3.1.10 \(\int \frac {a+b \sinh ^{-1}(c x)}{(d+e x)^3} \, dx\) [10]

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

[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 = 0.06, 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 = {5828, 745, 739, 212} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*c*Sqrt[1 + c^2*x^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*A
rcTanh[(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 212

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

Rule 739

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 745

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 5828

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 ) \text {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 = 0.26, size = 166, normalized size = 1.30 \begin {gather*} \frac {1}{2} \left (-\frac {a}{e (d+e x)^2}-\frac {b c \sqrt {1+c^2 x^2}}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac {b \sinh ^{-1}(c x)}{e (d+e x)^2}+\frac {b c^3 d \log (d+e x)}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac {b c^3 d \log \left (e-c^2 d x+\sqrt {c^2 d^2+e^2} \sqrt {1+c^2 x^2}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}\right ) \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(282\) vs. \(2(117)=234\).
time = 4.52, size = 283, normalized size = 2.21

method	result	size
derivativedivides	\(\frac {-\frac {a \,c^{3}}{2 \left (c e x +c d \right )^{2} e}-\frac {b \,c^{3} \arcsinh \left (c x \right )}{2 \left (c e x +c d \right )^{2} e}-\frac {b \,c^{3} \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 d c \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{2 e \left (c^{2} d^{2}+e^{2}\right ) \left (c x +\frac {c d}{e}\right )}-\frac {b \,c^{4} d \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 d c \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 d c \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{2 e^{2} \left (c^{2} d^{2}+e^{2}\right ) \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}}{c}\)	\(283\)
default	\(\frac {-\frac {a \,c^{3}}{2 \left (c e x +c d \right )^{2} e}-\frac {b \,c^{3} \arcsinh \left (c x \right )}{2 \left (c e x +c d \right )^{2} e}-\frac {b \,c^{3} \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 d c \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{2 e \left (c^{2} d^{2}+e^{2}\right ) \left (c x +\frac {c d}{e}\right )}-\frac {b \,c^{4} d \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 d c \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 d c \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{2 e^{2} \left (c^{2} d^{2}+e^{2}\right ) \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}}{c}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-1/2*a*c^3/(c*e*x+c*d)^2/e-1/2*b*c^3/(c*e*x+c*d)^2/e*arcsinh(c*x)-1/2*b*c^3/e/(c^2*d^2+e^2)/(c*x+c*d/e)*(
(c*x+c*d/e)^2-2*d*c/e*(c*x+c*d/e)+(c^2*d^2+e^2)/e^2)^(1/2)-1/2*b*c^4/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*d*c/e*(c*x+c*d/e)+2*((c^2*d^2+e^2)/e^2)^(1/2)*((c*x+c*d/e)^2-2*d*c/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 = 0.30, size = 151, normalized size = 1.18 \begin {gather*} \frac {1}{2} \, {\left ({\left (\frac {c^{2} d \operatorname {arsinh}\left (\frac {c d x e^{\left (-1\right )}}{{\left | d e^{\left (-1\right )} + x \right |}} - \frac {1}{c {\left | d e^{\left (-1\right )} + x \right |}}\right ) e^{\left (-4\right )}}{{\left (c^{2} d^{2} e^{\left (-2\right )} + 1\right )}^{\frac {3}{2}}} - \frac {\sqrt {c^{2} x^{2} + 1}}{c^{2} d^{2} x e + c^{2} d^{3} + x e^{3} + d e^{2}}\right )} c - \frac {\operatorname {arsinh}\left (c x\right )}{x^{2} e^{3} + 2 \, d x e^{2} + d^{2} e}\right )} b - \frac {a}{2 \, {\left (x^{2} e^{3} + 2 \, d x e^{2} + d^{2} e\right )}} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2021 vs. \(2 (114) = 228\).
time = 0.45, size = 2021, normalized size = 15.79 \begin {gather*} \text {Too large to display} \end {gather*}

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*(2*b*c^4*d^5*x*cosh(1) + (a + b)*c^4*d^6 + 2*b*c^2*d^3*x*cosh(1)^3 + (b*c^2*d^2*x^2 + a*d^2)*cosh(1)^4 +
(b*c^2*d^2*x^2 + a*d^2)*sinh(1)^4 + 2*(b*c^2*d^3*x + 2*(b*c^2*d^2*x^2 + a*d^2)*cosh(1))*sinh(1)^3 + (b*c^4*d^4
*x^2 + (2*a + b)*c^2*d^4)*cosh(1)^2 + (b*c^4*d^4*x^2 + 6*b*c^2*d^3*x*cosh(1) + (2*a + b)*c^2*d^4 + 6*(b*c^2*d^
2*x^2 + a*d^2)*cosh(1)^2)*sinh(1)^2 - (b*c^3*d^3*x^2*cosh(1)^2 + b*c^3*d^3*x^2*sinh(1)^2 + 2*b*c^3*d^4*x*cosh(
1) + b*c^3*d^5 + 2*(b*c^3*d^3*x^2*cosh(1) + b*c^3*d^4*x)*sinh(1))*sqrt(((c^2*d^2 + 1)*cosh(1) - (c^2*d^2 - 1)*
sinh(1))/(cosh(1) - sinh(1)))*log(-(c^3*d^2*x - c*d*cosh(1) - c*d*sinh(1) + (c^2*d^2 + c*d*sqrt(((c^2*d^2 + 1)
*cosh(1) - (c^2*d^2 - 1)*sinh(1))/(cosh(1) - sinh(1))) + cosh(1)^2 + 2*cosh(1)*sinh(1) + sinh(1)^2)*sqrt(c^2*x
^2 + 1) + (c^2*d*x - cosh(1) - sinh(1))*sqrt(((c^2*d^2 + 1)*cosh(1) - (c^2*d^2 - 1)*sinh(1))/(cosh(1) - sinh(1
))))/(x*cosh(1) + x*sinh(1) + d)) - (b*c^4*d^4*x^2*cosh(1)^2 + 2*b*c^4*d^5*x*cosh(1) + 2*b*c^2*d^2*x^2*cosh(1)
^4 + 4*b*c^2*d^3*x*cosh(1)^3 + b*x^2*cosh(1)^6 + b*x^2*sinh(1)^6 + 2*b*d*x*cosh(1)^5 + 2*(3*b*x^2*cosh(1) + b*
d*x)*sinh(1)^5 + (2*b*c^2*d^2*x^2 + 15*b*x^2*cosh(1)^2 + 10*b*d*x*cosh(1))*sinh(1)^4 + 4*(2*b*c^2*d^2*x^2*cosh
(1) + b*c^2*d^3*x + 5*b*x^2*cosh(1)^3 + 5*b*d*x*cosh(1)^2)*sinh(1)^3 + (b*c^4*d^4*x^2 + 12*b*c^2*d^2*x^2*cosh(
1)^2 + 12*b*c^2*d^3*x*cosh(1) + 15*b*x^2*cosh(1)^4 + 20*b*d*x*cosh(1)^3)*sinh(1)^2 + 2*(b*c^4*d^4*x^2*cosh(1)
+ b*c^4*d^5*x + 4*b*c^2*d^2*x^2*cosh(1)^3 + 6*b*c^2*d^3*x*cosh(1)^2 + 3*b*x^2*cosh(1)^5 + 5*b*d*x*cosh(1)^4)*s
inh(1))*log(c*x + sqrt(c^2*x^2 + 1)) - (2*b*c^4*d^5*x*cosh(1) + b*c^4*d^6 + 4*b*c^2*d^3*x*cosh(1)^3 + b*x^2*co
sh(1)^6 + b*x^2*sinh(1)^6 + 2*b*d*x*cosh(1)^5 + 2*(3*b*x^2*cosh(1) + b*d*x)*sinh(1)^5 + (2*b*c^2*d^2*x^2 + b*d
^2)*cosh(1)^4 + (2*b*c^2*d^2*x^2 + 15*b*x^2*cosh(1)^2 + 10*b*d*x*cosh(1) + b*d^2)*sinh(1)^4 + 4*(b*c^2*d^3*x +
5*b*x^2*cosh(1)^3 + 5*b*d*x*cosh(1)^2 + (2*b*c^2*d^2*x^2 + b*d^2)*cosh(1))*sinh(1)^3 + (b*c^4*d^4*x^2 + 2*b*c
^2*d^4)*cosh(1)^2 + (b*c^4*d^4*x^2 + 12*b*c^2*d^3*x*cosh(1) + 2*b*c^2*d^4 + 15*b*x^2*cosh(1)^4 + 20*b*d*x*cosh
(1)^3 + 6*(2*b*c^2*d^2*x^2 + b*d^2)*cosh(1)^2)*sinh(1)^2 + 2*(b*c^4*d^5*x + 6*b*c^2*d^3*x*cosh(1)^2 + 3*b*x^2*
cosh(1)^5 + 5*b*d*x*cosh(1)^4 + 2*(2*b*c^2*d^2*x^2 + b*d^2)*cosh(1)^3 + (b*c^4*d^4*x^2 + 2*b*c^2*d^4)*cosh(1))
*sinh(1))*log(-c*x + sqrt(c^2*x^2 + 1)) + 2*(b*c^4*d^5*x + 3*b*c^2*d^3*x*cosh(1)^2 + 2*(b*c^2*d^2*x^2 + a*d^2)
*cosh(1)^3 + (b*c^4*d^4*x^2 + (2*a + b)*c^2*d^4)*cosh(1))*sinh(1) + (b*c^3*d^4*x*cosh(1)^2 + b*c^3*d^5*cosh(1)
+ b*c*d^2*x*cosh(1)^4 + b*c*d^2*x*sinh(1)^4 + b*c*d^3*cosh(1)^3 + (4*b*c*d^2*x*cosh(1) + b*c*d^3)*sinh(1)^3 +
(b*c^3*d^4*x + 6*b*c*d^2*x*cosh(1)^2 + 3*b*c*d^3*cosh(1))*sinh(1)^2 + (2*b*c^3*d^4*x*cosh(1) + b*c^3*d^5 + 4*
b*c*d^2*x*cosh(1)^3 + 3*b*c*d^3*cosh(1)^2)*sinh(1))*sqrt(c^2*x^2 + 1))/(2*c^4*d^7*x*cosh(1)^2 + c^4*d^8*cosh(1
) + 4*c^2*d^5*x*cosh(1)^4 + d^2*x^2*cosh(1)^7 + d^2*x^2*sinh(1)^7 + 2*d^3*x*cosh(1)^6 + (7*d^2*x^2*cosh(1) + 2
*d^3*x)*sinh(1)^6 + (2*c^2*d^4*x^2 + d^4)*cosh(1)^5 + (2*c^2*d^4*x^2 + 21*d^2*x^2*cosh(1)^2 + 12*d^3*x*cosh(1)
+ d^4)*sinh(1)^5 + (4*c^2*d^5*x + 35*d^2*x^2*cosh(1)^3 + 30*d^3*x*cosh(1)^2 + 5*(2*c^2*d^4*x^2 + d^4)*cosh(1)
)*sinh(1)^4 + (c^4*d^6*x^2 + 2*c^2*d^6)*cosh(1)^3 + (c^4*d^6*x^2 + 16*c^2*d^5*x*cosh(1) + 2*c^2*d^6 + 35*d^2*x
^2*cosh(1)^4 + 40*d^3*x*cosh(1)^3 + 10*(2*c^2*d^4*x^2 + d^4)*cosh(1)^2)*sinh(1)^3 + (2*c^4*d^7*x + 24*c^2*d^5*
x*cosh(1)^2 + 21*d^2*x^2*cosh(1)^5 + 30*d^3*x*cosh(1)^4 + 10*(2*c^2*d^4*x^2 + d^4)*cosh(1)^3 + 3*(c^4*d^6*x^2
+ 2*c^2*d^6)*cosh(1))*sinh(1)^2 + (4*c^4*d^7*x*cosh(1) + c^4*d^8 + 16*c^2*d^5*x*cosh(1)^3 + 7*d^2*x^2*cosh(1)^
6 + 12*d^3*x*cosh(1)^5 + 5*(2*c^2*d^4*x^2 + d^4)*cosh(1)^4 + 3*(c^4*d^6*x^2 + 2*c^2*d^6)*cosh(1)^2)*sinh(1))

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

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

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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