Integrand size = 20, antiderivative size = 111 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=-\frac {3 (a B-A b x) \sqrt {a+b x^2}}{2 x}-\frac {(A-B x) \left (a+b x^2\right )^{3/2}}{2 x^2}+\frac {3}{2} a \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {3}{2} \sqrt {a} A b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
-1/2*(-B*x+A)*(b*x^2+a)^(3/2)/x^2-3/2*A*b*arctanh((b*x^2+a)^(1/2)/a^(1/2)) *a^(1/2)+3/2*a*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)-3/2*(-A*b*x+B* a)*(b*x^2+a)^(1/2)/x
Time = 0.32 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=\frac {1}{2} \left (\frac {\sqrt {a+b x^2} \left (b x^2 (2 A+B x)-a (A+2 B x)\right )}{x^2}+6 \sqrt {a} A b \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-3 a \sqrt {b} B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right ) \]
((Sqrt[a + b*x^2]*(b*x^2*(2*A + B*x) - a*(A + 2*B*x)))/x^2 + 6*Sqrt[a]*A*b *ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] - 3*a*Sqrt[b]*B*Log[-(Sqrt [b]*x) + Sqrt[a + b*x^2]])/2
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {537, 25, 535, 27, 538, 224, 219, 243, 73, 221}
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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} (A+B x)}{x^3} \, dx\) |
| \(\Big \downarrow \) 537 |
| \(\displaystyle -\frac {3}{2} b \int -\frac {(A+2 B x) \sqrt {b x^2+a}}{x}dx-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3}{2} b \int \frac {(A+2 B x) \sqrt {b x^2+a}}{x}dx-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 535 |
| \(\displaystyle \frac {3}{2} b \left (\frac {1}{2} a \int \frac {2 (A+B x)}{x \sqrt {b x^2+a}}dx+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{2} b \left (a \int \frac {A+B x}{x \sqrt {b x^2+a}}dx+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {3}{2} b \left (a \left (A \int \frac {1}{x \sqrt {b x^2+a}}dx+B \int \frac {1}{\sqrt {b x^2+a}}dx\right )+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {3}{2} b \left (a \left (A \int \frac {1}{x \sqrt {b x^2+a}}dx+B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{2} b \left (a \left (A \int \frac {1}{x \sqrt {b x^2+a}}dx+\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {3}{2} b \left (a \left (\frac {1}{2} A \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {3}{2} b \left (a \left (\frac {A \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {3}{2} b \left (a \left (\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-\frac {A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\sqrt {a+b x^2} (A+B x)\right )-\frac {\left (a+b x^2\right )^{3/2} (A+2 B x)}{2 x^2}\) |
-1/2*((A + 2*B*x)*(a + b*x^2)^(3/2))/x^2 + (3*b*((A + B*x)*Sqrt[a + b*x^2] + a*((B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b] - (A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a])))/2
3.1.14.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p + 1) Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Time = 3.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(-\frac {a \sqrt {b \,x^{2}+a}\, \left (2 B x +A \right )}{2 x^{2}}+\frac {3 \sqrt {b}\, B a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2}+\frac {b B x \sqrt {b \,x^{2}+a}}{2}+b A \sqrt {b \,x^{2}+a}-\frac {3 b A \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2}\) | \(102\) |
| default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )\) | \(157\) |
-1/2*a*(b*x^2+a)^(1/2)*(2*B*x+A)/x^2+3/2*b^(1/2)*B*a*ln(x*b^(1/2)+(b*x^2+a )^(1/2))+1/2*b*B*x*(b*x^2+a)^(1/2)+b*A*(b*x^2+a)^(1/2)-3/2*b*A*a^(1/2)*ln( (2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
Time = 0.31 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.83 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 3 \, A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, -\frac {6 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, A \sqrt {a} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, \frac {6 \, A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 3 \, B a \sqrt {b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{4 \, x^{2}}, -\frac {3 \, B a \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, A \sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (B b x^{3} + 2 \, A b x^{2} - 2 \, B a x - A a\right )} \sqrt {b x^{2} + a}}{2 \, x^{2}}\right ] \]
[1/4*(3*B*a*sqrt(b)*x^2*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 3*A*sqrt(a)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2* (B*b*x^3 + 2*A*b*x^2 - 2*B*a*x - A*a)*sqrt(b*x^2 + a))/x^2, -1/4*(6*B*a*sq rt(-b)*x^2*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*A*sqrt(a)*b*x^2*log(-(b* x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(B*b*x^3 + 2*A*b*x^2 - 2*B *a*x - A*a)*sqrt(b*x^2 + a))/x^2, 1/4*(6*A*sqrt(-a)*b*x^2*arctan(sqrt(-a)/ sqrt(b*x^2 + a)) + 3*B*a*sqrt(b)*x^2*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt (b)*x - a) + 2*(B*b*x^3 + 2*A*b*x^2 - 2*B*a*x - A*a)*sqrt(b*x^2 + a))/x^2, -1/2*(3*B*a*sqrt(-b)*x^2*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*A*sqrt(-a )*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (B*b*x^3 + 2*A*b*x^2 - 2*B*a*x - A*a)*sqrt(b*x^2 + a))/x^2]
Time = 2.91 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.02 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=- \frac {3 A \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} + \frac {A a \sqrt {b}}{x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {3}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + B a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + B b \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \]
-3*A*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - A*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(2*x) + A*a*sqrt(b)/(x*sqrt(a/(b*x**2) + 1)) + A*b**(3/2)*x/sqrt(a/(b *x**2) + 1) - B*a**(3/2)/(x*sqrt(1 + b*x**2/a)) - B*sqrt(a)*b*x/sqrt(1 + b *x**2/a) + B*a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) + B*b*Piecewise((a*Piecewi se((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/ sqrt(b*x**2), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True ))
Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=\frac {3}{2} \, \sqrt {b x^{2} + a} B b x + \frac {3}{2} \, B a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {3}{2} \, A \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {3}{2} \, \sqrt {b x^{2} + a} A b + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{2 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{2 \, a x^{2}} \]
3/2*sqrt(b*x^2 + a)*B*b*x + 3/2*B*a*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - 3/2*A *sqrt(a)*b*arcsinh(a/(sqrt(a*b)*abs(x))) + 3/2*sqrt(b*x^2 + a)*A*b + 1/2*( b*x^2 + a)^(3/2)*A*b/a - (b*x^2 + a)^(3/2)*B/x - 1/2*(b*x^2 + a)^(5/2)*A/( a*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (88) = 176\).
Time = 0.31 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.72 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=\frac {3 \, A a b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3}{2} \, B a \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {1}{2} \, {\left (B b x + 2 \, A b\right )} \sqrt {b x^{2} + a} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A a b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a^{2} b - 2 \, B a^{3} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2}} \]
3*A*a*b*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/2*B*a *sqrt(b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a))) + 1/2*(B*b*x + 2*A*b)*sqrt (b*x^2 + a) + ((sqrt(b)*x - sqrt(b*x^2 + a))^3*A*a*b + 2*(sqrt(b)*x - sqrt (b*x^2 + a))^2*B*a^2*sqrt(b) + (sqrt(b)*x - sqrt(b*x^2 + a))*A*a^2*b - 2*B *a^3*sqrt(b))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^2
Time = 7.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^3} \, dx=A\,b\,\sqrt {b\,x^2+a}-\frac {A\,a\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {3\,A\,\sqrt {a}\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2}-\frac {B\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]