Integrand size = 25, antiderivative size = 187 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=-\frac {b}{2 c^5 d^2 \sqrt {1-c^2 x^2}}+\frac {b \sqrt {1-c^2 x^2}}{c^5 d^2}+\frac {3 x (a+b \arcsin (c x))}{2 c^4 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {3 i (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c^5 d^2}-\frac {3 i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{2 c^5 d^2}+\frac {3 i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{2 c^5 d^2} \] Output:
-1/2*b/c^5/d^2/(-c^2*x^2+1)^(1/2)+b*(-c^2*x^2+1)^(1/2)/c^5/d^2+3/2*x*(a+b* arcsin(c*x))/c^4/d^2+1/2*x^3*(a+b*arcsin(c*x))/c^2/d^2/(-c^2*x^2+1)+3*I*(a +b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c^5/d^2-3/2*I*b*polylog(2 ,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d^2+3/2*I*b*polylog(2,I*(I*c*x+(-c^2*x ^2+1)^(1/2)))/c^5/d^2
Time = 0.45 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.78 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {4 a c x+4 b \sqrt {1-c^2 x^2}+\frac {b \sqrt {1-c^2 x^2}}{-1+c x}-\frac {b \sqrt {1-c^2 x^2}}{1+c x}-\frac {2 a c x}{-1+c^2 x^2}+3 i b \pi \arcsin (c x)+4 b c x \arcsin (c x)+\frac {b \arcsin (c x)}{1-c x}-\frac {b \arcsin (c x)}{1+c x}-3 b \pi \log \left (1-i e^{i \arcsin (c x)}\right )-6 b \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )-3 b \pi \log \left (1+i e^{i \arcsin (c x)}\right )+6 b \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )+3 a \log (1-c x)-3 a \log (1+c x)+3 b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+3 b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-6 i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )+6 i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{4 c^5 d^2} \] Input:
Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^2,x]
Output:
(4*a*c*x + 4*b*Sqrt[1 - c^2*x^2] + (b*Sqrt[1 - c^2*x^2])/(-1 + c*x) - (b*S qrt[1 - c^2*x^2])/(1 + c*x) - (2*a*c*x)/(-1 + c^2*x^2) + (3*I)*b*Pi*ArcSin [c*x] + 4*b*c*x*ArcSin[c*x] + (b*ArcSin[c*x])/(1 - c*x) - (b*ArcSin[c*x])/ (1 + c*x) - 3*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 6*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 3*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 6*b*ArcSin[c* x]*Log[1 + I*E^(I*ArcSin[c*x])] + 3*a*Log[1 - c*x] - 3*a*Log[1 + c*x] + 3* b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 3*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x ])/4]] - (6*I)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (6*I)*b*PolyLog[2, I *E^(I*ArcSin[c*x])])/(4*c^5*d^2)
Time = 0.94 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {5206, 27, 243, 53, 2009, 5210, 241, 5164, 3042, 4669, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 5206 |
\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{d \left (1-c^2 x^2\right )}dx}{2 c^2 d}-\frac {b \int \frac {x^3}{\left (1-c^2 x^2\right )^{3/2}}dx}{2 c d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{1-c^2 x^2}dx}{2 c^2 d^2}-\frac {b \int \frac {x^3}{\left (1-c^2 x^2\right )^{3/2}}dx}{2 c d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{1-c^2 x^2}dx}{2 c^2 d^2}-\frac {b \int \frac {x^2}{\left (1-c^2 x^2\right )^{3/2}}dx^2}{4 c d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{1-c^2 x^2}dx}{2 c^2 d^2}-\frac {b \int \left (\frac {1}{c^2 \left (1-c^2 x^2\right )^{3/2}}-\frac {1}{c^2 \sqrt {1-c^2 x^2}}\right )dx^2}{4 c d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 \int \frac {x^2 (a+b \arcsin (c x))}{1-c^2 x^2}dx}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 5210 |
\(\displaystyle -\frac {3 \left (\frac {\int \frac {a+b \arcsin (c x)}{1-c^2 x^2}dx}{c^2}+\frac {b \int \frac {x}{\sqrt {1-c^2 x^2}}dx}{c}-\frac {x (a+b \arcsin (c x))}{c^2}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle -\frac {3 \left (\frac {\int \frac {a+b \arcsin (c x)}{1-c^2 x^2}dx}{c^2}-\frac {x (a+b \arcsin (c x))}{c^2}-\frac {b \sqrt {1-c^2 x^2}}{c^3}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 5164 |
\(\displaystyle -\frac {3 \left (\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}d\arcsin (c x)}{c^3}-\frac {x (a+b \arcsin (c x))}{c^2}-\frac {b \sqrt {1-c^2 x^2}}{c^3}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 \left (\frac {\int (a+b \arcsin (c x)) \csc \left (\arcsin (c x)+\frac {\pi }{2}\right )d\arcsin (c x)}{c^3}-\frac {x (a+b \arcsin (c x))}{c^2}-\frac {b \sqrt {1-c^2 x^2}}{c^3}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 4669 |
\(\displaystyle -\frac {3 \left (\frac {-b \int \log \left (1-i e^{i \arcsin (c x)}\right )d\arcsin (c x)+b \int \log \left (1+i e^{i \arcsin (c x)}\right )d\arcsin (c x)-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c^3}-\frac {x (a+b \arcsin (c x))}{c^2}-\frac {b \sqrt {1-c^2 x^2}}{c^3}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -\frac {3 \left (\frac {i b \int e^{-i \arcsin (c x)} \log \left (1-i e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-i b \int e^{-i \arcsin (c x)} \log \left (1+i e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c^3}-\frac {x (a+b \arcsin (c x))}{c^2}-\frac {b \sqrt {1-c^2 x^2}}{c^3}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -\frac {3 \left (\frac {-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))+i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^3}-\frac {x (a+b \arcsin (c x))}{c^2}-\frac {b \sqrt {1-c^2 x^2}}{c^3}\right )}{2 c^2 d^2}+\frac {x^3 (a+b \arcsin (c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \left (\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {2}{c^4 \sqrt {1-c^2 x^2}}\right )}{4 c d^2}\) |
Input:
Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^2,x]
Output:
-1/4*(b*(2/(c^4*Sqrt[1 - c^2*x^2]) + (2*Sqrt[1 - c^2*x^2])/c^4))/(c*d^2) + (x^3*(a + b*ArcSin[c*x]))/(2*c^2*d^2*(1 - c^2*x^2)) - (3*(-((b*Sqrt[1 - c ^2*x^2])/c^3) - (x*(a + b*ArcSin[c*x]))/c^2 + ((-2*I)*(a + b*ArcSin[c*x])* ArcTan[E^(I*ArcSin[c*x])] + I*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - I*b*P olyLog[2, I*E^(I*ArcSin[c*x])])/c^3))/(2*c^2*d^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol ] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si mp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x ))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> Simp[1/(c*d) Subst[Int[(a + b*x)^n*Sec[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + (-Simp[f^2*((m - 1)/(2*e*(p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Simp [b*f*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{ a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && IG tQ[m, 1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + S imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f* x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m , 1] && NeQ[m + 2*p + 1, 0]
Time = 0.47 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.39
method | result | size |
derivativedivides | \(\frac {\frac {a \left (c x -\frac {1}{4 \left (c x +1\right )}-\frac {3 \ln \left (c x +1\right )}{4}-\frac {1}{4 \left (c x -1\right )}+\frac {3 \ln \left (c x -1\right )}{4}\right )}{d^{2}}+\frac {b \sqrt {-c^{2} x^{2}+1}}{d^{2}}+\frac {b \arcsin \left (c x \right ) c x}{d^{2}}-\frac {b \arcsin \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 i b \operatorname {dilog}\left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {3 i b \operatorname {dilog}\left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}}{c^{5}}\) | \(259\) |
default | \(\frac {\frac {a \left (c x -\frac {1}{4 \left (c x +1\right )}-\frac {3 \ln \left (c x +1\right )}{4}-\frac {1}{4 \left (c x -1\right )}+\frac {3 \ln \left (c x -1\right )}{4}\right )}{d^{2}}+\frac {b \sqrt {-c^{2} x^{2}+1}}{d^{2}}+\frac {b \arcsin \left (c x \right ) c x}{d^{2}}-\frac {b \arcsin \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 i b \operatorname {dilog}\left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {3 i b \operatorname {dilog}\left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}}{c^{5}}\) | \(259\) |
parts | \(\frac {a \left (\frac {x}{c^{4}}-\frac {1}{4 c^{5} \left (c x -1\right )}+\frac {3 \ln \left (c x -1\right )}{4 c^{5}}-\frac {1}{4 c^{5} \left (c x +1\right )}-\frac {3 \ln \left (c x +1\right )}{4 c^{5}}\right )}{d^{2}}+\frac {b \sqrt {-c^{2} x^{2}+1}}{c^{5} d^{2}}+\frac {b \arcsin \left (c x \right ) x}{d^{2} c^{4}}-\frac {b \arcsin \left (c x \right ) x}{2 d^{2} c^{4} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} c^{5} \left (c^{2} x^{2}-1\right )}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2} c^{5}}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2} c^{5}}-\frac {3 i b \operatorname {dilog}\left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2} c^{5}}+\frac {3 i b \operatorname {dilog}\left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2} c^{5}}\) | \(291\) |
Input:
int(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x,method=_RETURNVERBOSE)
Output:
1/c^5*(a/d^2*(c*x-1/4/(c*x+1)-3/4*ln(c*x+1)-1/4/(c*x-1)+3/4*ln(c*x-1))+b/d ^2*(-c^2*x^2+1)^(1/2)+b/d^2*arcsin(c*x)*c*x-1/2*b/d^2/(c^2*x^2-1)*arcsin(c *x)*c*x+1/2*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+3/2*b/d^2*arcsin(c*x)*ln( 1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/2*b/d^2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2* x^2+1)^(1/2)))-3/2*I*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/2*I*b/d ^2*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2))))
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \] Input:
integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="fricas")
Output:
integral((b*x^4*arcsin(c*x) + a*x^4)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {\int \frac {a x^{4}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{4} \operatorname {asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \] Input:
integrate(x**4*(a+b*asin(c*x))/(-c**2*d*x**2+d)**2,x)
Output:
(Integral(a*x**4/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*x**4*asin( c*x)/(c**4*x**4 - 2*c**2*x**2 + 1), x))/d**2
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \] Input:
integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="maxima")
Output:
-1/4*a*(2*x/(c^6*d^2*x^2 - c^4*d^2) - 4*x/(c^4*d^2) + 3*log(c*x + 1)/(c^5* d^2) - 3*log(c*x - 1)/(c^5*d^2)) - 1/4*(3*(c^2*x^2 - 1)*arctan2(c*x, sqrt( c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(c^2*x^2 - 1)*arctan2(c*x, sqrt( c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(2*c^3*x^3 - 3*c*x)*arctan2(c*x , sqrt(c*x + 1)*sqrt(-c*x + 1)) + 4*(c^7*d^2*x^2 - c^5*d^2)*integrate(-1/4 *(4*c^3*x^3 - 6*c*x - 3*(c^2*x^2 - 1)*log(c*x + 1) + 3*(c^2*x^2 - 1)*log(- c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^8*d^2*x^4 - 2*c^6*d^2*x^2 + c^4* d^2), x))*b/(c^7*d^2*x^2 - c^5*d^2)
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} - d\right )}^{2}} \,d x } \] Input:
integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="giac")
Output:
integrate((b*arcsin(c*x) + a)*x^4/(c^2*d*x^2 - d)^2, x)
Timed out. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^2} \,d x \] Input:
int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2)^2,x)
Output:
int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2)^2, x)
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^2} \, dx=\frac {4 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{4}}{c^{4} x^{4}-2 c^{2} x^{2}+1}d x \right ) b \,c^{7} x^{2}-4 \left (\int \frac {\mathit {asin} \left (c x \right ) x^{4}}{c^{4} x^{4}-2 c^{2} x^{2}+1}d x \right ) b \,c^{5}+3 \,\mathrm {log}\left (c^{2} x -c \right ) a \,c^{2} x^{2}-3 \,\mathrm {log}\left (c^{2} x -c \right ) a -3 \,\mathrm {log}\left (c^{2} x +c \right ) a \,c^{2} x^{2}+3 \,\mathrm {log}\left (c^{2} x +c \right ) a +4 a \,c^{3} x^{3}-6 a c x}{4 c^{5} d^{2} \left (c^{2} x^{2}-1\right )} \] Input:
int(x^4*(a+b*asin(c*x))/(-c^2*d*x^2+d)^2,x)
Output:
(4*int((asin(c*x)*x**4)/(c**4*x**4 - 2*c**2*x**2 + 1),x)*b*c**7*x**2 - 4*i nt((asin(c*x)*x**4)/(c**4*x**4 - 2*c**2*x**2 + 1),x)*b*c**5 + 3*log(c**2*x - c)*a*c**2*x**2 - 3*log(c**2*x - c)*a - 3*log(c**2*x + c)*a*c**2*x**2 + 3*log(c**2*x + c)*a + 4*a*c**3*x**3 - 6*a*c*x)/(4*c**5*d**2*(c**2*x**2 - 1 ))