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

Optimal. Leaf size=128 \[ c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+b c^3 d^2 \sqrt{1-c^2 x^2}-\frac{b c d^2 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{11}{6} b c^3 d^2 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right ) \]

[Out]

b*c^3*d^2*Sqrt[1 - c^2*x^2] - (b*c*d^2*Sqrt[1 - c^2*x^2])/(6*x^2) - (d^2*(a + b*ArcSin[c*x]))/(3*x^3) + (2*c^2
*d^2*(a + b*ArcSin[c*x]))/x + c^4*d^2*x*(a + b*ArcSin[c*x]) + (11*b*c^3*d^2*ArcTanh[Sqrt[1 - c^2*x^2]])/6

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Rubi [A]  time = 0.161704, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {270, 4687, 12, 1251, 897, 1157, 388, 208} \[ c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+b c^3 d^2 \sqrt{1-c^2 x^2}-\frac{b c d^2 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{11}{6} b c^3 d^2 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*c^3*d^2*Sqrt[1 - c^2*x^2] - (b*c*d^2*Sqrt[1 - c^2*x^2])/(6*x^2) - (d^2*(a + b*ArcSin[c*x]))/(3*x^3) + (2*c^2
*d^2*(a + b*ArcSin[c*x]))/x + c^4*d^2*x*(a + b*ArcSin[c*x]) + (11*b*c^3*d^2*ArcTanh[Sqrt[1 - c^2*x^2]])/6

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 4687

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = I
ntHide[(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] && EqQ[c^2*d + e, 0] && IGtQ[p, 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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

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

Rubi steps

\begin{align*} \int \frac{\left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{d^2 \left (-1+6 c^2 x^2+3 c^4 x^4\right )}{3 x^3 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} \left (b c d^2\right ) \int \frac{-1+6 c^2 x^2+3 c^4 x^4}{x^3 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{-1+6 c^2 x+3 c^4 x^2}{x^2 \sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{8-12 x^2+3 x^4}{\left (\frac{1}{c^2}-\frac{x^2}{c^2}\right )^2} \, dx,x,\sqrt{1-c^2 x^2}\right )}{3 c}\\ &=-\frac{b c d^2 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} \left (b c d^2\right ) \operatorname{Subst}\left (\int \frac{-17+6 x^2}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )\\ &=b c^3 d^2 \sqrt{1-c^2 x^2}-\frac{b c d^2 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{6} \left (11 b c d^2\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 )\\ &=b c^3 d^2 \sqrt{1-c^2 x^2}-\frac{b c d^2 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{d^2 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{2 c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )}{x}+c^4 d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{11}{6} b c^3 d^2 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0932271, size = 136, normalized size = 1.06 \[ \frac{d^2 \left (6 a c^4 x^4+12 a c^2 x^2-2 a+6 b c^3 x^3 \sqrt{1-c^2 x^2}-b c x \sqrt{1-c^2 x^2}-11 b c^3 x^3 \log (x)+11 b c^3 x^3 \log \left (\sqrt{1-c^2 x^2}+1\right )+2 b \left (3 c^4 x^4+6 c^2 x^2-1\right ) \sin ^{-1}(c x)\right )}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(-2*a + 12*a*c^2*x^2 + 6*a*c^4*x^4 - b*c*x*Sqrt[1 - c^2*x^2] + 6*b*c^3*x^3*Sqrt[1 - c^2*x^2] + 2*b*(-1 +
6*c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x] - 11*b*c^3*x^3*Log[x] + 11*b*c^3*x^3*Log[1 + Sqrt[1 - c^2*x^2]]))/(6*x^3)

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Maple [A]  time = 0.01, size = 115, normalized size = 0.9 \begin{align*}{c}^{3} \left ({d}^{2}a \left ( cx+2\,{\frac{1}{cx}}-{\frac{1}{3\,{c}^{3}{x}^{3}}} \right ) +{d}^{2}b \left ( cx\arcsin \left ( cx \right ) +2\,{\frac{\arcsin \left ( cx \right ) }{cx}}-{\frac{\arcsin \left ( cx \right ) }{3\,{c}^{3}{x}^{3}}}+\sqrt{-{c}^{2}{x}^{2}+1}+{\frac{11}{6}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) }-{\frac{1}{6\,{c}^{2}{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(d^2*a*(c*x+2/c/x-1/3/c^3/x^3)+d^2*b*(c*x*arcsin(c*x)+2/c/x*arcsin(c*x)-1/3/c^3/x^3*arcsin(c*x)+(-c^2*x^2+
1)^(1/2)+11/6*arctanh(1/(-c^2*x^2+1)^(1/2))-1/6/c^2/x^2*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.58534, size = 230, normalized size = 1.8 \begin{align*} a c^{4} d^{2} x +{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} b c^{3} d^{2} + 2 \,{\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 c^{2} d^{2} - \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^{2} + \frac{2 \, a c^{2} d^{2}}{x} - \frac{a d^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.52897, size = 358, normalized size = 2.8 \begin{align*} \frac{12 \, a c^{4} d^{2} x^{4} + 11 \, b c^{3} d^{2} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} + 1\right ) - 11 \, b c^{3} d^{2} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} - 1\right ) + 24 \, a c^{2} d^{2} x^{2} - 4 \, a d^{2} + 4 \,{\left (3 \, b c^{4} d^{2} x^{4} + 6 \, b c^{2} d^{2} x^{2} - b d^{2}\right )} \arcsin \left (c x\right ) + 2 \,{\left (6 \, b c^{3} d^{2} x^{3} - b c d^{2} x\right )} \sqrt{-c^{2} x^{2} + 1}}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a*c^4*d^2*x^4 + 11*b*c^3*d^2*x^3*log(sqrt(-c^2*x^2 + 1) + 1) - 11*b*c^3*d^2*x^3*log(sqrt(-c^2*x^2 + 1
) - 1) + 24*a*c^2*d^2*x^2 - 4*a*d^2 + 4*(3*b*c^4*d^2*x^4 + 6*b*c^2*d^2*x^2 - b*d^2)*arcsin(c*x) + 2*(6*b*c^3*d
^2*x^3 - b*c*d^2*x)*sqrt(-c^2*x^2 + 1))/x^3

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Sympy [A]  time = 12.1423, size = 235, normalized size = 1.84 \begin{align*} a c^{4} d^{2} x + \frac{2 a c^{2} d^{2}}{x} - \frac{a d^{2}}{3 x^{3}} + b c^{4} d^{2} \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 ) - 2 b c^{3} d^{2} \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{2 b c^{2} d^{2} \operatorname{asin}{\left (c x \right )}}{x} + \frac{b c d^{2} \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} - \frac{b d^{2} \operatorname{asin}{\left (c x \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**4*d**2*x + 2*a*c**2*d**2/x - a*d**2/(3*x**3) + b*c**4*d**2*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-
c**2*x**2 + 1)/c, True)) - 2*b*c**3*d**2*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)),
True)) + 2*b*c**2*d**2*asin(c*x)/x + b*c*d**2*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*d**2*asin(c*x)/(3*x**3)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x