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

Optimal. Leaf size=66 \[ -\frac {d \left (a+b \sin ^{-1}(c x)\right )}{x}+e x \left (a+b \sin ^{-1}(c x)\right )-b c d \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )+\frac {b e \sqrt {1-c^2 x^2}}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 66, 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, 4731, 446, 80, 63, 208} \[ -\frac {d \left (a+b \sin ^{-1}(c x)\right )}{x}+e x \left (a+b \sin ^{-1}(c x)\right )-b c d \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )+\frac {b e \sqrt {1-c^2 x^2}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 71, normalized size = 1.08 \[ -\frac {a d}{x}+a e x-b c d \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )+\frac {b e \sqrt {1-c^2 x^2}}{c}-\frac {b d \sin ^{-1}(c x)}{x}+b e x \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 103, normalized size = 1.56 \[ -\frac {b c^{2} d x \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - b c^{2} d x \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) - 2 \, a c e x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b e x + 2 \, a c d - 2 \, {\left (b c e x^{2} - b c d\right )} \arcsin \left (c x\right )}{2 \, c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.71, size = 1036, normalized size = 15.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*c^6*d*x^4*arcsin(c*x)/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2
*x^2 + 1) + 1)^4) - 1/2*a*c^6*d*x^4/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sq
rt(-c^2*x^2 + 1) + 1)^4) + b*c^5*d*x^3*log(abs(c)*abs(x))/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-
c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^3) - b*c^5*d*x^3*log(sqrt(-c^2*x^2 + 1) + 1)/((c^4*x^3/(sqrt(-c^2*
x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^3) - b*c^4*d*x^2*arcsin(c*x)/((c^4*
x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^2) - a*c^4*d*x^2/((c
^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^2) + b*c^3*d*x*lo
g(abs(c)*abs(x))/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) +
1)) - b*c^3*d*x*log(sqrt(-c^2*x^2 + 1) + 1)/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) +
1))*(sqrt(-c^2*x^2 + 1) + 1)) - 1/2*b*c^2*d*arcsin(c*x)/(c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^
2*x^2 + 1) + 1)) - b*c^3*x^3*e/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c
^2*x^2 + 1) + 1)^3) + 2*b*c^2*x^2*arcsin(c*x)*e/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 +
1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^2) - 1/2*a*c^2*d/(c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2
+ 1) + 1)) + 2*a*c^2*x^2*e/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x
^2 + 1) + 1)^2) + b*c*x*e/((c^4*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^2*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^
2 + 1) + 1))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 79, normalized size = 1.20 \[ c \left (\frac {a \left (c e x -\frac {c d}{x}\right )}{c^{2}}+\frac {b \left (\arcsin \left (c x \right ) e c x -\frac {\arcsin \left (c x \right ) c d}{x}+e \sqrt {-c^{2} x^{2}+1}-c^{2} d \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 79, normalized size = 1.20 \[ -{\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b d + a e x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b e}{c} - \frac {a d}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.36, size = 70, normalized size = 1.06 \[ \frac {b\,e\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )}{c}-\frac {b\,d\,\mathrm {asin}\left (c\,x\right )}{x}-b\,c\,d\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-c^2\,x^2}}\right )-\frac {a\,\left (d-e\,x^2\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e*((1 - c^2*x^2)^(1/2) + c*x*asin(c*x)))/c - (b*d*asin(c*x))/x - b*c*d*atanh(1/(1 - c^2*x^2)^(1/2)) - (a*(d
- e*x^2))/x

________________________________________________________________________________________

sympy [A]  time = 3.94, size = 75, normalized size = 1.14 \[ - \frac {a d}{x} + a e x + b c d \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d \operatorname {asin}{\left (c x \right )}}{x} + b e \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*d/x + a*e*x + b*c*d*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)), True)) - b*d*asin(
c*x)/x + b*e*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True))

________________________________________________________________________________________