Integrand size = 19, antiderivative size = 75 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=-\frac {c \sqrt {b x^2+c x^4}}{x^2}-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6}+c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right ) \]
-1/3*(c*x^4+b*x^2)^(3/2)/x^6+c^(3/2)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/ 2))-c*(c*x^4+b*x^2)^(1/2)/x^2
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b+c x^2} \left (b+4 c x^2\right )+3 c^{3/2} x^3 \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{3 x^4 \sqrt {b+c x^2}} \]
-1/3*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b + c*x^2]*(b + 4*c*x^2) + 3*c^(3/2)*x^3 *Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]]))/(x^4*Sqrt[b + c*x^2])
Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1424, 1130, 1125, 25, 27, 1091, 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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^8}dx^2\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {1}{2} \left (c \int \frac {\sqrt {c x^4+b x^2}}{x^4}dx^2-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 1125 |
\(\displaystyle \frac {1}{2} \left (c \left (-\int -\frac {c}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (c \left (\int \frac {c}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (c \left (c \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (c \left (2 c \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (c \left (2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{x^2}\right )-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{3 x^6}\right )\) |
((-2*(b*x^2 + c*x^4)^(3/2))/(3*x^6) + c*((-2*Sqrt[b*x^2 + c*x^4])/x^2 + 2* Sqrt[c]*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]]))/2
3.3.43.3.1 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[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 2)*(d + e*x))), x] - Simp[e^(2*m + 2) Int[(1/Sqrt[a + b*x + c*x^2])*Expan dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x ), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[m, 0] && EqQ[m + p, -3/2]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97
method | result | size |
risch | \(-\frac {\left (4 c \,x^{2}+b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3 x^{4}}+\frac {c^{\frac {3}{2}} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\) | \(73\) |
pseudoelliptic | \(\frac {3 x^{4} \left (-\ln \left (2\right )+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right )\right ) c^{\frac {3}{2}}-2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (4 c \,x^{2}+b \right )}{6 x^{4}}\) | \(74\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (2 c^{\frac {5}{2}} \left (c \,x^{2}+b \right )^{\frac {3}{2}} x^{4}+3 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b}\, b \,x^{4}-2 c^{\frac {3}{2}} \left (c \,x^{2}+b \right )^{\frac {5}{2}} x^{2}+3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{2} c^{2} x^{3}-\left (c \,x^{2}+b \right )^{\frac {5}{2}} b \sqrt {c}\right )}{3 x^{6} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}}\) | \(129\) |
-1/3*(4*c*x^2+b)/x^4*(x^2*(c*x^2+b))^(1/2)+c^(3/2)*ln(x*c^(1/2)+(c*x^2+b)^ (1/2))*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\left [\frac {3 \, c^{\frac {3}{2}} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (4 \, c x^{2} + b\right )}}{6 \, x^{4}}, -\frac {3 \, \sqrt {-c} c x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (4 \, c x^{2} + b\right )}}{3 \, x^{4}}\right ] \]
[1/6*(3*c^(3/2)*x^4*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2* sqrt(c*x^4 + b*x^2)*(4*c*x^2 + b))/x^4, -1/3*(3*sqrt(-c)*c*x^4*arctan(sqrt (c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + sqrt(c*x^4 + b*x^2)*(4*c*x^2 + b)) /x^4]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{7}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.19 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\frac {1}{2} \, c^{\frac {3}{2}} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{4} + b x^{2}} c}{6 \, x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} b}{6 \, x^{4}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{6 \, x^{6}} \]
1/2*c^(3/2)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 7/6*sqrt(c* x^4 + b*x^2)*c/x^2 - 1/6*sqrt(c*x^4 + b*x^2)*b/x^4 - 1/6*(c*x^4 + b*x^2)^( 3/2)/x^6
Time = 0.42 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.63 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=-\frac {1}{2} \, c^{\frac {3}{2}} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {4 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{2} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 2 \, b^{3} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3}} \]
-1/2*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 4/3*(3*(sqrt(c) *x - sqrt(c*x^2 + b))^4*b*c^(3/2)*sgn(x) - 3*(sqrt(c)*x - sqrt(c*x^2 + b)) ^2*b^2*c^(3/2)*sgn(x) + 2*b^3*c^(3/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b ))^2 - b)^3
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^7} \,d x \]