Integrand size = 18, antiderivative size = 81 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=2 (A b+a B) \sqrt {a+b x}-\frac {a A \sqrt {a+b x}}{x}+\frac {2}{3} B (a+b x)^{3/2}-\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \] Output:
2*(A*b+B*a)*(b*x+a)^(1/2)-a*A*(b*x+a)^(1/2)/x+2/3*B*(b*x+a)^(3/2)-a^(1/2)* (3*A*b+2*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\frac {\sqrt {a+b x} (2 b x (3 A+B x)+a (-3 A+8 B x))}{3 x}-\sqrt {a} (3 A b+2 a B) \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \] Input:
Integrate[((a + b*x)^(3/2)*(A + B*x))/x^2,x]
Output:
(Sqrt[a + b*x]*(2*b*x*(3*A + B*x) + a*(-3*A + 8*B*x)))/(3*x) - Sqrt[a]*(3* A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {87, 60, 60, 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.
\(\displaystyle \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(2 a B+3 A b) \int \frac {(a+b x)^{3/2}}{x}dx}{2 a}-\frac {A (a+b x)^{5/2}}{a x}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(2 a B+3 A b) \left (a \int \frac {\sqrt {a+b x}}{x}dx+\frac {2}{3} (a+b x)^{3/2}\right )}{2 a}-\frac {A (a+b x)^{5/2}}{a x}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(2 a B+3 A b) \left (a \left (a \int \frac {1}{x \sqrt {a+b x}}dx+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )}{2 a}-\frac {A (a+b x)^{5/2}}{a x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(2 a B+3 A b) \left (a \left (\frac {2 a \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{b}+2 \sqrt {a+b x}\right )+\frac {2}{3} (a+b x)^{3/2}\right )}{2 a}-\frac {A (a+b x)^{5/2}}{a x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(2 a B+3 A b) \left (a \left (2 \sqrt {a+b x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )+\frac {2}{3} (a+b x)^{3/2}\right )}{2 a}-\frac {A (a+b x)^{5/2}}{a x}\) |
Input:
Int[((a + b*x)^(3/2)*(A + B*x))/x^2,x]
Output:
-((A*(a + b*x)^(5/2))/(a*x)) + ((3*A*b + 2*a*B)*((2*(a + b*x)^(3/2))/3 + a *(2*Sqrt[a + b*x] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])))/(2*a)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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]
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(-\frac {3 \left (a x \left (A b +\frac {2 B a}{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {2 \left (\left (\frac {4 B x}{3}-\frac {A}{2}\right ) a^{\frac {3}{2}}+b x \sqrt {a}\, \left (\frac {B x}{3}+A \right )\right ) \sqrt {b x +a}}{3}\right )}{\sqrt {a}\, x}\) | \(67\) |
risch | \(-\frac {a A \sqrt {b x +a}}{x}+\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-\sqrt {a}\, \left (3 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\) | \(74\) |
derivativedivides | \(\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-2 a \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (3 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\) | \(77\) |
default | \(\frac {2 B \left (b x +a \right )^{\frac {3}{2}}}{3}+2 A b \sqrt {b x +a}+2 B a \sqrt {b x +a}-2 a \left (\frac {A \sqrt {b x +a}}{2 x}+\frac {\left (3 A b +2 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\) | \(77\) |
Input:
int((b*x+a)^(3/2)*(B*x+A)/x^2,x,method=_RETURNVERBOSE)
Output:
-3*(a*x*(A*b+2/3*B*a)*arctanh((b*x+a)^(1/2)/a^(1/2))-2/3*((4/3*B*x-1/2*A)* a^(3/2)+b*x*a^(1/2)*(1/3*B*x+A))*(b*x+a)^(1/2))/a^(1/2)/x
Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\left [\frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (2 \, B b x^{2} - 3 \, A a + 2 \, {\left (4 \, B a + 3 \, A b\right )} x\right )} \sqrt {b x + a}}{6 \, x}, \frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {-a} x \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) + {\left (2 \, B b x^{2} - 3 \, A a + 2 \, {\left (4 \, B a + 3 \, A b\right )} x\right )} \sqrt {b x + a}}{3 \, x}\right ] \] Input:
integrate((b*x+a)^(3/2)*(B*x+A)/x^2,x, algorithm="fricas")
Output:
[1/6*(3*(2*B*a + 3*A*b)*sqrt(a)*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a )/x) + 2*(2*B*b*x^2 - 3*A*a + 2*(4*B*a + 3*A*b)*x)*sqrt(b*x + a))/x, 1/3*( 3*(2*B*a + 3*A*b)*sqrt(-a)*x*arctan(sqrt(-a)/sqrt(b*x + a)) + (2*B*b*x^2 - 3*A*a + 2*(4*B*a + 3*A*b)*x)*sqrt(b*x + a))/x]
Time = 14.03 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.10 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=- A \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x} + 1}}{\sqrt {x}} + A b \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + B a \left (\begin {cases} \frac {2 a \operatorname {atan}{\left (\frac {\sqrt {a + b x}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 \sqrt {a + b x} & \text {for}\: b \neq 0 \\\sqrt {a} \log {\left (x \right )} & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} \frac {2 \left (a + b x\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x+a)**(3/2)*(B*x+A)/x**2,x)
Output:
-A*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*sqrt(x))) - A*a*sqrt(b)*sqrt(a/(b*x) + 1)/sqrt(x) + A*b*Piecewise((2*a*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2 *sqrt(a + b*x), Ne(b, 0)), (sqrt(a)*log(x), True)) + B*a*Piecewise((2*a*at an(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) + 2*sqrt(a + b*x), Ne(b, 0)), (sqrt(a) *log(x), True)) + B*b*Piecewise((2*(a + b*x)**(3/2)/(3*b), Ne(b, 0)), (sqr t(a)*x, True))
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\frac {1}{6} \, {\left (\frac {3 \, {\left (2 \, B a + 3 \, A b\right )} \sqrt {a} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{b} - \frac {6 \, \sqrt {b x + a} A a}{b x} + \frac {4 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B + 3 \, {\left (B a + A b\right )} \sqrt {b x + a}\right )}}{b}\right )} b \] Input:
integrate((b*x+a)^(3/2)*(B*x+A)/x^2,x, algorithm="maxima")
Output:
1/6*(3*(2*B*a + 3*A*b)*sqrt(a)*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a ) + sqrt(a)))/b - 6*sqrt(b*x + a)*A*a/(b*x) + 4*((b*x + a)^(3/2)*B + 3*(B* a + A*b)*sqrt(b*x + a))/b)*b
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=-\frac {1}{3} \, b {\left (\frac {3 \, \sqrt {b x + a} A a}{b x} - \frac {3 \, {\left (2 \, B a^{2} + 3 \, A a b\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} b} - \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} B b^{2} + 3 \, \sqrt {b x + a} B a b^{2} + 3 \, \sqrt {b x + a} A b^{3}\right )}}{b^{3}}\right )} \] Input:
integrate((b*x+a)^(3/2)*(B*x+A)/x^2,x, algorithm="giac")
Output:
-1/3*b*(3*sqrt(b*x + a)*A*a/(b*x) - 3*(2*B*a^2 + 3*A*a*b)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*b) - 2*((b*x + a)^(3/2)*B*b^2 + 3*sqrt(b*x + a)*B *a*b^2 + 3*sqrt(b*x + a)*A*b^3)/b^3)
Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\left (2\,A\,b+2\,B\,a\right )\,\sqrt {a+b\,x}+\frac {2\,B\,{\left (a+b\,x\right )}^{3/2}}{3}+2\,\mathrm {atan}\left (\frac {2\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {-\frac {a}{4}}\,\sqrt {a+b\,x}}{2\,B\,a^2+3\,A\,b\,a}\right )\,\left (3\,A\,b+2\,B\,a\right )\,\sqrt {-\frac {a}{4}}-\frac {A\,a\,\sqrt {a+b\,x}}{x} \] Input:
int(((A + B*x)*(a + b*x)^(3/2))/x^2,x)
Output:
(2*A*b + 2*B*a)*(a + b*x)^(1/2) + (2*B*(a + b*x)^(3/2))/3 + 2*atan((2*(3*A *b + 2*B*a)*(-a/4)^(1/2)*(a + b*x)^(1/2))/(2*B*a^2 + 3*A*a*b))*(3*A*b + 2* B*a)*(-a/4)^(1/2) - (A*a*(a + b*x)^(1/2))/x
Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x)^{3/2} (A+B x)}{x^2} \, dx=\frac {-6 \sqrt {b x +a}\, a^{2}+28 \sqrt {b x +a}\, a b x +4 \sqrt {b x +a}\, b^{2} x^{2}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b x -15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b x}{6 x} \] Input:
int((b*x+a)^(3/2)*(B*x+A)/x^2,x)
Output:
( - 6*sqrt(a + b*x)*a**2 + 28*sqrt(a + b*x)*a*b*x + 4*sqrt(a + b*x)*b**2*x **2 + 15*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b*x - 15*sqrt(a)*log(sqrt( a + b*x) + sqrt(a))*a*b*x)/(6*x)