Integrand size = 23, antiderivative size = 117 \[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=\frac {4 b (e (c+d x))^{3/2} \sqrt {1-(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} (a+b \arcsin (c+d x))}{5 d e}+\frac {12 b e \sqrt {e (c+d x)} E\left (\left .\arcsin \left (\frac {\sqrt {1-c-d x}}{\sqrt {2}}\right )\right |2\right )}{25 d \sqrt {c+d x}} \]
2/5*(e*(d*x+c))^(5/2)*(a+b*arcsin(d*x+c))/d/e+12/25*b*e*EllipticE(1/2*(-d* x-c+1)^(1/2)*2^(1/2),2^(1/2))*(e*(d*x+c))^(1/2)/d/(d*x+c)^(1/2)+4/25*b*(e* (d*x+c))^(3/2)*(1-(d*x+c)^2)^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=\frac {2 (e (c+d x))^{3/2} \left (5 a c+5 a d x+2 b \sqrt {1-(c+d x)^2}+5 b c \arcsin (c+d x)+5 b d x \arcsin (c+d x)-2 b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},(c+d x)^2\right )\right )}{25 d} \]
(2*(e*(c + d*x))^(3/2)*(5*a*c + 5*a*d*x + 2*b*Sqrt[1 - (c + d*x)^2] + 5*b* c*ArcSin[c + d*x] + 5*b*d*x*ArcSin[c + d*x] - 2*b*Hypergeometric2F1[1/2, 3 /4, 7/4, (c + d*x)^2]))/(25*d)
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5304, 5138, 262, 261, 259, 327}
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.
\(\displaystyle \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx\) |
\(\Big \downarrow \) 5304 |
\(\displaystyle \frac {\int (e (c+d x))^{3/2} (a+b \arcsin (c+d x))d(c+d x)}{d}\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {\frac {2 (e (c+d x))^{5/2} (a+b \arcsin (c+d x))}{5 e}-\frac {2 b \int \frac {(e (c+d x))^{5/2}}{\sqrt {1-(c+d x)^2}}d(c+d x)}{5 e}}{d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {2 (e (c+d x))^{5/2} (a+b \arcsin (c+d x))}{5 e}-\frac {2 b \left (\frac {3}{5} e^2 \int \frac {\sqrt {e (c+d x)}}{\sqrt {1-(c+d x)^2}}d(c+d x)-\frac {2}{5} e \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}\right )}{5 e}}{d}\) |
\(\Big \downarrow \) 261 |
\(\displaystyle \frac {\frac {2 (e (c+d x))^{5/2} (a+b \arcsin (c+d x))}{5 e}-\frac {2 b \left (\frac {3 e^2 \sqrt {e (c+d x)} \int \frac {\sqrt {c+d x}}{\sqrt {1-(c+d x)^2}}d(c+d x)}{5 \sqrt {c+d x}}-\frac {2}{5} e \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}\right )}{5 e}}{d}\) |
\(\Big \downarrow \) 259 |
\(\displaystyle \frac {\frac {2 (e (c+d x))^{5/2} (a+b \arcsin (c+d x))}{5 e}-\frac {2 b \left (-\frac {6 e^2 \sqrt {e (c+d x)} \int \frac {\sqrt {c+d x}}{\sqrt {\frac {1}{2} (c+d x-1)+1}}d\frac {\sqrt {-c-d x+1}}{\sqrt {2}}}{5 \sqrt {c+d x}}-\frac {2}{5} e \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}\right )}{5 e}}{d}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {\frac {2 (e (c+d x))^{5/2} (a+b \arcsin (c+d x))}{5 e}-\frac {2 b \left (-\frac {6 e^2 \sqrt {e (c+d x)} E\left (\left .\arcsin \left (\frac {\sqrt {-c-d x+1}}{\sqrt {2}}\right )\right |2\right )}{5 \sqrt {c+d x}}-\frac {2}{5} e \sqrt {1-(c+d x)^2} (e (c+d x))^{3/2}\right )}{5 e}}{d}\) |
((2*(e*(c + d*x))^(5/2)*(a + b*ArcSin[c + d*x]))/(5*e) - (2*b*((-2*e*(e*(c + d*x))^(3/2)*Sqrt[1 - (c + d*x)^2])/5 - (6*e^2*Sqrt[e*(c + d*x)]*Ellipti cE[ArcSin[Sqrt[1 - c - d*x]/Sqrt[2]], 2])/(5*Sqrt[c + d*x])))/(5*e))/d
3.3.83.3.1 Defintions of rubi rules used
Int[Sqrt[x_]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[-2/(Sqrt[a]*(-b/a )^(3/4)) Subst[Int[Sqrt[1 - 2*x^2]/Sqrt[1 - x^2], x], x, Sqrt[1 - Sqrt[-b /a]*x]/Sqrt[2]], x] /; FreeQ[{a, b}, x] && GtQ[-b/a, 0] && GtQ[a, 0]
Int[Sqrt[(c_)*(x_)]/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[Sqrt[c*x]/ Sqrt[x] Int[Sqrt[x]/Sqrt[a + b*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ [-b/a, 0]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 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_) + (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]
Result contains complex when optimal does not.
Time = 2.45 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.66
method | result | size |
derivativedivides | \(\frac {\frac {2 a \left (d e x +c e \right )^{\frac {5}{2}}}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \arcsin \left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (-\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 e^{3} \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {1+\frac {d e x +c e}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )\right )}{5 \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e}\) | \(194\) |
default | \(\frac {\frac {2 a \left (d e x +c e \right )^{\frac {5}{2}}}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \arcsin \left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (-\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 e^{3} \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {1+\frac {d e x +c e}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )\right )}{5 \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e}\) | \(194\) |
parts | \(\frac {2 a \left (d e x +c e \right )^{\frac {5}{2}}}{5 d e}+\frac {2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \arcsin \left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (-\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 e^{3} \sqrt {1-\frac {d e x +c e}{e}}\, \sqrt {1+\frac {d e x +c e}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {1}{e}}, i\right )\right )}{5 \sqrt {\frac {1}{e}}\, \sqrt {-\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{e d}\) | \(199\) |
2/d/e*(1/5*a*(d*e*x+c*e)^(5/2)+b*(1/5*(d*e*x+c*e)^(5/2)*arcsin(1/e*(d*e*x+ c*e))-2/5/e*(-1/5*e^2*(d*e*x+c*e)^(3/2)*(-1/e^2*(d*e*x+c*e)^2+1)^(1/2)-3/5 *e^3/(1/e)^(1/2)*(1-1/e*(d*e*x+c*e))^(1/2)*(1+1/e*(d*e*x+c*e))^(1/2)/(-1/e ^2*(d*e*x+c*e)^2+1)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(1/e)^(1/2),I)-Elli pticE((d*e*x+c*e)^(1/2)*(1/e)^(1/2),I)))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.30 \[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=-\frac {2 \, {\left (6 \, \sqrt {-d^{3} e} b e {\rm weierstrassZeta}\left (\frac {4}{d^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right ) - {\left (5 \, a d^{3} e x^{2} + 10 \, a c d^{2} e x + 5 \, a c^{2} d e + 5 \, {\left (b d^{3} e x^{2} + 2 \, b c d^{2} e x + b c^{2} d e\right )} \arcsin \left (d x + c\right ) + 2 \, {\left (b d^{2} e x + b c d e\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}\right )} \sqrt {d e x + c e}\right )}}{25 \, d^{2}} \]
-2/25*(6*sqrt(-d^3*e)*b*e*weierstrassZeta(4/d^2, 0, weierstrassPInverse(4/ d^2, 0, (d*x + c)/d)) - (5*a*d^3*e*x^2 + 10*a*c*d^2*e*x + 5*a*c^2*d*e + 5* (b*d^3*e*x^2 + 2*b*c*d^2*e*x + b*c^2*d*e)*arcsin(d*x + c) + 2*(b*d^2*e*x + b*c*d*e)*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1))*sqrt(d*e*x + c*e))/d^2
\[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=\int \left (e \left (c + d x\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c + d x \right )}\right )\, dx \]
Exception generated. \[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=\int { {\left (d e x + c e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (d x + c\right ) + a\right )} \,d x } \]
Timed out. \[ \int (c e+d e x)^{3/2} (a+b \arcsin (c+d x)) \, dx=\int {\left (c\,e+d\,e\,x\right )}^{3/2}\,\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right ) \,d x \]