Integrand size = 17, antiderivative size = 97 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {a^2 \sqrt {a x+b x^2}}{8 b}+\frac {7}{12} a x \sqrt {a x+b x^2}+\frac {1}{3} b x^2 \sqrt {a x+b x^2}-\frac {a^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 b^{3/2}} \] Output:
1/8*a^2*(b*x^2+a*x)^(1/2)/b+7/12*a*x*(b*x^2+a*x)^(1/2)+1/3*b*x^2*(b*x^2+a* x)^(1/2)-1/8*a^3*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/b^(3/2)
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (3 a^2+14 a b x+8 b^2 x^2\right )+\frac {6 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{24 b^{3/2}} \] Input:
Integrate[(a*x + b*x^2)^(3/2)/x,x]
Output:
(Sqrt[x*(a + b*x)]*(Sqrt[b]*(3*a^2 + 14*a*b*x + 8*b^2*x^2) + (6*a^3*ArcTan h[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])])/(Sqrt[x]*Sqrt[a + b*x])))/ (24*b^(3/2))
Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1131, 1087, 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.
\(\displaystyle \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {1}{2} a \int \sqrt {b x^2+a x}dx+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{2} a \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{\sqrt {b x^2+a x}}dx}{8 b}\right )+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} a \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{4 b}\right )+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} a \left (\frac {(a+2 b x) \sqrt {a x+b x^2}}{4 b}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{4 b^{3/2}}\right )+\frac {1}{3} \left (a x+b x^2\right )^{3/2}\) |
Input:
Int[(a*x + b*x^2)^(3/2)/x,x]
Output:
(a*x + b*x^2)^(3/2)/3 + (a*(((a + 2*b*x)*Sqrt[a*x + b*x^2])/(4*b) - (a^2*A rcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/(4*b^(3/2))))/2
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_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.75
method | result | size |
default | \(\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2}\) | \(73\) |
risch | \(\frac {\left (8 b^{2} x^{2}+14 a b x +3 a^{2}\right ) x \left (b x +a \right )}{24 b \sqrt {x \left (b x +a \right )}}-\frac {a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 b^{\frac {3}{2}}}\) | \(73\) |
pseudoelliptic | \(\frac {8 b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, x^{2}+14 a \,b^{\frac {3}{2}} x \sqrt {x \left (b x +a \right )}+3 a^{2} \sqrt {b}\, \sqrt {x \left (b x +a \right )}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) a^{3}}{24 b^{\frac {3}{2}}}\) | \(79\) |
Input:
int((b*x^2+a*x)^(3/2)/x,x,method=_RETURNVERBOSE)
Output:
1/3*(b*x^2+a*x)^(3/2)+1/2*a*(1/4*(2*b*x+a)/b*(b*x^2+a*x)^(1/2)-1/8*a^2/b^( 3/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\left [\frac {3 \, a^{3} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (8 \, b^{3} x^{2} + 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{48 \, b^{2}}, \frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + {\left (8 \, b^{3} x^{2} + 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{24 \, b^{2}}\right ] \] Input:
integrate((b*x^2+a*x)^(3/2)/x,x, algorithm="fricas")
Output:
[1/48*(3*a^3*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(8*b ^3*x^2 + 14*a*b^2*x + 3*a^2*b)*sqrt(b*x^2 + a*x))/b^2, 1/24*(3*a^3*sqrt(-b )*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) + (8*b^3*x^2 + 14*a*b^2*x + 3*a^2*b)*sqrt(b*x^2 + a*x))/b^2]
Time = 0.95 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.31 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=a \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \left (\frac {a}{4 b} + \frac {x}{2}\right ) \sqrt {a x + b x^{2}} & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {3}{2}}}{3 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {a^{3} \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a x + b x^{2}} \left (- \frac {a^{2}}{8 b^{2}} + \frac {a x}{12 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (a x\right )^{\frac {5}{2}}}{5 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x**2+a*x)**(3/2)/x,x)
Output:
a*Piecewise((-a**2*Piecewise((log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x )/sqrt(b), Ne(a**2/b, 0)), ((a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(8*b) + (a/(4*b) + x/2)*sqrt(a*x + b*x**2), Ne(b, 0)), ( 2*(a*x)**(3/2)/(3*a), Ne(a, 0)), (0, True)) + b*Piecewise((a**3*Piecewise( (log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), (( a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(16*b**2) + sqrt(a*x + b*x**2)*(-a**2/(8*b**2) + a*x/(12*b) + x**2/3), Ne(b, 0)), (2* (a*x)**(5/2)/(5*a**2), Ne(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {1}{4} \, \sqrt {b x^{2} + a x} a x - \frac {a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {3}{2}}} + \frac {1}{3} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} + \frac {\sqrt {b x^{2} + a x} a^{2}}{8 \, b} \] Input:
integrate((b*x^2+a*x)^(3/2)/x,x, algorithm="maxima")
Output:
1/4*sqrt(b*x^2 + a*x)*a*x - 1/16*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*s qrt(b))/b^(3/2) + 1/3*(b*x^2 + a*x)^(3/2) + 1/8*sqrt(b*x^2 + a*x)*a^2/b
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {a^{3} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, b^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, b x + 7 \, a\right )} x + \frac {3 \, a^{2}}{b}\right )} \] Input:
integrate((b*x^2+a*x)^(3/2)/x,x, algorithm="giac")
Output:
1/16*a^3*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(3/2) + 1/24*sqrt(b*x^2 + a*x)*(2*(4*b*x + 7*a)*x + 3*a^2/b)
Timed out. \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{3/2}}{x} \,d x \] Input:
int((a*x + b*x^2)^(3/2)/x,x)
Output:
int((a*x + b*x^2)^(3/2)/x, x)
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a x+b x^2\right )^{3/2}}{x} \, dx=\frac {3 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b +14 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, b^{3} x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3}}{24 b^{2}} \] Input:
int((b*x^2+a*x)^(3/2)/x,x)
Output:
(3*sqrt(x)*sqrt(a + b*x)*a**2*b + 14*sqrt(x)*sqrt(a + b*x)*a*b**2*x + 8*sq rt(x)*sqrt(a + b*x)*b**3*x**2 - 3*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqr t(b))/sqrt(a))*a**3)/(24*b**2)