3.11 \(\int \frac {a+b \sinh ^{-1}(c x)}{(d+e x)^4} \, dx\)

Optimal. Leaf size=183 \[ -\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}-\frac {b c \sqrt {c^2 x^2+1}}{6 \left (c^2 d^2+e^2\right ) (d+e x)^2}-\frac {b c^3 d \sqrt {c^2 x^2+1}}{2 \left (c^2 d^2+e^2\right )^2 (d+e x)}-\frac {b c^3 \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{6 e \left (c^2 d^2+e^2\right )^{5/2}} \]

[Out]

1/3*(-a-b*arcsinh(c*x))/e/(e*x+d)^3-1/6*b*c^3*(2*c^2*d^2-e^2)*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)^(5/2)-1/6*b*c*(c^2*x^2+1)^(1/2)/(c^2*d^2+e^2)/(e*x+d)^2-1/2*b*c^3*d*(c^2*x^2+1)^(1
/2)/(c^2*d^2+e^2)^2/(e*x+d)

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Rubi [A]  time = 0.14, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5801, 745, 807, 725, 206} \[ -\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}-\frac {b c^3 d \sqrt {c^2 x^2+1}}{2 \left (c^2 d^2+e^2\right )^2 (d+e x)}-\frac {b c \sqrt {c^2 x^2+1}}{6 \left (c^2 d^2+e^2\right ) (d+e x)^2}-\frac {b c^3 \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}}\right )}{6 e \left (c^2 d^2+e^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(d + e*x)^4,x]

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(6*(c^2*d^2 + e^2)*(d + e*x)^2) - (b*c^3*d*Sqrt[1 + c^2*x^2])/(2*(c^2*d^2 + e^2)^2*(d
+ e*x)) - (a + b*ArcSinh[c*x])/(3*e*(d + e*x)^3) - (b*c^3*(2*c^2*d^2 - e^2)*ArcTanh[(e - c^2*d*x)/(Sqrt[c^2*d
^2 + e^2]*Sqrt[1 + c^2*x^2])])/(6*e*(c^2*d^2 + e^2)^(5/2))

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 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/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[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)^4} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}+\frac {(b c) \int \frac {1}{(d+e x)^3 \sqrt {1+c^2 x^2}} \, dx}{3 e}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 \left (c^2 d^2+e^2\right ) (d+e x)^2}-\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}-\frac {\left (b c^3\right ) \int \frac {-2 d+e x}{(d+e x)^2 \sqrt {1+c^2 x^2}} \, dx}{6 e \left (c^2 d^2+e^2\right )}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 \left (c^2 d^2+e^2\right ) (d+e x)^2}-\frac {b c^3 d \sqrt {1+c^2 x^2}}{2 \left (c^2 d^2+e^2\right )^2 (d+e x)}-\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}+\frac {\left (b c^3 \left (2 c^2 d^2-e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {1+c^2 x^2}} \, dx}{6 e \left (c^2 d^2+e^2\right )^2}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 \left (c^2 d^2+e^2\right ) (d+e x)^2}-\frac {b c^3 d \sqrt {1+c^2 x^2}}{2 \left (c^2 d^2+e^2\right )^2 (d+e x)}-\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}-\frac {\left (b c^3 \left (2 c^2 d^2-e^2\right )\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 )}{6 e \left (c^2 d^2+e^2\right )^2}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 \left (c^2 d^2+e^2\right ) (d+e x)^2}-\frac {b c^3 d \sqrt {1+c^2 x^2}}{2 \left (c^2 d^2+e^2\right )^2 (d+e x)}-\frac {a+b \sinh ^{-1}(c x)}{3 e (d+e x)^3}-\frac {b c^3 \left (2 c^2 d^2-e^2\right ) \tanh ^{-1}\left (\frac {e-c^2 d x}{\sqrt {c^2 d^2+e^2} \sqrt {1+c^2 x^2}}\right )}{6 e \left (c^2 d^2+e^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 205, normalized size = 1.12 \[ \frac {1}{6} \left (-\frac {2 a}{e (d+e x)^3}-\frac {b c \sqrt {c^2 x^2+1} \left (c^2 d (4 d+3 e x)+e^2\right )}{\left (c^2 d^2+e^2\right )^2 (d+e x)^2}+\frac {b c^3 \left (e^2-2 c^2 d^2\right ) \log \left (\sqrt {c^2 x^2+1} \sqrt {c^2 d^2+e^2}+c^2 (-d) x+e\right )}{e \left (c^2 d^2+e^2\right )^{5/2}}-\frac {b c^3 \left (e^2-2 c^2 d^2\right ) \log (d+e x)}{e \left (c^2 d^2+e^2\right )^{5/2}}-\frac {2 b \sinh ^{-1}(c x)}{e (d+e x)^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(d + e*x)^4,x]

[Out]

((-2*a)/(e*(d + e*x)^3) - (b*c*Sqrt[1 + c^2*x^2]*(e^2 + c^2*d*(4*d + 3*e*x)))/((c^2*d^2 + e^2)^2*(d + e*x)^2)
- (2*b*ArcSinh[c*x])/(e*(d + e*x)^3) - (b*c^3*(-2*c^2*d^2 + e^2)*Log[d + e*x])/(e*(c^2*d^2 + e^2)^(5/2)) + (b*
c^3*(-2*c^2*d^2 + e^2)*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)^(5/2)))/6

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fricas [B]  time = 1.67, size = 977, normalized size = 5.34 \[ -\frac {{\left (2 \, a + 3 \, b\right )} c^{6} d^{9} + 3 \, {\left (2 \, a + b\right )} c^{4} d^{7} e^{2} + 6 \, a c^{2} d^{5} e^{4} + 2 \, a d^{3} e^{6} + 3 \, {\left (b c^{6} d^{6} e^{3} + b c^{4} d^{4} e^{5}\right )} x^{3} + 9 \, {\left (b c^{6} d^{7} e^{2} + b c^{4} d^{5} e^{4}\right )} x^{2} + {\left (2 \, b c^{5} d^{8} - b c^{3} d^{6} e^{2} + {\left (2 \, b c^{5} d^{5} e^{3} - b c^{3} d^{3} e^{5}\right )} x^{3} + 3 \, {\left (2 \, b c^{5} d^{6} e^{2} - b c^{3} d^{4} e^{4}\right )} x^{2} + 3 \, {\left (2 \, b c^{5} d^{7} e - b c^{3} d^{5} e^{3}\right )} x\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 ) + 9 \, {\left (b c^{6} d^{8} e + b c^{4} d^{6} e^{3}\right )} x - 2 \, {\left ({\left (b c^{6} d^{6} e^{3} + 3 \, b c^{4} d^{4} e^{5} + 3 \, b c^{2} d^{2} e^{7} + b e^{9}\right )} x^{3} + 3 \, {\left (b c^{6} d^{7} e^{2} + 3 \, b c^{4} d^{5} e^{4} + 3 \, b c^{2} d^{3} e^{6} + b d e^{8}\right )} x^{2} + 3 \, {\left (b c^{6} d^{8} e + 3 \, b c^{4} d^{6} e^{3} + 3 \, b c^{2} d^{4} e^{5} + b d^{2} e^{7}\right )} x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (b c^{6} d^{9} + 3 \, b c^{4} d^{7} e^{2} + 3 \, b c^{2} d^{5} e^{4} + b d^{3} e^{6} + {\left (b c^{6} d^{6} e^{3} + 3 \, b c^{4} d^{4} e^{5} + 3 \, b c^{2} d^{2} e^{7} + b e^{9}\right )} x^{3} + 3 \, {\left (b c^{6} d^{7} e^{2} + 3 \, b c^{4} d^{5} e^{4} + 3 \, b c^{2} d^{3} e^{6} + b d e^{8}\right )} x^{2} + 3 \, {\left (b c^{6} d^{8} e + 3 \, b c^{4} d^{6} e^{3} + 3 \, b c^{2} d^{4} e^{5} + b d^{2} e^{7}\right )} x\right )} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (4 \, b c^{5} d^{8} e + 5 \, b c^{3} d^{6} e^{3} + b c d^{4} e^{5} + 3 \, {\left (b c^{5} d^{6} e^{3} + b c^{3} d^{4} e^{5}\right )} x^{2} + {\left (7 \, b c^{5} d^{7} e^{2} + 8 \, b c^{3} d^{5} e^{4} + b c d^{3} e^{6}\right )} x\right )} \sqrt {c^{2} x^{2} + 1}}{6 \, {\left (c^{6} d^{12} e + 3 \, c^{4} d^{10} e^{3} + 3 \, c^{2} d^{8} e^{5} + d^{6} e^{7} + {\left (c^{6} d^{9} e^{4} + 3 \, c^{4} d^{7} e^{6} + 3 \, c^{2} d^{5} e^{8} + d^{3} e^{10}\right )} x^{3} + 3 \, {\left (c^{6} d^{10} e^{3} + 3 \, c^{4} d^{8} e^{5} + 3 \, c^{2} d^{6} e^{7} + d^{4} e^{9}\right )} x^{2} + 3 \, {\left (c^{6} d^{11} e^{2} + 3 \, c^{4} d^{9} e^{4} + 3 \, c^{2} d^{7} e^{6} + d^{5} e^{8}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x+d)^4,x, algorithm="fricas")

[Out]

-1/6*((2*a + 3*b)*c^6*d^9 + 3*(2*a + b)*c^4*d^7*e^2 + 6*a*c^2*d^5*e^4 + 2*a*d^3*e^6 + 3*(b*c^6*d^6*e^3 + b*c^4
*d^4*e^5)*x^3 + 9*(b*c^6*d^7*e^2 + b*c^4*d^5*e^4)*x^2 + (2*b*c^5*d^8 - b*c^3*d^6*e^2 + (2*b*c^5*d^5*e^3 - b*c^
3*d^3*e^5)*x^3 + 3*(2*b*c^5*d^6*e^2 - b*c^3*d^4*e^4)*x^2 + 3*(2*b*c^5*d^7*e - b*c^3*d^5*e^3)*x)*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)) + 9*(b*c^6*d^8*e + b*c^4*d^6*e^3)*x - 2*((b*c^6*d^6*e^3 + 3*b*c^4*d^4*e^5 + 3*b*
c^2*d^2*e^7 + b*e^9)*x^3 + 3*(b*c^6*d^7*e^2 + 3*b*c^4*d^5*e^4 + 3*b*c^2*d^3*e^6 + b*d*e^8)*x^2 + 3*(b*c^6*d^8*
e + 3*b*c^4*d^6*e^3 + 3*b*c^2*d^4*e^5 + b*d^2*e^7)*x)*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(b*c^6*d^9 + 3*b*c^4*d^
7*e^2 + 3*b*c^2*d^5*e^4 + b*d^3*e^6 + (b*c^6*d^6*e^3 + 3*b*c^4*d^4*e^5 + 3*b*c^2*d^2*e^7 + b*e^9)*x^3 + 3*(b*c
^6*d^7*e^2 + 3*b*c^4*d^5*e^4 + 3*b*c^2*d^3*e^6 + b*d*e^8)*x^2 + 3*(b*c^6*d^8*e + 3*b*c^4*d^6*e^3 + 3*b*c^2*d^4
*e^5 + b*d^2*e^7)*x)*log(-c*x + sqrt(c^2*x^2 + 1)) + (4*b*c^5*d^8*e + 5*b*c^3*d^6*e^3 + b*c*d^4*e^5 + 3*(b*c^5
*d^6*e^3 + b*c^3*d^4*e^5)*x^2 + (7*b*c^5*d^7*e^2 + 8*b*c^3*d^5*e^4 + b*c*d^3*e^6)*x)*sqrt(c^2*x^2 + 1))/(c^6*d
^12*e + 3*c^4*d^10*e^3 + 3*c^2*d^8*e^5 + d^6*e^7 + (c^6*d^9*e^4 + 3*c^4*d^7*e^6 + 3*c^2*d^5*e^8 + d^3*e^10)*x^
3 + 3*(c^6*d^10*e^3 + 3*c^4*d^8*e^5 + 3*c^2*d^6*e^7 + d^4*e^9)*x^2 + 3*(c^6*d^11*e^2 + 3*c^4*d^9*e^4 + 3*c^2*d
^7*e^6 + d^5*e^8)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (e x + d\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/(e*x + d)^4, x)

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maple [B]  time = 0.01, size = 516, normalized size = 2.82 \[ -\frac {c^{3} a}{3 \left (c e x +c d \right )^{3} e}-\frac {c^{3} b \arcsinh \left (c x \right )}{3 \left (c e x +c d \right )^{3} e}-\frac {c^{3} b \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{6 e^{2} \left (c^{2} d^{2}+e^{2}\right ) \left (c x +\frac {c d}{e}\right )^{2}}-\frac {c^{4} b d \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{2 e \left (c^{2} d^{2}+e^{2}\right )^{2} \left (c x +\frac {c d}{e}\right )}-\frac {c^{5} b \,d^{2} \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{2 e^{2} \left (c^{2} d^{2}+e^{2}\right )^{2} \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}+\frac {c^{3} b \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{6 e^{2} \left (c^{2} d^{2}+e^{2}\right ) \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(e*x+d)^4,x)

[Out]

-1/3*c^3*a/(c*e*x+c*d)^3/e-1/3*c^3*b/(c*e*x+c*d)^3/e*arcsinh(c*x)-1/6*c^3*b/e^2/(c^2*d^2+e^2)/(c*x+c*d/e)^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)-1/2*c^4*b/e*d/(c^2*d^2+e^2)^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^5*b/e^2*d^2/(c^2*d^2+e^2)^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))+1/6*c^3*b/e^2/(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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, {\left (6 \, c \int \frac {1}{3 \, {\left (c^{3} e^{4} x^{6} + 3 \, c^{3} d e^{3} x^{5} + 3 \, c d^{2} e^{2} x^{2} + c d^{3} e x + {\left (3 \, c^{3} d^{2} e^{2} + c e^{4}\right )} x^{4} + {\left (c^{3} d^{3} e + 3 \, c d e^{3}\right )} x^{3} + {\left (c^{2} e^{4} x^{5} + 3 \, c^{2} d e^{3} x^{4} + 3 \, d^{2} e^{2} x + d^{3} e + {\left (3 \, c^{2} d^{2} e^{2} + e^{4}\right )} x^{3} + {\left (c^{2} d^{3} e + 3 \, d e^{3}\right )} x^{2}\right )} \sqrt {c^{2} x^{2} + 1}\right )}}\,{d x} - \frac {2 \, {\left (c^{6} d^{3} - 3 \, c^{4} d e^{2}\right )} \log \left (e x + d\right )}{c^{6} d^{6} e + 3 \, c^{4} d^{4} e^{3} + 3 \, c^{2} d^{2} e^{5} + e^{7}} + \frac {3 \, c^{6} d^{6} + 2 \, c^{4} d^{4} e^{2} - c^{2} d^{2} e^{4} + 2 \, {\left (c^{6} d^{4} e^{2} - c^{2} e^{6}\right )} x^{2} + {\left (5 \, c^{6} d^{5} e + 2 \, c^{4} d^{3} e^{3} - 3 \, c^{2} d e^{5}\right )} x + {\left (c^{6} d^{6} - 3 \, c^{4} d^{4} e^{2} + {\left (c^{6} d^{3} e^{3} - 3 \, c^{4} d e^{5}\right )} x^{3} + 3 \, {\left (c^{6} d^{4} e^{2} - 3 \, c^{4} d^{2} e^{4}\right )} x^{2} + 3 \, {\left (c^{6} d^{5} e - 3 \, c^{4} d^{3} e^{3}\right )} x\right )} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (c^{6} d^{6} + 3 \, c^{4} d^{4} e^{2} + 3 \, c^{2} d^{2} e^{4} + e^{6}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{9} e + 3 \, c^{4} d^{7} e^{3} + 3 \, c^{2} d^{5} e^{5} + d^{3} e^{7} + {\left (c^{6} d^{6} e^{4} + 3 \, c^{4} d^{4} e^{6} + 3 \, c^{2} d^{2} e^{8} + e^{10}\right )} x^{3} + 3 \, {\left (c^{6} d^{7} e^{3} + 3 \, c^{4} d^{5} e^{5} + 3 \, c^{2} d^{3} e^{7} + d e^{9}\right )} x^{2} + 3 \, {\left (c^{6} d^{8} e^{2} + 3 \, c^{4} d^{6} e^{4} + 3 \, c^{2} d^{4} e^{6} + d^{2} e^{8}\right )} x} - \frac {i \, {\left (3 \, c^{6} d^{2} - c^{4} e^{2}\right )} {\left (\log \left (i \, c x + 1\right ) - \log \left (-i \, c x + 1\right )\right )}}{{\left (c^{6} d^{6} + 3 \, c^{4} d^{4} e^{2} + 3 \, c^{2} d^{2} e^{4} + e^{6}\right )} c}\right )} b - \frac {a}{3 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*(6*c*integrate(1/3/(c^3*e^4*x^6 + 3*c^3*d*e^3*x^5 + 3*c*d^2*e^2*x^2 + c*d^3*e*x + (3*c^3*d^2*e^2 + c*e^4)*
x^4 + (c^3*d^3*e + 3*c*d*e^3)*x^3 + (c^2*e^4*x^5 + 3*c^2*d*e^3*x^4 + 3*d^2*e^2*x + d^3*e + (3*c^2*d^2*e^2 + e^
4)*x^3 + (c^2*d^3*e + 3*d*e^3)*x^2)*sqrt(c^2*x^2 + 1)), x) - 2*(c^6*d^3 - 3*c^4*d*e^2)*log(e*x + d)/(c^6*d^6*e
+ 3*c^4*d^4*e^3 + 3*c^2*d^2*e^5 + e^7) + (3*c^6*d^6 + 2*c^4*d^4*e^2 - c^2*d^2*e^4 + 2*(c^6*d^4*e^2 - c^2*e^6)
*x^2 + (5*c^6*d^5*e + 2*c^4*d^3*e^3 - 3*c^2*d*e^5)*x + (c^6*d^6 - 3*c^4*d^4*e^2 + (c^6*d^3*e^3 - 3*c^4*d*e^5)*
x^3 + 3*(c^6*d^4*e^2 - 3*c^4*d^2*e^4)*x^2 + 3*(c^6*d^5*e - 3*c^4*d^3*e^3)*x)*log(c^2*x^2 + 1) - 2*(c^6*d^6 + 3
*c^4*d^4*e^2 + 3*c^2*d^2*e^4 + e^6)*log(c*x + sqrt(c^2*x^2 + 1)))/(c^6*d^9*e + 3*c^4*d^7*e^3 + 3*c^2*d^5*e^5 +
d^3*e^7 + (c^6*d^6*e^4 + 3*c^4*d^4*e^6 + 3*c^2*d^2*e^8 + e^10)*x^3 + 3*(c^6*d^7*e^3 + 3*c^4*d^5*e^5 + 3*c^2*d
^3*e^7 + d*e^9)*x^2 + 3*(c^6*d^8*e^2 + 3*c^4*d^6*e^4 + 3*c^2*d^4*e^6 + d^2*e^8)*x) - I*(3*c^6*d^2 - c^4*e^2)*(
log(I*c*x + 1) - log(-I*c*x + 1))/((c^6*d^6 + 3*c^4*d^4*e^2 + 3*c^2*d^2*e^4 + e^6)*c))*b - 1/3*a/(e^4*x^3 + 3*
d*e^3*x^2 + 3*d^2*e^2*x + d^3*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/(d + e*x)^4,x)

[Out]

int((a + b*asinh(c*x))/(d + e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (d + e x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/(e*x+d)**4,x)

[Out]

Integral((a + b*asinh(c*x))/(d + e*x)**4, x)

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