Integrand size = 19, antiderivative size = 139 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}-\frac {3 c^4 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{5/2}} \]
-1/4*(c*x^2+b*x)^(3/2)/x^(11/2)-3/64*c^4*arctanh((c*x^2+b*x)^(1/2)/b^(1/2) /x^(1/2))/b^(5/2)-1/8*c*(c*x^2+b*x)^(1/2)/x^(7/2)-1/32*c^2*(c*x^2+b*x)^(1/ 2)/b/x^(5/2)+3/64*c^3*(c*x^2+b*x)^(1/2)/b^2/x^(3/2)
Time = 0.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.75 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} \sqrt {b+c x} \left (16 b^3+24 b^2 c x+2 b c^2 x^2-3 c^3 x^3\right )+3 c^4 x^4 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{64 b^{5/2} x^{9/2} \sqrt {b+c x}} \]
-1/64*(Sqrt[x*(b + c*x)]*(Sqrt[b]*Sqrt[b + c*x]*(16*b^3 + 24*b^2*c*x + 2*b *c^2*x^2 - 3*c^3*x^3) + 3*c^4*x^4*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(b^(5/2 )*x^(9/2)*Sqrt[b + c*x])
Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1130, 1130, 1135, 1135, 1136, 220}
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.
\(\displaystyle \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {3}{8} c \int \frac {\sqrt {c x^2+b x}}{x^{9/2}}dx-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {3}{8} c \left (\frac {1}{6} c \int \frac {1}{x^{5/2} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{3 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {3}{8} c \left (\frac {1}{6} c \left (-\frac {3 c \int \frac {1}{x^{3/2} \sqrt {c x^2+b x}}dx}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )-\frac {\sqrt {b x+c x^2}}{3 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {3}{8} c \left (\frac {1}{6} c \left (-\frac {3 c \left (-\frac {c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )-\frac {\sqrt {b x+c x^2}}{3 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {3}{8} c \left (\frac {1}{6} c \left (-\frac {3 c \left (-\frac {c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )-\frac {\sqrt {b x+c x^2}}{3 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {3}{8} c \left (\frac {1}{6} c \left (-\frac {3 c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )-\frac {\sqrt {b x+c x^2}}{3 x^{7/2}}\right )-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}\) |
-1/4*(b*x + c*x^2)^(3/2)/x^(11/2) + (3*c*(-1/3*Sqrt[b*x + c*x^2]/x^(7/2) + (c*(-1/2*Sqrt[b*x + c*x^2]/(b*x^(5/2)) - (3*c*(-(Sqrt[b*x + c*x^2]/(b*x^( 3/2))) + (c*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)))/(4*b)) )/6))/8
3.1.92.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
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[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e))) Int [(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I ntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Time = 2.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.67
method | result | size |
risch | \(-\frac {\left (c x +b \right ) \left (-3 c^{3} x^{3}+2 b \,c^{2} x^{2}+24 b^{2} c x +16 b^{3}\right )}{64 x^{\frac {7}{2}} b^{2} \sqrt {x \left (c x +b \right )}}-\frac {3 c^{4} \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{64 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\) | \(93\) |
default | \(-\frac {\sqrt {x \left (c x +b \right )}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-3 c^{3} x^{3} \sqrt {c x +b}\, \sqrt {b}+2 b^{\frac {3}{2}} c^{2} x^{2} \sqrt {c x +b}+24 b^{\frac {5}{2}} c x \sqrt {c x +b}+16 b^{\frac {7}{2}} \sqrt {c x +b}\right )}{64 b^{\frac {5}{2}} x^{\frac {9}{2}} \sqrt {c x +b}}\) | \(108\) |
-1/64*(c*x+b)*(-3*c^3*x^3+2*b*c^2*x^2+24*b^2*c*x+16*b^3)/x^(7/2)/b^2/(x*(c *x+b))^(1/2)-3/64*c^4/b^(5/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2) *x^(1/2)/(x*(c*x+b))^(1/2)
Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.41 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\left [\frac {3 \, \sqrt {b} c^{4} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (3 \, b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 24 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{128 \, b^{3} x^{5}}, \frac {3 \, \sqrt {-b} c^{4} x^{5} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 24 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{64 \, b^{3} x^{5}}\right ] \]
[1/128*(3*sqrt(b)*c^4*x^5*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b )*sqrt(x))/x^2) + 2*(3*b*c^3*x^3 - 2*b^2*c^2*x^2 - 24*b^3*c*x - 16*b^4)*sq rt(c*x^2 + b*x)*sqrt(x))/(b^3*x^5), 1/64*(3*sqrt(-b)*c^4*x^5*arctan(sqrt(- b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (3*b*c^3*x^3 - 2*b^2*c^2*x^2 - 24*b^3*c*x - 16*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^5)]
\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}{x^{\frac {13}{2}}}\, dx \]
\[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {13}{2}}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.71 \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\frac {\frac {3 \, c^{5} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (c x + b\right )}^{\frac {7}{2}} c^{5} - 11 \, {\left (c x + b\right )}^{\frac {5}{2}} b c^{5} - 11 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c^{5} + 3 \, \sqrt {c x + b} b^{3} c^{5}}{b^{2} c^{4} x^{4}}}{64 \, c} \]
1/64*(3*c^5*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x + b)^( 7/2)*c^5 - 11*(c*x + b)^(5/2)*b*c^5 - 11*(c*x + b)^(3/2)*b^2*c^5 + 3*sqrt( c*x + b)*b^3*c^5)/(b^2*c^4*x^4))/c
Timed out. \[ \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{13/2}} \,d x \]