3.195 \(\int (c e+d e x) (a+b \sinh ^{-1}(c+d x))^{5/2} \, dx\)

Optimal. Leaf size=262 \[ -\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} e e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{256 d}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} e e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{256 d}+\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}+\frac {15 b^2 e \sqrt {a+b \sinh ^{-1}(c+d x)}}{64 d}-\frac {5 b e (c+d x) \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d} \]

[Out]

1/4*e*(a+b*arcsinh(d*x+c))^(5/2)/d+1/2*e*(d*x+c)^2*(a+b*arcsinh(d*x+c))^(5/2)/d-15/512*b^(5/2)*e*exp(2*a/b)*er
f(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/d-15/512*b^(5/2)*e*erfi(2^(1/2)*(a+b*arcsinh(d*
x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/d/exp(2*a/b)-5/8*b*e*(d*x+c)*(a+b*arcsinh(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1
/2)/d+15/64*b^2*e*(a+b*arcsinh(d*x+c))^(1/2)/d+15/32*b^2*e*(d*x+c)^2*(a+b*arcsinh(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.73, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5865, 12, 5663, 5758, 5675, 5779, 3312, 3307, 2180, 2204, 2205} \[ -\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} e e^{\frac {2 a}{b}} \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{256 d}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} e e^{-\frac {2 a}{b}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{256 d}+\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}+\frac {15 b^2 e \sqrt {a+b \sinh ^{-1}(c+d x)}}{64 d}-\frac {5 b e (c+d x) \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*b^2*e*Sqrt[a + b*ArcSinh[c + d*x]])/(64*d) + (15*b^2*e*(c + d*x)^2*Sqrt[a + b*ArcSinh[c + d*x]])/(32*d) -
(5*b*e*(c + d*x)*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(8*d) + (e*(a + b*ArcSinh[c + d*x])^(5/
2))/(4*d) + (e*(c + d*x)^2*(a + b*ArcSinh[c + d*x])^(5/2))/(2*d) - (15*b^(5/2)*e*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(S
qrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(256*d) - (15*b^(5/2)*e*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*Arc
Sinh[c + d*x]])/Sqrt[b]])/(256*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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5663

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*(a + b*ArcSinh[c*x])^n)/
(m + 1), x] - Dist[(b*c*n)/(m + 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] && GtQ[n, 0]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || 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 (c e+d e x) \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int e x \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e \operatorname {Subst}\left (\int x \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {(5 b e) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}+\frac {(5 b e) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{8 d}+\frac {\left (15 b^2 e\right ) \operatorname {Subst}\left (\int x \sqrt {a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{16 d}\\ &=\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {\left (15 b^3 e\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2} \sqrt {a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{64 d}\\ &=\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {\left (15 b^3 e\right ) \operatorname {Subst}\left (\int \frac {\sinh ^2(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{64 d}\\ &=\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}+\frac {\left (15 b^3 e\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2 \sqrt {a+b x}}-\frac {\cosh (2 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{64 d}\\ &=\frac {15 b^2 e \sqrt {a+b \sinh ^{-1}(c+d x)}}{64 d}+\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {\left (15 b^3 e\right ) \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{128 d}\\ &=\frac {15 b^2 e \sqrt {a+b \sinh ^{-1}(c+d x)}}{64 d}+\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {\left (15 b^3 e\right ) \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{256 d}-\frac {\left (15 b^3 e\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{256 d}\\ &=\frac {15 b^2 e \sqrt {a+b \sinh ^{-1}(c+d x)}}{64 d}+\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {\left (15 b^2 e\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 )}{128 d}-\frac {\left (15 b^2 e\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 )}{128 d}\\ &=\frac {15 b^2 e \sqrt {a+b \sinh ^{-1}(c+d x)}}{64 d}+\frac {15 b^2 e (c+d x)^2 \sqrt {a+b \sinh ^{-1}(c+d x)}}{32 d}-\frac {5 b e (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{8 d}+\frac {e \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac {e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{2 d}-\frac {15 b^{5/2} e e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{256 d}-\frac {15 b^{5/2} e e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{256 d}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 126, normalized size = 0.48 \[ \frac {e e^{-\frac {2 a}{b}} \left (b^3 e^{\frac {4 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \Gamma \left (\frac {7}{2},\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-b^3 \sqrt {-\frac {a+b \sinh ^{-1}(c+d x)}{b}} \Gamma \left (\frac {7}{2},-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{32 \sqrt {2} d \sqrt {a+b \sinh ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(e*(-(b^3*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[7/2, (-2*(a + b*ArcSinh[c + d*x]))/b]) + b^3*E^((4*a)/b)*S
qrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (2*(a + b*ArcSinh[c + d*x]))/b]))/(32*Sqrt[2]*d*E^((2*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)*(a+b*arcsinh(d*x+c))^(5/2),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 \[ \int {\left (d e x + c e\right )} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right ) \left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (c\,e+d\,e\,x\right )\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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