Integrand size = 26, antiderivative size = 192 \[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {b x^2}{4 c^5 \pi ^{5/2}}-\frac {b}{6 c^7 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac {5 x^3 (a+b \text {arcsinh}(c x))}{3 c^4 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {5 x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{2 c^6 \pi ^3}-\frac {5 (a+b \text {arcsinh}(c x))^2}{4 b c^7 \pi ^{5/2}}-\frac {7 b \log \left (1+c^2 x^2\right )}{6 c^7 \pi ^{5/2}} \]
-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*arcsin h(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/(P i*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
Time = 0.60 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.05 \[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {60 a c x+80 a c^3 x^3+12 a c^5 x^5-7 b \sqrt {1+c^2 x^2}-9 b c^2 x^2 \sqrt {1+c^2 x^2}-6 b c^4 x^4 \sqrt {1+c^2 x^2}+4 \left (-15 a \left (1+c^2 x^2\right )^{3/2}+b c x \left (15+20 c^2 x^2+3 c^4 x^4\right )\right ) \text {arcsinh}(c x)-30 b \left (1+c^2 x^2\right )^{3/2} \text {arcsinh}(c x)^2-28 b \left (1+c^2 x^2\right )^{3/2} \log \left (1+c^2 x^2\right )}{24 c^7 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}} \]
(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))
Time = 1.11 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.31, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {6225, 243, 49, 2009, 6225, 243, 49, 2009, 6227, 15, 6198}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi c^2 x^2+\pi \right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6225 |
\(\displaystyle \frac {5 \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (c^2 \pi x^2+\pi \right )^{3/2}}dx}{3 \pi c^2}+\frac {b \int \frac {x^5}{\left (c^2 x^2+1\right )^2}dx}{3 \pi ^{5/2} c}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {5 \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (c^2 \pi x^2+\pi \right )^{3/2}}dx}{3 \pi c^2}+\frac {b \int \frac {x^4}{\left (c^2 x^2+1\right )^2}dx^2}{6 \pi ^{5/2} c}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {5 \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (c^2 \pi x^2+\pi \right )^{3/2}}dx}{3 \pi c^2}+\frac {b \int \left (\frac {1}{c^4}-\frac {2}{c^4 \left (c^2 x^2+1\right )}+\frac {1}{c^4 \left (c^2 x^2+1\right )^2}\right )dx^2}{6 \pi ^{5/2} c}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (c^2 \pi x^2+\pi \right )^{3/2}}dx}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 6225 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{\pi c^2}+\frac {b \int \frac {x^3}{c^2 x^2+1}dx}{\pi ^{3/2} c}-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}\right )}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{\pi c^2}+\frac {b \int \frac {x^2}{c^2 x^2+1}dx^2}{2 \pi ^{3/2} c}-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}\right )}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{\pi c^2}+\frac {b \int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^2+1\right )}\right )dx^2}{2 \pi ^{3/2} c}-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}\right )}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 \pi x^2+\pi }}dx}{\pi c^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {b \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{2 \pi ^{3/2} c}\right )}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 6227 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 \pi x^2+\pi }}dx}{2 c^2}-\frac {b \int xdx}{2 \sqrt {\pi } c}+\frac {x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{2 \pi c^2}\right )}{\pi c^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {b \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{2 \pi ^{3/2} c}\right )}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 \pi x^2+\pi }}dx}{2 c^2}+\frac {x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{2 \pi c^2}-\frac {b x^2}{4 \sqrt {\pi } c}\right )}{\pi c^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {b \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{2 \pi ^{3/2} c}\right )}{3 \pi c^2}-\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
\(\Big \downarrow \) 6198 |
\(\displaystyle -\frac {x^5 (a+b \text {arcsinh}(c x))}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac {5 \left (-\frac {x^3 (a+b \text {arcsinh}(c x))}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {3 \left (-\frac {(a+b \text {arcsinh}(c x))^2}{4 \sqrt {\pi } b c^3}+\frac {x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{2 \pi c^2}-\frac {b x^2}{4 \sqrt {\pi } c}\right )}{\pi c^2}+\frac {b \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )}{2 \pi ^{3/2} c}\right )}{3 \pi c^2}+\frac {b \left (\frac {x^2}{c^4}-\frac {1}{c^6 \left (c^2 x^2+1\right )}-\frac {2 \log \left (c^2 x^2+1\right )}{c^6}\right )}{6 \pi ^{5/2} c}\) |
-1/3*(x^5*(a + b*ArcSinh[c*x]))/(c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)) + (b*(x^2 /c^4 - 1/(c^6*(1 + c^2*x^2)) - (2*Log[1 + c^2*x^2])/c^6))/(6*c*Pi^(5/2)) + (5*(-((x^3*(a + b*ArcSinh[c*x]))/(c^2*Pi*Sqrt[Pi + c^2*Pi*x^2])) + (3*(-1 /4*(b*x^2)/(c*Sqrt[Pi]) + (x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/( 2*c^2*Pi) - (a + b*ArcSinh[c*x])^2/(4*b*c^3*Sqrt[Pi])))/(c^2*Pi) + (b*(x^2 /c^2 - Log[1 + c^2*x^2]/c^4))/(2*c*Pi^(3/2))))/(3*c^2*Pi)
3.1.100.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ Symbol] :> Simp[(1/(b*c*(n + 1)))*Simp[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]
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] + (-Simp[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] - S imp[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]) /; Fre eQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IG tQ[m, 1]
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] + (-Simp[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] - Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(969\) vs. \(2(164)=328\).
Time = 0.33 (sec) , antiderivative size = 970, normalized size of antiderivative = 5.05
method | result | size |
default | \(\text {Expression too large to display}\) | \(970\) |
parts | \(\text {Expression too large to display}\) | \(970\) |
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+1/2*b/c^6/Pi^(5/2)*(c^2*x^2+1)^(1 /2)*arcsinh(c*x)*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^6-4 9*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)/(6 3*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^7*arcsinh(c*x)+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*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/2*a*x ^5/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)-1/8*b/c^7/Pi^(5/2)-1/4*b*x^2/c^5/Pi^(5/2)- 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)+14/3*b/c^7/Pi^(5/2)*arcsinh(c*x)-49/6*b/Pi^(5/2)/(63*c^4*x^4+11 1*c^2*x^2+49)/(c^2*x^2+1)^2/c^7-147*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+4 9)/(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^4-1463/3*b/Pi^(5/2)/(63*c^4*x^4+1 11*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^(...
\[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{6}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}} \,d x } \]
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)
\[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\frac {\int \frac {a x^{6}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{6} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]
(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)
\[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{6}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}}} \,d x } \]
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*sq rt(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)
Exception generated. \[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {x^6 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx=\int \frac {x^6\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \]