3.85 \(\int \frac{(d+e x^2)^2 (a+b \sec ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=183 \[ -\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+\frac{b c \sqrt{c^2 x^2-1} \left (24 c^4 d^2+100 c^2 d e+225 e^2\right )}{225 \sqrt{c^2 x^2}}+\frac{b c d^2 \sqrt{c^2 x^2-1}}{25 x^4 \sqrt{c^2 x^2}}+\frac{2 b c d \sqrt{c^2 x^2-1} \left (6 c^2 d+25 e\right )}{225 x^2 \sqrt{c^2 x^2}} \]

[Out]

(b*c*(24*c^4*d^2 + 100*c^2*d*e + 225*e^2)*Sqrt[-1 + c^2*x^2])/(225*Sqrt[c^2*x^2]) + (b*c*d^2*Sqrt[-1 + c^2*x^2
])/(25*x^4*Sqrt[c^2*x^2]) + (2*b*c*d*(6*c^2*d + 25*e)*Sqrt[-1 + c^2*x^2])/(225*x^2*Sqrt[c^2*x^2]) - (d^2*(a +
b*ArcSec[c*x]))/(5*x^5) - (2*d*e*(a + b*ArcSec[c*x]))/(3*x^3) - (e^2*(a + b*ArcSec[c*x]))/x

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Rubi [A]  time = 0.157466, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {270, 5238, 12, 1265, 453, 264} \[ -\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+\frac{b c \sqrt{c^2 x^2-1} \left (24 c^4 d^2+100 c^2 d e+225 e^2\right )}{225 \sqrt{c^2 x^2}}+\frac{b c d^2 \sqrt{c^2 x^2-1}}{25 x^4 \sqrt{c^2 x^2}}+\frac{2 b c d \sqrt{c^2 x^2-1} \left (6 c^2 d+25 e\right )}{225 x^2 \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcSec[c*x]))/x^6,x]

[Out]

(b*c*(24*c^4*d^2 + 100*c^2*d*e + 225*e^2)*Sqrt[-1 + c^2*x^2])/(225*Sqrt[c^2*x^2]) + (b*c*d^2*Sqrt[-1 + c^2*x^2
])/(25*x^4*Sqrt[c^2*x^2]) + (2*b*c*d*(6*c^2*d + 25*e)*Sqrt[-1 + c^2*x^2])/(225*x^2*Sqrt[c^2*x^2]) - (d^2*(a +
b*ArcSec[c*x]))/(5*x^5) - (2*d*e*(a + b*ArcSec[c*x]))/(3*x^3) - (e^2*(a + b*ArcSec[c*x]))/x

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 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 )^2 \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx &=-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{(b c x) \int \frac{-3 d^2-10 d e x^2-15 e^2 x^4}{15 x^6 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{(b c x) \int \frac{-3 d^2-10 d e x^2-15 e^2 x^4}{x^6 \sqrt{-1+c^2 x^2}} \, dx}{15 \sqrt{c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1+c^2 x^2}}{25 x^4 \sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{(b c x) \int \frac{-2 d \left (6 c^2 d+25 e\right )-75 e^2 x^2}{x^4 \sqrt{-1+c^2 x^2}} \, dx}{75 \sqrt{c^2 x^2}}\\ &=\frac{b c d^2 \sqrt{-1+c^2 x^2}}{25 x^4 \sqrt{c^2 x^2}}+\frac{2 b c d \left (6 c^2 d+25 e\right ) \sqrt{-1+c^2 x^2}}{225 x^2 \sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}-\frac{\left (b c \left (-225 e^2-4 c^2 d \left (6 c^2 d+25 e\right )\right ) x\right ) \int \frac{1}{x^2 \sqrt{-1+c^2 x^2}} \, dx}{225 \sqrt{c^2 x^2}}\\ &=\frac{b c \left (225 e^2+4 c^2 d \left (6 c^2 d+25 e\right )\right ) \sqrt{-1+c^2 x^2}}{225 \sqrt{c^2 x^2}}+\frac{b c d^2 \sqrt{-1+c^2 x^2}}{25 x^4 \sqrt{c^2 x^2}}+\frac{2 b c d \left (6 c^2 d+25 e\right ) \sqrt{-1+c^2 x^2}}{225 x^2 \sqrt{c^2 x^2}}-\frac{d^2 \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac{2 d e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac{e^2 \left (a+b \sec ^{-1}(c x)\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.211026, size = 127, normalized size = 0.69 \[ \frac{-15 a \left (3 d^2+10 d e x^2+15 e^2 x^4\right )+b c x \sqrt{1-\frac{1}{c^2 x^2}} \left (3 d^2 \left (8 c^4 x^4+4 c^2 x^2+3\right )+50 d e x^2 \left (2 c^2 x^2+1\right )+225 e^2 x^4\right )-15 b \sec ^{-1}(c x) \left (3 d^2+10 d e x^2+15 e^2 x^4\right )}{225 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcSec[c*x]))/x^6,x]

[Out]

(-15*a*(3*d^2 + 10*d*e*x^2 + 15*e^2*x^4) + b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(225*e^2*x^4 + 50*d*e*x^2*(1 + 2*c^2*x^
2) + 3*d^2*(3 + 4*c^2*x^2 + 8*c^4*x^4)) - 15*b*(3*d^2 + 10*d*e*x^2 + 15*e^2*x^4)*ArcSec[c*x])/(225*x^5)

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Maple [A]  time = 0.177, size = 191, normalized size = 1. \begin{align*}{c}^{5} \left ({\frac{a}{{c}^{4}} \left ( -{\frac{{e}^{2}}{cx}}-{\frac{{d}^{2}}{5\,c{x}^{5}}}-{\frac{2\,de}{3\,c{x}^{3}}} \right ) }+{\frac{b}{{c}^{4}} \left ( -{\frac{{\rm arcsec} \left (cx\right ){e}^{2}}{cx}}-{\frac{{\rm arcsec} \left (cx\right ){d}^{2}}{5\,c{x}^{5}}}-{\frac{2\,{\rm arcsec} \left (cx\right )ed}{3\,c{x}^{3}}}+{\frac{ \left ({c}^{2}{x}^{2}-1 \right ) \left ( 24\,{c}^{8}{d}^{2}{x}^{4}+100\,{c}^{6}de{x}^{4}+12\,{c}^{6}{d}^{2}{x}^{2}+225\,{c}^{4}{e}^{2}{x}^{4}+50\,{c}^{4}de{x}^{2}+9\,{d}^{2}{c}^{4} \right ) }{225\,{c}^{6}{x}^{6}}{\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)^2*(a+b*arcsec(c*x))/x^6,x)

[Out]

c^5*(a/c^4*(-e^2/c/x-1/5*d^2/c/x^5-2/3/c*e*d/x^3)+b/c^4*(-arcsec(c*x)*e^2/c/x-1/5*arcsec(c*x)*d^2/c/x^5-2/3*ar
csec(c*x)/c*e*d/x^3+1/225*(c^2*x^2-1)*(24*c^8*d^2*x^4+100*c^6*d*e*x^4+12*c^6*d^2*x^2+225*c^4*e^2*x^4+50*c^4*d*
e*x^2+9*c^4*d^2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^6/x^6))

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Maxima [A]  time = 0.973861, size = 244, normalized size = 1.33 \begin{align*}{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} - \frac{\operatorname{arcsec}\left (c x\right )}{x}\right )} b e^{2} + \frac{1}{75} \, b d^{2}{\left (\frac{3 \, c^{6}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{5}{2}} - 10 \, c^{6}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 15 \, c^{6} \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c} - \frac{15 \, \operatorname{arcsec}\left (c x\right )}{x^{5}}\right )} - \frac{2}{9} \, b d e{\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^{2}}{x} - \frac{2 \, a d e}{3 \, x^{3}} - \frac{a d^{2}}{5 \, x^{5}} \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^6,x, algorithm="maxima")

[Out]

(c*sqrt(-1/(c^2*x^2) + 1) - arcsec(c*x)/x)*b*e^2 + 1/75*b*d^2*((3*c^6*(-1/(c^2*x^2) + 1)^(5/2) - 10*c^6*(-1/(c
^2*x^2) + 1)^(3/2) + 15*c^6*sqrt(-1/(c^2*x^2) + 1))/c - 15*arcsec(c*x)/x^5) - 2/9*b*d*e*((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^2/x - 2/3*a*d*e/x^3 - 1/5*a*d^2/x^5

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Fricas [A]  time = 1.67044, size = 302, normalized size = 1.65 \begin{align*} -\frac{225 \, a e^{2} x^{4} + 150 \, a d e x^{2} + 45 \, a d^{2} + 15 \,{\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \operatorname{arcsec}\left (c x\right ) -{\left ({\left (24 \, b c^{4} d^{2} + 100 \, b c^{2} d e + 225 \, b e^{2}\right )} x^{4} + 9 \, b d^{2} + 2 \,{\left (6 \, b c^{2} d^{2} + 25 \, b d e\right )} x^{2}\right )} \sqrt{c^{2} x^{2} - 1}}{225 \, x^{5}} \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^6,x, algorithm="fricas")

[Out]

-1/225*(225*a*e^2*x^4 + 150*a*d*e*x^2 + 45*a*d^2 + 15*(15*b*e^2*x^4 + 10*b*d*e*x^2 + 3*b*d^2)*arcsec(c*x) - ((
24*b*c^4*d^2 + 100*b*c^2*d*e + 225*b*e^2)*x^4 + 9*b*d^2 + 2*(6*b*c^2*d^2 + 25*b*d*e)*x^2)*sqrt(c^2*x^2 - 1))/x
^5

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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^{6}}\, 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**6,x)

[Out]

Integral((a + b*asec(c*x))*(d + e*x**2)**2/x**6, 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^{6}}\,{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^6,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arcsec(c*x) + a)/x^6, x)