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

Optimal. Leaf size=128 \[ -\frac {a+b \sinh ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c \sqrt {c^2 x^2+1}}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {b c^3 d \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{2 e \left (c^2 d^2+e^2\right )^{3/2}} \]

[Out]

1/2*(-a-b*arcsinh(c*x))/e/(e*x+d)^2-1/2*b*c^3*d*arctanh((-c^2*d*x+e)/(c^2*d^2+e^2)^(1/2)/(c^2*x^2+1)^(1/2))/e/
(c^2*d^2+e^2)^(3/2)-1/2*b*c*(c^2*x^2+1)^(1/2)/(c^2*d^2+e^2)/(e*x+d)

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Rubi [A]  time = 0.08, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5801, 731, 725, 206} \[ -\frac {a+b \sinh ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c \sqrt {c^2 x^2+1}}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {b c^3 d \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{2 e \left (c^2 d^2+e^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(2*(c^2*d^2 + e^2)*(d + e*x)) - (a + b*ArcSinh[c*x])/(2*e*(d + e*x)^2) - (b*c^3*d*Arc
Tanh[(e - c^2*d*x)/(Sqrt[c^2*d^2 + e^2]*Sqrt[1 + c^2*x^2])])/(2*e*(c^2*d^2 + e^2)^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 5801

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

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{(d+e x)^3} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 e (d+e x)^2}+\frac {(b c) \int \frac {1}{(d+e x)^2 \sqrt {1+c^2 x^2}} \, dx}{2 e}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \sinh ^{-1}(c x)}{2 e (d+e x)^2}+\frac {\left (b c^3 d\right ) \int \frac {1}{(d+e x) \sqrt {1+c^2 x^2}} \, dx}{2 e \left (c^2 d^2+e^2\right )}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \sinh ^{-1}(c x)}{2 e (d+e x)^2}-\frac {\left (b c^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{c^2 d^2+e^2-x^2} \, dx,x,\frac {e-c^2 d x}{\sqrt {1+c^2 x^2}}\right )}{2 e \left (c^2 d^2+e^2\right )}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{2 \left (c^2 d^2+e^2\right ) (d+e x)}-\frac {a+b \sinh ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c^3 d \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 d^2+e^2} \sqrt {1+c^2 x^2}}\right )}{2 e \left (c^2 d^2+e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 166, normalized size = 1.30 \[ \frac {1}{2} \left (-\frac {a}{e (d+e x)^2}-\frac {b c \sqrt {c^2 x^2+1}}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac {b c^3 d \log \left (\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}+c^2 (-d) x+e\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}+\frac {b c^3 d \log (d+e x)}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac {b \sinh ^{-1}(c x)}{e (d+e x)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a/(e*(d + e*x)^2)) - (b*c*Sqrt[1 + c^2*x^2])/((c^2*d^2 + e^2)*(d + e*x)) - (b*ArcSinh[c*x])/(e*(d + e*x)^2)
+ (b*c^3*d*Log[d + e*x])/(e*(c^2*d^2 + e^2)^(3/2)) - (b*c^3*d*Log[e - c^2*d*x + Sqrt[c^2*d^2 + e^2]*Sqrt[1 +
c^2*x^2]])/(e*(c^2*d^2 + e^2)^(3/2)))/2

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fricas [B]  time = 0.76, size = 566, normalized size = 4.42 \[ -\frac {{\left (a + b\right )} c^{4} d^{6} + {\left (2 \, a + b\right )} c^{2} d^{4} e^{2} + a d^{2} e^{4} + {\left (b c^{4} d^{4} e^{2} + b c^{2} d^{2} e^{4}\right )} x^{2} - {\left (b c^{3} d^{3} e^{2} x^{2} + 2 \, b c^{3} d^{4} e x + b c^{3} d^{5}\right )} \sqrt {c^{2} d^{2} + e^{2}} \log \left (-\frac {c^{3} d^{2} x - c d e + \sqrt {c^{2} d^{2} + e^{2}} {\left (c^{2} d x - e\right )} + {\left (c^{2} d^{2} + \sqrt {c^{2} d^{2} + e^{2}} c d + e^{2}\right )} \sqrt {c^{2} x^{2} + 1}}{e x + d}\right ) + 2 \, {\left (b c^{4} d^{5} e + b c^{2} d^{3} e^{3}\right )} x - {\left ({\left (b c^{4} d^{4} e^{2} + 2 \, b c^{2} d^{2} e^{4} + b e^{6}\right )} x^{2} + 2 \, {\left (b c^{4} d^{5} e + 2 \, b c^{2} d^{3} e^{3} + b d e^{5}\right )} x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (b c^{4} d^{6} + 2 \, b c^{2} d^{4} e^{2} + b d^{2} e^{4} + {\left (b c^{4} d^{4} e^{2} + 2 \, b c^{2} d^{2} e^{4} + b e^{6}\right )} x^{2} + 2 \, {\left (b c^{4} d^{5} e + 2 \, b c^{2} d^{3} e^{3} + b d e^{5}\right )} x\right )} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (b c^{3} d^{5} e + b c d^{3} e^{3} + {\left (b c^{3} d^{4} e^{2} + b c d^{2} e^{4}\right )} x\right )} \sqrt {c^{2} x^{2} + 1}}{2 \, {\left (c^{4} d^{8} e + 2 \, c^{2} d^{6} e^{3} + d^{4} e^{5} + {\left (c^{4} d^{6} e^{3} + 2 \, c^{2} d^{4} e^{5} + d^{2} e^{7}\right )} x^{2} + 2 \, {\left (c^{4} d^{7} e^{2} + 2 \, c^{2} d^{5} e^{4} + d^{3} e^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a + b)*c^4*d^6 + (2*a + b)*c^2*d^4*e^2 + a*d^2*e^4 + (b*c^4*d^4*e^2 + b*c^2*d^2*e^4)*x^2 - (b*c^3*d^3*e
^2*x^2 + 2*b*c^3*d^4*e*x + b*c^3*d^5)*sqrt(c^2*d^2 + e^2)*log(-(c^3*d^2*x - c*d*e + sqrt(c^2*d^2 + e^2)*(c^2*d
*x - e) + (c^2*d^2 + sqrt(c^2*d^2 + e^2)*c*d + e^2)*sqrt(c^2*x^2 + 1))/(e*x + d)) + 2*(b*c^4*d^5*e + b*c^2*d^3
*e^3)*x - ((b*c^4*d^4*e^2 + 2*b*c^2*d^2*e^4 + b*e^6)*x^2 + 2*(b*c^4*d^5*e + 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*log(
c*x + sqrt(c^2*x^2 + 1)) - (b*c^4*d^6 + 2*b*c^2*d^4*e^2 + b*d^2*e^4 + (b*c^4*d^4*e^2 + 2*b*c^2*d^2*e^4 + b*e^6
)*x^2 + 2*(b*c^4*d^5*e + 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*log(-c*x + sqrt(c^2*x^2 + 1)) + (b*c^3*d^5*e + b*c*d^3*
e^3 + (b*c^3*d^4*e^2 + b*c*d^2*e^4)*x)*sqrt(c^2*x^2 + 1))/(c^4*d^8*e + 2*c^2*d^6*e^3 + d^4*e^5 + (c^4*d^6*e^3
+ 2*c^2*d^4*e^5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 + 2*c^2*d^5*e^4 + d^3*e^6)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 279, normalized size = 2.18 \[ -\frac {c^{2} a}{2 \left (c e x +c d \right )^{2} e}-\frac {c^{2} b \arcsinh \left (c x \right )}{2 \left (c e x +c d \right )^{2} e}-\frac {c^{2} b \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{2 e \left (c^{2} d^{2}+e^{2}\right ) \left (c x +\frac {c d}{e}\right )}-\frac {c^{3} b d \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{2 e^{2} \left (c^{2} d^{2}+e^{2}\right ) \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a/(c*e*x+c*d)^2/e-1/2*c^2*b/(c*e*x+c*d)^2/e*arcsinh(c*x)-1/2*c^2*b/e/(c^2*d^2+e^2)/(c*x+c*d/e)*((c*x+
c*d/e)^2-2*c*d/e*(c*x+c*d/e)+(c^2*d^2+e^2)/e^2)^(1/2)-1/2*c^3*b/e^2*d/(c^2*d^2+e^2)/((c^2*d^2+e^2)/e^2)^(1/2)*
ln((2*(c^2*d^2+e^2)/e^2-2*c*d/e*(c*x+c*d/e)+2*((c^2*d^2+e^2)/e^2)^(1/2)*((c*x+c*d/e)^2-2*c*d/e*(c*x+c*d/e)+(c^
2*d^2+e^2)/e^2)^(1/2))/(c*x+c*d/e))

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maxima [A]  time = 0.39, size = 158, normalized size = 1.23 \[ -\frac {1}{2} \, {\left (c {\left (\frac {\sqrt {c^{2} x^{2} + 1}}{c^{2} d^{2} e x + c^{2} d^{3} + e^{3} x + d e^{2}} - \frac {c^{2} d \operatorname {arsinh}\left (\frac {c d x}{e {\left | x + \frac {d}{e} \right |}} - \frac {1}{c {\left | x + \frac {d}{e} \right |}}\right )}{{\left (\frac {c^{2} d^{2}}{e^{2}} + 1\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {\operatorname {arsinh}\left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} b - \frac {a}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(sqrt(c^2*x^2 + 1)/(c^2*d^2*e*x + c^2*d^3 + e^3*x + d*e^2) - c^2*d*arcsinh(c*d*x/(e*abs(x + d/e)) - 1/
(c*abs(x + d/e)))/((c^2*d^2/e^2 + 1)^(3/2)*e^4)) + arcsinh(c*x)/(e^3*x^2 + 2*d*e^2*x + d^2*e))*b - 1/2*a/(e^3*
x^2 + 2*d*e^2*x + d^2*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(d + e*x)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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