Integrand size = 19, antiderivative size = 139 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\frac {3 a^2 b c x^2 \sqrt {c \left (a+b x^2\right )^2}}{2 \left (a+b x^2\right )}+\frac {3 a b^2 c x^4 \sqrt {c \left (a+b x^2\right )^2}}{4 \left (a+b x^2\right )}+\frac {b^3 c x^6 \sqrt {c \left (a+b x^2\right )^2}}{6 \left (a+b x^2\right )}+\frac {a^3 c \sqrt {c \left (a+b x^2\right )^2} \log (x)}{a+b x^2} \]
3/2*a^2*b*c*x^2*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)+3/4*a*b^2*c*x^4*(c*(b*x^2+ a)^2)^(1/2)/(b*x^2+a)+1/6*b^3*c*x^6*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)+a^3*c* ln(x)*(c*(b*x^2+a)^2)^(1/2)/(b*x^2+a)
Time = 0.52 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.68 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\frac {c \left (b \left (2 b^2 x^4 \left (-\sqrt {b^2 c} x^2+\sqrt {c \left (a+b x^2\right )^2}\right )+a b x^2 \left (-9 \sqrt {b^2 c} x^2+7 \sqrt {c \left (a+b x^2\right )^2}\right )+a^2 \left (-18 \sqrt {b^2 c} x^2+11 \sqrt {c \left (a+b x^2\right )^2}\right )\right )+12 a^3 b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {b^2 c} x^2-\sqrt {c \left (a+b x^2\right )^2}}{a \sqrt {c}}\right )-6 a^3 \sqrt {b^2 c} \log \left (x^2 \left (a b c+b^2 c x^2-\sqrt {b^2 c} \sqrt {c \left (a+b x^2\right )^2}\right )\right )\right )}{24 b} \]
(c*(b*(2*b^2*x^4*(-(Sqrt[b^2*c]*x^2) + Sqrt[c*(a + b*x^2)^2]) + a*b*x^2*(- 9*Sqrt[b^2*c]*x^2 + 7*Sqrt[c*(a + b*x^2)^2]) + a^2*(-18*Sqrt[b^2*c]*x^2 + 11*Sqrt[c*(a + b*x^2)^2])) + 12*a^3*b*Sqrt[c]*ArcTanh[(Sqrt[b^2*c]*x^2 - S qrt[c*(a + b*x^2)^2])/(a*Sqrt[c])] - 6*a^3*Sqrt[b^2*c]*Log[x^2*(a*b*c + b^ 2*c*x^2 - Sqrt[b^2*c]*Sqrt[c*(a + b*x^2)^2])]))/(24*b)
Time = 0.22 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.49, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2045, 27, 243, 49, 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.
\(\displaystyle \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \frac {a^3 c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{a^3 x}dx}{a+b x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{x}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \frac {\left (b x^2+a\right )^3}{x^2}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \int \left (b^3 x^4+3 a b^2 x^2+3 a^2 b+\frac {a^3}{x^2}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \sqrt {c \left (a+b x^2\right )^2} \left (a^3 \log \left (x^2\right )+3 a^2 b x^2+\frac {3}{2} a b^2 x^4+\frac {b^3 x^6}{3}\right )}{2 \left (a+b x^2\right )}\) |
(c*Sqrt[c*(a + b*x^2)^2]*(3*a^2*b*x^2 + (3*a*b^2*x^4)/2 + (b^3*x^6)/3 + a^ 3*Log[x^2]))/(2*(a + b*x^2))
3.3.34.3.1 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_))^(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[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.42
method | result | size |
default | \(\frac {{\left (c \left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}} \left (2 b^{3} x^{6}+9 b^{2} x^{4} a +18 a^{2} b \,x^{2}+12 a^{3} \ln \left (x \right )\right )}{12 \left (b \,x^{2}+a \right )^{3}}\) | \(59\) |
pseudoelliptic | \(\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, \left (2 b^{3} x^{6}+9 b^{2} x^{4} a +6 a^{3} \ln \left (x^{2}\right )+18 a^{2} b \,x^{2}\right )}{12 b \,x^{2}+12 a}\) | \(64\) |
risch | \(\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, b \left (\frac {1}{6} b^{2} x^{6}+\frac {3}{4} a b \,x^{4}+\frac {3}{2} a^{2} x^{2}\right )}{b \,x^{2}+a}+\frac {a^{3} c \ln \left (x \right ) \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\) | \(80\) |
Time = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.53 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\frac {{\left (2 \, b^{3} c x^{6} + 9 \, a b^{2} c x^{4} + 18 \, a^{2} b c x^{2} + 12 \, a^{3} c \log \left (x\right )\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{12 \, {\left (b x^{2} + a\right )}} \]
1/12*(2*b^3*c*x^6 + 9*a*b^2*c*x^4 + 18*a^2*b*c*x^2 + 12*a^3*c*log(x))*sqrt (b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^2 + a)
\[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\int \frac {\left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x}\, dx \]
Time = 0.19 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.23 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\frac {1}{2} \, \left (-1\right )^{2 \, b^{2} c x^{2} + 2 \, a b c} a^{3} c^{\frac {3}{2}} \log \left (2 \, b^{2} c x^{2} + 2 \, a b c\right ) - \frac {1}{2} \, \left (-1\right )^{2 \, a b c x^{2} + 2 \, a^{2} c} a^{3} c^{\frac {3}{2}} \log \left (2 \, a b c + \frac {2 \, a^{2} c}{x^{2}}\right ) + \frac {1}{4} \, \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} a b c x^{2} + \frac {3}{4} \, \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} a^{2} c + \frac {1}{6} \, {\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )}^{\frac {3}{2}} \]
1/2*(-1)^(2*b^2*c*x^2 + 2*a*b*c)*a^3*c^(3/2)*log(2*b^2*c*x^2 + 2*a*b*c) - 1/2*(-1)^(2*a*b*c*x^2 + 2*a^2*c)*a^3*c^(3/2)*log(2*a*b*c + 2*a^2*c/x^2) + 1/4*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*a*b*c*x^2 + 3/4*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*a^2*c + 1/6*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)^(3/2)
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.53 \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\frac {1}{12} \, {\left (2 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 18 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 6 \, a^{3} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right )\right )} c^{\frac {3}{2}} \]
1/12*(2*b^3*x^6*sgn(b*x^2 + a) + 9*a*b^2*x^4*sgn(b*x^2 + a) + 18*a^2*b*x^2 *sgn(b*x^2 + a) + 6*a^3*log(x^2)*sgn(b*x^2 + a))*c^(3/2)
Timed out. \[ \int \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2}}{x} \, dx=\int \frac {{\left (c\,{\left (b\,x^2+a\right )}^2\right )}^{3/2}}{x} \,d x \]