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

Optimal. Leaf size=100 \[ \frac {e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac {b e^4 \left ((c+d x)^2+1\right )^{5/2}}{25 d}+\frac {2 b e^4 \left ((c+d x)^2+1\right )^{3/2}}{15 d}-\frac {b e^4 \sqrt {(c+d x)^2+1}}{5 d} \]

[Out]

2/15*b*e^4*(1+(d*x+c)^2)^(3/2)/d-1/25*b*e^4*(1+(d*x+c)^2)^(5/2)/d+1/5*e^4*(d*x+c)^5*(a+b*arcsinh(d*x+c))/d-1/5
*b*e^4*(1+(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.08, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5865, 12, 5661, 266, 43} \[ \frac {e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac {b e^4 \left ((c+d x)^2+1\right )^{5/2}}{25 d}+\frac {2 b e^4 \left ((c+d x)^2+1\right )^{3/2}}{15 d}-\frac {b e^4 \sqrt {(c+d x)^2+1}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e^4*Sqrt[1 + (c + d*x)^2])/(5*d) + (2*b*e^4*(1 + (c + d*x)^2)^(3/2))/(15*d) - (b*e^4*(1 + (c + d*x)^2)^(5/
2))/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x]))/(5*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5661

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

Rule 5865

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

Rubi steps

\begin {align*} \int (c e+d e x)^4 \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e^4 x^4 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int x^4 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac {\left (b e^4\right ) \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{5 d}\\ &=\frac {e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac {\left (b e^4\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x}} \, dx,x,(c+d x)^2\right )}{10 d}\\ &=\frac {e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}-\frac {\left (b e^4\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt {1+x}}-2 \sqrt {1+x}+(1+x)^{3/2}\right ) \, dx,x,(c+d x)^2\right )}{10 d}\\ &=-\frac {b e^4 \sqrt {1+(c+d x)^2}}{5 d}+\frac {2 b e^4 \left (1+(c+d x)^2\right )^{3/2}}{15 d}-\frac {b e^4 \left (1+(c+d x)^2\right )^{5/2}}{25 d}+\frac {e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 71, normalized size = 0.71 \[ \frac {e^4 \left (\frac {1}{5} (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )-\frac {1}{75} b \sqrt {(c+d x)^2+1} \left (-10 (c+d x)^2+3 \left ((c+d x)^2+1\right )^2+5\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*(-1/75*(b*Sqrt[1 + (c + d*x)^2]*(5 - 10*(c + d*x)^2 + 3*(1 + (c + d*x)^2)^2)) + ((c + d*x)^5*(a + b*ArcSi
nh[c + d*x]))/5))/d

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fricas [B]  time = 0.56, size = 279, normalized size = 2.79 \[ \frac {15 \, a d^{5} e^{4} x^{5} + 75 \, a c d^{4} e^{4} x^{4} + 150 \, a c^{2} d^{3} e^{4} x^{3} + 150 \, a c^{3} d^{2} e^{4} x^{2} + 75 \, a c^{4} d e^{4} x + 15 \, {\left (b d^{5} e^{4} x^{5} + 5 \, b c d^{4} e^{4} x^{4} + 10 \, b c^{2} d^{3} e^{4} x^{3} + 10 \, b c^{3} d^{2} e^{4} x^{2} + 5 \, b c^{4} d e^{4} x + b c^{5} e^{4}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (3 \, b d^{4} e^{4} x^{4} + 12 \, b c d^{3} e^{4} x^{3} + 2 \, {\left (9 \, b c^{2} - 2 \, b\right )} d^{2} e^{4} x^{2} + 4 \, {\left (3 \, b c^{3} - 2 \, b c\right )} d e^{4} x + {\left (3 \, b c^{4} - 4 \, b c^{2} + 8 \, b\right )} e^{4}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{75 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/75*(15*a*d^5*e^4*x^5 + 75*a*c*d^4*e^4*x^4 + 150*a*c^2*d^3*e^4*x^3 + 150*a*c^3*d^2*e^4*x^2 + 75*a*c^4*d*e^4*x
+ 15*(b*d^5*e^4*x^5 + 5*b*c*d^4*e^4*x^4 + 10*b*c^2*d^3*e^4*x^3 + 10*b*c^3*d^2*e^4*x^2 + 5*b*c^4*d*e^4*x + b*c
^5*e^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (3*b*d^4*e^4*x^4 + 12*b*c*d^3*e^4*x^3 + 2*(9*b*c^2
- 2*b)*d^2*e^4*x^2 + 4*(3*b*c^3 - 2*b*c)*d*e^4*x + (3*b*c^4 - 4*b*c^2 + 8*b)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2
+ 1))/d

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giac [B]  time = 1.93, size = 817, normalized size = 8.17 \[ \frac {1}{600} \, {\left (120 \, a d^{4} x^{5} + 600 \, a c d^{3} x^{4} + 1200 \, a c^{2} d^{2} x^{3} + 1200 \, a c^{3} d x^{2} - 600 \, {\left (d {\left (\frac {c \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d {\left | d \right |}} + \frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{d^{2}}\right )} - x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )\right )} b c^{4} + 600 \, {\left (2 \, x^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (\frac {x}{d^{2}} - \frac {3 \, c}{d^{3}}\right )} - \frac {{\left (2 \, c^{2} - 1\right )} \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d^{2} {\left | d \right |}}\right )} d\right )} b c^{3} d + 200 \, {\left (6 \, x^{3} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (x {\left (\frac {2 \, x}{d^{2}} - \frac {5 \, c}{d^{3}}\right )} + \frac {11 \, c^{2} d - 4 \, d}{d^{5}}\right )} + \frac {3 \, {\left (2 \, c^{3} - 3 \, c\right )} \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d^{3} {\left | d \right |}}\right )} d\right )} b c^{2} d^{2} + 25 \, {\left (24 \, x^{4} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left ({\left (2 \, x {\left (\frac {3 \, x}{d^{2}} - \frac {7 \, c}{d^{3}}\right )} + \frac {26 \, c^{2} d^{3} - 9 \, d^{3}}{d^{7}}\right )} x - \frac {5 \, {\left (10 \, c^{3} d^{2} - 11 \, c d^{2}\right )}}{d^{7}}\right )} - \frac {3 \, {\left (8 \, c^{4} - 24 \, c^{2} + 3\right )} \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d^{4} {\left | d \right |}}\right )} d\right )} b c d^{3} + {\left (120 \, x^{5} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left ({\left (2 \, {\left (3 \, x {\left (\frac {4 \, x}{d^{2}} - \frac {9 \, c}{d^{3}}\right )} + \frac {47 \, c^{2} d^{5} - 16 \, d^{5}}{d^{9}}\right )} x - \frac {7 \, {\left (22 \, c^{3} d^{4} - 23 \, c d^{4}\right )}}{d^{9}}\right )} x + \frac {274 \, c^{4} d^{3} - 607 \, c^{2} d^{3} + 64 \, d^{3}}{d^{9}}\right )} + \frac {15 \, {\left (8 \, c^{5} - 40 \, c^{3} + 15 \, c\right )} \log \left (-c d - {\left (x {\left | d \right |} - \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} {\left | d \right |}\right )}{d^{5} {\left | d \right |}}\right )} d\right )} b d^{4} + 600 \, a c^{4} x\right )} e^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/600*(120*a*d^4*x^5 + 600*a*c*d^3*x^4 + 1200*a*c^2*d^2*x^3 + 1200*a*c^3*d*x^2 - 600*(d*(c*log(-c*d - (x*abs(d
) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d))/(d*abs(d)) + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2) - x*log(d*x
+ c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))*b*c^4 + 600*(2*x^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))
- (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x/d^2 - 3*c/d^3) - (2*c^2 - 1)*log(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*
c*d*x + c^2 + 1))*abs(d))/(d^2*abs(d)))*d)*b*c^3*d + 200*(6*x^3*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1
)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x*(2*x/d^2 - 5*c/d^3) + (11*c^2*d - 4*d)/d^5) + 3*(2*c^3 - 3*c)*log(-
c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d))/(d^3*abs(d)))*d)*b*c^2*d^2 + 25*(24*x^4*log(d*x +
c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*((2*x*(3*x/d^2 - 7*c/d^3) + (26*c
^2*d^3 - 9*d^3)/d^7)*x - 5*(10*c^3*d^2 - 11*c*d^2)/d^7) - 3*(8*c^4 - 24*c^2 + 3)*log(-c*d - (x*abs(d) - sqrt(d
^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d))/(d^4*abs(d)))*d)*b*c*d^3 + (120*x^5*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x
+ c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*((2*(3*x*(4*x/d^2 - 9*c/d^3) + (47*c^2*d^5 - 16*d^5)/d^9)*x -
7*(22*c^3*d^4 - 23*c*d^4)/d^9)*x + (274*c^4*d^3 - 607*c^2*d^3 + 64*d^3)/d^9) + 15*(8*c^5 - 40*c^3 + 15*c)*log
(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d))/(d^5*abs(d)))*d)*b*d^4 + 600*a*c^4*x)*e^4

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maple [A]  time = 0.01, size = 93, normalized size = 0.93 \[ \frac {\frac {\left (d x +c \right )^{5} e^{4} a}{5}+e^{4} b \left (\frac {\left (d x +c \right )^{5} \arcsinh \left (d x +c \right )}{5}-\frac {\left (d x +c \right )^{4} \sqrt {1+\left (d x +c \right )^{2}}}{25}+\frac {4 \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{75}-\frac {8 \sqrt {1+\left (d x +c \right )^{2}}}{75}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/5*(d*x+c)^5*e^4*a+e^4*b*(1/5*(d*x+c)^5*arcsinh(d*x+c)-1/25*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+4/75*(d*x+c)^2
*(1+(d*x+c)^2)^(1/2)-8/75*(1+(d*x+c)^2)^(1/2)))

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maxima [B]  time = 0.48, size = 1231, normalized size = 12.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*d^4*e^4*x^5 + a*c*d^3*e^4*x^4 + 2*a*c^2*d^2*e^4*x^3 + 2*a*c^3*d*e^4*x^2 + (2*x^2*arcsinh(d*x + c) - d*(3
*c^2*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^3 + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x/d^2
- (c^2 + 1)*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^3 - 3*sqrt(d^2*x^2 + 2*c*d*x + c^2
+ 1)*c/d^3))*b*c^3*d*e^4 + 1/3*(6*x^3*arcsinh(d*x + c) - d*(2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x^2/d^2 - 15*c
^3*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^4 - 5*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c*x/d
^3 + 9*(c^2 + 1)*c*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^4 + 15*sqrt(d^2*x^2 + 2*c*d*x
+ c^2 + 1)*c^2/d^4 - 4*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)/d^4))*b*c^2*d^2*e^4 + 1/24*(24*x^4*arcsinh
(d*x + c) - (6*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x^3/d^2 - 14*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c*x^2/d^3 + 10
5*c^4*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^5 + 35*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c
^2*x/d^4 - 90*(c^2 + 1)*c^2*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^5 - 105*sqrt(d^2*x^2
+ 2*c*d*x + c^2 + 1)*c^3/d^5 - 9*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)*x/d^4 + 9*(c^2 + 1)^2*arcsinh(2*
(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^5 + 55*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)*c/d^5)*
d)*b*c*d^3*e^4 + 1/600*(120*x^5*arcsinh(d*x + c) - (24*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x^4/d^2 - 54*sqrt(d^2
*x^2 + 2*c*d*x + c^2 + 1)*c*x^3/d^3 + 126*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c^2*x^2/d^4 - 945*c^5*arcsinh(2*(d
^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^6 - 315*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c^3*x/d^5 - 32*sqr
t(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)*x^2/d^4 + 1050*(c^2 + 1)*c^3*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2
+ 4*(c^2 + 1)*d^2))/d^6 + 945*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c^4/d^6 + 161*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1
)*(c^2 + 1)*c*x/d^5 - 225*(c^2 + 1)^2*c*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^6 - 735*
sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)*c^2/d^6 + 64*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)^2/d^6)*d)
*b*d^4*e^4 + a*c^4*e^4*x + ((d*x + c)*arcsinh(d*x + c) - sqrt((d*x + c)^2 + 1))*b*c^4*e^4/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.50, size = 527, normalized size = 5.27 \[ \begin {cases} a c^{4} e^{4} x + 2 a c^{3} d e^{4} x^{2} + 2 a c^{2} d^{2} e^{4} x^{3} + a c d^{3} e^{4} x^{4} + \frac {a d^{4} e^{4} x^{5}}{5} + \frac {b c^{5} e^{4} \operatorname {asinh}{\left (c + d x \right )}}{5 d} + b c^{4} e^{4} x \operatorname {asinh}{\left (c + d x \right )} - \frac {b c^{4} e^{4} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25 d} + 2 b c^{3} d e^{4} x^{2} \operatorname {asinh}{\left (c + d x \right )} - \frac {4 b c^{3} e^{4} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + 2 b c^{2} d^{2} e^{4} x^{3} \operatorname {asinh}{\left (c + d x \right )} - \frac {6 b c^{2} d e^{4} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + \frac {4 b c^{2} e^{4} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75 d} + b c d^{3} e^{4} x^{4} \operatorname {asinh}{\left (c + d x \right )} - \frac {4 b c d^{2} e^{4} x^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + \frac {8 b c e^{4} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75} + \frac {b d^{4} e^{4} x^{5} \operatorname {asinh}{\left (c + d x \right )}}{5} - \frac {b d^{3} e^{4} x^{4} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{25} + \frac {4 b d e^{4} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75} - \frac {8 b e^{4} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{75 d} & \text {for}\: d \neq 0 \\c^{4} e^{4} x \left (a + b \operatorname {asinh}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*e**4*x + 2*a*c**3*d*e**4*x**2 + 2*a*c**2*d**2*e**4*x**3 + a*c*d**3*e**4*x**4 + a*d**4*e**4*x
**5/5 + b*c**5*e**4*asinh(c + d*x)/(5*d) + b*c**4*e**4*x*asinh(c + d*x) - b*c**4*e**4*sqrt(c**2 + 2*c*d*x + d*
*2*x**2 + 1)/(25*d) + 2*b*c**3*d*e**4*x**2*asinh(c + d*x) - 4*b*c**3*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 +
1)/25 + 2*b*c**2*d**2*e**4*x**3*asinh(c + d*x) - 6*b*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25
+ 4*b*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(75*d) + b*c*d**3*e**4*x**4*asinh(c + d*x) - 4*b*c*d**2*e
**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 8*b*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/75 + b*d*
*4*e**4*x**5*asinh(c + d*x)/5 - b*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 4*b*d*e**4*x**2*sqr
t(c**2 + 2*c*d*x + d**2*x**2 + 1)/75 - 8*b*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(75*d), Ne(d, 0)), (c**4*
e**4*x*(a + b*asinh(c)), True))

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