Integrand size = 22, antiderivative size = 119 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=-\frac {2 B c^2 \sqrt {b x+c x^2}}{x}-\frac {2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac {2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+2 B c^{5/2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \]
-2/3*B*c*(c*x^2+b*x)^(3/2)/x^3-2/5*B*(c*x^2+b*x)^(5/2)/x^5-2/7*A*(c*x^2+b* x)^(7/2)/b/x^7+2*B*c^(5/2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))-2*B*c^2*(c *x^2+b*x)^(1/2)/x
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=-\frac {2 \sqrt {x (b+c x)} \left (\sqrt {b+c x} \left (15 A (b+c x)^3+7 b B x \left (3 b^2+11 b c x+23 c^2 x^2\right )\right )+105 b B c^{5/2} x^{7/2} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{105 b x^4 \sqrt {b+c x}} \]
(-2*Sqrt[x*(b + c*x)]*(Sqrt[b + c*x]*(15*A*(b + c*x)^3 + 7*b*B*x*(3*b^2 + 11*b*c*x + 23*c^2*x^2)) + 105*b*B*c^(5/2)*x^(7/2)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]))/(105*b*x^4*Sqrt[b + c*x])
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1220, 1130, 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.
| \(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle B \int \frac {\left (c x^2+b x\right )^{5/2}}{x^6}dx-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle B \left (c \int \frac {\left (c x^2+b x\right )^{3/2}}{x^4}dx-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle B \left (c \left (c \int \frac {\sqrt {c x^2+b x}}{x^2}dx-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 1125 |
| \(\displaystyle B \left (c \left (c \left (-\int -\frac {c}{\sqrt {c x^2+b x}}dx-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle B \left (c \left (c \left (\int \frac {c}{\sqrt {c x^2+b x}}dx-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle B \left (c \left (c \left (c \int \frac {1}{\sqrt {c x^2+b x}}dx-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 1091 |
| \(\displaystyle B \left (c \left (c \left (2 c \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle B \left (c \left (c \left (2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )-\frac {2 \sqrt {b x+c x^2}}{x}\right )-\frac {2 \left (b x+c x^2\right )^{3/2}}{3 x^3}\right )-\frac {2 \left (b x+c x^2\right )^{5/2}}{5 x^5}\right )-\frac {2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}\) |
(-2*A*(b*x + c*x^2)^(7/2))/(7*b*x^7) + B*((-2*(b*x + c*x^2)^(5/2))/(5*x^5) + c*((-2*(b*x + c*x^2)^(3/2))/(3*x^3) + c*((-2*Sqrt[b*x + c*x^2])/x + 2*S qrt[c]*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])))
3.2.3.3.1 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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80
| method | result | size |
| pseudoelliptic | \(\frac {2 B b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) x^{4}-\frac {2 \left (\left (\frac {7 B x}{5}+A \right ) b^{3}+3 c \left (\frac {77 B x}{45}+A \right ) x \,b^{2}+3 c^{2} \left (\frac {161 B x}{45}+A \right ) x^{2} b +A \,c^{3} x^{3}\right ) \sqrt {x \left (c x +b \right )}}{7}}{b \,x^{4}}\) | \(95\) |
| risch | \(-\frac {2 \left (c x +b \right ) \left (15 A \,c^{3} x^{3}+161 B b \,c^{2} x^{3}+45 A b \,c^{2} x^{2}+77 B \,b^{2} c \,x^{2}+45 A \,b^{2} c x +21 B \,b^{3} x +15 A \,b^{3}\right )}{105 x^{3} \sqrt {x \left (c x +b \right )}\, b}+B \,c^{\frac {5}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )\) | \(114\) |
| default | \(-\frac {2 A \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{7 b \,x^{7}}+B \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{5 b \,x^{6}}+\frac {2 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{5}}+\frac {4 c \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{4}}+\frac {6 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{b \,x^{3}}-\frac {8 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{3 b \,x^{2}}-\frac {10 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{8 c}-\frac {3 b^{2} \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}\right )}{3 b}\right )}{b}\right )}{b}\right )}{3 b}\right )}{5 b}\right )\) | \(257\) |
2/7*(7*B*b*c^(5/2)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))*x^4-((7/5*B*x+A)*b ^3+3*c*(77/45*B*x+A)*x*b^2+3*c^2*(161/45*B*x+A)*x^2*b+A*c^3*x^3)*(x*(c*x+b ))^(1/2))/b/x^4
Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.00 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=\left [\frac {105 \, B b c^{\frac {5}{2}} x^{4} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (15 \, A b^{3} + {\left (161 \, B b c^{2} + 15 \, A c^{3}\right )} x^{3} + {\left (77 \, B b^{2} c + 45 \, A b c^{2}\right )} x^{2} + 3 \, {\left (7 \, B b^{3} + 15 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}}{105 \, b x^{4}}, -\frac {2 \, {\left (105 \, B b \sqrt {-c} c^{2} x^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (15 \, A b^{3} + {\left (161 \, B b c^{2} + 15 \, A c^{3}\right )} x^{3} + {\left (77 \, B b^{2} c + 45 \, A b c^{2}\right )} x^{2} + 3 \, {\left (7 \, B b^{3} + 15 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x}\right )}}{105 \, b x^{4}}\right ] \]
[1/105*(105*B*b*c^(5/2)*x^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(15*A*b^3 + (161*B*b*c^2 + 15*A*c^3)*x^3 + (77*B*b^2*c + 45*A*b*c^2)*x^ 2 + 3*(7*B*b^3 + 15*A*b^2*c)*x)*sqrt(c*x^2 + b*x))/(b*x^4), -2/105*(105*B* b*sqrt(-c)*c^2*x^4*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (15*A*b^3 + (161*B*b*c^2 + 15*A*c^3)*x^3 + (77*B*b^2*c + 45*A*b*c^2)*x^2 + 3*(7*B*b^3 + 15*A*b^2*c)*x)*sqrt(c*x^2 + b*x))/(b*x^4)]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{7}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (99) = 198\).
Time = 0.20 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.17 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=B c^{\frac {5}{2}} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - \frac {38 \, \sqrt {c x^{2} + b x} B c^{2}}{15 \, x} - \frac {2 \, \sqrt {c x^{2} + b x} A c^{3}}{7 \, b x} - \frac {7 \, \sqrt {c x^{2} + b x} B b c}{30 \, x^{2}} + \frac {\sqrt {c x^{2} + b x} A c^{2}}{7 \, x^{2}} + \frac {3 \, \sqrt {c x^{2} + b x} B b^{2}}{10 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B c}{3 \, x^{3}} - \frac {3 \, \sqrt {c x^{2} + b x} A b c}{28 \, x^{3}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B b}{2 \, x^{4}} - \frac {15 \, \sqrt {c x^{2} + b x} A b^{2}}{28 \, x^{4}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} B}{5 \, x^{5}} + \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} A b}{4 \, x^{5}} - \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} A}{x^{6}} \]
B*c^(5/2)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 38/15*sqrt(c*x^2 + b*x)*B*c^2/x - 2/7*sqrt(c*x^2 + b*x)*A*c^3/(b*x) - 7/30*sqrt(c*x^2 + b*x )*B*b*c/x^2 + 1/7*sqrt(c*x^2 + b*x)*A*c^2/x^2 + 3/10*sqrt(c*x^2 + b*x)*B*b ^2/x^3 - 1/3*(c*x^2 + b*x)^(3/2)*B*c/x^3 - 3/28*sqrt(c*x^2 + b*x)*A*b*c/x^ 3 - 1/2*(c*x^2 + b*x)^(3/2)*B*b/x^4 - 15/28*sqrt(c*x^2 + b*x)*A*b^2/x^4 - 1/5*(c*x^2 + b*x)^(5/2)*B/x^5 + 5/4*(c*x^2 + b*x)^(3/2)*A*b/x^5 - (c*x^2 + b*x)^(5/2)*A/x^6
Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (99) = 198\).
Time = 0.29 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.18 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=-B c^{\frac {5}{2}} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right ) + \frac {2 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} B b c^{2} + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{6} A c^{3} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} B b^{2} c^{\frac {3}{2}} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5} A b c^{\frac {5}{2}} + 245 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B b^{3} c + 525 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} A b^{2} c^{2} + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b^{4} \sqrt {c} + 525 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A b^{3} c^{\frac {3}{2}} + 21 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{5} + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b^{4} c + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{5} \sqrt {c} + 15 \, A b^{6}\right )}}{105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{7}} \]
-B*c^(5/2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b)) + 2/105 *(315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b*c^2 + 105*(sqrt(c)*x - sqrt(c* x^2 + b*x))^6*A*c^3 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c^(3/2) + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b*c^(5/2) + 245*(sqrt(c)*x - sqr t(c*x^2 + b*x))^4*B*b^3*c + 525*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^2*c^ 2 + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*sqrt(c) + 525*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^3*c^(3/2) + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2 *B*b^5 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*A*b^4*c + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^5*sqrt(c) + 15*A*b^6)/(sqrt(c)*x - sqrt(c*x^2 + b*x ))^7
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^7} \,d x \]