3.2.65 \(\int \frac {x^6 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [165]

Optimal. Leaf size=281 \[ -\frac {b}{6 c^7 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {b x^2 \sqrt {1+c^2 x^2}}{4 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 d^3}-\frac {5 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^7 d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 c^7 d^2 \sqrt {d+c^2 d x^2}} \]

[Out]

-1/3*x^5*(a+b*arcsinh(c*x))/c^2/d/(c^2*d*x^2+d)^(3/2)-5/3*x^3*(a+b*arcsinh(c*x))/c^4/d^2/(c^2*d*x^2+d)^(1/2)-1
/6*b/c^7/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-1/4*b*x^2*(c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*d*x^2+d)^(1/2)-5/4
*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/b/c^7/d^2/(c^2*d*x^2+d)^(1/2)-7/6*b*ln(c^2*x^2+1)*(c^2*x^2+1)^(1/2)/c^
7/d^2/(c^2*d*x^2+d)^(1/2)+5/2*x*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2)/c^6/d^3

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Rubi [A]
time = 0.28, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5810, 5812, 5783, 30, 272, 45} \begin {gather*} -\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {5 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^7 d^2 \sqrt {c^2 d x^2+d}}+\frac {5 x \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 d^3}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {c^2 d x^2+d}}-\frac {b}{6 c^7 d^2 \sqrt {c^2 x^2+1} \sqrt {c^2 d x^2+d}}-\frac {7 b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 c^7 d^2 \sqrt {c^2 d x^2+d}}-\frac {b x^2 \sqrt {c^2 x^2+1}}{4 c^5 d^2 \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

-1/6*b/(c^7*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) - (b*x^2*Sqrt[1 + c^2*x^2])/(4*c^5*d^2*Sqrt[d + c^2*d*x
^2]) - (x^5*(a + b*ArcSinh[c*x]))/(3*c^2*d*(d + c^2*d*x^2)^(3/2)) - (5*x^3*(a + b*ArcSinh[c*x]))/(3*c^4*d^2*Sq
rt[d + c^2*d*x^2]) + (5*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(2*c^6*d^3) - (5*Sqrt[1 + c^2*x^2]*(a + b*
ArcSinh[c*x])^2)/(4*b*c^7*d^2*Sqrt[d + c^2*d*x^2]) - (7*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*c^7*d^2*Sqrt[
d + c^2*d*x^2])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

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 272

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 5783

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

Rule 5810

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

Rule 5812

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

Rubi steps

\begin {align*} \int \frac {x^6 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {5 \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{3/2}} \, dx}{3 c^2 d}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {x^5}{\left (1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+c^2 d x^2}} \, dx}{c^4 d^2}+\frac {\left (5 b \sqrt {1+c^2 x^2}\right ) \int \frac {x^3}{1+c^2 x^2} \, dx}{3 c^3 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 c d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 d^3}-\frac {5 \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {d+c^2 d x^2}} \, dx}{2 c^6 d^2}-\frac {\left (5 b \sqrt {1+c^2 x^2}\right ) \int x \, dx}{2 c^5 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )}{6 c^3 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \left (\frac {1}{c^4}+\frac {1}{c^4 \left (1+c^2 x\right )^2}-\frac {2}{c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 c d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b}{6 c^7 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {13 b x^2 \sqrt {1+c^2 x^2}}{12 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 d^3}-\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c^7 d^2 \sqrt {d+c^2 d x^2}}-\frac {\left (5 \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{2 c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\left (5 b \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 c^3 d^2 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b}{6 c^7 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}-\frac {b x^2 \sqrt {1+c^2 x^2}}{4 c^5 d^2 \sqrt {d+c^2 d x^2}}-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 x \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 d^3}-\frac {5 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^7 d^2 \sqrt {d+c^2 d x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 c^7 d^2 \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.65, size = 222, normalized size = 0.79 \begin {gather*} \frac {4 a c d x \left (15+20 c^2 x^2+3 c^4 x^4\right )+b d \left (4 c x \left (15+20 c^2 x^2+3 c^4 x^4\right ) \sinh ^{-1}(c x)-30 \left (1+c^2 x^2\right )^{3/2} \sinh ^{-1}(c x)^2-\sqrt {1+c^2 x^2} \left (7+9 c^2 x^2+6 c^4 x^4+28 \left (1+c^2 x^2\right ) \log \left (1+c^2 x^2\right )\right )\right )-60 a \sqrt {d} \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{24 c^7 d^3 \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

(4*a*c*d*x*(15 + 20*c^2*x^2 + 3*c^4*x^4) + b*d*(4*c*x*(15 + 20*c^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] - 30*(1 + c^2
*x^2)^(3/2)*ArcSinh[c*x]^2 - Sqrt[1 + c^2*x^2]*(7 + 9*c^2*x^2 + 6*c^4*x^4 + 28*(1 + c^2*x^2)*Log[1 + c^2*x^2])
) - 60*a*Sqrt[d]*(1 + c^2*x^2)*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/(24*c^7*d^3*(1 +
c^2*x^2)*Sqrt[d + c^2*d*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1606\) vs. \(2(245)=490\).
time = 4.68, size = 1607, normalized size = 5.72

method	result	size
default	\(\text {Expression too large to display}\)	\(1607\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

5/6*a/c^4*x^3/d/(c^2*d*x^2+d)^(3/2)+5/2*a/c^6/d^2*x/(c^2*d*x^2+d)^(1/2)-5/2*a/c^6/d^2*ln(x*c^2*d/(c^2*d)^(1/2)
+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-1/8*b*(d*(c^2*x^2+1))^(1/2)/c^7/d^3/(c^2*x^2+1)^(1/2)-406*b*(d*(c^2*x^2+1)
)^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/c^3/d^3*(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^4+1/2*a*x
^5/c^2/d/(c^2*d*x^2+d)^(3/2)+147*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/d
^3*arcsinh(c*x)*x^7+14/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^7/d^3*arcsinh(c*x)+70/3*b*(d*(c^2*x^2+1))
^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/c^2/d^3*x^5+133/6*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x
^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/c^4/d^3*x^3+7*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c
^4*x^4+209*c^2*x^2+49)/c^6/d^3*x-49/6*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+
49)/c^7/d^3*(c^2*x^2+1)^(1/2)-7/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^7/d^3*ln(1+(c*x+(c^2*x^2+1)^(1/2
))^2)-5/4*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^7/d^3*arcsinh(c*x)^2-1/4*b*(d*(c^2*x^2+1))^(1/2)/c^5/d^3
/(c^2*x^2+1)^(1/2)*x^2-147*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/c/d^3*(
c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^6-1120/3*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x
^2+49)/c^5/d^3*(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^2-49/6*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4
*x^4+209*c^2*x^2+49)/c^2/d^3*(c^2*x^2+1)*x^5+385*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+2
09*c^2*x^2+49)/c^2/d^3*arcsinh(c*x)*x^5-21/2*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c
^2*x^2+49)/c^3/d^3*(c^2*x^2+1)^(1/2)*x^4-91/6*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*
c^2*x^2+49)/c^4/d^3*(c^2*x^2+1)*x^3+1009/3*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2
*x^2+49)/c^4/d^3*arcsinh(c*x)*x^3-37/2*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2
+49)/c^5/d^3*(c^2*x^2+1)^(1/2)*x^2-7*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+4
9)/c^6/d^3*(c^2*x^2+1)*x+98*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/c^6/d^
3*arcsinh(c*x)*x-343/3*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4+209*c^2*x^2+49)/c^7/d^3*(c^
2*x^2+1)^(1/2)*arcsinh(c*x)+1/2*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^3/(c^2*x^2+1)*arcsinh(c*x)*x^3+1/2*b*(d*(c^2*x^2
+1))^(1/2)/c^6/d^3/(c^2*x^2+1)*arcsinh(c*x)*x+49/6*b*(d*(c^2*x^2+1))^(1/2)/(63*c^8*x^8+237*c^6*x^6+334*c^4*x^4
+209*c^2*x^2+49)/d^3*x^7

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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(x^6*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/6*a*(3*x^5/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 5*x*(3*x^2/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 2/((c^2*d*x^2 + d)^(3/
2)*c^4*d))/c^2 + 5*x/(sqrt(c^2*d*x^2 + d)*c^6*d^2) - 15*arcsinh(c*x)/(c^7*d^(5/2))) + b*integrate(x^6*log(c*x
+ sqrt(c^2*x^2 + 1))/(c^2*d*x^2 + d)^(5/2), x)

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Fricas [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(x^6*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral((b*x^6*arcsinh(c*x) + a*x^6)*sqrt(c^2*d*x^2 + d)/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3),
x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**6*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^6\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2),x)

[Out]

int((x^6*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(5/2), x)

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