Integrand size = 21, antiderivative size = 88 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=-\frac {b \sqrt {1-(c+d x)^2}}{6 d e^4 (c+d x)^2}-\frac {a+b \arcsin (c+d x)}{3 d e^4 (c+d x)^3}-\frac {b \text {arctanh}\left (\sqrt {1-(c+d x)^2}\right )}{6 d e^4} \]
1/3*(-a-b*arcsin(d*x+c))/d/e^4/(d*x+c)^3-1/6*b*arctanh((1-(d*x+c)^2)^(1/2) )/d/e^4-1/6*b*(1-(d*x+c)^2)^(1/2)/d/e^4/(d*x+c)^2
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=-\frac {2 a+b (c+d x) \sqrt {1-(c+d x)^2}+2 b \arcsin (c+d x)+b (c+d x)^3 \text {arctanh}\left (\sqrt {1-(c+d x)^2}\right )}{6 d e^4 (c+d x)^3} \]
-1/6*(2*a + b*(c + d*x)*Sqrt[1 - (c + d*x)^2] + 2*b*ArcSin[c + d*x] + b*(c + d*x)^3*ArcTanh[Sqrt[1 - (c + d*x)^2]])/(d*e^4*(c + d*x)^3)
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5304, 27, 5138, 243, 52, 73, 219}
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.
| \(\displaystyle \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx\) |
| \(\Big \downarrow \) 5304 |
| \(\displaystyle \frac {\int \frac {a+b \arcsin (c+d x)}{e^4 (c+d x)^4}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a+b \arcsin (c+d x)}{(c+d x)^4}d(c+d x)}{d e^4}\) |
| \(\Big \downarrow \) 5138 |
| \(\displaystyle \frac {\frac {1}{3} b \int \frac {1}{(c+d x)^3 \sqrt {1-(c+d x)^2}}d(c+d x)-\frac {a+b \arcsin (c+d x)}{3 (c+d x)^3}}{d e^4}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {1}{6} b \int \frac {1}{\sqrt {-c-d x+1} (c+d x)^4}d(c+d x)^2-\frac {a+b \arcsin (c+d x)}{3 (c+d x)^3}}{d e^4}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {\frac {1}{6} b \left (\frac {1}{2} \int \frac {1}{\sqrt {-c-d x+1} (c+d x)^2}d(c+d x)^2-\frac {\sqrt {-c-d x+1}}{(c+d x)^2}\right )-\frac {a+b \arcsin (c+d x)}{3 (c+d x)^3}}{d e^4}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {1}{6} b \left (-\int \frac {1}{1-(c+d x)^4}d\sqrt {-c-d x+1}-\frac {\sqrt {-c-d x+1}}{(c+d x)^2}\right )-\frac {a+b \arcsin (c+d x)}{3 (c+d x)^3}}{d e^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{6} b \left (-\text {arctanh}\left (\sqrt {-c-d x+1}\right )-\frac {\sqrt {-c-d x+1}}{(c+d x)^2}\right )-\frac {a+b \arcsin (c+d x)}{3 (c+d x)^3}}{d e^4}\) |
(-1/3*(a + b*ArcSin[c + d*x])/(c + d*x)^3 + (b*(-(Sqrt[1 - c - d*x]/(c + d *x)^2) - ArcTanh[Sqrt[1 - c - d*x]]))/6)/(d*e^4)
3.2.85.3.1 Defintions of rubi rules used
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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[((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_) + (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 = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {-\frac {a}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\) | \(78\) |
| default | \(\frac {-\frac {a}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\) | \(78\) |
| parts | \(-\frac {a}{3 e^{4} \left (d x +c \right )^{3} d}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4} d}\) | \(80\) |
1/d*(-1/3*a/e^4/(d*x+c)^3+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))))
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (78) = 156\).
Time = 0.31 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.38 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=-\frac {4 \, b \arcsin \left (d x + c\right ) + {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} + 1\right ) - {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} - 1\right ) + 2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (b d x + b c\right )} + 4 \, a}{12 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]
-1/12*(4*b*arcsin(d*x + c) + (b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b* c^3)*log(sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1) + 1) - (b*d^3*x^3 + 3*b*c*d^2* x^2 + 3*b*c^2*d*x + b*c^3)*log(sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1) - 1) + 2 *sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(b*d*x + b*c) + 4*a)/(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)
\[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=\frac {\int \frac {a}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b \operatorname {asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
(Integral(a/(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*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
\[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=\int { \frac {b \arcsin \left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{4}} \,d x } \]
-1/3*(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)*inte grate(1/3*e^(1/2*log(d*x + c + 1) + 1/2*log(-d*x - c + 1))/(d^7*e^4*x^7 + 7*c*d^6*e^4*x^6 + (21*c^2 - 1)*d^5*e^4*x^5 + 5*(7*c^3 - c)*d^4*e^4*x^4 + 5 *(7*c^4 - 2*c^2)*d^3*e^4*x^3 + (21*c^5 - 10*c^3)*d^2*e^4*x^2 + (7*c^6 - 5* c^4)*d*e^4*x + (c^7 - c^5)*e^4 + (d^5*e^4*x^5 + 5*c*d^4*e^4*x^4 + (10*c^2 - 1)*d^3*e^4*x^3 + (10*c^3 - 3*c)*d^2*e^4*x^2 + (5*c^4 - 3*c^2)*d*e^4*x + (c^5 - c^3)*e^4)*e^(log(d*x + c + 1) + log(-d*x - c + 1))), x) + arctan2(d *x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))*b/(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*a/(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)
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (78) = 156\).
Time = 0.67 (sec) , antiderivative size = 388, normalized size of antiderivative = 4.41 \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=-\frac {{\left (d x + c\right )}^{3} b \arcsin \left (d x + c\right )}{24 \, d e^{4} {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3}} - \frac {{\left (d x + c\right )} b \arcsin \left (d x + c\right )}{8 \, d e^{4} {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac {b {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )} \arcsin \left (d x + c\right )}{8 \, {\left (d x + c\right )} d e^{4}} - \frac {b {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3} \arcsin \left (d x + c\right )}{24 \, {\left (d x + c\right )}^{3} d e^{4}} - \frac {b \log \left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}{6 \, d e^{4}} + \frac {b \log \left ({\left | d x + c \right |}\right )}{6 \, d e^{4}} - \frac {{\left (d x + c\right )}^{3} a}{24 \, d e^{4} {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3}} + \frac {{\left (d x + c\right )}^{2} b}{24 \, d e^{4} {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} - \frac {{\left (d x + c\right )} a}{8 \, d e^{4} {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac {a {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}}{8 \, {\left (d x + c\right )} d e^{4}} - \frac {b {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}}{24 \, {\left (d x + c\right )}^{2} d e^{4}} - \frac {a {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3}}{24 \, {\left (d x + c\right )}^{3} d e^{4}} \]
-1/24*(d*x + c)^3*b*arcsin(d*x + c)/(d*e^4*(sqrt(-(d*x + c)^2 + 1) + 1)^3) - 1/8*(d*x + c)*b*arcsin(d*x + c)/(d*e^4*(sqrt(-(d*x + c)^2 + 1) + 1)) - 1/8*b*(sqrt(-(d*x + c)^2 + 1) + 1)*arcsin(d*x + c)/((d*x + c)*d*e^4) - 1/2 4*b*(sqrt(-(d*x + c)^2 + 1) + 1)^3*arcsin(d*x + c)/((d*x + c)^3*d*e^4) - 1 /6*b*log(sqrt(-(d*x + c)^2 + 1) + 1)/(d*e^4) + 1/6*b*log(abs(d*x + c))/(d* e^4) - 1/24*(d*x + c)^3*a/(d*e^4*(sqrt(-(d*x + c)^2 + 1) + 1)^3) + 1/24*(d *x + c)^2*b/(d*e^4*(sqrt(-(d*x + c)^2 + 1) + 1)^2) - 1/8*(d*x + c)*a/(d*e^ 4*(sqrt(-(d*x + c)^2 + 1) + 1)) - 1/8*a*(sqrt(-(d*x + c)^2 + 1) + 1)/((d*x + c)*d*e^4) - 1/24*b*(sqrt(-(d*x + c)^2 + 1) + 1)^2/((d*x + c)^2*d*e^4) - 1/24*a*(sqrt(-(d*x + c)^2 + 1) + 1)^3/((d*x + c)^3*d*e^4)
Timed out. \[ \int \frac {a+b \arcsin (c+d x)}{(c e+d e x)^4} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]