3.3.28 \(\int (c e+d e x)^{7/2} (a+b \sinh ^{-1}(c+d x)) \, dx\) [228]

Optimal. Leaf size=298 \[ \frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}-\frac {28 b e^3 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{135 d (1+c+d x)}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^{7/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {1+(c+d x)^2}}-\frac {14 b e^{7/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {1+(c+d x)^2}} \]

[Out]

2/9*(e*(d*x+c))^(9/2)*(a+b*arcsinh(d*x+c))/d/e+28/405*b*e^2*(e*(d*x+c))^(3/2)*(1+(d*x+c)^2)^(1/2)/d-4/81*b*(e*
(d*x+c))^(7/2)*(1+(d*x+c)^2)^(1/2)/d-28/135*b*e^3*(e*(d*x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d/(d*x+c+1)+28/135*b*e
^(7/2)*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))^2)^(1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*E
llipticE(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/(1+(d*x+c)^
2)^(1/2)-14/135*b*e^(7/2)*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))^2)^(1/2)/cos(2*arctan((e*(d*x+c)
)^(1/2)/e^(1/2)))*EllipticF(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*((1+(d*x+c)^2)/(d*x+c+1)^2)^
(1/2)/d/(1+(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5859, 5776, 327, 335, 311, 226, 1210} \begin {gather*} \frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {14 b e^{7/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {(c+d x)^2+1}}+\frac {28 b e^{7/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {(c+d x)^2+1}}-\frac {28 b e^3 \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{135 d (c+d x+1)}+\frac {28 b e^2 \sqrt {(c+d x)^2+1} (e (c+d x))^{3/2}}{405 d}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{7/2}}{81 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(7/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(28*b*e^2*(e*(c + d*x))^(3/2)*Sqrt[1 + (c + d*x)^2])/(405*d) - (4*b*(e*(c + d*x))^(7/2)*Sqrt[1 + (c + d*x)^2])
/(81*d) - (28*b*e^3*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(135*d*(1 + c + d*x)) + (2*(e*(c + d*x))^(9/2)*(a
+ b*ArcSinh[c + d*x]))/(9*d*e) + (28*b*e^(7/2)*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*Elliptic
E[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(135*d*Sqrt[1 + (c + d*x)^2]) - (14*b*e^(7/2)*(1 + c + d*x)*Sqrt[
(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(135*d*Sqrt[1 + (c + d
*x)^2])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 5776

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 5859

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)^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\text {Subst}\left (\int (e x)^{7/2} \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {(2 b) \text {Subst}\left (\int \frac {(e x)^{9/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{9 d e}\\ &=-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}+\frac {(14 b e) \text {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{81 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (14 b e^3\right ) \text {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (28 b e^2\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}-\frac {\left (28 b e^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{135 d}+\frac {\left (28 b e^3\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{e}}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{135 d}\\ &=\frac {28 b e^2 (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{405 d}-\frac {4 b (e (c+d x))^{7/2} \sqrt {1+(c+d x)^2}}{81 d}-\frac {28 b e^3 \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{135 d (1+c+d x)}+\frac {2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d e}+\frac {28 b e^{7/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {1+(c+d x)^2}}-\frac {14 b e^{7/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{135 d \sqrt {1+(c+d x)^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.15, size = 113, normalized size = 0.38 \begin {gather*} \frac {2 (e (c+d x))^{7/2} \left (45 a (c+d x)^3+14 b \sqrt {1+(c+d x)^2}-10 b (c+d x)^2 \sqrt {1+(c+d x)^2}+45 b (c+d x)^3 \sinh ^{-1}(c+d x)-14 b \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-(c+d x)^2\right )\right )}{405 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(7/2)*(a + b*ArcSinh[c + d*x]),x]

[Out]

(2*(e*(c + d*x))^(7/2)*(45*a*(c + d*x)^3 + 14*b*Sqrt[1 + (c + d*x)^2] - 10*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]
+ 45*b*(c + d*x)^3*ArcSinh[c + d*x] - 14*b*Hypergeometric2F1[1/2, 3/4, 7/4, -(c + d*x)^2]))/(405*d*(c + d*x)^
2)

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Maple [C] Result contains complex when optimal does not.
time = 3.00, size = 238, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d e x +c e \right )^{\frac {9}{2}} a}{9}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {9}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{9}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {7}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{9}-\frac {7 e^{4} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{45}+\frac {7 i e^{5} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{15 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{9 e}\right )}{d e}\)	\(238\)
default	\(\frac {\frac {2 \left (d e x +c e \right )^{\frac {9}{2}} a}{9}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {9}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{9}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {7}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{9}-\frac {7 e^{4} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{45}+\frac {7 i e^{5} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{15 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{9 e}\right )}{d e}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/d/e*(1/9*(d*e*x+c*e)^(9/2)*a+b*(1/9*(d*e*x+c*e)^(9/2)*arcsinh((d*e*x+c*e)/e)-2/9/e*(1/9*e^2*(d*e*x+c*e)^(7/2
)*((d*e*x+c*e)^2/e^2+1)^(1/2)-7/45*e^4*(d*e*x+c*e)^(3/2)*((d*e*x+c*e)^2/e^2+1)^(1/2)+7/15*I*e^5/(I/e)^(1/2)*(1
-I/e*(d*e*x+c*e))^(1/2)*(1+I/e*(d*e*x+c*e))^(1/2)/((d*e*x+c*e)^2/e^2+1)^(1/2)*(EllipticF((d*e*x+c*e)^(1/2)*(I/
e)^(1/2),I)-EllipticE((d*e*x+c*e)^(1/2)*(I/e)^(1/2),I)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x, algorithm="maxima")

[Out]

2/9*(d*x*e + c*e)^(9/2)*a*e^(-1)/d + 1/810*(180*(d^4*x^4*e^(7/2) + 4*c*d^3*x^3*e^(7/2) + 6*c^2*d^2*x^2*e^(7/2)
+ 4*c^3*d*x*e^(7/2) + c^4*e^(7/2))*sqrt(d*x + c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d - (45*(I*
sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) +
1))*e^3 - I*sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) - 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2) - 2*sqr
t(d*x + c)) + 1))*e^3 - sqrt(2)*e^3*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) + sqrt(2)*e^3*log(d*x - sqrt(2)*s
qrt(d*x + c) + c + 1))*e^(1/2) + 40*e^(9/2*log(d*x + c) + 7/2) - 72*e^(5/2*log(d*x + c) + 7/2) + 360*e^(1/2*lo
g(d*x + c) + 7/2))/d - 810*integrate(2/9*(d^4*x^4*e^(7/2) + 4*c*d^3*x^3*e^(7/2) + 6*c^2*d^2*x^2*e^(7/2) + 4*c^
3*d*x*e^(7/2) + c^4*e^(7/2))*sqrt(d*x + c)/(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))*b

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 815, normalized size = 2.73 \begin {gather*} \frac {2 \, {\left (45 \, {\left ({\left (b d^{5} x^{4} + 4 \, b c d^{4} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c^{3} d^{2} x + b c^{4} d\right )} \cosh \left (1\right )^{3} + 3 \, {\left (b d^{5} x^{4} + 4 \, b c d^{4} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c^{3} d^{2} x + b c^{4} d\right )} \cosh \left (1\right )^{2} \sinh \left (1\right ) + 3 \, {\left (b d^{5} x^{4} + 4 \, b c d^{4} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c^{3} d^{2} x + b c^{4} d\right )} \cosh \left (1\right ) \sinh \left (1\right )^{2} + {\left (b d^{5} x^{4} + 4 \, b c d^{4} x^{3} + 6 \, b c^{2} d^{3} x^{2} + 4 \, b c^{3} d^{2} x + b c^{4} d\right )} \sinh \left (1\right )^{3}\right )} \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 42 \, \sqrt {d^{3} \cosh \left (1\right ) + d^{3} \sinh \left (1\right )} {\left (b \cosh \left (1\right )^{3} + 3 \, b \cosh \left (1\right )^{2} \sinh \left (1\right ) + 3 \, b \cosh \left (1\right ) \sinh \left (1\right )^{2} + b \sinh \left (1\right )^{3}\right )} {\rm weierstrassZeta}\left (-\frac {4}{d^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right )\right ) + {\left (45 \, {\left (a d^{5} x^{4} + 4 \, a c d^{4} x^{3} + 6 \, a c^{2} d^{3} x^{2} + 4 \, a c^{3} d^{2} x + a c^{4} d\right )} \cosh \left (1\right )^{3} + 135 \, {\left (a d^{5} x^{4} + 4 \, a c d^{4} x^{3} + 6 \, a c^{2} d^{3} x^{2} + 4 \, a c^{3} d^{2} x + a c^{4} d\right )} \cosh \left (1\right )^{2} \sinh \left (1\right ) + 135 \, {\left (a d^{5} x^{4} + 4 \, a c d^{4} x^{3} + 6 \, a c^{2} d^{3} x^{2} + 4 \, a c^{3} d^{2} x + a c^{4} d\right )} \cosh \left (1\right ) \sinh \left (1\right )^{2} + 45 \, {\left (a d^{5} x^{4} + 4 \, a c d^{4} x^{3} + 6 \, a c^{2} d^{3} x^{2} + 4 \, a c^{3} d^{2} x + a c^{4} d\right )} \sinh \left (1\right )^{3} - 2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left ({\left (5 \, b d^{4} x^{3} + 15 \, b c d^{3} x^{2} + {\left (15 \, b c^{2} - 7 \, b\right )} d^{2} x + {\left (5 \, b c^{3} - 7 \, b c\right )} d\right )} \cosh \left (1\right )^{3} + 3 \, {\left (5 \, b d^{4} x^{3} + 15 \, b c d^{3} x^{2} + {\left (15 \, b c^{2} - 7 \, b\right )} d^{2} x + {\left (5 \, b c^{3} - 7 \, b c\right )} d\right )} \cosh \left (1\right )^{2} \sinh \left (1\right ) + 3 \, {\left (5 \, b d^{4} x^{3} + 15 \, b c d^{3} x^{2} + {\left (15 \, b c^{2} - 7 \, b\right )} d^{2} x + {\left (5 \, b c^{3} - 7 \, b c\right )} d\right )} \cosh \left (1\right ) \sinh \left (1\right )^{2} + {\left (5 \, b d^{4} x^{3} + 15 \, b c d^{3} x^{2} + {\left (15 \, b c^{2} - 7 \, b\right )} d^{2} x + {\left (5 \, b c^{3} - 7 \, b c\right )} d\right )} \sinh \left (1\right )^{3}\right )}\right )} \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )}\right )}}{405 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x, algorithm="fricas")

[Out]

2/405*(45*((b*d^5*x^4 + 4*b*c*d^4*x^3 + 6*b*c^2*d^3*x^2 + 4*b*c^3*d^2*x + b*c^4*d)*cosh(1)^3 + 3*(b*d^5*x^4 +
4*b*c*d^4*x^3 + 6*b*c^2*d^3*x^2 + 4*b*c^3*d^2*x + b*c^4*d)*cosh(1)^2*sinh(1) + 3*(b*d^5*x^4 + 4*b*c*d^4*x^3 +
6*b*c^2*d^3*x^2 + 4*b*c^3*d^2*x + b*c^4*d)*cosh(1)*sinh(1)^2 + (b*d^5*x^4 + 4*b*c*d^4*x^3 + 6*b*c^2*d^3*x^2 +
4*b*c^3*d^2*x + b*c^4*d)*sinh(1)^3)*sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1))*log(d*x + c + sqrt(d^2*x^2 + 2
*c*d*x + c^2 + 1)) + 42*sqrt(d^3*cosh(1) + d^3*sinh(1))*(b*cosh(1)^3 + 3*b*cosh(1)^2*sinh(1) + 3*b*cosh(1)*sin
h(1)^2 + b*sinh(1)^3)*weierstrassZeta(-4/d^2, 0, weierstrassPInverse(-4/d^2, 0, (d*x + c)/d)) + (45*(a*d^5*x^4
+ 4*a*c*d^4*x^3 + 6*a*c^2*d^3*x^2 + 4*a*c^3*d^2*x + a*c^4*d)*cosh(1)^3 + 135*(a*d^5*x^4 + 4*a*c*d^4*x^3 + 6*a
*c^2*d^3*x^2 + 4*a*c^3*d^2*x + a*c^4*d)*cosh(1)^2*sinh(1) + 135*(a*d^5*x^4 + 4*a*c*d^4*x^3 + 6*a*c^2*d^3*x^2 +
4*a*c^3*d^2*x + a*c^4*d)*cosh(1)*sinh(1)^2 + 45*(a*d^5*x^4 + 4*a*c*d^4*x^3 + 6*a*c^2*d^3*x^2 + 4*a*c^3*d^2*x
+ a*c^4*d)*sinh(1)^3 - 2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*((5*b*d^4*x^3 + 15*b*c*d^3*x^2 + (15*b*c^2 - 7*b)*d
^2*x + (5*b*c^3 - 7*b*c)*d)*cosh(1)^3 + 3*(5*b*d^4*x^3 + 15*b*c*d^3*x^2 + (15*b*c^2 - 7*b)*d^2*x + (5*b*c^3 -
7*b*c)*d)*cosh(1)^2*sinh(1) + 3*(5*b*d^4*x^3 + 15*b*c*d^3*x^2 + (15*b*c^2 - 7*b)*d^2*x + (5*b*c^3 - 7*b*c)*d)*
cosh(1)*sinh(1)^2 + (5*b*d^4*x^3 + 15*b*c*d^3*x^2 + (15*b*c^2 - 7*b)*d^2*x + (5*b*c^3 - 7*b*c)*d)*sinh(1)^3))*
sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1)))/d^2

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(7/2)*(a+b*asinh(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3876 deep

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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((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(7/2)*(b*arcsinh(d*x + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^{7/2}\,\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^(7/2)*(a + b*asinh(c + d*x)),x)

[Out]

int((c*e + d*e*x)^(7/2)*(a + b*asinh(c + d*x)), x)

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