Integrand size = 47, antiderivative size = 255 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=\frac {2 b \sqrt {e g+f g x}}{d f g}+\frac {(b c-a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {e g+f g x}}{\sqrt [4]{c} \sqrt {g}}\right )}{\sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}}-\frac {(b c-a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {e g+f g x}}{\sqrt [4]{c} \sqrt {g}}\right )}{\sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {e g+f g x}}{\sqrt {c} \sqrt {g}+\sqrt {d} \sqrt {g} (e+f x)}\right )}{\sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}} \] Output:
2*b*(f*g*x+e*g)^(1/2)/d/f/g+1/2*(-a*d+b*c)*arctan(1-2^(1/2)*d^(1/4)*(f*g*x +e*g)^(1/2)/c^(1/4)/g^(1/2))*2^(1/2)/c^(3/4)/d^(5/4)/f/g^(1/2)-1/2*(-a*d+b *c)*arctan(1+2^(1/2)*d^(1/4)*(f*g*x+e*g)^(1/2)/c^(1/4)/g^(1/2))*2^(1/2)/c^ (3/4)/d^(5/4)/f/g^(1/2)-1/2*(-a*d+b*c)*arctanh(2^(1/2)*c^(1/4)*d^(1/4)*(f* g*x+e*g)^(1/2)/(c^(1/2)*g^(1/2)+d^(1/2)*g^(1/2)*(f*x+e)))*2^(1/2)/c^(3/4)/ d^(5/4)/f/g^(1/2)
Time = 0.01 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.72 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=\frac {4 b c^{3/4} \sqrt [4]{d} (e+f x)+\sqrt {2} (b c-a d) \sqrt {e+f x} \arctan \left (\frac {\sqrt {c}-\sqrt {d} (e+f x)}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {e+f x}}\right )-\sqrt {2} (b c-a d) \sqrt {e+f x} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {e+f x}}{\sqrt {c}+\sqrt {d} (e+f x)}\right )}{2 c^{3/4} d^{5/4} f \sqrt {g (e+f x)}} \] Input:
Integrate[(a + b*(e + f*x)^2)/(Sqrt[e*g + f*g*x]*(c + d*e^2 + 2*d*e*f*x + d*f^2*x^2)),x]
Output:
(4*b*c^(3/4)*d^(1/4)*(e + f*x) + Sqrt[2]*(b*c - a*d)*Sqrt[e + f*x]*ArcTan[ (Sqrt[c] - Sqrt[d]*(e + f*x))/(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[e + f*x])] - S qrt[2]*(b*c - a*d)*Sqrt[e + f*x]*ArcTanh[(Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[e + f*x])/(Sqrt[c] + Sqrt[d]*(e + f*x))])/(2*c^(3/4)*d^(5/4)*f*Sqrt[g*(e + f* x)])
Time = 1.62 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.35, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {2084, 2159, 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.
\(\displaystyle \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx\) |
\(\Big \downarrow \) 2084 |
\(\displaystyle \int \frac {a+b e^2+2 b e f x+b f^2 x^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )}dx\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \int \left (\frac {a-\frac {b c}{d}}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )}+\frac {b}{d \sqrt {e g+f g x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(b c-a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {g (e+f x)}}{\sqrt [4]{c} \sqrt {g}}\right )}{\sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}}-\frac {(b c-a d) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {g (e+f x)}}{\sqrt [4]{c} \sqrt {g}}+1\right )}{\sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}}+\frac {(b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {g (e+f x)}+\sqrt {c} \sqrt {g}+\sqrt {d} \sqrt {g} (e+f x)\right )}{2 \sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}}-\frac {(b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {g (e+f x)}+\sqrt {c} \sqrt {g}+\sqrt {d} \sqrt {g} (e+f x)\right )}{2 \sqrt {2} c^{3/4} d^{5/4} f \sqrt {g}}+\frac {2 b \sqrt {e g+f g x}}{d f g}\) |
Input:
Int[(a + b*(e + f*x)^2)/(Sqrt[e*g + f*g*x]*(c + d*e^2 + 2*d*e*f*x + d*f^2* x^2)),x]
Output:
(2*b*Sqrt[e*g + f*g*x])/(d*f*g) + ((b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4) *Sqrt[g*(e + f*x)])/(c^(1/4)*Sqrt[g])])/(Sqrt[2]*c^(3/4)*d^(5/4)*f*Sqrt[g] ) - ((b*c - a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[g*(e + f*x)])/(c^(1/4)*S qrt[g])])/(Sqrt[2]*c^(3/4)*d^(5/4)*f*Sqrt[g]) + ((b*c - a*d)*Log[Sqrt[c]*S qrt[g] + Sqrt[d]*Sqrt[g]*(e + f*x) - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[g*(e + f *x)]])/(2*Sqrt[2]*c^(3/4)*d^(5/4)*f*Sqrt[g]) - ((b*c - a*d)*Log[Sqrt[c]*Sq rt[g] + Sqrt[d]*Sqrt[g]*(e + f*x) + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[g*(e + f* x)]])/(2*Sqrt[2]*c^(3/4)*d^(5/4)*f*Sqrt[g])
Int[(u_)^(p_.)*(v_)^(q_.)*(z_)^(m_.), x_Symbol] :> Int[ExpandToSum[z, x]^m* ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{m, p, q}, x] && Linea rQ[z, x] && QuadraticQ[{u, v}, x] && !(LinearMatchQ[z, x] && QuadraticMatc hQ[{u, v}, x]) && !MatchQ[z^m*u^p*v^q, ((d_.) + (e_.)*x)^m*((f_.) + (g_.)* x)^2*((a_.) + (b_.)*x + (c_.)*x^2)^(t_.) /; FreeQ[{a, b, c, d, e, f, g, t}, x]] && !MatchQ[z^m*u^p*v^q, ((d_.) + (e_.)*x)^m*((f_.) + (g_.)*x)^2*((a_. ) + (c_.)*x^2)^(t_.) /; FreeQ[{a, c, d, e, f, g, t}, x]]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.26 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {\frac {2 b \sqrt {f g x +e g}}{d}+\frac {\left (a d -c b \right ) \left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {f g x +e g +\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {f g x +e g}\, \sqrt {2}+\sqrt {\frac {c \,g^{2}}{d}}}{f g x +e g -\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {f g x +e g}\, \sqrt {2}+\sqrt {\frac {c \,g^{2}}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {f g x +e g}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {f g x +e g}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d c}}{f g}\) | \(203\) |
default | \(\frac {\frac {2 b \sqrt {f g x +e g}}{d}+\frac {\left (a d -c b \right ) \left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {f g x +e g +\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {f g x +e g}\, \sqrt {2}+\sqrt {\frac {c \,g^{2}}{d}}}{f g x +e g -\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {f g x +e g}\, \sqrt {2}+\sqrt {\frac {c \,g^{2}}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {f g x +e g}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {f g x +e g}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d c}}{f g}\) | \(203\) |
risch | \(\frac {2 b \left (f x +e \right )}{d f \sqrt {g \left (f x +e \right )}}+\frac {\left (a d -c b \right ) \left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {f g x +e g +\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {f g x +e g}\, \sqrt {2}+\sqrt {\frac {c \,g^{2}}{d}}}{f g x +e g -\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {f g x +e g}\, \sqrt {2}+\sqrt {\frac {c \,g^{2}}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {f g x +e g}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {f g x +e g}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d f g c}\) | \(209\) |
pseudoelliptic | \(\frac {\left (\ln \left (\frac {-\sqrt {\frac {c \,g^{2}}{d}}-\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {g \left (f x +e \right )}\, \sqrt {2}+\left (-f x -e \right ) g}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}} \sqrt {g \left (f x +e \right )}\, \sqrt {2}-\sqrt {\frac {c \,g^{2}}{d}}+\left (-f x -e \right ) g}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {g \left (f x +e \right )}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}-1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {g \left (f x +e \right )}+\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}{\left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}}\right )\right ) \left (a d -c b \right ) \sqrt {2}\, \left (\frac {c \,g^{2}}{d}\right )^{\frac {1}{4}}+8 b \sqrt {g \left (f x +e \right )}\, c}{4 d f g c}\) | \(215\) |
Input:
int((a+b*(f*x+e)^2)/(f*g*x+e*g)^(1/2)/(d*f^2*x^2+2*d*e*f*x+d*e^2+c),x,meth od=_RETURNVERBOSE)
Output:
2/g/f*(b/d*(f*g*x+e*g)^(1/2)+1/8*(a*d-b*c)/d*(c*g^2/d)^(1/4)/c*2^(1/2)*(ln ((f*g*x+e*g+(c*g^2/d)^(1/4)*(f*g*x+e*g)^(1/2)*2^(1/2)+(c*g^2/d)^(1/2))/(f* g*x+e*g-(c*g^2/d)^(1/4)*(f*g*x+e*g)^(1/2)*2^(1/2)+(c*g^2/d)^(1/2)))+2*arct an(2^(1/2)/(c*g^2/d)^(1/4)*(f*g*x+e*g)^(1/2)+1)-2*arctan(-2^(1/2)/(c*g^2/d )^(1/4)*(f*g*x+e*g)^(1/2)+1)))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 675, normalized size of antiderivative = 2.65 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=\frac {d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} \log \left (c d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} - \sqrt {f g x + e g} {\left (b c - a d\right )}\right ) + i \, d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} \log \left (i \, c d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} - \sqrt {f g x + e g} {\left (b c - a d\right )}\right ) - i \, d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} \log \left (-i \, c d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} - \sqrt {f g x + e g} {\left (b c - a d\right )}\right ) - d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} \log \left (-c d f g \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5} f^{4} g^{2}}\right )^{\frac {1}{4}} - \sqrt {f g x + e g} {\left (b c - a d\right )}\right ) + 4 \, \sqrt {f g x + e g} b}{2 \, d f g} \] Input:
integrate((a+b*(f*x+e)^2)/(f*g*x+e*g)^(1/2)/(d*f^2*x^2+2*d*e*f*x+d*e^2+c), x, algorithm="fricas")
Output:
1/2*(d*f*g*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5*f^4*g^2))^(1/4)*log(c*d*f*g*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5*f^4*g^2))^(1/4) - sqrt(f*g*x + e*g)*(b*c - a*d)) + I*d*f*g*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a ^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5*f^4*g^2))^(1/4)*log(I*c *d*f*g*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^ 4*d^4)/(c^3*d^5*f^4*g^2))^(1/4) - sqrt(f*g*x + e*g)*(b*c - a*d)) - I*d*f*g *(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4) /(c^3*d^5*f^4*g^2))^(1/4)*log(-I*c*d*f*g*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^ 2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5*f^4*g^2))^(1/4) - sqrt(f *g*x + e*g)*(b*c - a*d)) - d*f*g*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^ 2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5*f^4*g^2))^(1/4)*log(-c*d*f*g*(-( b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^ 3*d^5*f^4*g^2))^(1/4) - sqrt(f*g*x + e*g)*(b*c - a*d)) + 4*sqrt(f*g*x + e* g)*b)/(d*f*g)
\[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=\int \frac {a + b e^{2} + 2 b e f x + b f^{2} x^{2}}{\sqrt {g \left (e + f x\right )} \left (c + d e^{2} + 2 d e f x + d f^{2} x^{2}\right )}\, dx \] Input:
integrate((a+b*(f*x+e)**2)/(f*g*x+e*g)**(1/2)/(d*f**2*x**2+2*d*e*f*x+d*e** 2+c),x)
Output:
Integral((a + b*e**2 + 2*b*e*f*x + b*f**2*x**2)/(sqrt(g*(e + f*x))*(c + d* e**2 + 2*d*e*f*x + d*f**2*x**2)), x)
Time = 0.12 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.14 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=-\frac {\frac {{\left (\frac {\sqrt {2} g^{2} \log \left (\sqrt {2} \left (c g^{2}\right )^{\frac {1}{4}} \sqrt {f g x + e g} d^{\frac {1}{4}} + {\left (f g x + e g\right )} \sqrt {d} + \sqrt {c} g\right )}{\left (c g^{2}\right )^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} g^{2} \log \left (-\sqrt {2} \left (c g^{2}\right )^{\frac {1}{4}} \sqrt {f g x + e g} d^{\frac {1}{4}} + {\left (f g x + e g\right )} \sqrt {d} + \sqrt {c} g\right )}{\left (c g^{2}\right )^{\frac {3}{4}} d^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} g \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (c g^{2}\right )^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {f g x + e g} \sqrt {d}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d} g}}\right )}{\sqrt {\sqrt {c} \sqrt {d} g} \sqrt {c}} + \frac {2 \, \sqrt {2} g \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (c g^{2}\right )^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {f g x + e g} \sqrt {d}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d} g}}\right )}{\sqrt {\sqrt {c} \sqrt {d} g} \sqrt {c}}\right )} {\left (b c - a d\right )}}{d} - \frac {8 \, \sqrt {f g x + e g} b}{d}}{4 \, f g} \] Input:
integrate((a+b*(f*x+e)^2)/(f*g*x+e*g)^(1/2)/(d*f^2*x^2+2*d*e*f*x+d*e^2+c), x, algorithm="maxima")
Output:
-1/4*((sqrt(2)*g^2*log(sqrt(2)*(c*g^2)^(1/4)*sqrt(f*g*x + e*g)*d^(1/4) + ( f*g*x + e*g)*sqrt(d) + sqrt(c)*g)/((c*g^2)^(3/4)*d^(1/4)) - sqrt(2)*g^2*lo g(-sqrt(2)*(c*g^2)^(1/4)*sqrt(f*g*x + e*g)*d^(1/4) + (f*g*x + e*g)*sqrt(d) + sqrt(c)*g)/((c*g^2)^(3/4)*d^(1/4)) + 2*sqrt(2)*g*arctan(1/2*sqrt(2)*(sq rt(2)*(c*g^2)^(1/4)*d^(1/4) + 2*sqrt(f*g*x + e*g)*sqrt(d))/sqrt(sqrt(c)*sq rt(d)*g))/(sqrt(sqrt(c)*sqrt(d)*g)*sqrt(c)) + 2*sqrt(2)*g*arctan(-1/2*sqrt (2)*(sqrt(2)*(c*g^2)^(1/4)*d^(1/4) - 2*sqrt(f*g*x + e*g)*sqrt(d))/sqrt(sqr t(c)*sqrt(d)*g))/(sqrt(sqrt(c)*sqrt(d)*g)*sqrt(c)))*(b*c - a*d)/d - 8*sqrt (f*g*x + e*g)*b/d)/(f*g)
Time = 0.14 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.47 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=\frac {2 \, \sqrt {f g x + e g} b}{d f g} - \frac {\sqrt {2} {\left (\left (c d^{3} g^{2}\right )^{\frac {1}{4}} b c - \left (c d^{3} g^{2}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c g^{2}}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {f g x + e g}\right )}}{2 \, \left (\frac {c g^{2}}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c d^{2} f g} - \frac {\sqrt {2} {\left (\left (c d^{3} g^{2}\right )^{\frac {1}{4}} b c - \left (c d^{3} g^{2}\right )^{\frac {1}{4}} a d\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c g^{2}}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {f g x + e g}\right )}}{2 \, \left (\frac {c g^{2}}{d}\right )^{\frac {1}{4}}}\right )}{2 \, c d^{2} f g} - \frac {\sqrt {2} {\left (\left (c d^{3} g^{2}\right )^{\frac {1}{4}} b c - \left (c d^{3} g^{2}\right )^{\frac {1}{4}} a d\right )} \log \left (f g x + e g + \sqrt {2} \sqrt {f g x + e g} \left (\frac {c g^{2}}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c g^{2}}{d}}\right )}{4 \, c d^{2} f g} + \frac {\sqrt {2} {\left (\left (c d^{3} g^{2}\right )^{\frac {1}{4}} b c - \left (c d^{3} g^{2}\right )^{\frac {1}{4}} a d\right )} \log \left (f g x + e g - \sqrt {2} \sqrt {f g x + e g} \left (\frac {c g^{2}}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c g^{2}}{d}}\right )}{4 \, c d^{2} f g} \] Input:
integrate((a+b*(f*x+e)^2)/(f*g*x+e*g)^(1/2)/(d*f^2*x^2+2*d*e*f*x+d*e^2+c), x, algorithm="giac")
Output:
2*sqrt(f*g*x + e*g)*b/(d*f*g) - 1/2*sqrt(2)*((c*d^3*g^2)^(1/4)*b*c - (c*d^ 3*g^2)^(1/4)*a*d)*arctan(1/2*sqrt(2)*(sqrt(2)*(c*g^2/d)^(1/4) + 2*sqrt(f*g *x + e*g))/(c*g^2/d)^(1/4))/(c*d^2*f*g) - 1/2*sqrt(2)*((c*d^3*g^2)^(1/4)*b *c - (c*d^3*g^2)^(1/4)*a*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c*g^2/d)^(1/4) - 2*sqrt(f*g*x + e*g))/(c*g^2/d)^(1/4))/(c*d^2*f*g) - 1/4*sqrt(2)*((c*d^3*g ^2)^(1/4)*b*c - (c*d^3*g^2)^(1/4)*a*d)*log(f*g*x + e*g + sqrt(2)*sqrt(f*g* x + e*g)*(c*g^2/d)^(1/4) + sqrt(c*g^2/d))/(c*d^2*f*g) + 1/4*sqrt(2)*((c*d^ 3*g^2)^(1/4)*b*c - (c*d^3*g^2)^(1/4)*a*d)*log(f*g*x + e*g - sqrt(2)*sqrt(f *g*x + e*g)*(c*g^2/d)^(1/4) + sqrt(c*g^2/d))/(c*d^2*f*g)
Time = 0.09 (sec) , antiderivative size = 1044, normalized size of antiderivative = 4.09 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx =\text {Too large to display} \] Input:
int((a + b*(e + f*x)^2)/((e*g + f*g*x)^(1/2)*(c + d*e^2 + d*f^2*x^2 + 2*d* e*f*x)),x)
Output:
(2*b*(e*g + f*g*x)^(1/2))/(d*f*g) - (atan((((a*d - b*c)*((16*(e*g + f*g*x) ^(1/2)*(a^2*d^3*g^2 + b^2*c^2*d*g^2 - 2*a*b*c*d^2*g^2))/f^2 - (16*(b*c^2*d ^2*g^3 - a*c*d^3*g^3)*(a*d - b*c))/((-c)^(3/4)*d^(5/4)*f^2*g^(1/2)))*1i)/( 2*(-c)^(3/4)*d^(5/4)*f*g^(1/2)) + ((a*d - b*c)*((16*(e*g + f*g*x)^(1/2)*(a ^2*d^3*g^2 + b^2*c^2*d*g^2 - 2*a*b*c*d^2*g^2))/f^2 + (16*(b*c^2*d^2*g^3 - a*c*d^3*g^3)*(a*d - b*c))/((-c)^(3/4)*d^(5/4)*f^2*g^(1/2)))*1i)/(2*(-c)^(3 /4)*d^(5/4)*f*g^(1/2)))/(((a*d - b*c)*((16*(e*g + f*g*x)^(1/2)*(a^2*d^3*g^ 2 + b^2*c^2*d*g^2 - 2*a*b*c*d^2*g^2))/f^2 - (16*(b*c^2*d^2*g^3 - a*c*d^3*g ^3)*(a*d - b*c))/((-c)^(3/4)*d^(5/4)*f^2*g^(1/2))))/(2*(-c)^(3/4)*d^(5/4)* f*g^(1/2)) - ((a*d - b*c)*((16*(e*g + f*g*x)^(1/2)*(a^2*d^3*g^2 + b^2*c^2* d*g^2 - 2*a*b*c*d^2*g^2))/f^2 + (16*(b*c^2*d^2*g^3 - a*c*d^3*g^3)*(a*d - b *c))/((-c)^(3/4)*d^(5/4)*f^2*g^(1/2))))/(2*(-c)^(3/4)*d^(5/4)*f*g^(1/2)))) *(a*d - b*c)*1i)/((-c)^(3/4)*d^(5/4)*f*g^(1/2)) - (atan((((a*d - b*c)*((16 *(e*g + f*g*x)^(1/2)*(a^2*d^3*g^2 + b^2*c^2*d*g^2 - 2*a*b*c*d^2*g^2))/f^2 - ((b*c^2*d^2*g^3 - a*c*d^3*g^3)*(a*d - b*c)*16i)/((-c)^(3/4)*d^(5/4)*f^2* g^(1/2))))/(2*(-c)^(3/4)*d^(5/4)*f*g^(1/2)) + ((a*d - b*c)*((16*(e*g + f*g *x)^(1/2)*(a^2*d^3*g^2 + b^2*c^2*d*g^2 - 2*a*b*c*d^2*g^2))/f^2 + ((b*c^2*d ^2*g^3 - a*c*d^3*g^3)*(a*d - b*c)*16i)/((-c)^(3/4)*d^(5/4)*f^2*g^(1/2))))/ (2*(-c)^(3/4)*d^(5/4)*f*g^(1/2)))/(((a*d - b*c)*((16*(e*g + f*g*x)^(1/2)*( a^2*d^3*g^2 + b^2*c^2*d*g^2 - 2*a*b*c*d^2*g^2))/f^2 - ((b*c^2*d^2*g^3 -...
Time = 0.22 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.40 \[ \int \frac {a+b (e+f x)^2}{\sqrt {e g+f g x} \left (c+d e^2+2 d e f x+d f^2 x^2\right )} \, dx=\frac {\sqrt {g}\, \left (2 d^{\frac {7}{4}} c^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {d}\, \sqrt {f x +e}-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) a -2 d^{\frac {3}{4}} c^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {d}\, \sqrt {f x +e}-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) b +2 d^{\frac {7}{4}} c^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {d}\, \sqrt {f x +e}+d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) a -2 d^{\frac {3}{4}} c^{\frac {5}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {d}\, \sqrt {f x +e}+d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) b +8 \sqrt {f x +e}\, b c d -d^{\frac {7}{4}} c^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {f x +e}\, \sqrt {2}+\sqrt {c}+\sqrt {d}\, e +\sqrt {d}\, f x \right ) a +d^{\frac {3}{4}} c^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {f x +e}\, \sqrt {2}+\sqrt {c}+\sqrt {d}\, e +\sqrt {d}\, f x \right ) b +d^{\frac {7}{4}} c^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {f x +e}\, \sqrt {2}+\sqrt {c}+\sqrt {d}\, e +\sqrt {d}\, f x \right ) a -d^{\frac {3}{4}} c^{\frac {5}{4}} \sqrt {2}\, \mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {f x +e}\, \sqrt {2}+\sqrt {c}+\sqrt {d}\, e +\sqrt {d}\, f x \right ) b \right )}{4 c \,d^{2} f g} \] Input:
int((a+b*(f*x+e)^2)/(f*g*x+e*g)^(1/2)/(d*f^2*x^2+2*d*e*f*x+d*e^2+c),x)
Output:
(sqrt(g)*(2*d**(3/4)*c**(1/4)*sqrt(2)*atan((2*sqrt(d)*sqrt(e + f*x) - d**( 1/4)*c**(1/4)*sqrt(2))/(d**(1/4)*c**(1/4)*sqrt(2)))*a*d - 2*d**(3/4)*c**(1 /4)*sqrt(2)*atan((2*sqrt(d)*sqrt(e + f*x) - d**(1/4)*c**(1/4)*sqrt(2))/(d* *(1/4)*c**(1/4)*sqrt(2)))*b*c + 2*d**(3/4)*c**(1/4)*sqrt(2)*atan((2*sqrt(d )*sqrt(e + f*x) + d**(1/4)*c**(1/4)*sqrt(2))/(d**(1/4)*c**(1/4)*sqrt(2)))* a*d - 2*d**(3/4)*c**(1/4)*sqrt(2)*atan((2*sqrt(d)*sqrt(e + f*x) + d**(1/4) *c**(1/4)*sqrt(2))/(d**(1/4)*c**(1/4)*sqrt(2)))*b*c + 8*sqrt(e + f*x)*b*c* d - d**(3/4)*c**(1/4)*sqrt(2)*log( - d**(1/4)*c**(1/4)*sqrt(e + f*x)*sqrt( 2) + sqrt(c) + sqrt(d)*e + sqrt(d)*f*x)*a*d + d**(3/4)*c**(1/4)*sqrt(2)*lo g( - d**(1/4)*c**(1/4)*sqrt(e + f*x)*sqrt(2) + sqrt(c) + sqrt(d)*e + sqrt( d)*f*x)*b*c + d**(3/4)*c**(1/4)*sqrt(2)*log(d**(1/4)*c**(1/4)*sqrt(e + f*x )*sqrt(2) + sqrt(c) + sqrt(d)*e + sqrt(d)*f*x)*a*d - d**(3/4)*c**(1/4)*sqr t(2)*log(d**(1/4)*c**(1/4)*sqrt(e + f*x)*sqrt(2) + sqrt(c) + sqrt(d)*e + s qrt(d)*f*x)*b*c))/(4*c*d**2*f*g)