Integrand size = 23, antiderivative size = 223 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=-\frac {4 b \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{d e (1+c+d x)}+\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{d e}+\frac {4 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}}-\frac {2 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{d \sqrt {e} \sqrt {1+(c+d x)^2}} \]
2*(a+b*arcsinh(d*x+c))*(e*(d*x+c))^(1/2)/d/e-4*b*(e*(d*x+c))^(1/2)*(1+(d*x +c)^2)^(1/2)/d/e/(d*x+c+1)+4*b*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2)/e ^(1/2)))^2)^(1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*EllipticE(sin(2 *arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*((1+(d*x+c)^2)/(d*x+c+1)^ 2)^(1/2)/d/e^(1/2)/(1+(d*x+c)^2)^(1/2)-2*b*(d*x+c+1)*(cos(2*arctan((e*(d*x +c))^(1/2)/e^(1/2)))^2)^(1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*Ell ipticF(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*((1+(d*x+c)^2 )/(d*x+c+1)^2)^(1/2)/d/e^(1/2)/(1+(d*x+c)^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.27 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=-\frac {2 \sqrt {e (c+d x)} \left (-3 (a+b \text {arcsinh}(c+d x))+2 b (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-(c+d x)^2\right )\right )}{3 d e} \]
(-2*Sqrt[e*(c + d*x)]*(-3*(a + b*ArcSinh[c + d*x]) + 2*b*(c + d*x)*Hyperge ometric2F1[1/2, 3/4, 7/4, -(c + d*x)^2]))/(3*d*e)
Time = 0.41 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6274, 6191, 266, 834, 27, 761, 1510}
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.
| \(\displaystyle \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx\) |
| \(\Big \downarrow \) 6274 |
| \(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {e (c+d x)}}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 6191 |
| \(\displaystyle \frac {\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{e}-\frac {2 b \int \frac {\sqrt {e (c+d x)}}{\sqrt {(c+d x)^2+1}}d(c+d x)}{e}}{d}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{e}-\frac {4 b \int \frac {e (c+d x)}{\sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}}{e^2}}{d}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{e}-\frac {4 b \left (e \int \frac {1}{\sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}-e \int \frac {e-e (c+d x)}{e \sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}\right )}{e^2}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{e}-\frac {4 b \left (e \int \frac {1}{\sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}-\int \frac {e-e (c+d x)}{\sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}\right )}{e^2}}{d}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{e}-\frac {4 b \left (\frac {\sqrt {e} (e (c+d x)+e) \sqrt {\frac {e^2 (c+d x)^2+e^2}{(e (c+d x)+e)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{2 \sqrt {(c+d x)^2+1}}-\int \frac {e-e (c+d x)}{\sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}\right )}{e^2}}{d}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {\frac {2 \sqrt {e (c+d x)} (a+b \text {arcsinh}(c+d x))}{e}-\frac {4 b \left (\frac {\sqrt {e} (e (c+d x)+e) \sqrt {\frac {e^2 (c+d x)^2+e^2}{(e (c+d x)+e)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{2 \sqrt {(c+d x)^2+1}}-\frac {\sqrt {e} (e (c+d x)+e) \sqrt {\frac {e^2 (c+d x)^2+e^2}{(e (c+d x)+e)^2}} E\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{\sqrt {(c+d x)^2+1}}+\frac {e^2 \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{e (c+d x)+e}\right )}{e^2}}{d}\) |
((2*Sqrt[e*(c + d*x)]*(a + b*ArcSinh[c + d*x]))/e - (4*b*((e^2*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(e + e*(c + d*x)) - (Sqrt[e]*(e + e*(c + d*x ))*Sqrt[(e^2 + e^2*(c + d*x)^2)/(e + e*(c + d*x))^2]*EllipticE[2*ArcTan[Sq rt[e*(c + d*x)]/Sqrt[e]], 1/2])/Sqrt[1 + (c + d*x)^2] + (Sqrt[e]*(e + e*(c + d*x))*Sqrt[(e^2 + e^2*(c + d*x)^2)/(e + e*(c + d*x))^2]*EllipticF[2*Arc Tan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(2*Sqrt[1 + (c + d*x)^2])))/e^2)/d
3.3.32.3.1 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^n/(d*(m + 1))), x] - Simp[b*c* (n/(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSinh[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_.) + ArcSinh[(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* ArcSinh[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 = 0.56 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.72
| method | result | size |
| derivativedivides | \(\frac {2 a \sqrt {d e x +c e}+2 b \left (\sqrt {d e x +c e}\, \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )-\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{\sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\) | \(161\) |
| default | \(\frac {2 a \sqrt {d e x +c e}+2 b \left (\sqrt {d e x +c e}\, \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )-\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{\sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\) | \(161\) |
| parts | \(\frac {2 a \sqrt {d e x +c e}}{d e}+\frac {2 b \left (\sqrt {d e x +c e}\, \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )-\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{\sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{d e}\) | \(167\) |
2/d/e*(a*(d*e*x+c*e)^(1/2)+b*((d*e*x+c*e)^(1/2)*arcsinh(1/e*(d*e*x+c*e))-2 *I/(I/e)^(1/2)*(1-I/e*(d*e*x+c*e))^(1/2)*(1+I/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)*(I/e)^(1/2),I)-Ellipti cE((d*e*x+c*e)^(1/2)*(I/e)^(1/2),I))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.42 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=\frac {2 \, {\left (\sqrt {d e x + c e} b d \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + \sqrt {d e x + c e} a d + 2 \, \sqrt {d^{3} e} b {\rm weierstrassZeta}\left (-\frac {4}{d^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right )\right )}}{d^{2} e} \]
2*(sqrt(d*e*x + c*e)*b*d*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + sqrt(d*e*x + c*e)*a*d + 2*sqrt(d^3*e)*b*weierstrassZeta(-4/d^2, 0, weier strassPInverse(-4/d^2, 0, (d*x + c)/d)))/(d^2*e)
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c + d x \right )}}{\sqrt {e \left (c + d x\right )}}\, dx \]
Exception generated. \[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=\int { \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{\sqrt {d e x + c e}} \,d x } \]
Timed out. \[ \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {c e+d e x}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{\sqrt {c\,e+d\,e\,x}} \,d x \]