Integrand size = 23, antiderivative size = 187 \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=-\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{3 d e^4 (c+d x)^2}-\frac {(a+b \arcsin (c+d x))^2}{3 d e^4 (c+d x)^3}-\frac {2 b (a+b \arcsin (c+d x)) \text {arctanh}\left (e^{i \arcsin (c+d x)}\right )}{3 d e^4}+\frac {i b^2 \operatorname {PolyLog}\left (2,-e^{i \arcsin (c+d x)}\right )}{3 d e^4}-\frac {i b^2 \operatorname {PolyLog}\left (2,e^{i \arcsin (c+d x)}\right )}{3 d e^4} \]
-1/3*b^2/d/e^4/(d*x+c)-1/3*(a+b*arcsin(d*x+c))^2/d/e^4/(d*x+c)^3-2/3*b*(a+ b*arcsin(d*x+c))*arctanh(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^4+1/3*I*b^2*po lylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))/d/e^4-1/3*I*b^2*polylog(2,I*(d*x+c )+(1-(d*x+c)^2)^(1/2))/d/e^4-1/3*b*(a+b*arcsin(d*x+c))*(1-(d*x+c)^2)^(1/2) /d/e^4/(d*x+c)^2
Time = 2.53 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=-\frac {4 a^2+8 a b \arcsin (c+d x)-4 i b^2 (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{i \arcsin (c+d x)}\right )+2 a b \sin (2 \arcsin (c+d x))+a b \left (\log \left (\cos \left (\frac {1}{2} \arcsin (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} \arcsin (c+d x)\right )\right )\right ) (3 (c+d x)-\sin (3 \arcsin (c+d x)))+b^2 \left (4 (c+d x)^2+4 \arcsin (c+d x)^2+4 i (c+d x)^3 \operatorname {PolyLog}\left (2,e^{i \arcsin (c+d x)}\right )+\arcsin (c+d x) \left (2 \sin (2 \arcsin (c+d x))+\left (\log \left (1-e^{i \arcsin (c+d x)}\right )-\log \left (1+e^{i \arcsin (c+d x)}\right )\right ) (-3 (c+d x)+\sin (3 \arcsin (c+d x)))\right )\right )}{12 d e^4 (c+d x)^3} \]
-1/12*(4*a^2 + 8*a*b*ArcSin[c + d*x] - (4*I)*b^2*(c + d*x)^3*PolyLog[2, -E ^(I*ArcSin[c + d*x])] + 2*a*b*Sin[2*ArcSin[c + d*x]] + a*b*(Log[Cos[ArcSin [c + d*x]/2]] - Log[Sin[ArcSin[c + d*x]/2]])*(3*(c + d*x) - Sin[3*ArcSin[c + d*x]]) + b^2*(4*(c + d*x)^2 + 4*ArcSin[c + d*x]^2 + (4*I)*(c + d*x)^3*P olyLog[2, E^(I*ArcSin[c + d*x])] + ArcSin[c + d*x]*(2*Sin[2*ArcSin[c + d*x ]] + (Log[1 - E^(I*ArcSin[c + d*x])] - Log[1 + E^(I*ArcSin[c + d*x])])*(-3 *(c + d*x) + Sin[3*ArcSin[c + d*x]]))))/(d*e^4*(c + d*x)^3)
Time = 0.68 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.80, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5304, 27, 5138, 5204, 15, 5218, 3042, 4671, 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.
\(\displaystyle \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx\) |
\(\Big \downarrow \) 5304 |
\(\displaystyle \frac {\int \frac {(a+b \arcsin (c+d x))^2}{e^4 (c+d x)^4}d(c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(a+b \arcsin (c+d x))^2}{(c+d x)^4}d(c+d x)}{d e^4}\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {\frac {2}{3} b \int \frac {a+b \arcsin (c+d x)}{(c+d x)^3 \sqrt {1-(c+d x)^2}}d(c+d x)-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}}{d e^4}\) |
\(\Big \downarrow \) 5204 |
\(\displaystyle \frac {\frac {2}{3} b \left (\frac {1}{2} \int \frac {a+b \arcsin (c+d x)}{(c+d x) \sqrt {1-(c+d x)^2}}d(c+d x)+\frac {1}{2} b \int \frac {1}{(c+d x)^2}d(c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}\right )-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}}{d e^4}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\frac {2}{3} b \left (\frac {1}{2} \int \frac {a+b \arcsin (c+d x)}{(c+d x) \sqrt {1-(c+d x)^2}}d(c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}-\frac {b}{2 (c+d x)}\right )-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}}{d e^4}\) |
\(\Big \downarrow \) 5218 |
\(\displaystyle \frac {\frac {2}{3} b \left (\frac {1}{2} \int \frac {a+b \arcsin (c+d x)}{c+d x}d\arcsin (c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}-\frac {b}{2 (c+d x)}\right )-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}}{d e^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2}{3} b \left (\frac {1}{2} \int (a+b \arcsin (c+d x)) \csc (\arcsin (c+d x))d\arcsin (c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}-\frac {b}{2 (c+d x)}\right )-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}}{d e^4}\) |
\(\Big \downarrow \) 4671 |
\(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}+\frac {2}{3} b \left (\frac {1}{2} \left (-b \int \log \left (1-e^{i \arcsin (c+d x)}\right )d\arcsin (c+d x)+b \int \log \left (1+e^{i \arcsin (c+d x)}\right )d\arcsin (c+d x)-2 \text {arctanh}\left (e^{i \arcsin (c+d x)}\right ) (a+b \arcsin (c+d x))\right )-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}-\frac {b}{2 (c+d x)}\right )}{d e^4}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}+\frac {2}{3} b \left (\frac {1}{2} \left (-i b \int e^{-i \arcsin (c+d x)} \log \left (1+e^{i \arcsin (c+d x)}\right )de^{i \arcsin (c+d x)}+i b \int e^{-i \arcsin (c+d x)} \log (-c-d x+1)de^{i \arcsin (c+d x)}-2 \text {arctanh}\left (e^{i \arcsin (c+d x)}\right ) (a+b \arcsin (c+d x))\right )-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}-\frac {b}{2 (c+d x)}\right )}{d e^4}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^2}{3 (c+d x)^3}+\frac {2}{3} b \left (\frac {1}{2} \left (-2 \text {arctanh}\left (e^{i \arcsin (c+d x)}\right ) (a+b \arcsin (c+d x))-i b \operatorname {PolyLog}\left (2,e^{i \arcsin (c+d x)}\right )+i b \operatorname {PolyLog}(2,-c-d x)\right )-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{2 (c+d x)^2}-\frac {b}{2 (c+d x)}\right )}{d e^4}\) |
(-1/3*(a + b*ArcSin[c + d*x])^2/(c + d*x)^3 + (2*b*(-1/2*b/(c + d*x) - (Sq rt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(2*(c + d*x)^2) + (-2*(a + b* ArcSin[c + d*x])*ArcTanh[E^(I*ArcSin[c + d*x])] - I*b*PolyLog[2, E^(I*ArcS in[c + d*x])] + I*b*PolyLog[2, -c - d*x])/2))/3)/(d*e^4)
3.2.96.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x )^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG tQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b *ArcSin[c*x])^n/(d*f*(m + 1))), x] + (Simp[c^2*((m + 2*p + 3)/(f^2*(m + 1)) ) Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Simp[b* c*(n/(f*(m + 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] && ILtQ[m, -1]
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)* (x_)^2], x_Symbol] :> Simp[(1/c^(m + 1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e* x^2]] Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a , b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.99 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.40
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2} \left (-\frac {\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}\, \left (d x +c \right )+\arcsin \left (d x +c \right )^{2}+\left (d x +c \right )^{2}}{3 \left (d x +c \right )^{3}}-\frac {\arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}+\frac {i \operatorname {polylog}\left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}+\frac {\arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}-\frac {i \operatorname {polylog}\left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}\right )}{e^{4}}+\frac {2 a b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\) | \(262\) |
default | \(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b^{2} \left (-\frac {\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}\, \left (d x +c \right )+\arcsin \left (d x +c \right )^{2}+\left (d x +c \right )^{2}}{3 \left (d x +c \right )^{3}}-\frac {\arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}+\frac {i \operatorname {polylog}\left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}+\frac {\arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}-\frac {i \operatorname {polylog}\left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}\right )}{e^{4}}+\frac {2 a b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\) | \(262\) |
parts | \(-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3} d}+\frac {b^{2} \left (-\frac {\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}\, \left (d x +c \right )+\arcsin \left (d x +c \right )^{2}+\left (d x +c \right )^{2}}{3 \left (d x +c \right )^{3}}-\frac {\arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}+\frac {i \operatorname {polylog}\left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}+\frac {\arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}-\frac {i \operatorname {polylog}\left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{3}\right )}{e^{4} d}+\frac {2 a b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4} d}\) | \(267\) |
1/d*(-1/3*a^2/e^4/(d*x+c)^3+b^2/e^4*(-1/3*(arcsin(d*x+c)*(1-(d*x+c)^2)^(1/ 2)*(d*x+c)+arcsin(d*x+c)^2+(d*x+c)^2)/(d*x+c)^3-1/3*arcsin(d*x+c)*ln(1+I*( d*x+c)+(1-(d*x+c)^2)^(1/2))+1/3*I*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2) )+1/3*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-1/3*I*polylog(2,I* (d*x+c)+(1-(d*x+c)^2)^(1/2)))+2*a*b/e^4*(-1/3/(d*x+c)^3*arcsin(d*x+c)-1/6/ (d*x+c)^2*(1-(d*x+c)^2)^(1/2)-1/6*arctanh(1/(1-(d*x+c)^2)^(1/2))))
\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}} \,d x } \]
integral((b^2*arcsin(d*x + c)^2 + 2*a*b*arcsin(d*x + c) + a^2)/(d^4*e^4*x^ 4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)
\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=\frac {\int \frac {a^{2}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {2 a b \operatorname {asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
(Integral(a**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d** 4*x**4), x) + Integral(b**2*asin(c + d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d **2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(2*a*b*asin(c + d*x)/( c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e** 4
\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}} \,d x } \]
-1/3*a^2/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 1 /3*(b^2*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 + 3*(d^4* e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)*integrate(2/3*((b ^2*d*x + b^2*c)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*arctan2(d*x + c, sqrt (d*x + c + 1)*sqrt(-d*x - c + 1)) - 3*(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 - a*b)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^6*e^4*x ^6 + 6*c*d^5*e^4*x^5 + (15*c^2 - 1)*d^4*e^4*x^4 + 4*(5*c^3 - c)*d^3*e^4*x^ 3 + 3*(5*c^4 - 2*c^2)*d^2*e^4*x^2 + 2*(3*c^5 - 2*c^3)*d*e^4*x + (c^6 - c^4 )*e^4), x))/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)
\[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c+d x))^2}{(c e+d e x)^4} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]