Integrand size = 26, antiderivative size = 245 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=-\frac {8 c d \left (c d^2+a e^2\right ) \sqrt {f+g x}}{e^5}-\frac {\left (c d^2+a e^2\right )^2 \sqrt {f+g x}}{e^5 (d+e x)}+\frac {2 c \left (2 a e^2 g^2+c \left (e^2 f^2+2 d e f g+3 d^2 g^2\right )\right ) (f+g x)^{3/2}}{3 e^4 g^3}-\frac {4 c^2 (e f+d g) (f+g x)^{5/2}}{5 e^3 g^3}+\frac {2 c^2 (f+g x)^{7/2}}{7 e^2 g^3}-\frac {\left (c d^2+a e^2\right ) \left (a e^2 g-c d (8 e f-9 d g)\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{e^{11/2} \sqrt {e f-d g}} \] Output:
-8*c*d*(a*e^2+c*d^2)*(g*x+f)^(1/2)/e^5-(a*e^2+c*d^2)^2*(g*x+f)^(1/2)/e^5/( e*x+d)+2/3*c*(2*a*e^2*g^2+c*(3*d^2*g^2+2*d*e*f*g+e^2*f^2))*(g*x+f)^(3/2)/e ^4/g^3-4/5*c^2*(d*g+e*f)*(g*x+f)^(5/2)/e^3/g^3+2/7*c^2*(g*x+f)^(7/2)/e^2/g ^3-(a*e^2+c*d^2)*(a*e^2*g-c*d*(-9*d*g+8*e*f))*arctanh(e^(1/2)*(g*x+f)^(1/2 )/(-d*g+e*f)^(1/2))/e^(11/2)/(-d*g+e*f)^(1/2)
Time = 0.88 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\frac {\sqrt {f+g x} \left (-105 a^2 e^4 g^3-70 a c e^2 g^2 \left (15 d^2 g-2 d e (f-5 g x)-2 e^2 x (f+g x)\right )+c^2 \left (-945 d^4 g^3+210 d^3 e g^2 (f-3 g x)+14 d^2 e^2 g \left (4 f^2+13 f g x+9 g^2 x^2\right )+2 d e^3 \left (8 f^3+24 f^2 g x-11 f g^2 x^2-27 g^3 x^3\right )+2 e^4 x \left (8 f^3-4 f^2 g x+3 f g^2 x^2+15 g^3 x^3\right )\right )\right )}{105 e^5 g^3 (d+e x)}+\frac {\left (c d^2+a e^2\right ) \left (a e^2 g+c d (-8 e f+9 d g)\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{e^{11/2} \sqrt {-e f+d g}} \] Input:
Integrate[(Sqrt[f + g*x]*(a + c*x^2)^2)/(d + e*x)^2,x]
Output:
(Sqrt[f + g*x]*(-105*a^2*e^4*g^3 - 70*a*c*e^2*g^2*(15*d^2*g - 2*d*e*(f - 5 *g*x) - 2*e^2*x*(f + g*x)) + c^2*(-945*d^4*g^3 + 210*d^3*e*g^2*(f - 3*g*x) + 14*d^2*e^2*g*(4*f^2 + 13*f*g*x + 9*g^2*x^2) + 2*d*e^3*(8*f^3 + 24*f^2*g *x - 11*f*g^2*x^2 - 27*g^3*x^3) + 2*e^4*x*(8*f^3 - 4*f^2*g*x + 3*f*g^2*x^2 + 15*g^3*x^3))))/(105*e^5*g^3*(d + e*x)) + ((c*d^2 + a*e^2)*(a*e^2*g + c* d*(-8*e*f + 9*d*g))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/(e ^(11/2)*Sqrt[-(e*f) + d*g])
Time = 0.72 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {649, 1580, 25, 2341, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^2 \sqrt {f+g x}}{(d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 649 |
| \(\displaystyle \frac {2 \int \frac {(f+g x) \left (c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2\right )^2}{(e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{g^3}\) |
| \(\Big \downarrow \) 1580 |
| \(\displaystyle \frac {2 \left (\frac {\int -\frac {\left (c d^2+a e^2\right )^2 g^4+2 c^2 e^4 (f+g x)^4-2 c^2 e^3 (3 e f+d g) (f+g x)^3+2 c e^2 \left (2 a e^2 g^2+c \left (3 e^2 f^2+2 d e g f+d^2 g^2\right )\right ) (f+g x)^2-2 c e (e f+d g) \left (c e^2 f^2+c d^2 g^2+2 a e^2 g^2\right ) (f+g x)}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{2 e^5}+\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{2 e^5 (-d g-e (f+g x)+e f)}\right )}{g^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{2 e^5 (-d g-e (f+g x)+e f)}-\frac {\int \frac {\left (c d^2+a e^2\right )^2 g^4+2 c^2 e^4 (f+g x)^4-2 c^2 e^3 (3 e f+d g) (f+g x)^3+2 c e^2 \left (2 a e^2 g^2+c \left (3 e^2 f^2+2 d e g f+d^2 g^2\right )\right ) (f+g x)^2-2 c e (e f+d g) \left (c e^2 f^2+c d^2 g^2+2 a e^2 g^2\right ) (f+g x)}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{2 e^5}\right )}{g^3}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {2 \left (\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{2 e^5 (-d g-e (f+g x)+e f)}-\frac {\int \left (8 c d \left (c d^2+a e^2\right ) g^3-2 c^2 e^3 (f+g x)^3+4 c^2 e^2 (e f+d g) (f+g x)^2-2 c e \left (2 a e^2 g^2+c \left (e^2 f^2+2 d e g f+3 d^2 g^2\right )\right ) (f+g x)+\frac {9 c^2 d^4 g^4+a^2 e^4 g^4+10 a c d^2 e^2 g^4-8 a c d e^3 f g^3-8 c^2 d^3 e f g^3}{e f-d g-e (f+g x)}\right )d\sqrt {f+g x}}{2 e^5}\right )}{g^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {g^4 \sqrt {f+g x} \left (a e^2+c d^2\right )^2}{2 e^5 (-d g-e (f+g x)+e f)}-\frac {\frac {g^3 \left (a e^2+c d^2\right ) \left (a e^2 g-c d (8 e f-9 d g)\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e} \sqrt {e f-d g}}-\frac {2}{3} c e (f+g x)^{3/2} \left (2 a e^2 g^2+c \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )+8 c d g^3 \sqrt {f+g x} \left (a e^2+c d^2\right )+\frac {4}{5} c^2 e^2 (f+g x)^{5/2} (d g+e f)-\frac {2}{7} c^2 e^3 (f+g x)^{7/2}}{2 e^5}\right )}{g^3}\) |
Input:
Int[(Sqrt[f + g*x]*(a + c*x^2)^2)/(d + e*x)^2,x]
Output:
(2*(((c*d^2 + a*e^2)^2*g^4*Sqrt[f + g*x])/(2*e^5*(e*f - d*g - e*(f + g*x)) ) - (8*c*d*(c*d^2 + a*e^2)*g^3*Sqrt[f + g*x] - (2*c*e*(2*a*e^2*g^2 + c*(e^ 2*f^2 + 2*d*e*f*g + 3*d^2*g^2))*(f + g*x)^(3/2))/3 + (4*c^2*e^2*(e*f + d*g )*(f + g*x)^(5/2))/5 - (2*c^2*e^3*(f + g*x)^(7/2))/7 + ((c*d^2 + a*e^2)*g^ 3*(a*e^2*g - c*d*(8*e*f - 9*d*g))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(Sqrt[e]*Sqrt[e*f - d*g]))/(2*e^5)))/g^3
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[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) ^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* (q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b *d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e }, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 1.14 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.12
| method | result | size |
| pseudoelliptic | \(\frac {-\left (\left (\left (\frac {18}{35} d \,e^{3} x^{3}-\frac {6}{5} d^{2} e^{2} x^{2}+6 d^{3} e x +9 d^{4}-\frac {2}{7} e^{4} x^{4}\right ) g^{3}-2 e f \left (\frac {1}{35} e^{2} x^{2}-\frac {2}{15} d e x +d^{2}\right ) \left (e x +d \right ) g^{2}-\frac {8 e^{2} \left (-\frac {e x}{7}+d \right ) f^{2} \left (e x +d \right ) g}{15}-\frac {16 e^{3} f^{3} \left (e x +d \right )}{105}\right ) c^{2}+10 e^{2} g^{2} \left (\left (-\frac {2}{15} e^{2} x^{2}+\frac {2}{3} d e x +d^{2}\right ) g -\frac {2 e f \left (e x +d \right )}{15}\right ) a c +a^{2} e^{4} g^{3}\right ) \sqrt {\left (d g -e f \right ) e}\, \sqrt {g x +f}+\left (a \,e^{2}+c \,d^{2}\right ) g^{3} \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right ) \left (e x +d \right ) \left (\left (9 d^{2} g -8 d e f \right ) c +a \,e^{2} g \right )}{\sqrt {\left (d g -e f \right ) e}\, g^{3} e^{5} \left (e x +d \right )}\) | \(274\) |
| risch | \(-\frac {2 c \left (-15 x^{3} c \,e^{3} g^{3}+42 c d \,e^{2} g^{3} x^{2}-3 c \,e^{3} f \,g^{2} x^{2}-70 a \,e^{3} g^{3} x -105 c \,d^{2} e \,g^{3} x +14 c d \,e^{2} f \,g^{2} x +4 c \,e^{3} f^{2} g x +420 a d \,e^{2} g^{3}-70 a \,e^{3} f \,g^{2}+420 c \,d^{3} g^{3}-105 c \,d^{2} e f \,g^{2}-28 c d \,e^{2} f^{2} g -8 c \,e^{3} f^{3}\right ) \sqrt {g x +f}}{105 g^{3} e^{5}}+\frac {\left (2 a \,e^{2}+2 c \,d^{2}\right ) \left (\frac {\left (-\frac {1}{2} a \,e^{2} g -\frac {1}{2} c \,d^{2} g \right ) \sqrt {g x +f}}{e \left (g x +f \right )+d g -e f}+\frac {\left (a \,e^{2} g +9 c \,d^{2} g -8 c d e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \sqrt {\left (d g -e f \right ) e}}\right )}{e^{5}}\) | \(275\) |
| derivativedivides | \(\frac {-\frac {2 c \left (-\frac {c \left (g x +f \right )^{\frac {7}{2}} e^{3}}{7}+\frac {2 c d \,e^{2} g \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 c \,e^{3} f \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {2 a \,e^{3} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}-c \,d^{2} e \,g^{2} \left (g x +f \right )^{\frac {3}{2}}-\frac {2 c d \,e^{2} f g \left (g x +f \right )^{\frac {3}{2}}}{3}-\frac {c \,e^{3} f^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}+4 a d \,e^{2} g^{3} \sqrt {g x +f}+4 c \,d^{3} g^{3} \sqrt {g x +f}\right )}{e^{5}}+\frac {2 g^{3} \left (\frac {\left (-\frac {1}{2} a^{2} e^{4} g -a c \,d^{2} e^{2} g -\frac {1}{2} c^{2} d^{4} g \right ) \sqrt {g x +f}}{e \left (g x +f \right )+d g -e f}+\frac {\left (a^{2} e^{4} g +10 a c \,d^{2} e^{2} g -8 a c d \,e^{3} f +9 c^{2} d^{4} g -8 c^{2} d^{3} e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \sqrt {\left (d g -e f \right ) e}}\right )}{e^{5}}}{g^{3}}\) | \(301\) |
| default | \(\frac {-\frac {2 c \left (-\frac {c \left (g x +f \right )^{\frac {7}{2}} e^{3}}{7}+\frac {2 c d \,e^{2} g \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 c \,e^{3} f \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {2 a \,e^{3} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}-c \,d^{2} e \,g^{2} \left (g x +f \right )^{\frac {3}{2}}-\frac {2 c d \,e^{2} f g \left (g x +f \right )^{\frac {3}{2}}}{3}-\frac {c \,e^{3} f^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}+4 a d \,e^{2} g^{3} \sqrt {g x +f}+4 c \,d^{3} g^{3} \sqrt {g x +f}\right )}{e^{5}}+\frac {2 g^{3} \left (\frac {\left (-\frac {1}{2} a^{2} e^{4} g -a c \,d^{2} e^{2} g -\frac {1}{2} c^{2} d^{4} g \right ) \sqrt {g x +f}}{e \left (g x +f \right )+d g -e f}+\frac {\left (a^{2} e^{4} g +10 a c \,d^{2} e^{2} g -8 a c d \,e^{3} f +9 c^{2} d^{4} g -8 c^{2} d^{3} e f \right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{2 \sqrt {\left (d g -e f \right ) e}}\right )}{e^{5}}}{g^{3}}\) | \(301\) |
Input:
int((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
(-(((18/35*d*e^3*x^3-6/5*d^2*e^2*x^2+6*d^3*e*x+9*d^4-2/7*e^4*x^4)*g^3-2*e* f*(1/35*e^2*x^2-2/15*d*e*x+d^2)*(e*x+d)*g^2-8/15*e^2*(-1/7*e*x+d)*f^2*(e*x +d)*g-16/105*e^3*f^3*(e*x+d))*c^2+10*e^2*g^2*((-2/15*e^2*x^2+2/3*d*e*x+d^2 )*g-2/15*e*f*(e*x+d))*a*c+a^2*e^4*g^3)*((d*g-e*f)*e)^(1/2)*(g*x+f)^(1/2)+( a*e^2+c*d^2)*g^3*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))*(e*x+d)*((9*d ^2*g-8*d*e*f)*c+a*e^2*g))/((d*g-e*f)*e)^(1/2)/g^3/e^5/(e*x+d)
Leaf count of result is larger than twice the leaf count of optimal. 608 vs. \(2 (219) = 438\).
Time = 0.11 (sec) , antiderivative size = 1229, normalized size of antiderivative = 5.02 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\text {Too large to display} \] Input:
integrate((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^2,x, algorithm="fricas")
Output:
[-1/210*(105*(8*(c^2*d^4*e + a*c*d^2*e^3)*f*g^3 - (9*c^2*d^5 + 10*a*c*d^3* e^2 + a^2*d*e^4)*g^4 + (8*(c^2*d^3*e^2 + a*c*d*e^4)*f*g^3 - (9*c^2*d^4*e + 10*a*c*d^2*e^3 + a^2*e^5)*g^4)*x)*sqrt(e^2*f - d*e*g)*log((e*g*x + 2*e*f - d*g - 2*sqrt(e^2*f - d*e*g)*sqrt(g*x + f))/(e*x + d)) - 2*(16*c^2*d*e^5* f^4 + 40*c^2*d^2*e^4*f^3*g + 14*(11*c^2*d^3*e^3 + 10*a*c*d*e^5)*f^2*g^2 - 35*(33*c^2*d^4*e^2 + 34*a*c*d^2*e^4 + 3*a^2*e^6)*f*g^3 + 105*(9*c^2*d^5*e + 10*a*c*d^3*e^3 + a^2*d*e^5)*g^4 + 30*(c^2*e^6*f*g^3 - c^2*d*e^5*g^4)*x^4 + 6*(c^2*e^6*f^2*g^2 - 10*c^2*d*e^5*f*g^3 + 9*c^2*d^2*e^4*g^4)*x^3 - 2*(4 *c^2*e^6*f^3*g + 7*c^2*d*e^5*f^2*g^2 - 2*(37*c^2*d^2*e^4 + 35*a*c*e^6)*f*g ^3 + 7*(9*c^2*d^3*e^3 + 10*a*c*d*e^5)*g^4)*x^2 + 2*(8*c^2*e^6*f^4 + 16*c^2 *d*e^5*f^3*g + (67*c^2*d^2*e^4 + 70*a*c*e^6)*f^2*g^2 - 14*(29*c^2*d^3*e^3 + 30*a*c*d*e^5)*f*g^3 + 35*(9*c^2*d^4*e^2 + 10*a*c*d^2*e^4)*g^4)*x)*sqrt(g *x + f))/(d*e^7*f*g^3 - d^2*e^6*g^4 + (e^8*f*g^3 - d*e^7*g^4)*x), -1/105*( 105*(8*(c^2*d^4*e + a*c*d^2*e^3)*f*g^3 - (9*c^2*d^5 + 10*a*c*d^3*e^2 + a^2 *d*e^4)*g^4 + (8*(c^2*d^3*e^2 + a*c*d*e^4)*f*g^3 - (9*c^2*d^4*e + 10*a*c*d ^2*e^3 + a^2*e^5)*g^4)*x)*sqrt(-e^2*f + d*e*g)*arctan(sqrt(-e^2*f + d*e*g) *sqrt(g*x + f)/(e*g*x + e*f)) - (16*c^2*d*e^5*f^4 + 40*c^2*d^2*e^4*f^3*g + 14*(11*c^2*d^3*e^3 + 10*a*c*d*e^5)*f^2*g^2 - 35*(33*c^2*d^4*e^2 + 34*a*c* d^2*e^4 + 3*a^2*e^6)*f*g^3 + 105*(9*c^2*d^5*e + 10*a*c*d^3*e^3 + a^2*d*e^5 )*g^4 + 30*(c^2*e^6*f*g^3 - c^2*d*e^5*g^4)*x^4 + 6*(c^2*e^6*f^2*g^2 - 1...
Timed out. \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\text {Timed out} \] Input:
integrate((g*x+f)**(1/2)*(c*x**2+a)**2/(e*x+d)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^2,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 = 0.12 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx=-\frac {{\left (8 \, c^{2} d^{3} e f + 8 \, a c d e^{3} f - 9 \, c^{2} d^{4} g - 10 \, a c d^{2} e^{2} g - a^{2} e^{4} g\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{\sqrt {-e^{2} f + d e g} e^{5}} - \frac {\sqrt {g x + f} c^{2} d^{4} g + 2 \, \sqrt {g x + f} a c d^{2} e^{2} g + \sqrt {g x + f} a^{2} e^{4} g}{{\left ({\left (g x + f\right )} e - e f + d g\right )} e^{5}} + \frac {2 \, {\left (15 \, {\left (g x + f\right )}^{\frac {7}{2}} c^{2} e^{12} g^{18} - 42 \, {\left (g x + f\right )}^{\frac {5}{2}} c^{2} e^{12} f g^{18} + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} e^{12} f^{2} g^{18} - 42 \, {\left (g x + f\right )}^{\frac {5}{2}} c^{2} d e^{11} g^{19} + 70 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} d e^{11} f g^{19} + 105 \, {\left (g x + f\right )}^{\frac {3}{2}} c^{2} d^{2} e^{10} g^{20} + 70 \, {\left (g x + f\right )}^{\frac {3}{2}} a c e^{12} g^{20} - 420 \, \sqrt {g x + f} c^{2} d^{3} e^{9} g^{21} - 420 \, \sqrt {g x + f} a c d e^{11} g^{21}\right )}}{105 \, e^{14} g^{21}} \] Input:
integrate((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^2,x, algorithm="giac")
Output:
-(8*c^2*d^3*e*f + 8*a*c*d*e^3*f - 9*c^2*d^4*g - 10*a*c*d^2*e^2*g - a^2*e^4 *g)*arctan(sqrt(g*x + f)*e/sqrt(-e^2*f + d*e*g))/(sqrt(-e^2*f + d*e*g)*e^5 ) - (sqrt(g*x + f)*c^2*d^4*g + 2*sqrt(g*x + f)*a*c*d^2*e^2*g + sqrt(g*x + f)*a^2*e^4*g)/(((g*x + f)*e - e*f + d*g)*e^5) + 2/105*(15*(g*x + f)^(7/2)* c^2*e^12*g^18 - 42*(g*x + f)^(5/2)*c^2*e^12*f*g^18 + 35*(g*x + f)^(3/2)*c^ 2*e^12*f^2*g^18 - 42*(g*x + f)^(5/2)*c^2*d*e^11*g^19 + 70*(g*x + f)^(3/2)* c^2*d*e^11*f*g^19 + 105*(g*x + f)^(3/2)*c^2*d^2*e^10*g^20 + 70*(g*x + f)^( 3/2)*a*c*e^12*g^20 - 420*sqrt(g*x + f)*c^2*d^3*e^9*g^21 - 420*sqrt(g*x + f )*a*c*d*e^11*g^21)/(e^14*g^21)
Time = 0.09 (sec) , antiderivative size = 552, normalized size of antiderivative = 2.25 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx={\left (f+g\,x\right )}^{3/2}\,\left (\frac {12\,c^2\,f^2+4\,a\,c\,g^2}{3\,e^2\,g^3}+\frac {2\,\left (d\,g-e\,f\right )\,\left (\frac {8\,c^2\,f}{e^2\,g^3}+\frac {4\,c^2\,\left (d\,g-e\,f\right )}{e^3\,g^3}\right )}{3\,e}-\frac {2\,c^2\,{\left (d\,g-e\,f\right )}^2}{3\,e^4\,g^3}\right )-{\left (f+g\,x\right )}^{5/2}\,\left (\frac {8\,c^2\,f}{5\,e^2\,g^3}+\frac {4\,c^2\,\left (d\,g-e\,f\right )}{5\,e^3\,g^3}\right )-\sqrt {f+g\,x}\,\left (\frac {8\,c^2\,f^3+8\,a\,c\,f\,g^2}{e^2\,g^3}-\frac {{\left (d\,g-e\,f\right )}^2\,\left (\frac {8\,c^2\,f}{e^2\,g^3}+\frac {4\,c^2\,\left (d\,g-e\,f\right )}{e^3\,g^3}\right )}{e^2}+\frac {2\,\left (d\,g-e\,f\right )\,\left (\frac {12\,c^2\,f^2+4\,a\,c\,g^2}{e^2\,g^3}+\frac {2\,\left (d\,g-e\,f\right )\,\left (\frac {8\,c^2\,f}{e^2\,g^3}+\frac {4\,c^2\,\left (d\,g-e\,f\right )}{e^3\,g^3}\right )}{e}-\frac {2\,c^2\,{\left (d\,g-e\,f\right )}^2}{e^4\,g^3}\right )}{e}\right )-\frac {\sqrt {f+g\,x}\,\left (g\,a^2\,e^4+2\,g\,a\,c\,d^2\,e^2+g\,c^2\,d^4\right )}{e^6\,\left (f+g\,x\right )-e^6\,f+d\,e^5\,g}+\frac {2\,c^2\,{\left (f+g\,x\right )}^{7/2}}{7\,e^2\,g^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}\,\left (c\,d^2+a\,e^2\right )\,\left (9\,c\,g\,d^2-8\,c\,f\,d\,e+a\,g\,e^2\right )}{\sqrt {d\,g-e\,f}\,\left (g\,a^2\,e^4+10\,g\,a\,c\,d^2\,e^2-8\,f\,a\,c\,d\,e^3+9\,g\,c^2\,d^4-8\,f\,c^2\,d^3\,e\right )}\right )\,\left (c\,d^2+a\,e^2\right )\,\left (9\,c\,g\,d^2-8\,c\,f\,d\,e+a\,g\,e^2\right )}{e^{11/2}\,\sqrt {d\,g-e\,f}} \] Input:
int(((f + g*x)^(1/2)*(a + c*x^2)^2)/(d + e*x)^2,x)
Output:
(f + g*x)^(3/2)*((12*c^2*f^2 + 4*a*c*g^2)/(3*e^2*g^3) + (2*(d*g - e*f)*((8 *c^2*f)/(e^2*g^3) + (4*c^2*(d*g - e*f))/(e^3*g^3)))/(3*e) - (2*c^2*(d*g - e*f)^2)/(3*e^4*g^3)) - (f + g*x)^(5/2)*((8*c^2*f)/(5*e^2*g^3) + (4*c^2*(d* g - e*f))/(5*e^3*g^3)) - (f + g*x)^(1/2)*((8*c^2*f^3 + 8*a*c*f*g^2)/(e^2*g ^3) - ((d*g - e*f)^2*((8*c^2*f)/(e^2*g^3) + (4*c^2*(d*g - e*f))/(e^3*g^3)) )/e^2 + (2*(d*g - e*f)*((12*c^2*f^2 + 4*a*c*g^2)/(e^2*g^3) + (2*(d*g - e*f )*((8*c^2*f)/(e^2*g^3) + (4*c^2*(d*g - e*f))/(e^3*g^3)))/e - (2*c^2*(d*g - e*f)^2)/(e^4*g^3)))/e) - ((f + g*x)^(1/2)*(a^2*e^4*g + c^2*d^4*g + 2*a*c* d^2*e^2*g))/(e^6*(f + g*x) - e^6*f + d*e^5*g) + (2*c^2*(f + g*x)^(7/2))/(7 *e^2*g^3) + (atan((e^(1/2)*(f + g*x)^(1/2)*(a*e^2 + c*d^2)*(a*e^2*g + 9*c* d^2*g - 8*c*d*e*f))/((d*g - e*f)^(1/2)*(a^2*e^4*g + 9*c^2*d^4*g - 8*c^2*d^ 3*e*f + 10*a*c*d^2*e^2*g - 8*a*c*d*e^3*f)))*(a*e^2 + c*d^2)*(a*e^2*g + 9*c *d^2*g - 8*c*d*e*f))/(e^(11/2)*(d*g - e*f)^(1/2))
Time = 0.26 (sec) , antiderivative size = 1117, normalized size of antiderivative = 4.56 \[ \int \frac {\sqrt {f+g x} \left (a+c x^2\right )^2}{(d+e x)^2} \, dx =\text {Too large to display} \] Input:
int((g*x+f)^(1/2)*(c*x^2+a)^2/(e*x+d)^2,x)
Output:
(105*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e* f)))*a**2*d*e**4*g**4 + 105*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e) /(sqrt(e)*sqrt(d*g - e*f)))*a**2*e**5*g**4*x + 1050*sqrt(e)*sqrt(d*g - e*f )*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d**3*e**2*g**4 - 8 40*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f) ))*a*c*d**2*e**3*f*g**3 + 1050*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x) *e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d**2*e**3*g**4*x - 840*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*a*c*d*e**4*f*g** 3*x + 945*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c**2*d**5*g**4 - 840*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x) *e)/(sqrt(e)*sqrt(d*g - e*f)))*c**2*d**4*e*f*g**3 + 945*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c**2*d**4*e*g**4*x - 840*sqrt(e)*sqrt(d*g - e*f)*atan((sqrt(f + g*x)*e)/(sqrt(e)*sqrt(d*g - e*f)))*c**2*d**3*e**2*f*g**3*x - 105*sqrt(f + g*x)*a**2*d*e**5*g**4 + 105* sqrt(f + g*x)*a**2*e**6*f*g**3 - 1050*sqrt(f + g*x)*a*c*d**3*e**3*g**4 + 1 190*sqrt(f + g*x)*a*c*d**2*e**4*f*g**3 - 700*sqrt(f + g*x)*a*c*d**2*e**4*g **4*x - 140*sqrt(f + g*x)*a*c*d*e**5*f**2*g**2 + 840*sqrt(f + g*x)*a*c*d*e **5*f*g**3*x + 140*sqrt(f + g*x)*a*c*d*e**5*g**4*x**2 - 140*sqrt(f + g*x)* a*c*e**6*f**2*g**2*x - 140*sqrt(f + g*x)*a*c*e**6*f*g**3*x**2 - 945*sqrt(f + g*x)*c**2*d**5*e*g**4 + 1155*sqrt(f + g*x)*c**2*d**4*e**2*f*g**3 - 6...