Integrand size = 17, antiderivative size = 97 \[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 x^3}{c \sqrt {b x+c x^2}}-\frac {15 b \sqrt {b x+c x^2}}{4 c^3}+\frac {5 x \sqrt {b x+c x^2}}{2 c^2}+\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}} \]
15/4*b^2*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)-2*x^3/c/(c*x^2+b*x)^ (1/2)-15/4*b*(c*x^2+b*x)^(1/2)/c^3+5/2*x*(c*x^2+b*x)^(1/2)/c^2
Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c} x \left (-15 b^2-5 b c x+2 c^2 x^2\right )+30 b^2 \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{4 c^{7/2} \sqrt {x (b+c x)}} \]
(Sqrt[c]*x*(-15*b^2 - 5*b*c*x + 2*c^2*x^2) + 30*b^2*Sqrt[x]*Sqrt[b + c*x]* ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(4*c^(7/2)*Sqrt[x*( b + c*x)])
Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1124, 2192, 27, 1160, 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.
\(\displaystyle \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1124 |
\(\displaystyle \frac {\int \frac {b^2-c x b+c^2 x^2}{\sqrt {c x^2+b x}}dx}{c^3}-\frac {2 b^2 x}{c^3 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {\frac {\int \frac {b c (4 b-7 c x)}{2 \sqrt {c x^2+b x}}dx}{2 c}+\frac {1}{2} c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b^2 x}{c^3 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{4} b \int \frac {4 b-7 c x}{\sqrt {c x^2+b x}}dx+\frac {1}{2} c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b^2 x}{c^3 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\frac {1}{4} b \left (\frac {15}{2} b \int \frac {1}{\sqrt {c x^2+b x}}dx-7 \sqrt {b x+c x^2}\right )+\frac {1}{2} c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b^2 x}{c^3 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {\frac {1}{4} b \left (15 b \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}-7 \sqrt {b x+c x^2}\right )+\frac {1}{2} c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b^2 x}{c^3 \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{4} b \left (\frac {15 b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{\sqrt {c}}-7 \sqrt {b x+c x^2}\right )+\frac {1}{2} c x \sqrt {b x+c x^2}}{c^3}-\frac {2 b^2 x}{c^3 \sqrt {b x+c x^2}}\) |
(-2*b^2*x)/(c^3*Sqrt[b*x + c*x^2]) + ((c*x*Sqrt[b*x + c*x^2])/2 + (b*(-7*S qrt[b*x + c*x^2] + (15*b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/Sqrt[c])) /4)/c^3
3.1.52.3.1 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)^(3/2), x _Symbol] :> Simp[-2*e*(2*c*d - b*e)^(m - 2)*((d + e*x)/(c^(m - 1)*Sqrt[a + b*x + c*x^2])), x] + Simp[e^2/c^(m - 1) Int[(1/Sqrt[a + b*x + c*x^2])*Exp andToSum[((2*c*d - b*e)^(m - 1) - c^(m - 1)*(d + e*x)^(m - 1))/(c*d - b*e - c*e*x), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e ^2, 0] && IGtQ[m, 0]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 1.87 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(\frac {2 c^{\frac {5}{2}} x^{3}-5 c^{\frac {3}{2}} b \,x^{2}-15 x \,b^{2} \sqrt {c}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) \sqrt {x \left (c x +b \right )}\, b^{2}}{4 c^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}\) | \(73\) |
risch | \(-\frac {\left (-2 c x +7 b \right ) x \left (c x +b \right )}{4 c^{3} \sqrt {x \left (c x +b \right )}}+\frac {15 b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {7}{2}}}-\frac {2 b^{2} \sqrt {c \left (\frac {b}{c}+x \right )^{2}-b \left (\frac {b}{c}+x \right )}}{c^{4} \left (\frac {b}{c}+x \right )}\) | \(103\) |
default | \(\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\) | \(145\) |
1/4*(2*c^(5/2)*x^3-5*c^(3/2)*b*x^2-15*x*b^2*c^(1/2)+15*arctanh((x*(c*x+b)) ^(1/2)/x/c^(1/2))*(x*(c*x+b))^(1/2)*b^2)/c^(7/2)/(x*(c*x+b))^(1/2)
Time = 0.26 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.88 \[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{2} c x + b^{3}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, c^{3} x^{2} - 5 \, b c^{2} x - 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (c^{5} x + b c^{4}\right )}}, -\frac {15 \, {\left (b^{2} c x + b^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, c^{3} x^{2} - 5 \, b c^{2} x - 15 \, b^{2} c\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (c^{5} x + b c^{4}\right )}}\right ] \]
[1/8*(15*(b^2*c*x + b^3)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt( c)) + 2*(2*c^3*x^2 - 5*b*c^2*x - 15*b^2*c)*sqrt(c*x^2 + b*x))/(c^5*x + b*c ^4), -1/4*(15*(b^2*c*x + b^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/( c*x)) - (2*c^3*x^2 - 5*b*c^2*x - 15*b^2*c)*sqrt(c*x^2 + b*x))/(c^5*x + b*c ^4)]
\[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, b x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} - \frac {15 \, b^{2} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} \]
1/2*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*b*x^2/(sqrt(c*x^2 + b*x)*c^2) - 15/4*b ^2*x/(sqrt(c*x^2 + b*x)*c^3) + 15/8*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x )*sqrt(c))/c^(7/2)
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, x}{c^{2}} - \frac {7 \, b}{c^{3}}\right )} - \frac {15 \, b^{2} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, b^{3}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b\right )} c^{\frac {7}{2}}} \]
1/4*sqrt(c*x^2 + b*x)*(2*x/c^2 - 7*b/c^3) - 15/8*b^2*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2) - 2*b^3/(((sqrt(c)*x - sqrt(c* x^2 + b*x))*sqrt(c) + b)*c^(7/2))
Timed out. \[ \int \frac {x^4}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]