Integrand size = 22, antiderivative size = 113 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\frac {2 \left (b c^2+a d^2\right ) \sqrt {e x}}{d^3}-\frac {2 b c (e x)^{3/2}}{3 d^2 e}+\frac {2 b (e x)^{5/2}}{5 d e^2}-\frac {2 \sqrt {c} \left (b c^2+a d^2\right ) \sqrt {e} \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {c} \sqrt {e}}\right )}{d^{7/2}} \] Output:
2*(a*d^2+b*c^2)*(e*x)^(1/2)/d^3-2/3*b*c*(e*x)^(3/2)/d^2/e+2/5*b*(e*x)^(5/2 )/d/e^2-2*c^(1/2)*(a*d^2+b*c^2)*e^(1/2)*arctan(d^(1/2)*(e*x)^(1/2)/c^(1/2) /e^(1/2))/d^(7/2)
Time = 0.13 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\frac {2 \sqrt {e x} \left (\sqrt {d} \left (15 a d^2+b \left (15 c^2-5 c d x+3 d^2 x^2\right )\right )-\frac {15 \sqrt {c} \left (b c^2+a d^2\right ) \arctan \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )}{\sqrt {x}}\right )}{15 d^{7/2}} \] Input:
Integrate[(Sqrt[e*x]*(a + b*x^2))/(c + d*x),x]
Output:
(2*Sqrt[e*x]*(Sqrt[d]*(15*a*d^2 + b*(15*c^2 - 5*c*d*x + 3*d^2*x^2)) - (15* Sqrt[c]*(b*c^2 + a*d^2)*ArcTan[(Sqrt[d]*Sqrt[x])/Sqrt[c]])/Sqrt[x]))/(15*d ^(7/2))
Time = 0.45 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {518, 1585, 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.
\(\displaystyle \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx\) |
\(\Big \downarrow \) 518 |
\(\displaystyle \frac {2 \int \frac {e x \left (b x^2 e^2+a e^2\right )}{c e+d x e}d\sqrt {e x}}{e^2}\) |
\(\Big \downarrow \) 1585 |
\(\displaystyle \frac {2 \int \left (-\frac {\left (b c^3+a d^2 c\right ) e^3}{d^3 (c e+d x e)}+\frac {b x^2 e^2}{d}+\frac {\left (b c^2+a d^2\right ) e^2}{d^3}-\frac {b c x e^2}{d^2}\right )d\sqrt {e x}}{e^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {\sqrt {c} e^{5/2} \left (a d^2+b c^2\right ) \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {c} \sqrt {e}}\right )}{d^{7/2}}+\frac {e^2 \sqrt {e x} \left (a d^2+b c^2\right )}{d^3}-\frac {b c e (e x)^{3/2}}{3 d^2}+\frac {b (e x)^{5/2}}{5 d}\right )}{e^2}\) |
Input:
Int[(Sqrt[e*x]*(a + b*x^2))/(c + d*x),x]
Output:
(2*(((b*c^2 + a*d^2)*e^2*Sqrt[e*x])/d^3 - (b*c*e*(e*x)^(3/2))/(3*d^2) + (b *(e*x)^(5/2))/(5*d) - (Sqrt[c]*(b*c^2 + a*d^2)*e^(5/2)*ArcTan[(Sqrt[d]*Sqr t[e*x])/(Sqrt[c]*Sqrt[e])])/d^(7/2)))/e^2
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^(2*m + 1)*(e*c + d*x^2)^ n*(a*e^2 + b*x^4)^p, x], x, Sqrt[e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {2 \left (c e \left (a \,d^{2}+b \,c^{2}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )-\left (\left (\frac {b \,x^{2}}{5}+a \right ) d^{2}-\frac {b c d x}{3}+b \,c^{2}\right ) \sqrt {e x}\, \sqrt {d e c}\right )}{\sqrt {d e c}\, d^{3}}\) | \(78\) |
risch | \(\frac {2 \left (3 b \,x^{2} d^{2}-5 b c d x +15 a \,d^{2}+15 b \,c^{2}\right ) x e}{15 d^{3} \sqrt {e x}}-\frac {2 c \left (a \,d^{2}+b \,c^{2}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right ) e}{d^{3} \sqrt {d e c}}\) | \(80\) |
derivativedivides | \(\frac {\frac {2 \left (\frac {b \,d^{2} \left (e x \right )^{\frac {5}{2}}}{5}-\frac {b c d e \left (e x \right )^{\frac {3}{2}}}{3}+a \,d^{2} e^{2} \sqrt {e x}+b \,c^{2} e^{2} \sqrt {e x}\right )}{d^{3}}-\frac {2 c \,e^{3} \left (a \,d^{2}+b \,c^{2}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{d^{3} \sqrt {d e c}}}{e^{2}}\) | \(100\) |
default | \(\frac {\frac {2 \left (\frac {b \,d^{2} \left (e x \right )^{\frac {5}{2}}}{5}-\frac {b c d e \left (e x \right )^{\frac {3}{2}}}{3}+a \,d^{2} e^{2} \sqrt {e x}+b \,c^{2} e^{2} \sqrt {e x}\right )}{d^{3}}-\frac {2 c \,e^{3} \left (a \,d^{2}+b \,c^{2}\right ) \arctan \left (\frac {d \sqrt {e x}}{\sqrt {d e c}}\right )}{d^{3} \sqrt {d e c}}}{e^{2}}\) | \(100\) |
Input:
int((e*x)^(1/2)*(b*x^2+a)/(d*x+c),x,method=_RETURNVERBOSE)
Output:
-2/(d*e*c)^(1/2)*(c*e*(a*d^2+b*c^2)*arctan(d*(e*x)^(1/2)/(d*e*c)^(1/2))-(( 1/5*b*x^2+a)*d^2-1/3*b*c*d*x+b*c^2)*(e*x)^(1/2)*(d*e*c)^(1/2))/d^3
Time = 0.25 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.62 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\left [\frac {15 \, {\left (b c^{2} + a d^{2}\right )} \sqrt {-\frac {c e}{d}} \log \left (\frac {d e x - 2 \, \sqrt {e x} d \sqrt {-\frac {c e}{d}} - c e}{d x + c}\right ) + 2 \, {\left (3 \, b d^{2} x^{2} - 5 \, b c d x + 15 \, b c^{2} + 15 \, a d^{2}\right )} \sqrt {e x}}{15 \, d^{3}}, -\frac {2 \, {\left (15 \, {\left (b c^{2} + a d^{2}\right )} \sqrt {\frac {c e}{d}} \arctan \left (\frac {\sqrt {e x} d \sqrt {\frac {c e}{d}}}{c e}\right ) - {\left (3 \, b d^{2} x^{2} - 5 \, b c d x + 15 \, b c^{2} + 15 \, a d^{2}\right )} \sqrt {e x}\right )}}{15 \, d^{3}}\right ] \] Input:
integrate((e*x)^(1/2)*(b*x^2+a)/(d*x+c),x, algorithm="fricas")
Output:
[1/15*(15*(b*c^2 + a*d^2)*sqrt(-c*e/d)*log((d*e*x - 2*sqrt(e*x)*d*sqrt(-c* e/d) - c*e)/(d*x + c)) + 2*(3*b*d^2*x^2 - 5*b*c*d*x + 15*b*c^2 + 15*a*d^2) *sqrt(e*x))/d^3, -2/15*(15*(b*c^2 + a*d^2)*sqrt(c*e/d)*arctan(sqrt(e*x)*d* sqrt(c*e/d)/(c*e)) - (3*b*d^2*x^2 - 5*b*c*d*x + 15*b*c^2 + 15*a*d^2)*sqrt( e*x))/d^3]
Time = 1.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\begin {cases} \frac {2 \left (- \frac {b c \left (e x\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {b \left (e x\right )^{\frac {5}{2}}}{5 d e} - \frac {c e^{2} \left (a d^{2} + b c^{2}\right ) \operatorname {atan}{\left (\frac {\sqrt {e x}}{\sqrt {\frac {c e}{d}}} \right )}}{d^{4} \sqrt {\frac {c e}{d}}} + \frac {\sqrt {e x} \left (a d^{2} e + b c^{2} e\right )}{d^{3}}\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate((e*x)**(1/2)*(b*x**2+a)/(d*x+c),x)
Output:
Piecewise((2*(-b*c*(e*x)**(3/2)/(3*d**2) + b*(e*x)**(5/2)/(5*d*e) - c*e**2 *(a*d**2 + b*c**2)*atan(sqrt(e*x)/sqrt(c*e/d))/(d**4*sqrt(c*e/d)) + sqrt(e *x)*(a*d**2*e + b*c**2*e)/d**3)/e, Ne(e, 0)), (0, True))
Exception generated. \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x)^(1/2)*(b*x^2+a)/(d*x+c),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=-\frac {2 \, {\left (b c^{3} e + a c d^{2} e\right )} \arctan \left (\frac {\sqrt {e x} d}{\sqrt {c d e}}\right )}{\sqrt {c d e} d^{3}} + \frac {2 \, {\left (3 \, \sqrt {e x} b d^{4} e^{10} x^{2} - 5 \, \sqrt {e x} b c d^{3} e^{10} x + 15 \, \sqrt {e x} b c^{2} d^{2} e^{10} + 15 \, \sqrt {e x} a d^{4} e^{10}\right )}}{15 \, d^{5} e^{10}} \] Input:
integrate((e*x)^(1/2)*(b*x^2+a)/(d*x+c),x, algorithm="giac")
Output:
-2*(b*c^3*e + a*c*d^2*e)*arctan(sqrt(e*x)*d/sqrt(c*d*e))/(sqrt(c*d*e)*d^3) + 2/15*(3*sqrt(e*x)*b*d^4*e^10*x^2 - 5*sqrt(e*x)*b*c*d^3*e^10*x + 15*sqrt (e*x)*b*c^2*d^2*e^10 + 15*sqrt(e*x)*a*d^4*e^10)/(d^5*e^10)
Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\left (\frac {2\,a}{d}+\frac {2\,b\,c^2}{d^3}\right )\,\sqrt {e\,x}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {c\,e}\,\sqrt {e\,x}\,\left (b\,c^2+a\,d^2\right )}{b\,e\,c^3+a\,e\,c\,d^2}\right )\,\sqrt {c\,e}\,\left (b\,c^2+a\,d^2\right )}{d^{7/2}}+\frac {2\,b\,{\left (e\,x\right )}^{5/2}}{5\,d\,e^2}-\frac {2\,b\,c\,{\left (e\,x\right )}^{3/2}}{3\,d^2\,e} \] Input:
int(((e*x)^(1/2)*(a + b*x^2))/(c + d*x),x)
Output:
((2*a)/d + (2*b*c^2)/d^3)*(e*x)^(1/2) - (2*atan((d^(1/2)*(c*e)^(1/2)*(e*x) ^(1/2)*(a*d^2 + b*c^2))/(b*c^3*e + a*c*d^2*e))*(c*e)^(1/2)*(a*d^2 + b*c^2) )/d^(7/2) + (2*b*(e*x)^(5/2))/(5*d*e^2) - (2*b*c*(e*x)^(3/2))/(3*d^2*e)
Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{c+d x} \, dx=\frac {2 \sqrt {e}\, \left (-15 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) a \,d^{2}-15 \sqrt {d}\, \sqrt {c}\, \mathit {atan} \left (\frac {\sqrt {x}\, d}{\sqrt {d}\, \sqrt {c}}\right ) b \,c^{2}+15 \sqrt {x}\, a \,d^{3}+15 \sqrt {x}\, b \,c^{2} d -5 \sqrt {x}\, b c \,d^{2} x +3 \sqrt {x}\, b \,d^{3} x^{2}\right )}{15 d^{4}} \] Input:
int((e*x)^(1/2)*(b*x^2+a)/(d*x+c),x)
Output:
(2*sqrt(e)*( - 15*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt(c)))*a*d* *2 - 15*sqrt(d)*sqrt(c)*atan((sqrt(x)*d)/(sqrt(d)*sqrt(c)))*b*c**2 + 15*sq rt(x)*a*d**3 + 15*sqrt(x)*b*c**2*d - 5*sqrt(x)*b*c*d**2*x + 3*sqrt(x)*b*d* *3*x**2))/(15*d**4)