3.204 \(\int \frac {(c e+d e x)^4}{\sqrt {a+b \sinh ^{-1}(c+d x)}} \, dx\)

Optimal. Leaf size=326 \[ \frac {\sqrt {\pi } e^4 e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 \sqrt {b} d}-\frac {\sqrt {3 \pi } e^4 e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{5}} e^4 e^{\frac {5 a}{b}} \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\pi } e^4 e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 \sqrt {b} d}-\frac {\sqrt {3 \pi } e^4 e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{5}} e^4 e^{-\frac {5 a}{b}} \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d} \]

[Out]

1/160*e^4*exp(5*a/b)*erf(5^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/d/b^(1/2)+1/160*e^4*erfi
(5^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/d/exp(5*a/b)/b^(1/2)+1/16*e^4*exp(a/b)*erf((a+b*
arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d/b^(1/2)+1/16*e^4*erfi((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d
/exp(a/b)/b^(1/2)-1/32*e^4*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/d/b^(1/
2)-1/32*e^4*erfi(3^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/d/exp(3*a/b)/b^(1/2)

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Rubi [A]  time = 0.64, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {5865, 12, 5669, 5448, 3307, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } e^4 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 \sqrt {b} d}-\frac {\sqrt {3 \pi } e^4 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{5}} e^4 e^{\frac {5 a}{b}} \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\pi } e^4 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 \sqrt {b} d}-\frac {\sqrt {3 \pi } e^4 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{5}} e^4 e^{-\frac {5 a}{b}} \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/Sqrt[a + b*ArcSinh[c + d*x]],x]

[Out]

(e^4*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(16*Sqrt[b]*d) - (e^4*E^((3*a)/b)*Sqrt[3*Pi]*
Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(32*Sqrt[b]*d) + (e^4*E^((5*a)/b)*Sqrt[Pi/5]*Erf[(Sqrt[5]
*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(32*Sqrt[b]*d) + (e^4*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt
[b]])/(16*Sqrt[b]*d*E^(a/b)) - (e^4*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(32*Sqrt[
b]*d*E^((3*a)/b)) + (e^4*Sqrt[Pi/5]*Erfi[(Sqrt[5]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(32*Sqrt[b]*d*E^((5*
a)/b))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5865

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*} \int \frac {(c e+d e x)^4}{\sqrt {a+b \sinh ^{-1}(c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh ^4(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \left (\frac {\cosh (x)}{8 \sqrt {a+b x}}-\frac {3 \cosh (3 x)}{16 \sqrt {a+b x}}+\frac {\cosh (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {\cosh (5 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\cosh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\cosh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {e^{5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{32 d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int e^{\frac {5 a}{b}-\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{16 b d}+\frac {e^4 \operatorname {Subst}\left (\int e^{-\frac {5 a}{b}+\frac {5 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{16 b d}+\frac {e^4 \operatorname {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b d}+\frac {e^4 \operatorname {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{8 b d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{16 b d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{16 b d}\\ &=\frac {e^4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 \sqrt {b} d}-\frac {e^4 e^{\frac {3 a}{b}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {e^4 e^{\frac {5 a}{b}} \sqrt {\frac {\pi }{5}} \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {e^4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 \sqrt {b} d}-\frac {e^4 e^{-\frac {3 a}{b}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}+\frac {e^4 e^{-\frac {5 a}{b}} \sqrt {\frac {\pi }{5}} \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{32 \sqrt {b} d}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 320, normalized size = 0.98 \[ \frac {e^4 e^{-\frac {5 a}{b}} \left (-10 e^{\frac {6 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt {5} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-5 \sqrt {3} e^{\frac {2 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+10 e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )+5 \sqrt {3} e^{\frac {8 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-\sqrt {5} e^{\frac {10 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{160 d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^4/Sqrt[a + b*ArcSinh[c + d*x]],x]

[Out]

(e^4*(-10*E^((6*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[1/2, a/b + ArcSinh[c + d*x]] + Sqrt[5]*Sqrt[-((a + b*
ArcSinh[c + d*x])/b)]*Gamma[1/2, (-5*(a + b*ArcSinh[c + d*x]))/b] - 5*Sqrt[3]*E^((2*a)/b)*Sqrt[-((a + b*ArcSin
h[c + d*x])/b)]*Gamma[1/2, (-3*(a + b*ArcSinh[c + d*x]))/b] + 10*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b
)]*Gamma[1/2, -((a + b*ArcSinh[c + d*x])/b)] + 5*Sqrt[3]*E^((8*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[1/2, (
3*(a + b*ArcSinh[c + d*x]))/b] - Sqrt[5]*E^((10*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[1/2, (5*(a + b*ArcSin
h[c + d*x]))/b]))/(160*d*E^((5*a)/b)*Sqrt[a + b*ArcSinh[c + d*x]])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{\sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^4/sqrt(b*arcsinh(d*x + c) + a), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d e x +c e \right )^{4}}{\sqrt {a +b \arcsinh \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(1/2),x)

[Out]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{\sqrt {b \operatorname {arsinh}\left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^4/sqrt(b*arcsinh(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{\sqrt {a+b\,\mathrm {asinh}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^(1/2),x)

[Out]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{4} \left (\int \frac {c^{4}}{\sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}}\, dx + \int \frac {d^{4} x^{4}}{\sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}}\, dx + \int \frac {4 c d^{3} x^{3}}{\sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{\sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}}\, dx + \int \frac {4 c^{3} d x}{\sqrt {a + b \operatorname {asinh}{\left (c + d x \right )}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asinh(d*x+c))**(1/2),x)

[Out]

e**4*(Integral(c**4/sqrt(a + b*asinh(c + d*x)), x) + Integral(d**4*x**4/sqrt(a + b*asinh(c + d*x)), x) + Integ
ral(4*c*d**3*x**3/sqrt(a + b*asinh(c + d*x)), x) + Integral(6*c**2*d**2*x**2/sqrt(a + b*asinh(c + d*x)), x) +
Integral(4*c**3*d*x/sqrt(a + b*asinh(c + d*x)), x))

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