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

Optimal. Leaf size=152 \[ -\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}+\frac {b c \sqrt {c^2 x^2-1} \left (12 c^2 d+25 e\right )}{225 x^2 \sqrt {c^2 x^2}}+\frac {b c d \sqrt {c^2 x^2-1}}{25 x^4 \sqrt {c^2 x^2}}+\frac {2 b c^3 \sqrt {c^2 x^2-1} \left (12 c^2 d+25 e\right )}{225 \sqrt {c^2 x^2}} \]

[Out]

-1/5*d*(a+b*arcsec(c*x))/x^5-1/3*e*(a+b*arcsec(c*x))/x^3+2/225*b*c^3*(12*c^2*d+25*e)*(c^2*x^2-1)^(1/2)/(c^2*x^
2)^(1/2)+1/25*b*c*d*(c^2*x^2-1)^(1/2)/x^4/(c^2*x^2)^(1/2)+1/225*b*c*(12*c^2*d+25*e)*(c^2*x^2-1)^(1/2)/x^2/(c^2
*x^2)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {14, 5238, 12, 453, 271, 264} \[ -\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}+\frac {2 b c^3 \sqrt {c^2 x^2-1} \left (12 c^2 d+25 e\right )}{225 \sqrt {c^2 x^2}}+\frac {b c \sqrt {c^2 x^2-1} \left (12 c^2 d+25 e\right )}{225 x^2 \sqrt {c^2 x^2}}+\frac {b c d \sqrt {c^2 x^2-1}}{25 x^4 \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*c^3*(12*c^2*d + 25*e)*Sqrt[-1 + c^2*x^2])/(225*Sqrt[c^2*x^2]) + (b*c*d*Sqrt[-1 + c^2*x^2])/(25*x^4*Sqrt[c
^2*x^2]) + (b*c*(12*c^2*d + 25*e)*Sqrt[-1 + c^2*x^2])/(225*x^2*Sqrt[c^2*x^2]) - (d*(a + b*ArcSec[c*x]))/(5*x^5
) - (e*(a + b*ArcSec[c*x]))/(3*x^3)

Rule 12

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

Rule 271

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

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {(b c x) \int \frac {-3 d-5 e x^2}{15 x^6 \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {(b c x) \int \frac {-3 d-5 e x^2}{x^6 \sqrt {-1+c^2 x^2}} \, dx}{15 \sqrt {c^2 x^2}}\\ &=\frac {b c d \sqrt {-1+c^2 x^2}}{25 x^4 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {\left (b c \left (-12 c^2 d-25 e\right ) x\right ) \int \frac {1}{x^4 \sqrt {-1+c^2 x^2}} \, dx}{75 \sqrt {c^2 x^2}}\\ &=\frac {b c d \sqrt {-1+c^2 x^2}}{25 x^4 \sqrt {c^2 x^2}}+\frac {b c \left (12 c^2 d+25 e\right ) \sqrt {-1+c^2 x^2}}{225 x^2 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}-\frac {\left (2 b c^3 \left (-12 c^2 d-25 e\right ) x\right ) \int \frac {1}{x^2 \sqrt {-1+c^2 x^2}} \, dx}{225 \sqrt {c^2 x^2}}\\ &=\frac {2 b c^3 \left (12 c^2 d+25 e\right ) \sqrt {-1+c^2 x^2}}{225 \sqrt {c^2 x^2}}+\frac {b c d \sqrt {-1+c^2 x^2}}{25 x^4 \sqrt {c^2 x^2}}+\frac {b c \left (12 c^2 d+25 e\right ) \sqrt {-1+c^2 x^2}}{225 x^2 \sqrt {c^2 x^2}}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{5 x^5}-\frac {e \left (a+b \sec ^{-1}(c x)\right )}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 94, normalized size = 0.62 \[ \frac {-15 a \left (3 d+5 e x^2\right )+b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (25 e x^2 \left (2 c^2 x^2+1\right )+3 d \left (8 c^4 x^4+4 c^2 x^2+3\right )\right )-15 b \sec ^{-1}(c x) \left (3 d+5 e x^2\right )}{225 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a*(3*d + 5*e*x^2) + b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(25*e*x^2*(1 + 2*c^2*x^2) + 3*d*(3 + 4*c^2*x^2 + 8*c^4*x^
4)) - 15*b*(3*d + 5*e*x^2)*ArcSec[c*x])/(225*x^5)

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fricas [A]  time = 1.07, size = 89, normalized size = 0.59 \[ -\frac {75 \, a e x^{2} + 45 \, a d + 15 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \operatorname {arcsec}\left (c x\right ) - {\left (2 \, {\left (12 \, b c^{4} d + 25 \, b c^{2} e\right )} x^{4} + {\left (12 \, b c^{2} d + 25 \, b e\right )} x^{2} + 9 \, b d\right )} \sqrt {c^{2} x^{2} - 1}}{225 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/225*(75*a*e*x^2 + 45*a*d + 15*(5*b*e*x^2 + 3*b*d)*arcsec(c*x) - (2*(12*b*c^4*d + 25*b*c^2*e)*x^4 + (12*b*c^
2*d + 25*b*e)*x^2 + 9*b*d)*sqrt(c^2*x^2 - 1))/x^5

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giac [A]  time = 0.16, size = 162, normalized size = 1.07 \[ \frac {1}{225} \, {\left (24 \, b c^{4} d \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 50 \, b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e + \frac {12 \, b c^{2} d \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{2}} + \frac {25 \, b \sqrt {-\frac {1}{c^{2} x^{2}} + 1} e}{x^{2}} - \frac {75 \, b \arccos \left (\frac {1}{c x}\right ) e}{c x^{3}} + \frac {9 \, b d \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{x^{4}} - \frac {75 \, a e}{c x^{3}} - \frac {45 \, b d \arccos \left (\frac {1}{c x}\right )}{c x^{5}} - \frac {45 \, a d}{c x^{5}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(24*b*c^4*d*sqrt(-1/(c^2*x^2) + 1) + 50*b*c^2*sqrt(-1/(c^2*x^2) + 1)*e + 12*b*c^2*d*sqrt(-1/(c^2*x^2) +
1)/x^2 + 25*b*sqrt(-1/(c^2*x^2) + 1)*e/x^2 - 75*b*arccos(1/(c*x))*e/(c*x^3) + 9*b*d*sqrt(-1/(c^2*x^2) + 1)/x^4
- 75*a*e/(c*x^3) - 45*b*d*arccos(1/(c*x))/(c*x^5) - 45*a*d/(c*x^5))*c

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maple [A]  time = 0.06, size = 140, normalized size = 0.92 \[ c^{5} \left (\frac {a \left (-\frac {e}{3 c^{3} x^{3}}-\frac {d}{5 c^{3} x^{5}}\right )}{c^{2}}+\frac {b \left (-\frac {\mathrm {arcsec}\left (c x \right ) e}{3 c^{3} x^{3}}-\frac {\mathrm {arcsec}\left (c x \right ) d}{5 c^{3} x^{5}}+\frac {\left (c^{2} x^{2}-1\right ) \left (24 x^{4} c^{6} d +50 c^{4} e \,x^{4}+12 c^{4} d \,x^{2}+25 c^{2} e \,x^{2}+9 c^{2} d \right )}{225 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )}{c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^5*(a/c^2*(-1/3*e/c^3/x^3-1/5/c^3*d/x^5)+b/c^2*(-1/3*arcsec(c*x)*e/c^3/x^3-1/5*arcsec(c*x)/c^3*d/x^5+1/225*(c
^2*x^2-1)*(24*c^6*d*x^4+50*c^4*e*x^4+12*c^4*d*x^2+25*c^2*e*x^2+9*c^2*d)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c^6/x^6))

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maxima [A]  time = 0.32, size = 137, normalized size = 0.90 \[ \frac {1}{75} \, b d {\left (\frac {3 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{6} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {15 \, \operatorname {arcsec}\left (c x\right )}{x^{5}}\right )} - \frac {1}{9} \, b e {\left (\frac {c^{4} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} + \frac {3 \, \operatorname {arcsec}\left (c x\right )}{x^{3}}\right )} - \frac {a e}{3 \, x^{3}} - \frac {a d}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/75*b*d*((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) - 1/9*b*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) - 1/3*a*e/x^3 - 1/5*a*d/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)*(a + b*acos(1/(c*x))))/x^6,x)

[Out]

int(((d + e*x^2)*(a + b*acos(1/(c*x))))/x^6, x)

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sympy [A]  time = 9.29, size = 279, normalized size = 1.84 \[ - \frac {a d}{5 x^{5}} - \frac {a e}{3 x^{3}} - \frac {b d \operatorname {asec}{\left (c x \right )}}{5 x^{5}} - \frac {b e \operatorname {asec}{\left (c x \right )}}{3 x^{3}} + \frac {b d \left (\begin {cases} \frac {8 c^{5} \sqrt {c^{2} x^{2} - 1}}{15 x} + \frac {4 c^{3} \sqrt {c^{2} x^{2} - 1}}{15 x^{3}} + \frac {c \sqrt {c^{2} x^{2} - 1}}{5 x^{5}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {8 i c^{5} \sqrt {- c^{2} x^{2} + 1}}{15 x} + \frac {4 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{15 x^{3}} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{5 x^{5}} & \text {otherwise} \end {cases}\right )}{5 c} + \frac {b e \left (\begin {cases} \frac {2 c^{3} \sqrt {c^{2} x^{2} - 1}}{3 x} + \frac {c \sqrt {c^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {2 i c^{3} \sqrt {- c^{2} x^{2} + 1}}{3 x} + \frac {i c \sqrt {- c^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right )}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d/(5*x**5) - a*e/(3*x**3) - b*d*asec(c*x)/(5*x**5) - b*e*asec(c*x)/(3*x**3) + b*d*Piecewise((8*c**5*sqrt(c*
*2*x**2 - 1)/(15*x) + 4*c**3*sqrt(c**2*x**2 - 1)/(15*x**3) + c*sqrt(c**2*x**2 - 1)/(5*x**5), Abs(c**2*x**2) >
1), (8*I*c**5*sqrt(-c**2*x**2 + 1)/(15*x) + 4*I*c**3*sqrt(-c**2*x**2 + 1)/(15*x**3) + I*c*sqrt(-c**2*x**2 + 1)
/(5*x**5), True))/(5*c) + b*e*Piecewise((2*c**3*sqrt(c**2*x**2 - 1)/(3*x) + c*sqrt(c**2*x**2 - 1)/(3*x**3), Ab
s(c**2*x**2) > 1), (2*I*c**3*sqrt(-c**2*x**2 + 1)/(3*x) + I*c*sqrt(-c**2*x**2 + 1)/(3*x**3), True))/(3*c)

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