Integrand size = 29, antiderivative size = 151 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {a c-d \sqrt {a+b x^2}}{2 a \left (a c^2-d^2\right ) x^2}-\frac {b d \left (3 a c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{3/2} \left (a c^2-d^2\right )^2}-\frac {b c^3 \log (x)}{\left (a c^2-d^2\right )^2}+\frac {b c^3 \log \left (d+c \sqrt {a+b x^2}\right )}{\left (a c^2-d^2\right )^2} \]
-1/2*b*d*(3*a*c^2-d^2)*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(3/2)/(a*c^2-d^2 )^2-b*c^3*ln(x)/(a*c^2-d^2)^2+b*c^3*ln(d+c*(b*x^2+a)^(1/2))/(a*c^2-d^2)^2+ 1/2*(-a*c+d*(b*x^2+a)^(1/2))/a/(a*c^2-d^2)/x^2
Time = 0.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {b d \left (-3 a c^2+d^2\right ) x^2 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\sqrt {a} \left (-\left (\left (a c^2-d^2\right ) \left (a c-d \sqrt {a+b x^2}\right )\right )-a b c^3 x^2 \log \left (b x^2\right )+2 a b c^3 x^2 \log \left (d+c \sqrt {a+b x^2}\right )\right )}{2 a^{3/2} \left (-a c^2+d^2\right )^2 x^2} \]
(b*d*(-3*a*c^2 + d^2)*x^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]] + Sqrt[a]*(-((a *c^2 - d^2)*(a*c - d*Sqrt[a + b*x^2])) - a*b*c^3*x^2*Log[b*x^2] + 2*a*b*c^ 3*x^2*Log[d + c*Sqrt[a + b*x^2]]))/(2*a^(3/2)*(-(a*c^2) + d^2)^2*x^2)
Time = 0.56 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2586, 7267, 496, 25, 657, 2009}
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.
| \(\displaystyle \int \frac {1}{x^3 \left (d \sqrt {a+b x^2}+a c+b c x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2586 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b c x^2+a c+d \sqrt {b x^2+a}\right )}dx^2\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle b \int \frac {1}{\left (a-x^4\right )^2 \left (\sqrt {b x^2+a} c+d\right )}d\sqrt {b x^2+a}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle b \left (\frac {a c-d \sqrt {a+b x^2}}{2 a \left (a-x^4\right ) \left (a c^2-d^2\right )}-\frac {\int -\frac {2 a c^2-d \sqrt {b x^2+a} c-d^2}{\left (a-x^4\right ) \left (\sqrt {b x^2+a} c+d\right )}d\sqrt {b x^2+a}}{2 a \left (a c^2-d^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b \left (\frac {\int \frac {2 a c^2-d \sqrt {b x^2+a} c-d^2}{\left (a-x^4\right ) \left (\sqrt {b x^2+a} c+d\right )}d\sqrt {b x^2+a}}{2 a \left (a c^2-d^2\right )}+\frac {a c-d \sqrt {a+b x^2}}{2 a \left (a-x^4\right ) \left (a c^2-d^2\right )}\right )\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle b \left (\frac {\int \left (\frac {2 a c^4}{\left (a c^2-d^2\right ) \left (\sqrt {b x^2+a} c+d\right )}+\frac {2 a \sqrt {b x^2+a} c^3-3 a d c^2+d^3}{\left (a c^2-d^2\right ) \left (a-x^4\right )}\right )d\sqrt {b x^2+a}}{2 a \left (a c^2-d^2\right )}+\frac {a c-d \sqrt {a+b x^2}}{2 a \left (a-x^4\right ) \left (a c^2-d^2\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b \left (\frac {-\frac {d \left (3 a c^2-d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a} \left (a c^2-d^2\right )}+\frac {2 a c^3 \log \left (c \sqrt {a+b x^2}+d\right )}{a c^2-d^2}-\frac {a c^3 \log \left (a-x^4\right )}{a c^2-d^2}}{2 a \left (a c^2-d^2\right )}+\frac {a c-d \sqrt {a+b x^2}}{2 a \left (a-x^4\right ) \left (a c^2-d^2\right )}\right )\) |
b*((a*c - d*Sqrt[a + b*x^2])/(2*a*(a*c^2 - d^2)*(a - x^4)) + (-((d*(3*a*c^ 2 - d^2)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(Sqrt[a]*(a*c^2 - d^2))) - (a*c ^3*Log[a - x^4])/(a*c^2 - d^2) + (2*a*c^3*Log[d + c*Sqrt[a + b*x^2]])/(a*c ^2 - d^2))/(2*a*(a*c^2 - d^2)))
3.6.48.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] ), x_Symbol] :> Simp[1/n Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* d, 0] && IntegerQ[(m + 1)/n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(1909\) vs. \(2(137)=274\).
Time = 0.10 (sec) , antiderivative size = 1910, normalized size of antiderivative = 12.65
a*c*(1/2*b*c^4/(a*c^2-d^2)^2/d^2*ln(b*c^2*x^2+a*c^2-d^2)-1/2/a/(a*c^2-d^2) /x^2-b*(2*a*c^2-d^2)/a^2/(a*c^2-d^2)^2*ln(x)-1/2*b/a^2/d^2*ln(b*x^2+a))+b* c*(-1/2*c^2/(a*c^2-d^2)/d^2*ln(b*c^2*x^2+a*c^2-d^2)+1/a/(a*c^2-d^2)*ln(x)+ 1/2/a/d^2*ln(b*x^2+a))-d*(1/a/(a*c^2-d^2)*(-1/2/a/x^2*(b*x^2+a)^(3/2)+1/2* b/a*((b*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)))-b*(2* a*c^2-d^2)/a^2/(a*c^2-d^2)^2*((b*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(b *x^2+a)^(1/2))/x))+1/2*b^2*c^2/a^2/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^ (1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*((b*(x-1/b*(-a*b)^(1/ 2))^2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2)+(-a*b)^(1/2)*ln(((x-1/b*( -a*b)^(1/2))*b+(-a*b)^(1/2))/b^(1/2)+(b*(x-1/b*(-a*b)^(1/2))^2+2*(-a*b)^(1 /2)*(x-1/b*(-a*b)^(1/2)))^(1/2))/b^(1/2))+1/2*b^2*c^2/a^2/((-a*b)^(1/2)*c^ 2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2) )*((b*(x+1/b*(-a*b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/2)))^(1/2)-(- a*b)^(1/2)*ln(((x+1/b*(-a*b)^(1/2))*b-(-a*b)^(1/2))/b^(1/2)+(b*(x+1/b*(-a* b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/2)))^(1/2))/b^(1/2))-1/2*b^2*c ^6/(a*c^2-d^2)^2/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/ 2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*((b*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2 )^2-2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+ d^2/c^2)^(1/2)-(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*ln((-(-(a*c^2-d^2)*b*c^2)^(1 /2)/c^2+(x+(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)*b)/b^(1/2)+(b*(x+(-(a*c^2-...
Time = 0.74 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.51 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\left [\frac {2 \, a^{2} b c^{3} x^{2} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a^{2} b c^{3} x^{2} \log \left (x\right ) + a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} - {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {a} x^{2} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{2}}, \frac {2 \, a^{2} b c^{3} x^{2} \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a^{2} b c^{3} x^{2} \log \left (x\right ) + a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a^{2} b c^{3} x^{2} \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - 2 \, a^{3} c^{3} + 2 \, a^{2} c d^{2} + 2 \, {\left (3 \, a b c^{2} d - b d^{3}\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 2 \, {\left (a^{2} c^{2} d - a d^{3}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4}\right )} x^{2}}\right ] \]
[1/4*(2*a^2*b*c^3*x^2*log(b*c^2*x^2 + a*c^2 - d^2) - 4*a^2*b*c^3*x^2*log(x ) + a^2*b*c^3*x^2*log(-(b*c^2*x^2 + a*c^2 + 2*sqrt(b*x^2 + a)*c*d + d^2)/x ^2) - a^2*b*c^3*x^2*log(-(b*c^2*x^2 + a*c^2 - 2*sqrt(b*x^2 + a)*c*d + d^2) /x^2) - 2*a^3*c^3 + 2*a^2*c*d^2 - (3*a*b*c^2*d - b*d^3)*sqrt(a)*x^2*log(-( b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(a^2*c^2*d - a*d^3)*sqrt (b*x^2 + a))/((a^4*c^4 - 2*a^3*c^2*d^2 + a^2*d^4)*x^2), 1/4*(2*a^2*b*c^3*x ^2*log(b*c^2*x^2 + a*c^2 - d^2) - 4*a^2*b*c^3*x^2*log(x) + a^2*b*c^3*x^2*l og(-(b*c^2*x^2 + a*c^2 + 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) - a^2*b*c^3*x^2 *log(-(b*c^2*x^2 + a*c^2 - 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) - 2*a^3*c^3 + 2*a^2*c*d^2 + 2*(3*a*b*c^2*d - b*d^3)*sqrt(-a)*x^2*arctan(sqrt(-a)/sqrt(b *x^2 + a)) + 2*(a^2*c^2*d - a*d^3)*sqrt(b*x^2 + a))/((a^4*c^4 - 2*a^3*c^2* d^2 + a^2*d^4)*x^2)]
\[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\int \frac {1}{x^{3} \left (a c + b c x^{2} + d \sqrt {a + b x^{2}}\right )}\, dx \]
\[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\int { \frac {1}{{\left (b c x^{2} + a c + \sqrt {b x^{2} + a} d\right )} x^{3}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {b c^{4} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{a^{2} c^{5} - 2 \, a c^{3} d^{2} + c d^{4}} - \frac {b c^{3} \log \left (-b x^{2}\right )}{2 \, {\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )}} + \frac {{\left (3 \, a b c^{2} d - b d^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, {\left (a^{3} c^{4} - 2 \, a^{2} c^{2} d^{2} + a d^{4}\right )} \sqrt {-a}} - \frac {a^{2} b c^{3} - a b c d^{2} - {\left (a b c^{2} d - b d^{3}\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a c^{2} - d^{2}\right )}^{2} a b x^{2}} \]
b*c^4*log(abs(sqrt(b*x^2 + a)*c + d))/(a^2*c^5 - 2*a*c^3*d^2 + c*d^4) - 1/ 2*b*c^3*log(-b*x^2)/(a^2*c^4 - 2*a*c^2*d^2 + d^4) + 1/2*(3*a*b*c^2*d - b*d ^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/((a^3*c^4 - 2*a^2*c^2*d^2 + a*d^4)*sq rt(-a)) - 1/2*(a^2*b*c^3 - a*b*c*d^2 - (a*b*c^2*d - b*d^3)*sqrt(b*x^2 + a) )/((a*c^2 - d^2)^2*a*b*x^2)
Time = 19.54 (sec) , antiderivative size = 4602, normalized size of antiderivative = 30.48 \[ \int \frac {1}{x^3 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\text {Too large to display} \]
(atan(((((((12*a^2*b*c^6*d^9 - 28*a^3*b*c^8*d^7 + 32*a^4*b*c^10*d^5 - 18*a ^5*b*c^12*d^3 - 2*a*b*c^4*d^11 + 4*a^6*b*c^14*d)/(16*(a^5*c^6 - a^2*d^6 + 3*a^3*c^2*d^4 - 3*a^4*c^4*d^2)) - ((a + b*x^2)^(1/2)*(16*(b^2*d^6 - 6*a*b^ 2*c^2*d^4 + 9*a^2*b^2*c^4*d^2)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5* c^4*d^4 - 4*a^6*c^6*d^2))^(1/2)*(16*a^7*c^14 + 16*a^2*c^4*d^10 - 48*a^3*c^ 6*d^8 + 32*a^4*c^8*d^6 + 32*a^5*c^10*d^4 - 48*a^6*c^12*d^2))/(512*(a^4*c^4 + a^2*d^4 - 2*a^3*c^2*d^2)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4 *d^4 - 4*a^6*c^6*d^2)))*(16*(b^2*d^6 - 6*a*b^2*c^2*d^4 + 9*a^2*b^2*c^4*d^2 )*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4*d^4 - 4*a^6*c^6*d^2))^(1/ 2))/(16*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4*d^4 - 4*a^6*c^6*d^2 )) + ((a + b*x^2)^(1/2)*(b^2*c^6*d^6 - 6*a*b^2*c^8*d^4 + 13*a^2*b^2*c^10*d ^2))/(32*(a^4*c^4 + a^2*d^4 - 2*a^3*c^2*d^2)))*(16*(b^2*d^6 - 6*a*b^2*c^2* d^4 + 9*a^2*b^2*c^4*d^2)*(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c^4*d^ 4 - 4*a^6*c^6*d^2))^(1/2)*1i)/(a^7*c^8 + a^3*d^8 - 4*a^4*c^2*d^6 + 6*a^5*c ^4*d^4 - 4*a^6*c^6*d^2) - (((((12*a^2*b*c^6*d^9 - 28*a^3*b*c^8*d^7 + 32*a^ 4*b*c^10*d^5 - 18*a^5*b*c^12*d^3 - 2*a*b*c^4*d^11 + 4*a^6*b*c^14*d)/(16*(a ^5*c^6 - a^2*d^6 + 3*a^3*c^2*d^4 - 3*a^4*c^4*d^2)) + ((a + b*x^2)^(1/2)*(1 6*(b^2*d^6 - 6*a*b^2*c^2*d^4 + 9*a^2*b^2*c^4*d^2)*(a^7*c^8 + a^3*d^8 - 4*a ^4*c^2*d^6 + 6*a^5*c^4*d^4 - 4*a^6*c^6*d^2))^(1/2)*(16*a^7*c^14 + 16*a^2*c ^4*d^10 - 48*a^3*c^6*d^8 + 32*a^4*c^8*d^6 + 32*a^5*c^10*d^4 - 48*a^6*c^...