3.100 \(\int \frac {x^6 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\)

Optimal. Leaf size=192 \[ -\frac {5 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 \pi ^{5/2} b c^7}-\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {5 x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^3 c^6}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 c^4 \sqrt {\pi c^2 x^2+\pi }}-\frac {b x^2}{4 \pi ^{5/2} c^5}-\frac {b}{6 \pi ^{5/2} c^7 \left (c^2 x^2+1\right )}-\frac {7 b \log \left (c^2 x^2+1\right )}{6 \pi ^{5/2} c^7} \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 256, normalized size of antiderivative = 1.33, 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 = {5751, 5758, 5675, 30, 266, 43} \[ -\frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 c^4 \sqrt {\pi c^2 x^2+\pi }}+\frac {5 x \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^3 c^6}-\frac {5 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 \pi ^{5/2} b c^7}-\frac {b x^2 \sqrt {c^2 x^2+1}}{4 \pi ^2 c^5 \sqrt {\pi c^2 x^2+\pi }}-\frac {b}{6 \pi ^2 c^7 \sqrt {c^2 x^2+1} \sqrt {\pi c^2 x^2+\pi }}-\frac {7 b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 \pi ^2 c^7 \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b/(6*c^7*Pi^2*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^2*Pi*x^2]) - (b*x^2*Sqrt[1 + c^2*x^2])/(4*c^5*Pi^2*Sqrt[Pi + c^2*
Pi*x^2]) - (x^5*(a + b*ArcSinh[c*x]))/(3*c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)) - (5*x^3*(a + b*ArcSinh[c*x]))/(3*c^4
*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (5*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(2*c^6*Pi^3) - (5*(a + b*ArcSi
nh[c*x])^2)/(4*b*c^7*Pi^(5/2)) - (7*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*c^7*Pi^2*Sqrt[Pi + c^2*Pi*x^2])

Rule 30

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

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 5675

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

Rule 5751

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*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[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] && Gt
Q[m, 1]

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*Sqrt[d + e*x^2]
), Int[(f*x)^(m - 1)*(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] && GtQ[m, 1] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 202, normalized size = 1.05 \[ \frac {4 \sinh ^{-1}(c x) \left (b c x \left (3 c^4 x^4+20 c^2 x^2+15\right )-15 a \left (c^2 x^2+1\right )^{3/2}\right )+12 a c^5 x^5+80 a c^3 x^3+60 a c x-9 b c^2 x^2 \sqrt {c^2 x^2+1}-7 b \sqrt {c^2 x^2+1}-28 b \left (c^2 x^2+1\right )^{3/2} \log \left (c^2 x^2+1\right )-30 b \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)^2-6 b c^4 x^4 \sqrt {c^2 x^2+1}}{24 \pi ^{5/2} c^7 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*c*x + 80*a*c^3*x^3 + 12*a*c^5*x^5 - 7*b*Sqrt[1 + c^2*x^2] - 9*b*c^2*x^2*Sqrt[1 + c^2*x^2] - 6*b*c^4*x^4*
Sqrt[1 + c^2*x^2] + 4*(-15*a*(1 + c^2*x^2)^(3/2) + b*c*x*(15 + 20*c^2*x^2 + 3*c^4*x^4))*ArcSinh[c*x] - 30*b*(1
+ c^2*x^2)^(3/2)*ArcSinh[c*x]^2 - 28*b*(1 + c^2*x^2)^(3/2)*Log[1 + c^2*x^2])/(24*c^7*Pi^(5/2)*(1 + c^2*x^2)^(
3/2))

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b x^{6} \operatorname {arsinh}\left (c x\right ) + a x^{6}\right )}}{\pi ^{3} c^{6} x^{6} + 3 \, \pi ^{3} c^{4} x^{4} + 3 \, \pi ^{3} c^{2} x^{2} + \pi ^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.51, size = 970, normalized size = 5.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

1/2*b/Pi^(5/2)/c^6*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-49/6*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2*
c*x^8-98/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c*x^6-49*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49
)/(c^2*x^2+1)^2/c^3*x^4-98/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^5*x^2-343/3*b/Pi^(5/2)/(63
*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^7*arcsinh(c*x)+49/6*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1
)/c*x^6+14*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)/c^3*x^4+6*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)
/(c^2*x^2+1)/c^5*x^2+147*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x^7-1/4*b*x^2/c
^5/Pi^(5/2)+14/3*b/c^7/Pi^(5/2)*arcsinh(c*x)-5/4*b/c^7/Pi^(5/2)*arcsinh(c*x)^2-7/3*b/c^7/Pi^(5/2)*ln(1+(c*x+(c
^2*x^2+1)^(1/2))^2)+1/2*a*x^5/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)-1463/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*
x^2+1)^2/c^5*arcsinh(c*x)*x^2+385*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3/2)/c^2*arcsinh(c*x)*x^
5+1009/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3/2)/c^4*arcsinh(c*x)*x^3+98*b/Pi^(5/2)/(63*c^4*x
^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3/2)/c^6*arcsinh(c*x)*x-147*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)
^2*c*arcsinh(c*x)*x^8-553*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c*arcsinh(c*x)*x^6-2338/3*b/Pi^
(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^3*arcsinh(c*x)*x^4+5/6*a/c^4*x^3/Pi/(Pi*c^2*x^2+Pi)^(3/2)+5/
2*a/c^6/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2)-5/2*a/c^6/Pi^2*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)
^(1/2)-49/6*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^7-1/8*b/Pi^(5/2)/c^7

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a {\left (\frac {3 \, x^{5}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {5 \, x {\left (\frac {3 \, x^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {2}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{4}}\right )}}{c^{2}} + \frac {5 \, x}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}} c^{6}} - \frac {15 \, \operatorname {arsinh}\left (c x\right )}{\pi ^{\frac {5}{2}} c^{7}}\right )} + b \int \frac {x^{6} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{6}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{6} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**6/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) +
Integral(b*x**6*asinh(c*x)/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2
+ 1)), x))/pi**(5/2)

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