3.223 \(\int \frac {(c e+d e x)^3}{(a+b \sinh ^{-1}(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=420 \[ \frac {16 \sqrt {\pi } e^3 e^{\frac {4 a}{b}} \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {4 \sqrt {2 \pi } e^3 e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {16 \sqrt {\pi } e^3 e^{-\frac {4 a}{b}} \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {4 \sqrt {2 \pi } e^3 e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {128 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 \sqrt {(c+d x)^2+1} (c+d x)}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {2 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]

[Out]

-4/5*e^3*(d*x+c)^2/b^2/d/(a+b*arcsinh(d*x+c))^(3/2)-16/15*e^3*(d*x+c)^4/b^2/d/(a+b*arcsinh(d*x+c))^(3/2)+16/15
*e^3*exp(4*a/b)*erf(2*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/d+16/15*e^3*erfi(2*(a+b*arcsinh(d*x
+c))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/d/exp(4*a/b)-4/15*e^3*exp(2*a/b)*erf(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b
^(1/2))*2^(1/2)*Pi^(1/2)/b^(7/2)/d-4/15*e^3*erfi(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/
b^(7/2)/d/exp(2*a/b)-2/5*e^3*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d*x+c))^(5/2)-16/5*e^3*(d*x+c)*(1+
(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsinh(d*x+c))^(1/2)-128/15*e^3*(d*x+c)^3*(1+(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsinh(
d*x+c))^(1/2)

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Rubi [A]  time = 1.10, antiderivative size = 420, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {5865, 12, 5667, 5774, 5665, 3307, 2180, 2204, 2205} \[ \frac {16 \sqrt {\pi } e^3 e^{\frac {4 a}{b}} \text {Erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {4 \sqrt {2 \pi } e^3 e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {16 \sqrt {\pi } e^3 e^{-\frac {4 a}{b}} \text {Erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {4 \sqrt {2 \pi } e^3 e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {128 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 \sqrt {(c+d x)^2+1} (c+d x)}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {2 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSinh[c + d*x])^(7/2),x]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(5*b*d*(a + b*ArcSinh[c + d*x])^(5/2)) - (4*e^3*(c + d*x)^2)/(5*b^2
*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (16*e^3*(c + d*x)^4)/(15*b^2*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (16*e^3*
(c + d*x)*Sqrt[1 + (c + d*x)^2])/(5*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (128*e^3*(c + d*x)^3*Sqrt[1 + (c + d
*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) + (16*e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d
*x]])/Sqrt[b]])/(15*b^(7/2)*d) - (4*e^3*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt
[b]])/(15*b^(7/2)*d) + (16*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d*E^((4*a)
/b)) - (4*e^3*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d*E^((2*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 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 5667

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[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c
^2*x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 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)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}+\frac {\left (6 e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}+\frac {\left (8 e^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac {\left (8 e^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{5 b^2 d}+\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {128 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (16 e^3\right ) \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{5 b^3 d}+\frac {\left (128 e^3\right ) \operatorname {Subst}\left (\int \left (-\frac {\cosh (2 x)}{2 \sqrt {a+b x}}+\frac {\cosh (4 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {128 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (8 e^3\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{5 b^3 d}+\frac {\left (8 e^3\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{5 b^3 d}-\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}+\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {128 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {\left (16 e^3\right ) \operatorname {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{5 b^4 d}+\frac {\left (16 e^3\right ) \operatorname {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{5 b^4 d}+\frac {\left (32 e^3\right ) \operatorname {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}-\frac {\left (32 e^3\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}-\frac {\left (32 e^3\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}+\frac {\left (32 e^3\right ) \operatorname {Subst}\left (\int \frac {e^{4 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {128 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {4 e^3 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{5 b^{7/2} d}+\frac {4 e^3 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{5 b^{7/2} d}+\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int e^{\frac {4 a}{b}-\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{15 b^4 d}-\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int e^{\frac {2 a}{b}-\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{15 b^4 d}-\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int e^{-\frac {2 a}{b}+\frac {2 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{15 b^4 d}+\frac {\left (64 e^3\right ) \operatorname {Subst}\left (\int e^{-\frac {4 a}{b}+\frac {4 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c+d x)}\right )}{15 b^4 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {16 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{5 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {128 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {16 e^3 e^{\frac {4 a}{b}} \sqrt {\pi } \text {erf}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {4 e^3 e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {16 e^3 e^{-\frac {4 a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {4 e^3 e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}\\ \end {align*}

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Mathematica [A]  time = 2.11, size = 429, normalized size = 1.02 \[ \frac {e^3 \left (4 \left (a+b \sinh ^{-1}(c+d x)\right ) \left (e^{2 \sinh ^{-1}(c+d x)} \left (4 a+4 b \sinh ^{-1}(c+d x)+b\right )+4 \sqrt {2} b e^{-\frac {2 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+4 \sqrt {2} e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-4 a e^{-2 \sinh ^{-1}(c+d x)}+b e^{-2 \sinh ^{-1}(c+d x)}-4 b e^{-2 \sinh ^{-1}(c+d x)} \sinh ^{-1}(c+d x)\right )-4 \left (a+b \sinh ^{-1}(c+d x)\right ) \left (e^{4 \sinh ^{-1}(c+d x)} \left (8 a+8 b \sinh ^{-1}(c+d x)+b\right )+16 b e^{-\frac {4 a}{b}} \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+16 e^{\frac {4 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \Gamma \left (\frac {1}{2},\frac {4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-8 a e^{-4 \sinh ^{-1}(c+d x)}+b e^{-4 \sinh ^{-1}(c+d x)} \left (1-8 \sinh ^{-1}(c+d x)\right )\right )+6 b^2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )-3 b^2 \sinh \left (4 \sinh ^{-1}(c+d x)\right )\right )}{60 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSinh[c + d*x])^(7/2),x]

[Out]

(e^3*(4*(a + b*ArcSinh[c + d*x])*((-4*a)/E^(2*ArcSinh[c + d*x]) + b/E^(2*ArcSinh[c + d*x]) - (4*b*ArcSinh[c +
d*x])/E^(2*ArcSinh[c + d*x]) + E^(2*ArcSinh[c + d*x])*(4*a + b + 4*b*ArcSinh[c + d*x]) + (4*Sqrt[2]*b*(-((a +
b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-2*(a + b*ArcSinh[c + d*x]))/b])/E^((2*a)/b) + 4*Sqrt[2]*E^((2*a)/b)
*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamma[1/2, (2*(a + b*ArcSinh[c + d*x]))/b]) - 4*(a + b*
ArcSinh[c + d*x])*((-8*a)/E^(4*ArcSinh[c + d*x]) + (b*(1 - 8*ArcSinh[c + d*x]))/E^(4*ArcSinh[c + d*x]) + E^(4*
ArcSinh[c + d*x])*(8*a + b + 8*b*ArcSinh[c + d*x]) + (16*b*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-
4*(a + b*ArcSinh[c + d*x]))/b])/E^((4*a)/b) + 16*E^((4*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d
*x])*Gamma[1/2, (4*(a + b*ArcSinh[c + d*x]))/b]) + 6*b^2*Sinh[2*ArcSinh[c + d*x]] - 3*b^2*Sinh[4*ArcSinh[c + d
*x]]))/(60*b^3*d*(a + b*ArcSinh[c + d*x])^(5/2))

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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)^3/(a+b*arcsinh(d*x+c))^(7/2),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 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsinh(d*x + c) + a)^(7/2), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d e x +c e \right )^{3}}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(7/2),x)

[Out]

int((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d e x + c e\right )}^{3}}{{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsinh(d*x + c) + a)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^3/(a + b*asinh(c + d*x))^(7/2),x)

[Out]

int((c*e + d*e*x)^3/(a + b*asinh(c + d*x))^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asinh(d*x+c))**(7/2),x)

[Out]

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

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