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]

-(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))

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Rubi [A]  time = 0.0837947, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.330422, size = 166, normalized size = 1.3 \[ \frac{1}{2} \left (-\frac{a}{e (d+e x)^2}-\frac{b c \sqrt{c^2 x^2+1}}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac{b c^3 d \log \left (\sqrt{c^2 x^2+1} \sqrt{c^2 d^2+e^2}+c^2 (-d) x+e\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}+\frac{b c^3 d \log (d+e x)}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b \sinh ^{-1}(c x)}{e (d+e x)^2}\right ) \]

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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Maple [B]  time = 0.03, size = 279, normalized size = 2.2 \begin{align*} -{\frac{{c}^{2}a}{2\, \left ( cex+cd \right ) ^{2}e}}-{\frac{{c}^{2}b{\it Arcsinh} \left ( cx \right ) }{2\, \left ( cex+cd \right ) ^{2}e}}-{\frac{{c}^{2}b}{2\,e \left ({c}^{2}{d}^{2}+{e}^{2} \right ) }\sqrt{ \left ( cx+{\frac{cd}{e}} \right ) ^{2}-2\,{\frac{cd}{e} \left ( cx+{\frac{cd}{e}} \right ) }+{\frac{{c}^{2}{d}^{2}+{e}^{2}}{{e}^{2}}}} \left ( cx+{\frac{cd}{e}} \right ) ^{-1}}-{\frac{b{c}^{3}d}{2\,{e}^{2} \left ({c}^{2}{d}^{2}+{e}^{2} \right ) }\ln \left ({ \left ( 2\,{\frac{{c}^{2}{d}^{2}+{e}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ( cx+{\frac{cd}{e}} \right ) }+2\,\sqrt{{\frac{{c}^{2}{d}^{2}+{e}^{2}}{{e}^{2}}}}\sqrt{ \left ( cx+{\frac{cd}{e}} \right ) ^{2}-2\,{\frac{cd}{e} \left ( cx+{\frac{cd}{e}} \right ) }+{\frac{{c}^{2}{d}^{2}+{e}^{2}}{{e}^{2}}}} \right ) \left ( cx+{\frac{cd}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{d}^{2}+{e}^{2}}{{e}^{2}}}}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 3.44087, size = 1146, normalized size = 8.95 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asinh}{\left (c x \right )}}{\left (d + e x\right )^{3}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

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)