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

Optimal. Leaf size=115 \[ -\frac {24 b^3 \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac {12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac {4 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+24 b^4 x \]

[Out]

24*b^4*x+12*b^2*(d*x+c)*(a+b*arcsinh(d*x+c))^2/d+(d*x+c)*(a+b*arcsinh(d*x+c))^4/d-24*b^3*(a+b*arcsinh(d*x+c))*
(1+(d*x+c)^2)^(1/2)/d-4*b*(a+b*arcsinh(d*x+c))^3*(1+(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.16, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5863, 5653, 5717, 8} \[ -\frac {24 b^3 \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac {12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac {4 b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+24 b^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])^4,x]

[Out]

24*b^4*x - (24*b^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/d + (12*b^2*(c + d*x)*(a + b*ArcSinh[c + d*
x])^2)/d - (4*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^3)/d + ((c + d*x)*(a + b*ArcSinh[c + d*x])^4)/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5717

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

Rule 5863

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSinh[x])^n, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \left (a+b \sinh ^{-1}(c+d x)\right )^4 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^4 \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {x \left (a+b \sinh ^{-1}(x)\right )^3}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {4 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+\frac {\left (12 b^2\right ) \operatorname {Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac {4 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}-\frac {\left (24 b^3\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {24 b^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac {12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac {4 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}+\frac {\left (24 b^4\right ) \operatorname {Subst}(\int 1 \, dx,x,c+d x)}{d}\\ &=24 b^4 x-\frac {24 b^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{d}+\frac {12 b^2 (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d}-\frac {4 b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^4}{d}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 226, normalized size = 1.97 \[ \frac {-4 a b \left (a^2+6 b^2\right ) \sqrt {(c+d x)^2+1}+6 b^2 \sinh ^{-1}(c+d x)^2 \left (a^2 (c+d x)-2 a b \sqrt {(c+d x)^2+1}+2 b^2 (c+d x)\right )+\left (a^4+12 a^2 b^2+24 b^4\right ) (c+d x)-4 b \sinh ^{-1}(c+d x) \left (-\left (a^3 (c+d x)\right )+3 a^2 b \sqrt {(c+d x)^2+1}-6 a b^2 (c+d x)+6 b^3 \sqrt {(c+d x)^2+1}\right )-4 b^3 \sinh ^{-1}(c+d x)^3 \left (b \sqrt {(c+d x)^2+1}-a (c+d x)\right )+b^4 (c+d x) \sinh ^{-1}(c+d x)^4}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^4 + 12*a^2*b^2 + 24*b^4)*(c + d*x) - 4*a*b*(a^2 + 6*b^2)*Sqrt[1 + (c + d*x)^2] - 4*b*(-(a^3*(c + d*x)) - 6
*a*b^2*(c + d*x) + 3*a^2*b*Sqrt[1 + (c + d*x)^2] + 6*b^3*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] + 6*b^2*(a^2*
(c + d*x) + 2*b^2*(c + d*x) - 2*a*b*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x]^2 - 4*b^3*(-(a*(c + d*x)) + b*Sqrt
[1 + (c + d*x)^2])*ArcSinh[c + d*x]^3 + b^4*(c + d*x)*ArcSinh[c + d*x]^4)/d

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fricas [B]  time = 0.54, size = 344, normalized size = 2.99 \[ \frac {{\left (b^{4} d x + b^{4} c\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{4} + 4 \, {\left (a b^{3} d x + a b^{3} c - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} b^{4}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + {\left (a^{4} + 12 \, a^{2} b^{2} + 24 \, b^{4}\right )} d x - 6 \, {\left (2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} a b^{3} - {\left (a^{2} b^{2} + 2 \, b^{4}\right )} d x - {\left (a^{2} b^{2} + 2 \, b^{4}\right )} c\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 4 \, {\left ({\left (a^{3} b + 6 \, a b^{3}\right )} d x + {\left (a^{3} b + 6 \, a b^{3}\right )} c - 3 \, {\left (a^{2} b^{2} + 2 \, b^{4}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 4 \, {\left (a^{3} b + 6 \, a b^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b^4*d*x + b^4*c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4 + 4*(a*b^3*d*x + a*b^3*c - sqrt(d^2*x^2
+ 2*c*d*x + c^2 + 1)*b^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3 + (a^4 + 12*a^2*b^2 + 24*b^4)*d*x
- 6*(2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*a*b^3 - (a^2*b^2 + 2*b^4)*d*x - (a^2*b^2 + 2*b^4)*c)*log(d*x + c + s
qrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2 + 4*((a^3*b + 6*a*b^3)*d*x + (a^3*b + 6*a*b^3)*c - 3*(a^2*b^2 + 2*b^4)*sqr
t(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - 4*(a^3*b + 6*a*b^3)*sqrt(d^
2*x^2 + 2*c*d*x + c^2 + 1))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 245, normalized size = 2.13 \[ \frac {\left (d x +c \right ) a^{4}+b^{4} \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )^{4}-4 \arcsinh \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}+12 \left (d x +c \right ) \arcsinh \left (d x +c \right )^{2}-24 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+24 d x +24 c \right )+4 a \,b^{3} \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )^{3}-3 \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}+6 \left (d x +c \right ) \arcsinh \left (d x +c \right )-6 \sqrt {1+\left (d x +c \right )^{2}}\right )+6 a^{2} b^{2} \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )^{2}-2 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}+2 d x +2 c \right )+4 a^{3} b \left (\left (d x +c \right ) \arcsinh \left (d x +c \right )-\sqrt {1+\left (d x +c \right )^{2}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*((d*x+c)*a^4+b^4*((d*x+c)*arcsinh(d*x+c)^4-4*arcsinh(d*x+c)^3*(1+(d*x+c)^2)^(1/2)+12*(d*x+c)*arcsinh(d*x+c
)^2-24*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+24*d*x+24*c)+4*a*b^3*((d*x+c)*arcsinh(d*x+c)^3-3*arcsinh(d*x+c)^2*(1
+(d*x+c)^2)^(1/2)+6*(d*x+c)*arcsinh(d*x+c)-6*(1+(d*x+c)^2)^(1/2))+6*a^2*b^2*((d*x+c)*arcsinh(d*x+c)^2-2*arcsin
h(d*x+c)*(1+(d*x+c)^2)^(1/2)+2*d*x+2*c)+4*a^3*b*((d*x+c)*arcsinh(d*x+c)-(1+(d*x+c)^2)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{4} x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{4} + a^{4} x + \frac {4 \, {\left ({\left (d x + c\right )} \operatorname {arsinh}\left (d x + c\right ) - \sqrt {{\left (d x + c\right )}^{2} + 1}\right )} a^{3} b}{d} + \int \frac {2 \, {\left (2 \, {\left ({\left (c^{3} + c\right )} a b^{3} + {\left (a b^{3} d^{3} - b^{4} d^{3}\right )} x^{3} + {\left (3 \, a b^{3} c d^{2} - 2 \, b^{4} c d^{2}\right )} x^{2} + {\left ({\left (3 \, c^{2} d + d\right )} a b^{3} - {\left (c^{2} d + d\right )} b^{4}\right )} x + {\left ({\left (c^{2} + 1\right )} a b^{3} + {\left (a b^{3} d^{2} - b^{4} d^{2}\right )} x^{2} + {\left (2 \, a b^{3} c d - b^{4} c d\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + 3 \, {\left (a^{2} b^{2} d^{3} x^{3} + 3 \, a^{2} b^{2} c d^{2} x^{2} + {\left (3 \, c^{2} d + d\right )} a^{2} b^{2} x + {\left (c^{3} + c\right )} a^{2} b^{2} + {\left (a^{2} b^{2} d^{2} x^{2} + 2 \, a^{2} b^{2} c d x + {\left (c^{2} + 1\right )} a^{2} b^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2}\right )}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^4*x*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^4 + a^4*x + 4*((d*x + c)*arcsinh(d*x + c) - sqrt((d*x +
c)^2 + 1))*a^3*b/d + integrate(2*(2*((c^3 + c)*a*b^3 + (a*b^3*d^3 - b^4*d^3)*x^3 + (3*a*b^3*c*d^2 - 2*b^4*c*d
^2)*x^2 + ((3*c^2*d + d)*a*b^3 - (c^2*d + d)*b^4)*x + ((c^2 + 1)*a*b^3 + (a*b^3*d^2 - b^4*d^2)*x^2 + (2*a*b^3*
c*d - b^4*c*d)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3 + 3*(a
^2*b^2*d^3*x^3 + 3*a^2*b^2*c*d^2*x^2 + (3*c^2*d + d)*a^2*b^2*x + (c^3 + c)*a^2*b^2 + (a^2*b^2*d^2*x^2 + 2*a^2*
b^2*c*d*x + (c^2 + 1)*a^2*b^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 +
1))^2)/(d^3*x^3 + 3*c*d^2*x^2 + c^3 + (3*c^2*d + d)*x + (d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c + d*x))^4,x)

[Out]

int((a + b*asinh(c + d*x))^4, x)

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sympy [A]  time = 1.69, size = 444, normalized size = 3.86 \[ \begin {cases} a^{4} x + \frac {4 a^{3} b c \operatorname {asinh}{\left (c + d x \right )}}{d} + 4 a^{3} b x \operatorname {asinh}{\left (c + d x \right )} - \frac {4 a^{3} b \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac {6 a^{2} b^{2} c \operatorname {asinh}^{2}{\left (c + d x \right )}}{d} + 6 a^{2} b^{2} x \operatorname {asinh}^{2}{\left (c + d x \right )} + 12 a^{2} b^{2} x - \frac {12 a^{2} b^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{d} + \frac {4 a b^{3} c \operatorname {asinh}^{3}{\left (c + d x \right )}}{d} + \frac {24 a b^{3} c \operatorname {asinh}{\left (c + d x \right )}}{d} + 4 a b^{3} x \operatorname {asinh}^{3}{\left (c + d x \right )} + 24 a b^{3} x \operatorname {asinh}{\left (c + d x \right )} - \frac {12 a b^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (c + d x \right )}}{d} - \frac {24 a b^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{d} + \frac {b^{4} c \operatorname {asinh}^{4}{\left (c + d x \right )}}{d} + \frac {12 b^{4} c \operatorname {asinh}^{2}{\left (c + d x \right )}}{d} + b^{4} x \operatorname {asinh}^{4}{\left (c + d x \right )} + 12 b^{4} x \operatorname {asinh}^{2}{\left (c + d x \right )} + 24 b^{4} x - \frac {4 b^{4} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (c + d x \right )}}{d} - \frac {24 b^{4} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {asinh}{\relax (c )}\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x + 4*a**3*b*c*asinh(c + d*x)/d + 4*a**3*b*x*asinh(c + d*x) - 4*a**3*b*sqrt(c**2 + 2*c*d*x + d
**2*x**2 + 1)/d + 6*a**2*b**2*c*asinh(c + d*x)**2/d + 6*a**2*b**2*x*asinh(c + d*x)**2 + 12*a**2*b**2*x - 12*a*
*2*b**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/d + 4*a*b**3*c*asinh(c + d*x)**3/d + 24*a*b**3*c*a
sinh(c + d*x)/d + 4*a*b**3*x*asinh(c + d*x)**3 + 24*a*b**3*x*asinh(c + d*x) - 12*a*b**3*sqrt(c**2 + 2*c*d*x +
d**2*x**2 + 1)*asinh(c + d*x)**2/d - 24*a*b**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/d + b**4*c*asinh(c + d*x)*
*4/d + 12*b**4*c*asinh(c + d*x)**2/d + b**4*x*asinh(c + d*x)**4 + 12*b**4*x*asinh(c + d*x)**2 + 24*b**4*x - 4*
b**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**3/d - 24*b**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*a
sinh(c + d*x)/d, Ne(d, 0)), (x*(a + b*asinh(c))**4, True))

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