Integrand size = 23, antiderivative size = 145 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=-\frac {4 b \sqrt {1+(c+d x)^2}}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 (a+b \text {arcsinh}(c+d x))}{5 d e (e (c+d x))^{5/2}}-\frac {2 b (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{15 d e^{7/2} \sqrt {1+(c+d x)^2}} \]
-2/5*(a+b*arcsinh(d*x+c))/d/e/(e*(d*x+c))^(5/2)-4/15*b*(1+(d*x+c)^2)^(1/2) /d/e^2/(e*(d*x+c))^(3/2)-2/15*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)))*EllipticF(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^(7/2)/(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 = 0.42 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=\frac {-6 (a+b \text {arcsinh}(c+d x))-4 b (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-(c+d x)^2\right )}{15 d e (e (c+d x))^{5/2}} \]
(-6*(a + b*ArcSinh[c + d*x]) - 4*b*(c + d*x)*Hypergeometric2F1[-3/4, 1/2, 1/4, -(c + d*x)^2])/(15*d*e*(e*(c + d*x))^(5/2))
Time = 0.34 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6274, 6191, 264, 266, 761}
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.
\(\displaystyle \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 6274 |
\(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c+d x)}{(e (c+d x))^{7/2}}d(c+d x)}{d}\) |
\(\Big \downarrow \) 6191 |
\(\displaystyle \frac {\frac {2 b \int \frac {1}{(e (c+d x))^{5/2} \sqrt {(c+d x)^2+1}}d(c+d x)}{5 e}-\frac {2 (a+b \text {arcsinh}(c+d x))}{5 e (e (c+d x))^{5/2}}}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\frac {2 b \left (-\frac {\int \frac {1}{\sqrt {e (c+d x)} \sqrt {(c+d x)^2+1}}d(c+d x)}{3 e^2}-\frac {2 \sqrt {(c+d x)^2+1}}{3 e (e (c+d x))^{3/2}}\right )}{5 e}-\frac {2 (a+b \text {arcsinh}(c+d x))}{5 e (e (c+d x))^{5/2}}}{d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {\frac {2 b \left (-\frac {2 \int \frac {1}{\sqrt {(c+d x)^2+1}}d\sqrt {e (c+d x)}}{3 e^3}-\frac {2 \sqrt {(c+d x)^2+1}}{3 e (e (c+d x))^{3/2}}\right )}{5 e}-\frac {2 (a+b \text {arcsinh}(c+d x))}{5 e (e (c+d x))^{5/2}}}{d}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\frac {2 b \left (-\frac {(e (c+d x)+e) \sqrt {\frac {e^2 (c+d x)^2+e^2}{(e (c+d x)+e)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{3 e^{7/2} \sqrt {(c+d x)^2+1}}-\frac {2 \sqrt {(c+d x)^2+1}}{3 e (e (c+d x))^{3/2}}\right )}{5 e}-\frac {2 (a+b \text {arcsinh}(c+d x))}{5 e (e (c+d x))^{5/2}}}{d}\) |
((-2*(a + b*ArcSinh[c + d*x]))/(5*e*(e*(c + d*x))^(5/2)) + (2*b*((-2*Sqrt[ 1 + (c + d*x)^2])/(3*e*(e*(c + d*x))^(3/2)) - ((e + e*(c + d*x))*Sqrt[(e^2 + e^2*(c + d*x)^2)/(e + e*(c + d*x))^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d* x)]/Sqrt[e]], 1/2])/(3*e^(7/2)*Sqrt[1 + (c + d*x)^2])))/(5*e))/d
3.3.35.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, 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[((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 = 0.66 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.21
method | result | size |
derivativedivides | \(\frac {-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+2 b \left (-\frac {\operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{15 \left (d e x +c e \right )^{\frac {3}{2}}}-\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{15 e^{2} \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e}\) | \(176\) |
default | \(\frac {-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+2 b \left (-\frac {\operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{15 \left (d e x +c e \right )^{\frac {3}{2}}}-\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{15 e^{2} \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e}\) | \(176\) |
parts | \(-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}} d e}+\frac {2 b \left (-\frac {\operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{15 \left (d e x +c e \right )^{\frac {3}{2}}}-\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{15 e^{2} \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e}\) | \(181\) |
2/d/e*(-1/5*a/(d*e*x+c*e)^(5/2)+b*(-1/5/(d*e*x+c*e)^(5/2)*arcsinh(1/e*(d*e *x+c*e))+2/5/e*(-1/3*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)/(d*e*x+c*e)^(3/2)-1/3/e ^2/(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))))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.43 \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {d e x + c e} b d^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 3 \, \sqrt {d e x + c e} a d^{2} + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \sqrt {d^{3} e} {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right ) + 2 \, {\left (b d^{3} x + b c d^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} \sqrt {d e x + c e}\right )}}{15 \, {\left (d^{6} e^{4} x^{3} + 3 \, c d^{5} e^{4} x^{2} + 3 \, c^{2} d^{4} e^{4} x + c^{3} d^{3} e^{4}\right )}} \]
-2/15*(3*sqrt(d*e*x + c*e)*b*d^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^ 2 + 1)) + 3*sqrt(d*e*x + c*e)*a*d^2 + 2*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c ^2*d*x + b*c^3)*sqrt(d^3*e)*weierstrassPInverse(-4/d^2, 0, (d*x + c)/d) + 2*(b*d^3*x + b*c*d^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*sqrt(d*e*x + c*e)) /(d^6*e^4*x^3 + 3*c*d^5*e^4*x^2 + 3*c^2*d^4*e^4*x + c^3*d^3*e^4)
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c + d x \right )}}{\left (e \left (c + d x\right )\right )^{\frac {7}{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \text {arcsinh}(c+d x)}{(c e+d e x)^{7/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^{7/2}} \,d x \]