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

Optimal. Leaf size=85 \[ -\frac {d \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{6} b c \left (c^2 d+6 e\right ) \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2} \]

[Out]

-1/3*d*(a+b*arcsin(c*x))/x^3-e*(a+b*arcsin(c*x))/x-1/6*b*c*(c^2*d+6*e)*arctanh((-c^2*x^2+1)^(1/2))-1/6*b*c*d*(
-c^2*x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {14, 4731, 12, 446, 78, 63, 208} \[ -\frac {d \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{6} b c \left (c^2 d+6 e\right ) \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d*Sqrt[1 - c^2*x^2])/(6*x^2) - (d*(a + b*ArcSin[c*x]))/(3*x^3) - (e*(a + b*ArcSin[c*x]))/x - (b*c*(c^2*d
+ 6*e)*ArcTanh[Sqrt[1 - c^2*x^2]])/6

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 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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 4731

Int[((a_.) + ArcSin[(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*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 109, normalized size = 1.28 \[ -\frac {a d}{3 x^3}-\frac {a e}{x}-\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2}-b c e \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-\frac {1}{6} b c^3 d \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-\frac {b d \sin ^{-1}(c x)}{3 x^3}-\frac {b e \sin ^{-1}(c x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*d)/x^3 - (a*e)/x - (b*c*d*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*d*ArcSin[c*x])/(3*x^3) - (b*e*ArcSin[c*x])/x
- (b*c^3*d*ArcTanh[Sqrt[1 - c^2*x^2]])/6 - b*c*e*ArcTanh[Sqrt[1 - c^2*x^2]]

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fricas [A]  time = 0.79, size = 115, normalized size = 1.35 \[ -\frac {{\left (b c^{3} d + 6 \, b c e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - {\left (b c^{3} d + 6 \, b c e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 2 \, \sqrt {-c^{2} x^{2} + 1} b c d x + 12 \, a e x^{2} + 4 \, a d + 4 \, {\left (3 \, b e x^{2} + b d\right )} \arcsin \left (c x\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 4.16, size = 430, normalized size = 5.06 \[ -\frac {b c^{6} d x^{3} \arcsin \left (c x\right )}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} - \frac {a c^{6} d x^{3}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} + \frac {b c^{5} d x^{2}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}} - \frac {b c^{4} d x \arcsin \left (c x\right )}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {a c^{4} d x}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} + \frac {1}{6} \, b c^{3} d \log \left ({\left | c \right |} {\left | x \right |}\right ) - \frac {1}{6} \, b c^{3} d \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \frac {b c^{2} d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} \arcsin \left (c x\right )}{8 \, x} - \frac {b c^{2} x \arcsin \left (c x\right ) e}{2 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {a c^{2} d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}}{8 \, x} - \frac {a c^{2} x e}{2 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} + b c e \log \left ({\left | c \right |} {\left | x \right |}\right ) - b c e \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \frac {b c d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}}{24 \, x^{2}} - \frac {b d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3} \arcsin \left (c x\right )}{24 \, x^{3}} - \frac {b {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} \arcsin \left (c x\right ) e}{2 \, x} - \frac {a d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}}{24 \, x^{3}} - \frac {a {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} e}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*b*c^6*d*x^3*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^3 - 1/24*a*c^6*d*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + 1/24*
b*c^5*d*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 - 1/8*b*c^4*d*x*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1) - 1/8*a*c^4*d*x/(s
qrt(-c^2*x^2 + 1) + 1) + 1/6*b*c^3*d*log(abs(c)*abs(x)) - 1/6*b*c^3*d*log(sqrt(-c^2*x^2 + 1) + 1) - 1/8*b*c^2*
d*(sqrt(-c^2*x^2 + 1) + 1)*arcsin(c*x)/x - 1/2*b*c^2*x*arcsin(c*x)*e/(sqrt(-c^2*x^2 + 1) + 1) - 1/8*a*c^2*d*(s
qrt(-c^2*x^2 + 1) + 1)/x - 1/2*a*c^2*x*e/(sqrt(-c^2*x^2 + 1) + 1) + b*c*e*log(abs(c)*abs(x)) - b*c*e*log(sqrt(
-c^2*x^2 + 1) + 1) - 1/24*b*c*d*(sqrt(-c^2*x^2 + 1) + 1)^2/x^2 - 1/24*b*d*(sqrt(-c^2*x^2 + 1) + 1)^3*arcsin(c*
x)/x^3 - 1/2*b*(sqrt(-c^2*x^2 + 1) + 1)*arcsin(c*x)*e/x - 1/24*a*d*(sqrt(-c^2*x^2 + 1) + 1)^3/x^3 - 1/2*a*(sqr
t(-c^2*x^2 + 1) + 1)*e/x

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maple [A]  time = 0.01, size = 120, normalized size = 1.41 \[ c^{3} \left (\frac {a \left (-\frac {d}{3 c \,x^{3}}-\frac {e}{c x}\right )}{c^{2}}+\frac {b \left (-\frac {\arcsin \left (c x \right ) d}{3 c \,x^{3}}-\frac {\arcsin \left (c x \right ) e}{c x}-e \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )+\frac {c^{2} d \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}\right )}{c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(a/c^2*(-1/3/c*d/x^3-e/c/x)+b/c^2*(-1/3*arcsin(c*x)/c*d/x^3-arcsin(c*x)*e/c/x-e*arctanh(1/(-c^2*x^2+1)^(1/
2))+1/3*c^2*d*(-1/2/c^2/x^2*(-c^2*x^2+1)^(1/2)-1/2*arctanh(1/(-c^2*x^2+1)^(1/2)))))

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maxima [A]  time = 1.21, size = 119, normalized size = 1.40 \[ -\frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b d - {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c*x)/x^3)*b*d -
(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*b*e - a*e/x - 1/3*a*d/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (e\,x^2+d\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.87, size = 170, normalized size = 2.00 \[ - \frac {a d}{3 x^{3}} - \frac {a e}{x} + \frac {b c d \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} - \frac {c \sqrt {-1 + \frac {1}{c^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c}{2 x \sqrt {1 - \frac {1}{c^{2} x^{2}}}} + \frac {i}{2 c x^{3} \sqrt {1 - \frac {1}{c^{2} x^{2}}}} & \text {otherwise} \end {cases}\right )}{3} + b c e \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d \operatorname {asin}{\left (c x \right )}}{3 x^{3}} - \frac {b e \operatorname {asin}{\left (c x \right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d/(3*x**3) - a*e/x + b*c*d*Piecewise((-c**2*acosh(1/(c*x))/2 - c*sqrt(-1 + 1/(c**2*x**2))/(2*x), 1/Abs(c**2
*x**2) > 1), (I*c**2*asin(1/(c*x))/2 - I*c/(2*x*sqrt(1 - 1/(c**2*x**2))) + I/(2*c*x**3*sqrt(1 - 1/(c**2*x**2))
), True))/3 + b*c*e*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)), True)) - b*d*asin(c*x
)/(3*x**3) - b*e*asin(c*x)/x

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