\(\int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx\) [229]

Optimal result

Integrand size = 23, antiderivative size = 177 \[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\frac {20 b e^2 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{147 d}-\frac {4 b (e (c+d x))^{5/2} \sqrt {1+(c+d x)^2}}{49 d}+\frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}-\frac {10 b e^{5/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{147 d \sqrt {1+(c+d x)^2}} \]

[Out]

2/7*(e*(d*x+c))^(7/2)*(a+b*arcsinh(d*x+c))/d/e-4/49*b*(e*(d*x+c))^(5/2)*(1+(d*x+c)^2)^(1/2)/d+20/147*b*e^2*(e*
(d*x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d-10/147*b*e^(5/2)*(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/(1+(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5859, 5776, 327, 335, 226} \[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}-\frac {10 b e^{5/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right ),\frac {1}{2}\right )}{147 d \sqrt {(c+d x)^2+1}}+\frac {20 b e^2 \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{147 d}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{5/2}}{49 d} \]

[In]

Int[(c*e + d*e*x)^(5/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(20*b*e^2*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(147*d) - (4*b*(e*(c + d*x))^(5/2)*Sqrt[1 + (c + d*x)^2])/(
49*d) + (2*(e*(c + d*x))^(7/2)*(a + b*ArcSinh[c + d*x]))/(7*d*e) - (10*b*e^(5/2)*(1 + c + d*x)*Sqrt[(1 + (c +
d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(147*d*Sqrt[1 + (c + d*x)^2])

Rule 226

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]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[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]

Rule 5859

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[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
]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (e x)^{5/2} (a+b \text {arcsinh}(x)) \, dx,x,c+d x\right )}{d} \\ & = \frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}-\frac {(2 b) \text {Subst}\left (\int \frac {(e x)^{7/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{7 d e} \\ & = -\frac {4 b (e (c+d x))^{5/2} \sqrt {1+(c+d x)^2}}{49 d}+\frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}+\frac {(10 b e) \text {Subst}\left (\int \frac {(e x)^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{49 d} \\ & = \frac {20 b e^2 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{147 d}-\frac {4 b (e (c+d x))^{5/2} \sqrt {1+(c+d x)^2}}{49 d}+\frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}-\frac {\left (10 b e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x^2}} \, dx,x,c+d x\right )}{147 d} \\ & = \frac {20 b e^2 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{147 d}-\frac {4 b (e (c+d x))^{5/2} \sqrt {1+(c+d x)^2}}{49 d}+\frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}-\frac {\left (20 b e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{147 d} \\ & = \frac {20 b e^2 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{147 d}-\frac {4 b (e (c+d x))^{5/2} \sqrt {1+(c+d x)^2}}{49 d}+\frac {2 (e (c+d x))^{7/2} (a+b \text {arcsinh}(c+d x))}{7 d e}-\frac {10 b e^{5/2} (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 )}{147 d \sqrt {1+(c+d x)^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.64 \[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\frac {2 (e (c+d x))^{5/2} \left (21 a (c+d x)^3+10 b \sqrt {1+(c+d x)^2}-6 b (c+d x)^2 \sqrt {1+(c+d x)^2}+21 b (c+d x)^3 \text {arcsinh}(c+d x)-10 b \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-(c+d x)^2\right )\right )}{147 d (c+d x)^2} \]

[In]

Integrate[(c*e + d*e*x)^(5/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(2*(e*(c + d*x))^(5/2)*(21*a*(c + d*x)^3 + 10*b*Sqrt[1 + (c + d*x)^2] - 6*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]
+ 21*b*(c + d*x)^3*ArcSinh[c + d*x] - 10*b*Hypergeometric2F1[1/4, 1/2, 5/4, -(c + d*x)^2]))/(147*d*(c + d*x)^2
)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.29 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d e x +c e \right )^{\frac {7}{2}} a}{7}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {7}{2}} \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{7}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {5}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{7}-\frac {5 e^{4} \sqrt {d e x +c e}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{21}+\frac {5 e^{4} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{21 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{7 e}\right )}{d e}\)	\(212\)
default	\(\frac {\frac {2 \left (d e x +c e \right )^{\frac {7}{2}} a}{7}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {7}{2}} \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{7}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {5}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{7}-\frac {5 e^{4} \sqrt {d e x +c e}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{21}+\frac {5 e^{4} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{21 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{7 e}\right )}{d e}\)	\(212\)
parts	\(\frac {2 a \left (d e x +c e \right )^{\frac {7}{2}}}{7 d e}+\frac {2 b \left (\frac {\left (d e x +c e \right )^{\frac {7}{2}} \operatorname {arcsinh}\left (\frac {d e x +c e}{e}\right )}{7}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {5}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{7}-\frac {5 e^{4} \sqrt {d e x +c e}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{21}+\frac {5 e^{4} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \operatorname {EllipticF}\left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{21 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{7 e}\right )}{d e}\)	\(217\)

[In]

int((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/d/e*(1/7*(d*e*x+c*e)^(7/2)*a+b*(1/7*(d*e*x+c*e)^(7/2)*arcsinh(1/e*(d*e*x+c*e))-2/7/e*(1/7*e^2*(d*e*x+c*e)^(5
/2)*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)-5/21*e^4*(d*e*x+c*e)^(1/2)*(1/e^2*(d*e*x+c*e)^2+1)^(1/2)+5/21*e^4/(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))))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.44 \[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=-\frac {2 \, {\left (10 \, \sqrt {d^{3} e} b e^{2} {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right ) - 21 \, {\left (b d^{5} e^{2} x^{3} + 3 \, b c d^{4} e^{2} x^{2} + 3 \, b c^{2} d^{3} e^{2} x + b c^{3} d^{2} e^{2}\right )} \sqrt {d e x + c e} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 2 \, {\left (3 \, b d^{4} e^{2} x^{2} + 6 \, b c d^{3} e^{2} x + {\left (3 \, b c^{2} - 5 \, b\right )} d^{2} e^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} \sqrt {d e x + c e} - 21 \, {\left (a d^{5} e^{2} x^{3} + 3 \, a c d^{4} e^{2} x^{2} + 3 \, a c^{2} d^{3} e^{2} x + a c^{3} d^{2} e^{2}\right )} \sqrt {d e x + c e}\right )}}{147 \, d^{3}} \]

[In]

integrate((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c)),x, algorithm="fricas")

[Out]

-2/147*(10*sqrt(d^3*e)*b*e^2*weierstrassPInverse(-4/d^2, 0, (d*x + c)/d) - 21*(b*d^5*e^2*x^3 + 3*b*c*d^4*e^2*x
^2 + 3*b*c^2*d^3*e^2*x + b*c^3*d^2*e^2)*sqrt(d*e*x + c*e)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) + 2
*(3*b*d^4*e^2*x^2 + 6*b*c*d^3*e^2*x + (3*b*c^2 - 5*b)*d^2*e^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*sqrt(d*e*x +
c*e) - 21*(a*d^5*e^2*x^3 + 3*a*c*d^4*e^2*x^2 + 3*a*c^2*d^3*e^2*x + a*c^3*d^2*e^2)*sqrt(d*e*x + c*e))/d^3

Sympy [F]

\[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\int \left (e \left (c + d x\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )\, dx \]

[In]

integrate((d*e*x+c*e)**(5/2)*(a+b*asinh(d*x+c)),x)

[Out]

Integral((e*(c + d*x))**(5/2)*(a + b*asinh(c + d*x)), x)

Maxima [F(-2)]

Exception generated. \[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\int { {\left (d e x + c e\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )} \,d x } \]

[In]

integrate((d*e*x+c*e)^(5/2)*(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(5/2)*(b*arcsinh(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int (c e+d e x)^{5/2} (a+b \text {arcsinh}(c+d x)) \, dx=\int {\left (c\,e+d\,e\,x\right )}^{5/2}\,\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right ) \,d x \]

[In]

int((c*e + d*e*x)^(5/2)*(a + b*asinh(c + d*x)),x)

[Out]

int((c*e + d*e*x)^(5/2)*(a + b*asinh(c + d*x)), x)