Integrand size = 25, antiderivative size = 256 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=-\frac {8 b e^4 n \sqrt {d+e x^2}}{315 d^3 x}-\frac {8 b e^3 n \left (d+e x^2\right )^{3/2}}{945 d^3 x^3}-\frac {8 b e^2 n \left (d+e x^2\right )^{5/2}}{1575 d^3 x^5}-\frac {b n \left (d+e x^2\right )^{7/2}}{81 d^2 x^9}+\frac {50 b e n \left (d+e x^2\right )^{7/2}}{3969 d^3 x^7}+\frac {8 b e^{9/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{315 d^3}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5} \]
-8/945*b*e^3*n*(e*x^2+d)^(3/2)/d^3/x^3-8/1575*b*e^2*n*(e*x^2+d)^(5/2)/d^3/ x^5-1/81*b*n*(e*x^2+d)^(7/2)/d^2/x^9+50/3969*b*e*n*(e*x^2+d)^(7/2)/d^3/x^7 +8/315*b*e^(9/2)*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/d^3-1/9*(e*x^2+d)^(5 /2)*(a+b*ln(c*x^n))/d/x^9+4/63*e*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/d^2/x^7-8 /315*e^2*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/d^3/x^5-8/315*b*e^4*n*(e*x^2+d)^( 1/2)/d^3/x
Time = 0.22 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.70 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=-\frac {\sqrt {d+e x^2} \left (315 a \left (d+e x^2\right )^2 \left (35 d^2-20 d e x^2+8 e^2 x^4\right )+b n \left (1225 d^4+2425 d^3 e x^2+429 d^2 e^2 x^4-677 d e^3 x^6+2614 e^4 x^8\right )\right )+315 b \left (d+e x^2\right )^{5/2} \left (35 d^2-20 d e x^2+8 e^2 x^4\right ) \log \left (c x^n\right )-2520 b e^{9/2} n x^9 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{99225 d^3 x^9} \]
-1/99225*(Sqrt[d + e*x^2]*(315*a*(d + e*x^2)^2*(35*d^2 - 20*d*e*x^2 + 8*e^ 2*x^4) + b*n*(1225*d^4 + 2425*d^3*e*x^2 + 429*d^2*e^2*x^4 - 677*d*e^3*x^6 + 2614*e^4*x^8)) + 315*b*(d + e*x^2)^(5/2)*(35*d^2 - 20*d*e*x^2 + 8*e^2*x^ 4)*Log[c*x^n] - 2520*b*e^(9/2)*n*x^9*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/( d^3*x^9)
Time = 0.48 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2792, 27, 1588, 27, 358, 247, 247, 247, 224, 219}
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.
\(\displaystyle \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle -b n \int -\frac {\left (e x^2+d\right )^{5/2} \left (8 e^2 x^4-20 d e x^2+35 d^2\right )}{315 d^3 x^{10}}dx-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b n \int \frac {\left (e x^2+d\right )^{5/2} \left (8 e^2 x^4-20 d e x^2+35 d^2\right )}{x^{10}}dx}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 1588 |
\(\displaystyle \frac {b n \left (-\frac {\int \frac {2 d e \left (125 d-36 e x^2\right ) \left (e x^2+d\right )^{5/2}}{x^8}dx}{9 d}-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b n \left (-\frac {2}{9} e \int \frac {\left (125 d-36 e x^2\right ) \left (e x^2+d\right )^{5/2}}{x^8}dx-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 358 |
\(\displaystyle \frac {b n \left (-\frac {2}{9} e \left (-36 e \int \frac {\left (e x^2+d\right )^{5/2}}{x^6}dx-\frac {125 \left (d+e x^2\right )^{7/2}}{7 x^7}\right )-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {b n \left (-\frac {2}{9} e \left (-36 e \left (e \int \frac {\left (e x^2+d\right )^{3/2}}{x^4}dx-\frac {\left (d+e x^2\right )^{5/2}}{5 x^5}\right )-\frac {125 \left (d+e x^2\right )^{7/2}}{7 x^7}\right )-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {b n \left (-\frac {2}{9} e \left (-36 e \left (e \left (e \int \frac {\sqrt {e x^2+d}}{x^2}dx-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (d+e x^2\right )^{5/2}}{5 x^5}\right )-\frac {125 \left (d+e x^2\right )^{7/2}}{7 x^7}\right )-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {b n \left (-\frac {2}{9} e \left (-36 e \left (e \left (e \left (e \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (d+e x^2\right )^{5/2}}{5 x^5}\right )-\frac {125 \left (d+e x^2\right )^{7/2}}{7 x^7}\right )-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {b n \left (-\frac {2}{9} e \left (-36 e \left (e \left (e \left (e \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (d+e x^2\right )^{5/2}}{5 x^5}\right )-\frac {125 \left (d+e x^2\right )^{7/2}}{7 x^7}\right )-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}-\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {8 e^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{315 d^3 x^5}+\frac {4 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{63 d^2 x^7}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{9 d x^9}+\frac {b n \left (-\frac {2}{9} e \left (-36 e \left (e \left (e \left (\sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {\left (d+e x^2\right )^{5/2}}{5 x^5}\right )-\frac {125 \left (d+e x^2\right )^{7/2}}{7 x^7}\right )-\frac {35 d \left (d+e x^2\right )^{7/2}}{9 x^9}\right )}{315 d^3}\) |
(b*n*((-35*d*(d + e*x^2)^(7/2))/(9*x^9) - (2*e*((-125*(d + e*x^2)^(7/2))/( 7*x^7) - 36*e*(-1/5*(d + e*x^2)^(5/2)/x^5 + e*(-1/3*(d + e*x^2)^(3/2)/x^3 + e*(-(Sqrt[d + e*x^2]/x) + Sqrt[e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])) )))/9))/(315*d^3) - ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(9*d*x^9) + (4* e*(d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(63*d^2*x^7) - (8*e^2*(d + e*x^2)^ (5/2)*(a + b*Log[c*x^n]))/(315*d^3*x^5)
3.3.74.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x_ Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + S imp[d/e^2 Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, -1]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f ^2*(m + 1)) Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x ) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x] }, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2] ) || InverseFunctionFreeQ[u, x]] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x ] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])
\[\int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{10}}d x\]
Time = 0.41 (sec) , antiderivative size = 526, normalized size of antiderivative = 2.05 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=\left [\frac {1260 \, b e^{\frac {9}{2}} n x^{9} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - {\left (2 \, {\left (1307 \, b e^{4} n + 1260 \, a e^{4}\right )} x^{8} - {\left (677 \, b d e^{3} n + 1260 \, a d e^{3}\right )} x^{6} + 1225 \, b d^{4} n + 11025 \, a d^{4} + 3 \, {\left (143 \, b d^{2} e^{2} n + 315 \, a d^{2} e^{2}\right )} x^{4} + 25 \, {\left (97 \, b d^{3} e n + 630 \, a d^{3} e\right )} x^{2} + 315 \, {\left (8 \, b e^{4} x^{8} - 4 \, b d e^{3} x^{6} + 3 \, b d^{2} e^{2} x^{4} + 50 \, b d^{3} e x^{2} + 35 \, b d^{4}\right )} \log \left (c\right ) + 315 \, {\left (8 \, b e^{4} n x^{8} - 4 \, b d e^{3} n x^{6} + 3 \, b d^{2} e^{2} n x^{4} + 50 \, b d^{3} e n x^{2} + 35 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{99225 \, d^{3} x^{9}}, -\frac {2520 \, b \sqrt {-e} e^{4} n x^{9} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (2 \, {\left (1307 \, b e^{4} n + 1260 \, a e^{4}\right )} x^{8} - {\left (677 \, b d e^{3} n + 1260 \, a d e^{3}\right )} x^{6} + 1225 \, b d^{4} n + 11025 \, a d^{4} + 3 \, {\left (143 \, b d^{2} e^{2} n + 315 \, a d^{2} e^{2}\right )} x^{4} + 25 \, {\left (97 \, b d^{3} e n + 630 \, a d^{3} e\right )} x^{2} + 315 \, {\left (8 \, b e^{4} x^{8} - 4 \, b d e^{3} x^{6} + 3 \, b d^{2} e^{2} x^{4} + 50 \, b d^{3} e x^{2} + 35 \, b d^{4}\right )} \log \left (c\right ) + 315 \, {\left (8 \, b e^{4} n x^{8} - 4 \, b d e^{3} n x^{6} + 3 \, b d^{2} e^{2} n x^{4} + 50 \, b d^{3} e n x^{2} + 35 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{99225 \, d^{3} x^{9}}\right ] \]
[1/99225*(1260*b*e^(9/2)*n*x^9*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (2*(1307*b*e^4*n + 1260*a*e^4)*x^8 - (677*b*d*e^3*n + 1260*a*d*e^3) *x^6 + 1225*b*d^4*n + 11025*a*d^4 + 3*(143*b*d^2*e^2*n + 315*a*d^2*e^2)*x^ 4 + 25*(97*b*d^3*e*n + 630*a*d^3*e)*x^2 + 315*(8*b*e^4*x^8 - 4*b*d*e^3*x^6 + 3*b*d^2*e^2*x^4 + 50*b*d^3*e*x^2 + 35*b*d^4)*log(c) + 315*(8*b*e^4*n*x^ 8 - 4*b*d*e^3*n*x^6 + 3*b*d^2*e^2*n*x^4 + 50*b*d^3*e*n*x^2 + 35*b*d^4*n)*l og(x))*sqrt(e*x^2 + d))/(d^3*x^9), -1/99225*(2520*b*sqrt(-e)*e^4*n*x^9*arc tan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (2*(1307*b*e^4*n + 1260*a*e^4)*x^8 - (67 7*b*d*e^3*n + 1260*a*d*e^3)*x^6 + 1225*b*d^4*n + 11025*a*d^4 + 3*(143*b*d^ 2*e^2*n + 315*a*d^2*e^2)*x^4 + 25*(97*b*d^3*e*n + 630*a*d^3*e)*x^2 + 315*( 8*b*e^4*x^8 - 4*b*d*e^3*x^6 + 3*b*d^2*e^2*x^4 + 50*b*d^3*e*x^2 + 35*b*d^4) *log(c) + 315*(8*b*e^4*n*x^8 - 4*b*d*e^3*n*x^6 + 3*b*d^2*e^2*n*x^4 + 50*b* d^3*e*n*x^2 + 35*b*d^4*n)*log(x))*sqrt(e*x^2 + d))/(d^3*x^9)]
Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{10}} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^{10}} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^{10}} \,d x \]