Integrand size = 24, antiderivative size = 131 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\frac {2 \left (c d^2+a e^2\right ) \sqrt {f+g x}}{e^3}-\frac {2 c (e f+d g) (f+g x)^{3/2}}{3 e^2 g^2}+\frac {2 c (f+g x)^{5/2}}{5 e g^2}-\frac {2 \left (c d^2+a e^2\right ) \sqrt {e f-d g} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{7/2}} \] Output:
2*(a*e^2+c*d^2)*(g*x+f)^(1/2)/e^3-2/3*c*(d*g+e*f)*(g*x+f)^(3/2)/e^2/g^2+2/ 5*c*(g*x+f)^(5/2)/e/g^2-2*(a*e^2+c*d^2)*(-d*g+e*f)^(1/2)*arctanh(e^(1/2)*( g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(7/2)
Time = 0.45 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\frac {2 \sqrt {f+g x} \left (15 a e^2 g^2+c \left (15 d^2 g^2-5 d e g (f+g x)+e^2 \left (-2 f^2+f g x+3 g^2 x^2\right )\right )\right )}{15 e^3 g^2}-\frac {2 \left (c d^2+a e^2\right ) \sqrt {-e f+d g} \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{7/2}} \] Input:
Integrate[(Sqrt[f + g*x]*(a + c*x^2))/(d + e*x),x]
Output:
(2*Sqrt[f + g*x]*(15*a*e^2*g^2 + c*(15*d^2*g^2 - 5*d*e*g*(f + g*x) + e^2*( -2*f^2 + f*g*x + 3*g^2*x^2))))/(15*e^3*g^2) - (2*(c*d^2 + a*e^2)*Sqrt[-(e* f) + d*g]*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/e^(7/2)
Time = 0.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {649, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right ) \sqrt {f+g x}}{d+e x} \, dx\) |
\(\Big \downarrow \) 649 |
\(\displaystyle \frac {2 \int -\frac {(f+g x) \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{g^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 \int \frac {(f+g x) \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{g^2}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle -\frac {2 \int \left (-\frac {\left (c d^2+a e^2\right ) g^2}{e^3}-\frac {c (f+g x)^2}{e}+\frac {c (e f+d g) (f+g x)}{e^2}+\frac {a f g^2 e^3-a d g^3 e^2+c d^2 f g^2 e-c d^3 g^3}{e^3 (e f-d g-e (f+g x))}\right )d\sqrt {f+g x}}{g^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-\frac {g^2 \left (a e^2+c d^2\right ) \sqrt {e f-d g} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{7/2}}+\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{e^3}-\frac {c (f+g x)^{3/2} (d g+e f)}{3 e^2}+\frac {c (f+g x)^{5/2}}{5 e}\right )}{g^2}\) |
Input:
Int[(Sqrt[f + g*x]*(a + c*x^2))/(d + e*x),x]
Output:
(2*(((c*d^2 + a*e^2)*g^2*Sqrt[f + g*x])/e^3 - (c*(e*f + d*g)*(f + g*x)^(3/ 2))/(3*e^2) + (c*(f + g*x)^(5/2))/(5*e) - ((c*d^2 + a*e^2)*g^2*Sqrt[e*f - d*g]*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/e^(7/2)))/g^2
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^(2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x ]], x] /; FreeQ[{a, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && Integ erQ[m + 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 = 2.18 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.01
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-\left (\left (-\frac {2 \left (-\frac {3 g x}{2}+f \right ) \left (g x +f \right ) c}{15}+a \,g^{2}\right ) e^{2}-\frac {c d e g \left (g x +f \right )}{3}+c \,d^{2} g^{2}\right ) \sqrt {\left (d g -e f \right ) e}\, \sqrt {g x +f}+g^{2} \left (d g -e f \right ) \left (a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )\right )}{\sqrt {\left (d g -e f \right ) e}\, g^{2} e^{3}}\) | \(132\) |
risch | \(\frac {2 \left (3 c \,e^{2} g^{2} x^{2}-5 c d e \,g^{2} x +c \,e^{2} f g x +15 a \,e^{2} g^{2}+15 c \,d^{2} g^{2}-5 c d e f g -2 c \,e^{2} f^{2}\right ) \sqrt {g x +f}}{15 g^{2} e^{3}}-\frac {2 \left (a d \,e^{2} g -a \,e^{3} f +c \,d^{3} g -c \,d^{2} e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{3} \sqrt {\left (d g -e f \right ) e}}\) | \(149\) |
derivativedivides | \(\frac {\frac {2 \left (\frac {c \left (g x +f \right )^{\frac {5}{2}} e^{2}}{5}-\frac {c d e g \left (g x +f \right )^{\frac {3}{2}}}{3}-\frac {c \,e^{2} f \left (g x +f \right )^{\frac {3}{2}}}{3}+a \,e^{2} g^{2} \sqrt {g x +f}+c \,d^{2} g^{2} \sqrt {g x +f}\right )}{e^{3}}-\frac {2 g^{2} \left (a d \,e^{2} g -a \,e^{3} f +c \,d^{3} g -c \,d^{2} e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{3} \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\) | \(153\) |
default | \(\frac {\frac {2 \left (\frac {c \left (g x +f \right )^{\frac {5}{2}} e^{2}}{5}-\frac {c d e g \left (g x +f \right )^{\frac {3}{2}}}{3}-\frac {c \,e^{2} f \left (g x +f \right )^{\frac {3}{2}}}{3}+a \,e^{2} g^{2} \sqrt {g x +f}+c \,d^{2} g^{2} \sqrt {g x +f}\right )}{e^{3}}-\frac {2 g^{2} \left (a d \,e^{2} g -a \,e^{3} f +c \,d^{3} g -c \,d^{2} e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{e^{3} \sqrt {\left (d g -e f \right ) e}}}{g^{2}}\) | \(153\) |
Input:
int((g*x+f)^(1/2)*(c*x^2+a)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
-2*(-((-2/15*(-3/2*g*x+f)*(g*x+f)*c+a*g^2)*e^2-1/3*c*d*e*g*(g*x+f)+c*d^2*g ^2)*((d*g-e*f)*e)^(1/2)*(g*x+f)^(1/2)+g^2*(d*g-e*f)*(a*e^2+c*d^2)*arctan(e *(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)))/((d*g-e*f)*e)^(1/2)/g^2/e^3
Time = 0.10 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.34 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\left [\frac {15 \, {\left (c d^{2} + a e^{2}\right )} g^{2} \sqrt {\frac {e f - d g}{e}} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {g x + f} e \sqrt {\frac {e f - d g}{e}}}{e x + d}\right ) + 2 \, {\left (3 \, c e^{2} g^{2} x^{2} - 2 \, c e^{2} f^{2} - 5 \, c d e f g + 15 \, {\left (c d^{2} + a e^{2}\right )} g^{2} + {\left (c e^{2} f g - 5 \, c d e g^{2}\right )} x\right )} \sqrt {g x + f}}{15 \, e^{3} g^{2}}, -\frac {2 \, {\left (15 \, {\left (c d^{2} + a e^{2}\right )} g^{2} \sqrt {-\frac {e f - d g}{e}} \arctan \left (-\frac {\sqrt {g x + f} e \sqrt {-\frac {e f - d g}{e}}}{e f - d g}\right ) - {\left (3 \, c e^{2} g^{2} x^{2} - 2 \, c e^{2} f^{2} - 5 \, c d e f g + 15 \, {\left (c d^{2} + a e^{2}\right )} g^{2} + {\left (c e^{2} f g - 5 \, c d e g^{2}\right )} x\right )} \sqrt {g x + f}\right )}}{15 \, e^{3} g^{2}}\right ] \] Input:
integrate((g*x+f)^(1/2)*(c*x^2+a)/(e*x+d),x, algorithm="fricas")
Output:
[1/15*(15*(c*d^2 + a*e^2)*g^2*sqrt((e*f - d*g)/e)*log((e*g*x + 2*e*f - d*g - 2*sqrt(g*x + f)*e*sqrt((e*f - d*g)/e))/(e*x + d)) + 2*(3*c*e^2*g^2*x^2 - 2*c*e^2*f^2 - 5*c*d*e*f*g + 15*(c*d^2 + a*e^2)*g^2 + (c*e^2*f*g - 5*c*d* e*g^2)*x)*sqrt(g*x + f))/(e^3*g^2), -2/15*(15*(c*d^2 + a*e^2)*g^2*sqrt(-(e *f - d*g)/e)*arctan(-sqrt(g*x + f)*e*sqrt(-(e*f - d*g)/e)/(e*f - d*g)) - ( 3*c*e^2*g^2*x^2 - 2*c*e^2*f^2 - 5*c*d*e*f*g + 15*(c*d^2 + a*e^2)*g^2 + (c* e^2*f*g - 5*c*d*e*g^2)*x)*sqrt(g*x + f))/(e^3*g^2)]
Time = 5.58 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.36 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\begin {cases} \frac {2 \left (\frac {c \left (f + g x\right )^{\frac {5}{2}}}{5 e g} + \frac {\left (f + g x\right )^{\frac {3}{2}} \left (- c d g - c e f\right )}{3 e^{2} g} + \frac {\sqrt {f + g x} \left (a e^{2} g + c d^{2} g\right )}{e^{3}} - \frac {g \left (a e^{2} + c d^{2}\right ) \left (d g - e f\right ) \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{4} \sqrt {\frac {d g - e f}{e}}}\right )}{g} & \text {for}\: g \neq 0 \\\sqrt {f} \left (- \frac {c d x}{e^{2}} + \frac {c x^{2}}{2 e} + \frac {\left (a e^{2} + c d^{2}\right ) \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((g*x+f)**(1/2)*(c*x**2+a)/(e*x+d),x)
Output:
Piecewise((2*(c*(f + g*x)**(5/2)/(5*e*g) + (f + g*x)**(3/2)*(-c*d*g - c*e* f)/(3*e**2*g) + sqrt(f + g*x)*(a*e**2*g + c*d**2*g)/e**3 - g*(a*e**2 + c*d **2)*(d*g - e*f)*atan(sqrt(f + g*x)/sqrt((d*g - e*f)/e))/(e**4*sqrt((d*g - e*f)/e)))/g, Ne(g, 0)), (sqrt(f)*(-c*d*x/e**2 + c*x**2/(2*e) + (a*e**2 + c*d**2)*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e**2), True))
Exception generated. \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((g*x+f)^(1/2)*(c*x^2+a)/(e*x+d),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*(d*g-e*f)>0)', see `assume?` f or more de
Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\frac {2 \, {\left (c d^{2} e f + a e^{3} f - c d^{3} g - a d e^{2} g\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{\sqrt {-e^{2} f + d e g} e^{3}} + \frac {2 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} c e^{4} g^{8} - 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c e^{4} f g^{8} - 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c d e^{3} g^{9} + 15 \, \sqrt {g x + f} c d^{2} e^{2} g^{10} + 15 \, \sqrt {g x + f} a e^{4} g^{10}\right )}}{15 \, e^{5} g^{10}} \] Input:
integrate((g*x+f)^(1/2)*(c*x^2+a)/(e*x+d),x, algorithm="giac")
Output:
2*(c*d^2*e*f + a*e^3*f - c*d^3*g - a*d*e^2*g)*arctan(sqrt(g*x + f)*e/sqrt( -e^2*f + d*e*g))/(sqrt(-e^2*f + d*e*g)*e^3) + 2/15*(3*(g*x + f)^(5/2)*c*e^ 4*g^8 - 5*(g*x + f)^(3/2)*c*e^4*f*g^8 - 5*(g*x + f)^(3/2)*c*d*e^3*g^9 + 15 *sqrt(g*x + f)*c*d^2*e^2*g^10 + 15*sqrt(g*x + f)*a*e^4*g^10)/(e^5*g^10)
Time = 5.75 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\sqrt {f+g\,x}\,\left (\frac {2\,c\,f^2+2\,a\,g^2}{e\,g^2}+\frac {\left (\frac {2\,c\,\left (d\,g^3-e\,f\,g^2\right )}{e^2\,g^4}+\frac {4\,c\,f}{e\,g^2}\right )\,\left (d\,g^3-e\,f\,g^2\right )}{e\,g^2}\right )-{\left (f+g\,x\right )}^{3/2}\,\left (\frac {2\,c\,\left (d\,g^3-e\,f\,g^2\right )}{3\,e^2\,g^4}+\frac {4\,c\,f}{3\,e\,g^2}\right )+\frac {2\,c\,{\left (f+g\,x\right )}^{5/2}}{5\,e\,g^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}\,1{}\mathrm {i}}{\sqrt {e\,f-d\,g}}\right )\,\left (c\,d^2+a\,e^2\right )\,\sqrt {e\,f-d\,g}\,2{}\mathrm {i}}{e^{7/2}} \] Input:
int(((f + g*x)^(1/2)*(a + c*x^2))/(d + e*x),x)
Output:
(f + g*x)^(1/2)*((2*a*g^2 + 2*c*f^2)/(e*g^2) + (((2*c*(d*g^3 - e*f*g^2))/( e^2*g^4) + (4*c*f)/(e*g^2))*(d*g^3 - e*f*g^2))/(e*g^2)) - (f + g*x)^(3/2)* ((2*c*(d*g^3 - e*f*g^2))/(3*e^2*g^4) + (4*c*f)/(3*e*g^2)) + (atan((e^(1/2) *(f + g*x)^(1/2)*1i)/(e*f - d*g)^(1/2))*(a*e^2 + c*d^2)*(e*f - d*g)^(1/2)* 2i)/e^(7/2) + (2*c*(f + g*x)^(5/2))/(5*e*g^2)
Time = 0.24 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.58 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )}{d+e x} \, dx=\frac {-2 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) a \,e^{2} g^{2}-2 \sqrt {e}\, \sqrt {d g -e f}\, \mathit {atan} \left (\frac {\sqrt {g x +f}\, e}{\sqrt {e}\, \sqrt {d g -e f}}\right ) c \,d^{2} g^{2}+2 \sqrt {g x +f}\, a \,e^{3} g^{2}+2 \sqrt {g x +f}\, c \,d^{2} e \,g^{2}-\frac {2 \sqrt {g x +f}\, c d \,e^{2} f g}{3}-\frac {2 \sqrt {g x +f}\, c d \,e^{2} g^{2} x}{3}-\frac {4 \sqrt {g x +f}\, c \,e^{3} f^{2}}{15}+\frac {2 \sqrt {g x +f}\, c \,e^{3} f g x}{15}+\frac {2 \sqrt {g x +f}\, c \,e^{3} g^{2} x^{2}}{5}}{e^{4} g^{2}} \] Input:
int((g*x+f)^(1/2)*(c*x^2+a)/(e*x+d),x)
Output:
(2*( - 15*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*e**2*g**2 - 15*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/ (sqrt(e)*sqrt(d*g - e*f)))*c*d**2*g**2 + 15*sqrt(f + g*x)*a*e**3*g**2 + 15 *sqrt(f + g*x)*c*d**2*e*g**2 - 5*sqrt(f + g*x)*c*d*e**2*f*g - 5*sqrt(f + g *x)*c*d*e**2*g**2*x - 2*sqrt(f + g*x)*c*e**3*f**2 + sqrt(f + g*x)*c*e**3*f *g*x + 3*sqrt(f + g*x)*c*e**3*g**2*x**2))/(15*e**4*g**2)