3.210 \(\int \frac {(c e+d e x)^4}{(a+b \sinh ^{-1}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=367 \[ -\frac {\sqrt {\pi } e^4 e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}+\frac {3 \sqrt {3 \pi } e^4 e^{\frac {3 a}{b}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {\sqrt {5 \pi } e^4 e^{\frac {5 a}{b}} \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {\sqrt {\pi } e^4 e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}-\frac {3 \sqrt {3 \pi } e^4 e^{-\frac {3 a}{b}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {\sqrt {5 \pi } e^4 e^{-\frac {5 a}{b}} \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {2 e^4 (c+d x)^4 \sqrt {(c+d x)^2+1}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

[Out]

-1/8*e^4*exp(a/b)*erf((a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/d+1/8*e^4*erfi((a+b*arcsinh(d*x+c))
^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/d/exp(a/b)+3/16*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)/b^(3/2)/d-3/16*e^4*erfi(3^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/b^(3/2
)/d/exp(3*a/b)-1/16*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)/b^(3/2)/d+
1/16*e^4*erfi(5^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*5^(1/2)*Pi^(1/2)/b^(3/2)/d/exp(5*a/b)-2*e^4*(d*x+c)^
4*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^(1/2)

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Rubi [A]  time = 0.65, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5865, 12, 5665, 3308, 2180, 2204, 2205} \[ -\frac {\sqrt {\pi } e^4 e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}+\frac {3 \sqrt {3 \pi } e^4 e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {\sqrt {5 \pi } e^4 e^{\frac {5 a}{b}} \text {Erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {\sqrt {\pi } e^4 e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}-\frac {3 \sqrt {3 \pi } e^4 e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}+\frac {\sqrt {5 \pi } e^4 e^{-\frac {5 a}{b}} \text {Erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}-\frac {2 e^4 (c+d x)^4 \sqrt {(c+d x)^2+1}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^(3/2),x]

[Out]

(-2*e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])/(b*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (e^4*E^(a/b)*Sqrt[Pi]*Erf[Sqrt
[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(8*b^(3/2)*d) + (3*e^4*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSi
nh[c + d*x]])/Sqrt[b]])/(16*b^(3/2)*d) - (e^4*E^((5*a)/b)*Sqrt[5*Pi]*Erf[(Sqrt[5]*Sqrt[a + b*ArcSinh[c + d*x]]
)/Sqrt[b]])/(16*b^(3/2)*d) + (e^4*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(8*b^(3/2)*d*E^(a/b)) -
(3*e^4*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16*b^(3/2)*d*E^((3*a)/b)) + (e^4*Sqr
t[5*Pi]*Erfi[(Sqrt[5]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16*b^(3/2)*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 3308

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

Rule 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

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}{\left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{8 \sqrt {a+b x}}-\frac {9 \sinh (3 x)}{16 \sqrt {a+b x}}+\frac {5 \sinh (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4 b d}+\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}-\frac {\left (9 e^4\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^4 \operatorname {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8 b d}-\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{-5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}+\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{5 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}+\frac {\left (9 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}-\frac {\left (9 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\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 )}{4 b^2 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 )}{4 b^2 d}-\frac {\left (5 e^4\right ) \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 )}{8 b^2 d}+\frac {\left (5 e^4\right ) \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 )}{8 b^2 d}+\frac {\left (9 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 )}{8 b^2 d}-\frac {\left (9 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 )}{8 b^2 d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1+(c+d x)^2}}{b d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {e^4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 b^{3/2} d}+\frac {3 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 )}{16 b^{3/2} d}-\frac {e^4 e^{\frac {5 a}{b}} \sqrt {5 \pi } \text {erf}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} 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 )}{8 b^{3/2} d}-\frac {3 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 )}{16 b^{3/2} d}+\frac {e^4 e^{-\frac {5 a}{b}} \sqrt {5 \pi } \text {erfi}\left (\frac {\sqrt {5} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{16 b^{3/2} d}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 490, normalized size = 1.34 \[ \frac {e^4 e^{-5 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \left (3 e^{\frac {5 a}{b}+2 \sinh ^{-1}(c+d x)}-2 e^{\frac {5 a}{b}+4 \sinh ^{-1}(c+d x)}-2 e^{\frac {5 a}{b}+6 \sinh ^{-1}(c+d x)}+3 e^{\frac {5 a}{b}+8 \sinh ^{-1}(c+d x)}-e^{\frac {5 a}{b}+10 \sinh ^{-1}(c+d x)}+2 e^{\frac {6 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt {5} e^{5 \sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-3 \sqrt {3} e^{\frac {2 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+2 e^{\frac {4 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )-3 \sqrt {3} e^{\frac {8 a}{b}+5 \sinh ^{-1}(c+d x)} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+\sqrt {5} e^{5 \left (\frac {2 a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {1}{2},\frac {5 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-e^{\frac {5 a}{b}}\right )}{16 b d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSinh[c + d*x])^(3/2),x]

[Out]

(e^4*(-E^((5*a)/b) + 3*E^((5*a)/b + 2*ArcSinh[c + d*x]) - 2*E^((5*a)/b + 4*ArcSinh[c + d*x]) - 2*E^((5*a)/b +
6*ArcSinh[c + d*x]) + 3*E^((5*a)/b + 8*ArcSinh[c + d*x]) - E^((5*a)/b + 10*ArcSinh[c + d*x]) + 2*E^((6*a)/b +
5*ArcSinh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[1/2, a/b + ArcSinh[c + d*x]] + Sqrt[5]*E^(5*ArcSinh[c +
d*x])*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[1/2, (-5*(a + b*ArcSinh[c + d*x]))/b] - 3*Sqrt[3]*E^((2*a)/b
+ 5*ArcSinh[c + d*x])*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[1/2, (-3*(a + b*ArcSinh[c + d*x]))/b] + 2*E^((
4*a)/b + 5*ArcSinh[c + d*x])*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[1/2, -((a + b*ArcSinh[c + d*x])/b)] - 3
*Sqrt[3]*E^((8*a)/b + 5*ArcSinh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[1/2, (3*(a + b*ArcSinh[c + d*x]))
/b] + Sqrt[5]*E^(5*((2*a)/b + ArcSinh[c + d*x]))*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[1/2, (5*(a + b*ArcSinh[c +
d*x]))/b]))/(16*b*d*E^(5*(a/b + ArcSinh[c + d*x]))*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))^(3/2),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 \[ \int \frac {{\left (d e x + c e\right )}^{4}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^4/(b*arcsinh(d*x + c) + a)^(3/2), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d e x +c e \right )^{4}}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(3/2),x)

[Out]

int((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(3/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsinh(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^4/(b*arcsinh(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^(3/2),x)

[Out]

int((c*e + d*e*x)^4/(a + b*asinh(c + d*x))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asinh(d*x+c))**(3/2),x)

[Out]

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

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