Integrand size = 20, antiderivative size = 108 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{2} (2 a B+3 A b x) \sqrt {a+b x^2}-\frac {(3 A-B x) \left (a+b x^2\right )^{3/2}}{3 x}+\frac {3}{2} a A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-a^{3/2} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
-1/3*(-B*x+3*A)*(b*x^2+a)^(3/2)/x-a^(3/2)*B*arctanh((b*x^2+a)^(1/2)/a^(1/2 ))+3/2*a*A*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)+1/2*(3*A*b*x+2*B*a)* (b*x^2+a)^(1/2)
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (b x^2 (3 A+2 B x)+a (-6 A+8 B x)\right )}{6 x}+2 a^{3/2} B \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {3}{2} a A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]
(Sqrt[a + b*x^2]*(b*x^2*(3*A + 2*B*x) + a*(-6*A + 8*B*x)))/(6*x) + 2*a^(3/ 2)*B*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] - (3*a*A*Sqrt[b]*Log[- (Sqrt[b]*x) + Sqrt[a + b*x^2]])/2
Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {536, 535, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} (A+B x)}{x^2} \, dx\) |
| \(\Big \downarrow \) 536 |
| \(\displaystyle \int \frac {(a B+3 A b x) \sqrt {b x^2+a}}{x}dx-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}\) |
| \(\Big \downarrow \) 535 |
| \(\displaystyle \frac {1}{2} a \int \frac {2 a B+3 A b x}{x \sqrt {b x^2+a}}dx-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {1}{2} a \left (3 A b \int \frac {1}{\sqrt {b x^2+a}}dx+2 a B \int \frac {1}{x \sqrt {b x^2+a}}dx\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{2} a \left (3 A b \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+2 a B \int \frac {1}{x \sqrt {b x^2+a}}dx\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} a \left (2 a B \int \frac {1}{x \sqrt {b x^2+a}}dx+3 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{2} a \left (a B \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+3 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} a \left (\frac {2 a B \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+3 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} a \left (3 A \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-2 \sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )-\frac {\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac {1}{2} \sqrt {a+b x^2} (2 a B+3 A b x)\) |
((2*a*B + 3*A*b*x)*Sqrt[a + b*x^2])/2 - ((3*A - B*x)*(a + b*x^2)^(3/2))/(3 *x) + (a*(3*A*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]] - 2*Sqrt[a]*B*A rcTanh[Sqrt[a + b*x^2]/Sqrt[a]]))/2
3.1.13.3.1 Defintions of rubi rules used
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[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer Q[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 = 3.41 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05
| method | result | size |
| risch | \(-\frac {a A \sqrt {b \,x^{2}+a}}{x}+\frac {B b \,x^{2} \sqrt {b \,x^{2}+a}}{3}+\frac {4 a B \sqrt {b \,x^{2}+a}}{3}+\frac {b A x \sqrt {b \,x^{2}+a}}{2}+\frac {3 a \sqrt {b}\, A \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2}-a^{\frac {3}{2}} B \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\) | \(113\) |
| default | \(B \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )\) | \(133\) |
-a*A*(b*x^2+a)^(1/2)/x+1/3*B*b*x^2*(b*x^2+a)^(1/2)+4/3*a*B*(b*x^2+a)^(1/2) +1/2*b*A*x*(b*x^2+a)^(1/2)+3/2*a*b^(1/2)*A*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-a ^(3/2)*B*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)
Time = 0.30 (sec) , antiderivative size = 411, normalized size of antiderivative = 3.81 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\left [\frac {9 \, A a \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 6 \, B a^{\frac {3}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x}, -\frac {9 \, A a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 3 \, B a^{\frac {3}{2}} x \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x}, \frac {12 \, B \sqrt {-a} a x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 9 \, A a \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x}, -\frac {9 \, A a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 6 \, B \sqrt {-a} a x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x}\right ] \]
[1/12*(9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 6 *B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B* b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqrt( -b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*B*a^(3/2)*x*log(-(b*x^2 - 2*s qrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6* A*a)*sqrt(b*x^2 + a))/x, 1/12*(12*B*sqrt(-a)*a*x*arctan(sqrt(-a)/sqrt(b*x^ 2 + a)) + 9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9* A*a*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 6*B*sqrt(-a)*a*x*arcta n(sqrt(-a)/sqrt(b*x^2 + a)) - (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sq rt(b*x^2 + a))/x]
Time = 2.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=- \frac {A a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + A b \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) - B a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {B a^{2}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B a \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + B b \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
-A*a**(3/2)/(x*sqrt(1 + b*x**2/a)) - A*sqrt(a)*b*x/sqrt(1 + b*x**2/a) + A* a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) + A*b*Piecewise((a*Piecewise((log(2*sqr t(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)) - B*a**(3/ 2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**2/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + B*a*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + B*b*Piecewise((a*sqrt(a + b*x**2)/(3* b) + x**2*sqrt(a + b*x**2)/3, Ne(b, 0)), (sqrt(a)*x**2/2, True))
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {3}{2} \, \sqrt {b x^{2} + a} A b x + \frac {3}{2} \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - B a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B + \sqrt {b x^{2} + a} B a - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{x} \]
3/2*sqrt(b*x^2 + a)*A*b*x + 3/2*A*a*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - B*a^( 3/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b*x^2 + a)^(3/2)*B + sqrt(b*x^2 + a)*B*a - (b*x^2 + a)^(3/2)*A/x
Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.15 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {2 \, B a^{2} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {3}{2} \, A a \sqrt {b} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right ) + \frac {2 \, A a^{2} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{6} \, \sqrt {b x^{2} + a} {\left (8 \, B a + {\left (2 \, B b x + 3 \, A b\right )} x\right )} \]
2*B*a^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/2*A*a *sqrt(b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a))) + 2*A*a^2*sqrt(b)/((sqrt(b )*x - sqrt(b*x^2 + a))^2 - a) + 1/6*sqrt(b*x^2 + a)*(8*B*a + (2*B*b*x + 3* A*b)*x)
Time = 6.73 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x^2} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{3}-B\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+B\,a\,\sqrt {b\,x^2+a}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]