Integrand size = 20, antiderivative size = 106 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{8} a (8 A+3 B x) \sqrt {a+b x^2}+\frac {1}{12} (4 A+3 B x) \left (a+b x^2\right )^{3/2}+\frac {3 a^2 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}-a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]
1/12*(3*B*x+4*A)*(b*x^2+a)^(3/2)-a^(3/2)*A*arctanh((b*x^2+a)^(1/2)/a^(1/2) )+3/8*a^2*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(1/2)+1/8*a*(3*B*x+8*A)*( b*x^2+a)^(1/2)
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {1}{24} \left (\sqrt {a+b x^2} \left (32 a A+15 a B x+8 A b x^2+6 b B x^3\right )-\frac {9 a^2 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {b}}\right ) \]
2*a^(3/2)*A*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] + (Sqrt[a + b*x ^2]*(32*a*A + 15*a*B*x + 8*A*b*x^2 + 6*b*B*x^3) - (9*a^2*B*Log[-(Sqrt[b]*x ) + Sqrt[a + b*x^2]])/Sqrt[b])/24
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {535, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} (A+B x)}{x} \, dx\) |
| \(\Big \downarrow \) 535 |
| \(\displaystyle \frac {1}{4} a \int \frac {(4 A+3 B x) \sqrt {b x^2+a}}{x}dx+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 535 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \int \frac {8 A+3 B x}{x \sqrt {b x^2+a}}dx+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (8 A \int \frac {1}{x \sqrt {b x^2+a}}dx+3 B \int \frac {1}{\sqrt {b x^2+a}}dx\right )+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (8 A \int \frac {1}{x \sqrt {b x^2+a}}dx+3 B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (8 A \int \frac {1}{x \sqrt {b x^2+a}}dx+\frac {3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (4 A \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2+\frac {3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (\frac {8 A \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{4} a \left (\frac {1}{2} a \left (\frac {3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}-\frac {8 A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )+\frac {1}{2} \sqrt {a+b x^2} (8 A+3 B x)\right )+\frac {1}{12} \left (a+b x^2\right )^{3/2} (4 A+3 B x)\) |
((4*A + 3*B*x)*(a + b*x^2)^(3/2))/12 + (a*(((8*A + 3*B*x)*Sqrt[a + b*x^2]) /2 + (a*((3*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b] - (8*A*ArcTanh [Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/2))/4
3.1.12.3.1 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_))/((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 = 3.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.03
| method | result | size |
| default | \(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+A \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\) | \(109\) |
B*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b ^(1/2)+(b*x^2+a)^(1/2))))+A*(1/3*(b*x^2+a)^(3/2)+a*((b*x^2+a)^(1/2)-a^(1/2 )*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)))
Time = 0.31 (sec) , antiderivative size = 439, normalized size of antiderivative = 4.14 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\left [\frac {9 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 24 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b}, -\frac {9 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 12 \, A a^{\frac {3}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b}, \frac {48 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + 9 \, B a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{48 \, b}, -\frac {9 \, B a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - 24 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (6 \, B b^{2} x^{3} + 8 \, A b^{2} x^{2} + 15 \, B a b x + 32 \, A a b\right )} \sqrt {b x^{2} + a}}{24 \, b}\right ] \]
[1/48*(9*B*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2 4*A*a^(3/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*B *b^2*x^3 + 8*A*b^2*x^2 + 15*B*a*b*x + 32*A*a*b)*sqrt(b*x^2 + a))/b, -1/24* (9*B*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 12*A*a^(3/2)*b*log( -(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - (6*B*b^2*x^3 + 8*A*b^2*x ^2 + 15*B*a*b*x + 32*A*a*b)*sqrt(b*x^2 + a))/b, 1/48*(48*A*sqrt(-a)*a*b*ar ctan(sqrt(-a)/sqrt(b*x^2 + a)) + 9*B*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x ^2 + a)*sqrt(b)*x - a) + 2*(6*B*b^2*x^3 + 8*A*b^2*x^2 + 15*B*a*b*x + 32*A* a*b)*sqrt(b*x^2 + a))/b, -1/24*(9*B*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b* x^2 + a)) - 24*A*sqrt(-a)*a*b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (6*B*b^2* x^3 + 8*A*b^2*x^2 + 15*B*a*b*x + 32*A*a*b)*sqrt(b*x^2 + a))/b]
Time = 4.26 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.58 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=- A a^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )} + \frac {A a^{2}}{\sqrt {b} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A a \sqrt {b} x}{\sqrt {\frac {a}{b x^{2}} + 1}} + A b \left (\begin {cases} \frac {a \sqrt {a + b x^{2}}}{3 b} + \frac {x^{2} \sqrt {a + b x^{2}}}{3} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + B a \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a x \sqrt {a + b x^{2}}}{8 b} + \frac {x^{3} \sqrt {a + b x^{2}}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]
-A*a**(3/2)*asinh(sqrt(a)/(sqrt(b)*x)) + A*a**2/(sqrt(b)*x*sqrt(a/(b*x**2) + 1)) + A*a*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + A*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)) + B*a*Piecewise((a*Piecewise((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, N e(b, 0)), (sqrt(a)*x, True)) + B*b*Piecewise((-a**2*Piecewise((log(2*sqrt( b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), T rue))/(8*b) + a*x*sqrt(a + b*x**2)/(8*b) + x**3*sqrt(a + b*x**2)/4, Ne(b, 0)), (sqrt(a)*x**3/3, True))
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B x + \frac {3}{8} \, \sqrt {b x^{2} + a} B a x + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - A a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A + \sqrt {b x^{2} + a} A a \]
1/4*(b*x^2 + a)^(3/2)*B*x + 3/8*sqrt(b*x^2 + a)*B*a*x + 3/8*B*a^2*arcsinh( b*x/sqrt(a*b))/sqrt(b) - A*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/3*(b* x^2 + a)^(3/2)*A + sqrt(b*x^2 + a)*A*a
Exception generated. \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Time = 6.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B x) \left (a+b x^2\right )^{3/2}}{x} \, dx=\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{3}-A\,a^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )+A\,a\,\sqrt {b\,x^2+a}+\frac {B\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]