Integrand size = 19, antiderivative size = 142 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}-\frac {c \sqrt {b x+c x^2}}{24 b x^{7/2}}+\frac {5 c^2 \sqrt {b x+c x^2}}{96 b^2 x^{5/2}}-\frac {5 c^3 \sqrt {b x+c x^2}}{64 b^3 x^{3/2}}+\frac {5 c^4 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{7/2}} \]
5/64*c^4*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(7/2)-1/4*(c*x^2+b*x )^(1/2)/x^(9/2)-1/24*c*(c*x^2+b*x)^(1/2)/b/x^(7/2)+5/96*c^2*(c*x^2+b*x)^(1 /2)/b^2/x^(5/2)-5/64*c^3*(c*x^2+b*x)^(1/2)/b^3/x^(3/2)
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\frac {\sqrt {x (b+c x)} \left (-\sqrt {b} \sqrt {b+c x} \left (48 b^3+8 b^2 c x-10 b c^2 x^2+15 c^3 x^3\right )+15 c^4 x^4 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{192 b^{7/2} x^{9/2} \sqrt {b+c x}} \]
(Sqrt[x*(b + c*x)]*(-(Sqrt[b]*Sqrt[b + c*x]*(48*b^3 + 8*b^2*c*x - 10*b*c^2 *x^2 + 15*c^3*x^3)) + 15*c^4*x^4*ArcTanh[Sqrt[b + c*x]/Sqrt[b]]))/(192*b^( 7/2)*x^(9/2)*Sqrt[b + c*x])
Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1130, 1135, 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.
\(\displaystyle \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {1}{8} c \int \frac {1}{x^{7/2} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {1}{8} c \left (-\frac {5 c \int \frac {1}{x^{5/2} \sqrt {c x^2+b x}}dx}{6 b}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}\right )-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \int \frac {1}{x^{3/2} \sqrt {c x^2+b x}}dx}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )}{6 b}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}\right )-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}\) |
\(\Big \downarrow \) 1135 |
\(\displaystyle \frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (-\frac {c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )}{6 b}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}\right )-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}\) |
\(\Big \downarrow \) 1136 |
\(\displaystyle \frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (-\frac {c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )}{6 b}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}\right )-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )}{4 b}-\frac {\sqrt {b x+c x^2}}{2 b x^{5/2}}\right )}{6 b}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}\right )-\frac {\sqrt {b x+c x^2}}{4 x^{9/2}}\) |
-1/4*Sqrt[b*x + c*x^2]/x^(9/2) + (c*(-1/3*Sqrt[b*x + c*x^2]/(b*x^(7/2)) - (5*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*b)))/8
3.1.81.3.1 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 = 1.99 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.65
method | result | size |
risch | \(-\frac {\left (c x +b \right ) \left (15 c^{3} x^{3}-10 b \,c^{2} x^{2}+8 b^{2} c x +48 b^{3}\right )}{192 x^{\frac {7}{2}} b^{3} \sqrt {x \left (c x +b \right )}}+\frac {5 c^{4} \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{64 b^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}\) | \(93\) |
default | \(\frac {\sqrt {x \left (c x +b \right )}\, \left (15 \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-15 c^{3} x^{3} \sqrt {c x +b}\, \sqrt {b}+10 b^{\frac {3}{2}} c^{2} x^{2} \sqrt {c x +b}-8 b^{\frac {5}{2}} c x \sqrt {c x +b}-48 b^{\frac {7}{2}} \sqrt {c x +b}\right )}{192 b^{\frac {7}{2}} x^{\frac {9}{2}} \sqrt {c x +b}}\) | \(108\) |
-1/192*(c*x+b)*(15*c^3*x^3-10*b*c^2*x^2+8*b^2*c*x+48*b^3)/x^(7/2)/b^3/(x*( c*x+b))^(1/2)+5/64*c^4/b^(7/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2 )*x^(1/2)/(x*(c*x+b))^(1/2)
Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\left [\frac {15 \, \sqrt {b} c^{4} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) - 2 \, {\left (15 \, b c^{3} x^{3} - 10 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + 48 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{384 \, b^{4} x^{5}}, -\frac {15 \, \sqrt {-b} c^{4} x^{5} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, b c^{3} x^{3} - 10 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x + 48 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{192 \, b^{4} x^{5}}\right ] \]
[1/384*(15*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*(15*b*c^3*x^3 - 10*b^2*c^2*x^2 + 8*b^3*c*x + 48*b^4)* sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^5), -1/192*(15*sqrt(-b)*c^4*x^5*arctan(s qrt(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (15*b*c^3*x^3 - 10*b^2*c^2*x^2 + 8*b^ 3*c*x + 48*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^5)]
\[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )}}{x^{\frac {11}{2}}}\, dx \]
\[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x}}{x^{\frac {11}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=-\frac {\frac {15 \, c^{5} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} c^{5} - 55 \, {\left (c x + b\right )}^{\frac {5}{2}} b c^{5} + 73 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c^{5} + 15 \, \sqrt {c x + b} b^{3} c^{5}}{b^{3} c^{4} x^{4}}}{192 \, c} \]
-1/192*(15*c^5*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (15*(c*x + b)^(7/2)*c^5 - 55*(c*x + b)^(5/2)*b*c^5 + 73*(c*x + b)^(3/2)*b^2*c^5 + 15* sqrt(c*x + b)*b^3*c^5)/(b^3*c^4*x^4))/c
Timed out. \[ \int \frac {\sqrt {b x+c x^2}}{x^{11/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}}{x^{11/2}} \,d x \]