Integrand size = 20, antiderivative size = 72 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=2 a \sqrt {c+d x}-\frac {2 b c (c+d x)^{3/2}}{3 d^2}+\frac {2 b (c+d x)^{5/2}}{5 d^2}-2 a \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Output:
2*a*(d*x+c)^(1/2)-2/3*b*c*(d*x+c)^(3/2)/d^2+2/5*b*(d*x+c)^(5/2)/d^2-2*a*c^ (1/2)*arctanh((d*x+c)^(1/2)/c^(1/2))
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=\frac {2 \sqrt {c+d x} \left (15 a d^2+b \left (-2 c^2+c d x+3 d^2 x^2\right )\right )}{15 d^2}-2 a \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Input:
Integrate[(Sqrt[c + d*x]*(a + b*x^2))/x,x]
Output:
(2*Sqrt[c + d*x]*(15*a*d^2 + b*(-2*c^2 + c*d*x + 3*d^2*x^2)))/(15*d^2) - 2 *a*Sqrt[c]*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]
Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {517, 25, 1584, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \sqrt {c+d x}}{x} \, dx\) |
\(\Big \downarrow \) 517 |
\(\displaystyle \frac {2 \int \frac {(c+d x) \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )}{d x}d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 \int -\frac {(c+d x) \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )}{d x}d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle -\frac {2 \int \left (-a d^2-\frac {a c d}{x}-b (c+d x)^2+b c (c+d x)\right )d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-a \sqrt {c} d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+a d^2 \sqrt {c+d x}+\frac {1}{5} b (c+d x)^{5/2}-\frac {1}{3} b c (c+d x)^{3/2}\right )}{d^2}\) |
Input:
Int[(Sqrt[c + d*x]*(a + b*x^2))/x,x]
Output:
(2*(a*d^2*Sqrt[c + d*x] - (b*c*(c + d*x)^(3/2))/3 + (b*(c + d*x)^(5/2))/5 - a*Sqrt[c]*d^2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]))/d^2
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 b c \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} \sqrt {d x +c}-2 a \sqrt {c}\, d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{2}}\) | \(61\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {5}{2}}}{5}-\frac {2 b c \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a \,d^{2} \sqrt {d x +c}-2 a \sqrt {c}\, d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{2}}\) | \(61\) |
pseudoelliptic | \(\frac {-2 a \sqrt {c}\, d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+2 \sqrt {d x +c}\, \left (\left (\frac {b \,x^{2}}{5}+a \right ) d^{2}+\frac {b c d x}{15}-\frac {2 b \,c^{2}}{15}\right )}{d^{2}}\) | \(61\) |
Input:
int((d*x+c)^(1/2)*(b*x^2+a)/x,x,method=_RETURNVERBOSE)
Output:
2/d^2*(1/5*b*(d*x+c)^(5/2)-1/3*b*c*(d*x+c)^(3/2)+a*d^2*(d*x+c)^(1/2)-a*c^( 1/2)*d^2*arctanh((d*x+c)^(1/2)/c^(1/2)))
Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.97 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=\left [\frac {15 \, a \sqrt {c} d^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (3 \, b d^{2} x^{2} + b c d x - 2 \, b c^{2} + 15 \, a d^{2}\right )} \sqrt {d x + c}}{15 \, d^{2}}, \frac {2 \, {\left (15 \, a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (3 \, b d^{2} x^{2} + b c d x - 2 \, b c^{2} + 15 \, a d^{2}\right )} \sqrt {d x + c}\right )}}{15 \, d^{2}}\right ] \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/x,x, algorithm="fricas")
Output:
[1/15*(15*a*sqrt(c)*d^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*( 3*b*d^2*x^2 + b*c*d*x - 2*b*c^2 + 15*a*d^2)*sqrt(d*x + c))/d^2, 2/15*(15*a *sqrt(-c)*d^2*arctan(sqrt(-c)/sqrt(d*x + c)) + (3*b*d^2*x^2 + b*c*d*x - 2* b*c^2 + 15*a*d^2)*sqrt(d*x + c))/d^2]
Time = 2.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=\begin {cases} \frac {2 a c \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a \sqrt {c + d x} - \frac {2 b c \left (c + d x\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {2 b \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a \log {\left (b x^{2} \right )}}{2} + \frac {b x^{2}}{2}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**(1/2)*(b*x**2+a)/x,x)
Output:
Piecewise((2*a*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + 2*a*sqrt(c + d*x) - 2*b*c*(c + d*x)**(3/2)/(3*d**2) + 2*b*(c + d*x)**(5/2)/(5*d**2), Ne(d, 0)), (sqrt(c)*(a*log(b*x**2)/2 + b*x**2/2), True))
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=a \sqrt {c} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b c + 15 \, \sqrt {d x + c} a d^{2}\right )}}{15 \, d^{2}} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/x,x, algorithm="maxima")
Output:
a*sqrt(c)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c))) + 2/15* (3*(d*x + c)^(5/2)*b - 5*(d*x + c)^(3/2)*b*c + 15*sqrt(d*x + c)*a*d^2)/d^2
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=\frac {2 \, a c \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b d^{8} - 5 \, {\left (d x + c\right )}^{\frac {3}{2}} b c d^{8} + 15 \, \sqrt {d x + c} a d^{10}\right )}}{15 \, d^{10}} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/x,x, algorithm="giac")
Output:
2*a*c*arctan(sqrt(d*x + c)/sqrt(-c))/sqrt(-c) + 2/15*(3*(d*x + c)^(5/2)*b* d^8 - 5*(d*x + c)^(3/2)*b*c*d^8 + 15*sqrt(d*x + c)*a*d^10)/d^10
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=\left (\frac {2\,b\,c^2+2\,a\,d^2}{d^2}-\frac {2\,b\,c^2}{d^2}\right )\,\sqrt {c+d\,x}+\frac {2\,b\,{\left (c+d\,x\right )}^{5/2}}{5\,d^2}-\frac {2\,b\,c\,{\left (c+d\,x\right )}^{3/2}}{3\,d^2}+a\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,2{}\mathrm {i} \] Input:
int(((a + b*x^2)*(c + d*x)^(1/2))/x,x)
Output:
((2*a*d^2 + 2*b*c^2)/d^2 - (2*b*c^2)/d^2)*(c + d*x)^(1/2) + a*c^(1/2)*atan (((c + d*x)^(1/2)*1i)/c^(1/2))*2i + (2*b*(c + d*x)^(5/2))/(5*d^2) - (2*b*c *(c + d*x)^(3/2))/(3*d^2)
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{x} \, dx=\frac {30 \sqrt {d x +c}\, a \,d^{2}-4 \sqrt {d x +c}\, b \,c^{2}+2 \sqrt {d x +c}\, b c d x +6 \sqrt {d x +c}\, b \,d^{2} x^{2}+15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{2}-15 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{2}}{15 d^{2}} \] Input:
int((d*x+c)^(1/2)*(b*x^2+a)/x,x)
Output:
(30*sqrt(c + d*x)*a*d**2 - 4*sqrt(c + d*x)*b*c**2 + 2*sqrt(c + d*x)*b*c*d* x + 6*sqrt(c + d*x)*b*d**2*x**2 + 15*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))* a*d**2 - 15*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**2)/(15*d**2)