3.3.25 \(\int \frac {c e+d e x}{(a+b \sinh ^{-1}(c+d x))^{7/2}} \, dx\) [225]

Optimal. Leaf size=252 \[ -\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {32 e (c+d x) \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {8 e e^{\frac {2 a}{b}} \sqrt {2 \pi } \text {Erf}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {8 e e^{-\frac {2 a}{b}} \sqrt {2 \pi } \text {Erfi}\left (\frac {\sqrt {2} \sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d} \]

[Out]

-4/15*e/b^2/d/(a+b*arcsinh(d*x+c))^(3/2)-8/15*e*(d*x+c)^2/b^2/d/(a+b*arcsinh(d*x+c))^(3/2)+8/15*e*exp(2*a/b)*e
rf(2^(1/2)*(a+b*arcsinh(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(7/2)/d+8/15*e*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*(d*x+c)*(1+(d*x+c)^2)^(1/2)/b/d/(a+b*arcsinh(d
*x+c))^(5/2)-32/15*e*(d*x+c)*(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 = 0.36, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5859, 12, 5779, 5818, 5778, 3388, 2211, 2236, 2235, 5783} \begin {gather*} \frac {8 \sqrt {2 \pi } e 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 {8 \sqrt {2 \pi } e 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 {32 e \sqrt {(c+d x)^2+1} (c+d x)}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {2 e \sqrt {(c+d x)^2+1} (c+d x)}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e*(c + d*x)*Sqrt[1 + (c + d*x)^2])/(5*b*d*(a + b*ArcSinh[c + d*x])^(5/2)) - (4*e)/(15*b^2*d*(a + b*ArcSinh
[c + d*x])^(3/2)) - (8*e*(c + d*x)^2)/(15*b^2*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (32*e*(c + d*x)*Sqrt[1 + (c
+ d*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) + (8*e*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSi
nh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d) + (8*e*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 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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 2236

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 3388

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 5778

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^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rule 5779

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 5783

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

Rule 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/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, m}, x] && EqQ[e, c^2*d] && LtQ[n, -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*} \int \frac {c e+d e x}{\left (a+b \sinh ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {e x}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e \text {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}+\frac {(2 e) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}+\frac {(4 e) \text {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 {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac {(16 e) \text {Subst}\left (\int \frac {x}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {32 e (c+d x) \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {(32 e) \text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {32 e (c+d x) \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {(16 e) \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}+\frac {(16 e) \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {32 e (c+d x) \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {(32 e) \text {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 {(32 e) \text {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 {2 e (c+d x) \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 e}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 e (c+d x)^2}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {32 e (c+d x) \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}+\frac {8 e 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 {8 e 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 = 0.68, size = 235, normalized size = 0.93 \begin {gather*} -\frac {e \left (\left (a+b \sinh ^{-1}(c+d x)\right ) \left (e^{-\frac {2 a}{b}} \left (2 e^{2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \left (4 a+b+4 b \sinh ^{-1}(c+d x)\right )+8 \sqrt {2} 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 )\right )+e^{-2 \sinh ^{-1}(c+d x)} \left (-8 a+2 b-8 b \sinh ^{-1}(c+d x)+8 \sqrt {2} e^{2 \left (\frac {a}{b}+\sinh ^{-1}(c+d x)\right )} \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 )\right )\right )+3 b^2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )\right )}{15 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(e*((a + b*ArcSinh[c + d*x])*((2*E^(2*(a/b + ArcSinh[c + d*x]))*(4*a + b + 4*b*ArcSinh[c + d*x]) + 8*Sqr
t[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) + (-8*a
+ 2*b - 8*b*ArcSinh[c + d*x] + 8*Sqrt[2]*E^(2*(a/b + ArcSinh[c + d*x]))*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*Ar
cSinh[c + d*x])*Gamma[1/2, (2*(a + b*ArcSinh[c + d*x]))/b])/E^(2*ArcSinh[c + d*x])) + 3*b^2*Sinh[2*ArcSinh[c +
d*x]]))/(b^3*d*(a + b*ArcSinh[c + d*x])^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e \left (\int \frac {c}{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 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*(Integral(c/(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) + Integ
ral(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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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