt{c^2 x^2}}-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{\left (b c \left (-2 c^2 d-9 e\right ) x\right ) \int \frac{1}{x^2 \sqrt{-1+c^2 x^2}} \, dx}{9 \sqrt{c^2 x^2}}\\ &=\frac{b c \left (2 c^2 d+9 e\right ) \sqrt{-1+c^2 x^2}}{9 \sqrt{c^2 x^2}}+\frac{b c d \sqrt{-1+c^2 x^2}}{9 x^2 \sqrt{c^2 x^2}}-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \sec ^{-1}(c x)\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.0785472, size = 69, normalized size = 0.66 \[ \frac{-3 a \left (d+3 e x^2\right )+b c x \sqrt{1-\frac{1}{c^2 x^2}} \left (2 c^2 d x^2+d+9 e x^2\right )-3 b \sec ^{-1}(c x) \left (d+3 e x^2\right )}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcSec[c*x]))/x^4,x]

[Out]

(-3*a*(d + 3*e*x^2) + b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(d + 2*c^2*d*x^2 + 9*e*x^2) - 3*b*(d + 3*e*x^2)*ArcSec[c*x])
/(9*x^3)

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Maple [A]  time = 0.171, size = 121, normalized size = 1.2 \begin{align*}{c}^{3} \left ({\frac{a}{{c}^{2}} \left ( -{\frac{e}{cx}}-{\frac{d}{3\,c{x}^{3}}} \right ) }+{\frac{b}{{c}^{2}} \left ( -{\frac{{\rm arcsec} \left (cx\right )e}{cx}}-{\frac{{\rm arcsec} \left (cx\right )d}{3\,c{x}^{3}}}+{\frac{ \left ({c}^{2}{x}^{2}-1 \right ) \left ( 2\,{c}^{4}d{x}^{2}+9\,{c}^{2}e{x}^{2}+{c}^{2}d \right ) }{9\,{c}^{4}{x}^{4}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arcsec(c*x))/x^4,x)

[Out]

c^3*(a/c^2*(-e/c/x-1/3/c*d/x^3)+b/c^2*(-arcsec(c*x)*e/c/x-1/3*arcsec(c*x)/c*d/x^3+1/9*(c^2*x^2-1)*(2*c^4*d*x^2
+9*c^2*e*x^2+c^2*d)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^4/x^4))

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Maxima [A]  time = 0.983457, size = 127, normalized size = 1.21 \begin{align*}{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} - \frac{\operatorname{arcsec}\left (c x\right )}{x}\right )} b e - \frac{1}{9} \, b d{\left (\frac{c^{4}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 3 \, c^{4} \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c} + \frac{3 \, \operatorname{arcsec}\left (c x\right )}{x^{3}}\right )} - \frac{a e}{x} - \frac{a d}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsec(c*x))/x^4,x, algorithm="maxima")

[Out]

(c*sqrt(-1/(c^2*x^2) + 1) - arcsec(c*x)/x)*b*e - 1/9*b*d*((c^4*(-1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(-1/(c^2*x
^2) + 1))/c + 3*arcsec(c*x)/x^3) - a*e/x - 1/3*a*d/x^3

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Fricas [A]  time = 1.99869, size = 157, normalized size = 1.5 \begin{align*} -\frac{9 \, a e x^{2} + 3 \, a d + 3 \,{\left (3 \, b e x^{2} + b d\right )} \operatorname{arcsec}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1}{\left ({\left (2 \, b c^{2} d + 9 \, b e\right )} x^{2} + b d\right )}}{9 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsec(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/9*(9*a*e*x^2 + 3*a*d + 3*(3*b*e*x^2 + b*d)*arcsec(c*x) - sqrt(c^2*x^2 - 1)*((2*b*c^2*d + 9*b*e)*x^2 + b*d))
/x^3

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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 )}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*asec(c*x))/x**4,x)

[Out]

Integral((a + b*asec(c*x))*(d + e*x**2)/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsec(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsec(c*x) + a)/x^4, x)