Integrand size = 21, antiderivative size = 134 \[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {b}{a (b+a c) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}+\frac {\text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {b+a c}}\right )}{(b+a c)^{3/2}} \]
arctanh(((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/a^(1/2))/a^(3/2)-c^(3/2)*arctanh (c^(1/2)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*c+b)^(1/2))/(a*c+b)^(3/2)-b/ a/(a*c+b)/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)
Time = 0.41 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {b}{a (b+a c) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}-\frac {c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {-b-a c}}\right )}{(-b-a c)^{3/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {a}}\right )}{a^{3/2}} \]
-(b/(a*(b + a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])) - (c^(3/2)*ArcTan [(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[-b - a*c]])/(-b - a* c)^(3/2) + ArcTanh[Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]/Sqrt[a]]/a^(3/2)
Time = 0.39 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2057, 2053, 2052, 25, 27, 382, 397, 219, 221}
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.
| \(\displaystyle \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int \frac {1}{x \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (\frac {a d x^2+b+a c}{d x^2+c}\right )^{3/2}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int -\frac {1}{d x^4 \left (a-x^4\right ) \left (-c x^4+b+a c\right )}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b d \int \frac {1}{d x^4 \left (a-x^4\right ) \left (-c x^4+b+a c\right )}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle b \int \frac {1}{x^4 \left (a-x^4\right ) \left (-c x^4+b+a c\right )}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 382 |
| \(\displaystyle b \left (\frac {\int \frac {-c x^4+b+2 a c}{\left (a-x^4\right ) \left (-c x^4+b+a c\right )}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{a (a c+b)}-\frac {1}{a x^2 (a c+b)}\right )\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle b \left (\frac {\frac {(a c+b) \int \frac {1}{a-x^4}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{b}-\frac {a c^2 \int \frac {1}{-c x^4+b+a c}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{b}}{a (a c+b)}-\frac {1}{a x^2 (a c+b)}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle b \left (\frac {\frac {(a c+b) \text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {a c^2 \int \frac {1}{-c x^4+b+a c}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{b}}{a (a c+b)}-\frac {1}{a x^2 (a c+b)}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle b \left (\frac {\frac {(a c+b) \text {arctanh}\left (\frac {\sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {a c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{b \sqrt {a c+b}}}{a (a c+b)}-\frac {1}{a x^2 (a c+b)}\right )\) |
b*(-(1/(a*(b + a*c)*x^2)) + (((b + a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x^2)/( c + d*x^2)]/Sqrt[a]])/(Sqrt[a]*b) - (a*c^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[b + a*c]])/(b*Sqrt[b + a*c]))/(a*(b + a* c)))
3.4.58.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ (a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b* x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m + 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* ((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1014\) vs. \(2(116)=232\).
Time = 0.13 (sec) , antiderivative size = 1015, normalized size of antiderivative = 7.57
-1/2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)/a*(-ln(1/2*(2*a*d^2*x^2+2 *a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d )/(a*d^2)^(1/2))*a^3*c^2*d^2*x^2-2*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^ 4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a ^2*b*c*d^2*x^2+(a*c^2+b*c)^(1/2)*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*(a*c^2+ b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b*c)/x^2)*(a* d^2)^(1/2)*a^2*c*d*x^2-ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^ 2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a^3*c^3*d-ln( 1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2) *(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a*b^2*d^2*x^2-3*ln(1/2*(2*a*d^2*x^2+2*a *c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/ (a*d^2)^(1/2))*a^2*b*c^2*d+(a*c^2+b*c)^(1/2)*ln((2*a*c*d*x^2+b*d*x^2+2*a*c ^2+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b *c)/x^2)*(a*d^2)^(1/2)*a^2*c^2-3*ln(1/2*(2*a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+ 2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a*b ^2*c*d+(a*c^2+b*c)^(1/2)*ln((2*a*c*d*x^2+b*d*x^2+2*a*c^2+2*(a*c^2+b*c)^(1/ 2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b*c)/x^2)*(a*d^2)^(1/ 2)*a*b*c+2*((a*d*x^2+a*c+b)*(d*x^2+c))^(1/2)*(a*d^2)^(1/2)*a*b*c-ln(1/2*(2 *a*d^2*x^2+2*a*c*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*d^ 2)^(1/2)+b*d)/(a*d^2)^(1/2))*b^3*d+2*((a*d*x^2+a*c+b)*(d*x^2+c))^(1/2)*...
Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (116) = 232\).
Time = 0.47 (sec) , antiderivative size = 1477, normalized size of antiderivative = 11.02 \[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*((a^2*c^2 + (a^2*c + a*b)*d*x^2 + 2*a*b*c + b^2)*sqrt(a)*log(8*a^2*d^ 2*x^4 + 8*a^2*c^2 + 8*(2*a^2*c + a*b)*d*x^2 + 8*a*b*c + b^2 + 4*(2*a*d^2*x ^4 + (4*a*c + b)*d*x^2 + 2*a*c^2 + b*c)*sqrt(a)*sqrt((a*d*x^2 + a*c + b)/( d*x^2 + c))) + (a^3*c*d*x^2 + a^3*c^2 + a^2*b*c)*sqrt(c/(a*c + b))*log(((8 *a^2*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 8*a^2*c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8 *(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d*x^2 - 4*((2*a^2*c^2 + 3*a*b*c + b^2)*d^ 2*x^4 + 2*a^2*c^4 + 4*a*b*c^3 + 2*b^2*c^2 + (4*a^2*c^3 + 7*a*b*c^2 + 3*b^2 *c)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt(c/(a*c + b)))/x^4) - 4*(a*b*d*x^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*c^2 + 2 *a^3*b*c + a^2*b^2 + (a^4*c + a^3*b)*d*x^2), -1/4*(2*(a^2*c^2 + (a^2*c + a *b)*d*x^2 + 2*a*b*c + b^2)*sqrt(-a)*arctan(1/2*(2*a*d*x^2 + 2*a*c + b)*sqr t(-a)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))/(a^2*d*x^2 + a^2*c + a*b)) - ( a^3*c*d*x^2 + a^3*c^2 + a^2*b*c)*sqrt(c/(a*c + b))*log(((8*a^2*c^2 + 8*a*b *c + b^2)*d^2*x^4 + 8*a^2*c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8*(2*a^2*c^3 + 3* a*b*c^2 + b^2*c)*d*x^2 - 4*((2*a^2*c^2 + 3*a*b*c + b^2)*d^2*x^4 + 2*a^2*c^ 4 + 4*a*b*c^3 + 2*b^2*c^2 + (4*a^2*c^3 + 7*a*b*c^2 + 3*b^2*c)*d*x^2)*sqrt( (a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt(c/(a*c + b)))/x^4) + 4*(a*b*d*x^2 + a*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(a^4*c^2 + 2*a^3*b*c + a^2*b ^2 + (a^4*c + a^3*b)*d*x^2), 1/4*(2*(a^3*c*d*x^2 + a^3*c^2 + a^2*b*c)*sqrt (-c/(a*c + b))*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt((a...
\[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {1}{x \left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\frac {c^{2} \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{2 \, \sqrt {{\left (a c + b\right )} c} {\left (a c + b\right )}} - \frac {b}{{\left (a^{2} c + a b\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}} - \frac {\log \left (-\frac {\sqrt {a} - \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{\sqrt {a} + \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}\right )}{2 \, a^{\frac {3}{2}}} \]
1/2*c^2*log((c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) - sqrt((a*c + b)*c))/ (c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) + sqrt((a*c + b)*c)))/(sqrt((a*c + b)*c)*(a*c + b)) - b/((a^2*c + a*b)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c) )) - 1/2*log(-(sqrt(a) - sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/(sqrt(a) + sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))))/a^(3/2)
Exception generated. \[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {1}{x \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \,d x \]