Integrand size = 24, antiderivative size = 214 \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\frac {2 \left (c f^2+a g^2\right )}{(e f-d g)^3 \sqrt {f+g x}}-\frac {\left (c d^2+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {\left (7 a e^2 g+c d (8 e f-d g)\right ) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}-\frac {\left (15 a e^2 g^2+c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{3/2} (e f-d g)^{7/2}} \]
-1/4*(15*a*e^2*g^2+c*(-d^2*g^2+8*d*e*f*g+8*e^2*f^2))*arctanh(e^(1/2)*(g*x+ f)^(1/2)/(-d*g+e*f)^(1/2))/e^(3/2)/(-d*g+e*f)^(7/2)+2*(a*g^2+c*f^2)/(-d*g+ e*f)^3/(g*x+f)^(1/2)-1/2*(a*e^2+c*d^2)*(g*x+f)^(1/2)/e/(-d*g+e*f)^2/(e*x+d )^2+1/4*(7*a*e^2*g+c*d*(-d*g+8*e*f))*(g*x+f)^(1/2)/e/(-d*g+e*f)^3/(e*x+d)
Time = 0.91 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.07 \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\frac {\frac {\sqrt {e} \left (c \left (8 e^3 f^2 x^2+d^3 g (f+g x)+8 d e^2 f x (3 f+g x)+d^2 e \left (14 f^2+5 f g x-g^2 x^2\right )\right )+a e \left (8 d^2 g^2+d e g (9 f+25 g x)+e^2 \left (-2 f^2+5 f g x+15 g^2 x^2\right )\right )\right )}{(e f-d g)^3 (d+e x)^2 \sqrt {f+g x}}-\frac {\left (15 a e^2 g^2+c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )\right ) \arctan \left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{(-e f+d g)^{7/2}}}{4 e^{3/2}} \]
((Sqrt[e]*(c*(8*e^3*f^2*x^2 + d^3*g*(f + g*x) + 8*d*e^2*f*x*(3*f + g*x) + d^2*e*(14*f^2 + 5*f*g*x - g^2*x^2)) + a*e*(8*d^2*g^2 + d*e*g*(9*f + 25*g*x ) + e^2*(-2*f^2 + 5*f*g*x + 15*g^2*x^2))))/((e*f - d*g)^3*(d + e*x)^2*Sqrt [f + g*x]) - ((15*a*e^2*g^2 + c*(8*e^2*f^2 + 8*d*e*f*g - d^2*g^2))*ArcTan[ (Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/(-(e*f) + d*g)^(7/2))/(4*e^(3 /2))
Time = 0.50 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.21, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {649, 25, 1582, 25, 27, 361, 25, 359, 221}
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.
| \(\displaystyle \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 649 |
| \(\displaystyle 2 \int -\frac {c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2}{(f+g x) (e f-d g-e (f+g x))^3}d\sqrt {f+g x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {c f^2-2 c (f+g x) f+a g^2+c (f+g x)^2}{(f+g x) (e f-d g-e (f+g x))^3}d\sqrt {f+g x}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle 2 \left (\frac {\int -\frac {e \left (4 e (e f-d g) \left (c f^2+a g^2\right )+\left (3 a e^2 g^2-c \left (4 e^2 f^2-8 d e g f+d^2 g^2\right )\right ) (f+g x)\right )}{(f+g x) (e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{4 e^2 (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \left (-\frac {\int \frac {e \left (4 e (e f-d g) \left (c f^2+a g^2\right )+\left (3 a e^2 g^2-c \left (4 e^2 f^2-8 d e g f+d^2 g^2\right )\right ) (f+g x)\right )}{(f+g x) (e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{4 e^2 (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \left (-\frac {\int \frac {4 e (e f-d g) \left (c f^2+a g^2\right )+\left (3 a e^2 g^2-c \left (4 e^2 f^2-8 d e g f+d^2 g^2\right )\right ) (f+g x)}{(f+g x) (e f-d g-e (f+g x))^2}d\sqrt {f+g x}}{4 e (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle 2 \left (-\frac {\frac {g \sqrt {f+g x} \left (7 a e^2 g+c d (8 e f-d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}-\frac {1}{2} \int -\frac {8 e \left (c f^2+a g^2\right )+\frac {g \left (7 a g e^2+c d (8 e f-d g)\right ) (f+g x)}{e f-d g}}{(f+g x) (e f-d g-e (f+g x))}d\sqrt {f+g x}}{4 e (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} \int \frac {8 e \left (c f^2+a g^2\right )+\frac {g \left (7 a g e^2+c d (8 e f-d g)\right ) (f+g x)}{e f-d g}}{(f+g x) (e f-d g-e (f+g x))}d\sqrt {f+g x}+\frac {g \sqrt {f+g x} \left (7 a e^2 g+c d (8 e f-d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}}{4 e (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} \left (\frac {\left (15 a e^2 g^2+c \left (-d^2 g^2+8 d e f g+8 e^2 f^2\right )\right ) \int \frac {1}{e f-d g-e (f+g x)}d\sqrt {f+g x}}{e f-d g}-\frac {8 e \left (a g^2+c f^2\right )}{\sqrt {f+g x} (e f-d g)}\right )+\frac {g \sqrt {f+g x} \left (7 a e^2 g+c d (8 e f-d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}}{4 e (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} \left (\frac {\left (15 a e^2 g^2+c \left (-d^2 g^2+8 d e f g+8 e^2 f^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e} (e f-d g)^{3/2}}-\frac {8 e \left (a g^2+c f^2\right )}{\sqrt {f+g x} (e f-d g)}\right )+\frac {g \sqrt {f+g x} \left (7 a e^2 g+c d (8 e f-d g)\right )}{2 (e f-d g) (-d g-e (f+g x)+e f)}}{4 e (e f-d g)^2}-\frac {g^2 \sqrt {f+g x} \left (a e^2+c d^2\right )}{4 e (e f-d g)^2 (-d g-e (f+g x)+e f)^2}\right )\) |
2*(-1/4*((c*d^2 + a*e^2)*g^2*Sqrt[f + g*x])/(e*(e*f - d*g)^2*(e*f - d*g - e*(f + g*x))^2) - ((g*(7*a*e^2*g + c*d*(8*e*f - d*g))*Sqrt[f + g*x])/(2*(e *f - d*g)*(e*f - d*g - e*(f + g*x))) + ((-8*e*(c*f^2 + a*g^2))/((e*f - d*g )*Sqrt[f + g*x]) + ((15*a*e^2*g^2 + c*(8*e^2*f^2 + 8*d*e*f*g - d^2*g^2))*A rcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(Sqrt[e]*(e*f - d*g)^(3/2 )))/2)/(4*e*(e*f - d*g)^2))
3.7.2.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
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[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Time = 0.54 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.07
| method | result | size |
| derivativedivides | \(-\frac {2 \left (a \,g^{2}+c \,f^{2}\right )}{\left (d g -e f \right )^{3} \sqrt {g x +f}}-\frac {2 \left (\frac {\left (\frac {7}{8} a \,e^{2} g^{2}-\frac {1}{8} c \,d^{2} g^{2}+c d e f g \right ) \left (g x +f \right )^{\frac {3}{2}}+\frac {g \left (9 a d \,e^{2} g^{2}-9 a \,e^{3} f g +c \,d^{3} g^{2}+7 c \,d^{2} e f g -8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{8 e}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (15 a \,e^{2} g^{2}-c \,d^{2} g^{2}+8 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8 e \sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{3}}\) | \(230\) |
| default | \(-\frac {2 \left (a \,g^{2}+c \,f^{2}\right )}{\left (d g -e f \right )^{3} \sqrt {g x +f}}-\frac {2 \left (\frac {\left (\frac {7}{8} a \,e^{2} g^{2}-\frac {1}{8} c \,d^{2} g^{2}+c d e f g \right ) \left (g x +f \right )^{\frac {3}{2}}+\frac {g \left (9 a d \,e^{2} g^{2}-9 a \,e^{3} f g +c \,d^{3} g^{2}+7 c \,d^{2} e f g -8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{8 e}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (15 a \,e^{2} g^{2}-c \,d^{2} g^{2}+8 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8 e \sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{3}}\) | \(230\) |
| pseudoelliptic | \(-\frac {2 \left (\frac {15 \sqrt {g x +f}\, \left (\left (a \,g^{2}+\frac {8 c \,f^{2}}{15}\right ) e^{2}+\frac {8 c d e f g}{15}-\frac {c \,d^{2} g^{2}}{15}\right ) \left (e x +d \right )^{2} \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8}+\left (\left (\frac {15 a \,g^{2} x^{2}}{8}+\frac {5 a f g x}{8}-\frac {f^{2} \left (-4 c \,x^{2}+a \right )}{4}\right ) e^{3}+\frac {9 d \left (\frac {25 a \,g^{2} x}{9}+f \left (\frac {8 c \,x^{2}}{9}+a \right ) g +\frac {8 c \,f^{2} x}{3}\right ) e^{2}}{8}+\left (\left (-\frac {c \,x^{2}}{8}+a \right ) g^{2}+\frac {5 c f x g}{8}+\frac {7 c \,f^{2}}{4}\right ) d^{2} e +\frac {c \,d^{3} g \left (g x +f \right )}{8}\right ) \sqrt {\left (d g -e f \right ) e}\right )}{\sqrt {\left (d g -e f \right ) e}\, \sqrt {g x +f}\, \left (d g -e f \right )^{3} \left (e x +d \right )^{2} e}\) | \(235\) |
-2*(a*g^2+c*f^2)/(d*g-e*f)^3/(g*x+f)^(1/2)-2/(d*g-e*f)^3*(((7/8*a*e^2*g^2- 1/8*c*d^2*g^2+c*d*e*f*g)*(g*x+f)^(3/2)+1/8*g*(9*a*d*e^2*g^2-9*a*e^3*f*g+c* d^3*g^2+7*c*d^2*e*f*g-8*c*d*e^2*f^2)/e*(g*x+f)^(1/2))/(e*(g*x+f)+d*g-e*f)^ 2+1/8*(15*a*e^2*g^2-c*d^2*g^2+8*c*d*e*f*g+8*c*e^2*f^2)/e/((d*g-e*f)*e)^(1/ 2)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 763 vs. \(2 (192) = 384\).
Time = 0.34 (sec) , antiderivative size = 1539, normalized size of antiderivative = 7.19 \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\text {Too large to display} \]
[-1/8*((8*c*d^2*e^2*f^3 + 8*c*d^3*e*f^2*g - (c*d^4 - 15*a*d^2*e^2)*f*g^2 + (8*c*e^4*f^2*g + 8*c*d*e^3*f*g^2 - (c*d^2*e^2 - 15*a*e^4)*g^3)*x^3 + (8*c *e^4*f^3 + 24*c*d*e^3*f^2*g + 15*(c*d^2*e^2 + a*e^4)*f*g^2 - 2*(c*d^3*e - 15*a*d*e^3)*g^3)*x^2 + (16*c*d*e^3*f^3 + 24*c*d^2*e^2*f^2*g + 6*(c*d^3*e + 5*a*d*e^3)*f*g^2 - (c*d^4 - 15*a*d^2*e^2)*g^3)*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*(8*a*d^3*e^2*g^3 - 2*(7*c*d^2*e^3 - a*e^5)*f^3 + (13*c*d^3*e^2 - 11*a*d* e^4)*f^2*g + (c*d^4*e + a*d^2*e^3)*f*g^2 - (8*c*e^5*f^3 - 3*(3*c*d^2*e^3 - 5*a*e^5)*f*g^2 + (c*d^3*e^2 - 15*a*d*e^4)*g^3)*x^2 - (24*c*d*e^4*f^3 - (1 9*c*d^2*e^3 - 5*a*e^5)*f^2*g - 4*(c*d^3*e^2 - 5*a*d*e^4)*f*g^2 - (c*d^4*e + 25*a*d^2*e^3)*g^3)*x)*sqrt(g*x + f))/(d^2*e^6*f^5 - 4*d^3*e^5*f^4*g + 6* d^4*e^4*f^3*g^2 - 4*d^5*e^3*f^2*g^3 + d^6*e^2*f*g^4 + (e^8*f^4*g - 4*d*e^7 *f^3*g^2 + 6*d^2*e^6*f^2*g^3 - 4*d^3*e^5*f*g^4 + d^4*e^4*g^5)*x^3 + (e^8*f ^5 - 2*d*e^7*f^4*g - 2*d^2*e^6*f^3*g^2 + 8*d^3*e^5*f^2*g^3 - 7*d^4*e^4*f*g ^4 + 2*d^5*e^3*g^5)*x^2 + (2*d*e^7*f^5 - 7*d^2*e^6*f^4*g + 8*d^3*e^5*f^3*g ^2 - 2*d^4*e^4*f^2*g^3 - 2*d^5*e^3*f*g^4 + d^6*e^2*g^5)*x), 1/4*((8*c*d^2* e^2*f^3 + 8*c*d^3*e*f^2*g - (c*d^4 - 15*a*d^2*e^2)*f*g^2 + (8*c*e^4*f^2*g + 8*c*d*e^3*f*g^2 - (c*d^2*e^2 - 15*a*e^4)*g^3)*x^3 + (8*c*e^4*f^3 + 24*c* d*e^3*f^2*g + 15*(c*d^2*e^2 + a*e^4)*f*g^2 - 2*(c*d^3*e - 15*a*d*e^3)*g^3) *x^2 + (16*c*d*e^3*f^3 + 24*c*d^2*e^2*f^2*g + 6*(c*d^3*e + 5*a*d*e^3)*f...
Timed out. \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.27 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.72 \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\frac {{\left (8 \, c e^{2} f^{2} + 8 \, c d e f g - c d^{2} g^{2} + 15 \, a e^{2} g^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{4 \, {\left (e^{4} f^{3} - 3 \, d e^{3} f^{2} g + 3 \, d^{2} e^{2} f g^{2} - d^{3} e g^{3}\right )} \sqrt {-e^{2} f + d e g}} + \frac {2 \, {\left (c f^{2} + a g^{2}\right )}}{{\left (e^{3} f^{3} - 3 \, d e^{2} f^{2} g + 3 \, d^{2} e f g^{2} - d^{3} g^{3}\right )} \sqrt {g x + f}} + \frac {8 \, {\left (g x + f\right )}^{\frac {3}{2}} c d e^{2} f g - 8 \, \sqrt {g x + f} c d e^{2} f^{2} g - {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} e g^{2} + 7 \, {\left (g x + f\right )}^{\frac {3}{2}} a e^{3} g^{2} + 7 \, \sqrt {g x + f} c d^{2} e f g^{2} - 9 \, \sqrt {g x + f} a e^{3} f g^{2} + \sqrt {g x + f} c d^{3} g^{3} + 9 \, \sqrt {g x + f} a d e^{2} g^{3}}{4 \, {\left (e^{4} f^{3} - 3 \, d e^{3} f^{2} g + 3 \, d^{2} e^{2} f g^{2} - d^{3} e g^{3}\right )} {\left ({\left (g x + f\right )} e - e f + d g\right )}^{2}} \]
1/4*(8*c*e^2*f^2 + 8*c*d*e*f*g - c*d^2*g^2 + 15*a*e^2*g^2)*arctan(sqrt(g*x + f)*e/sqrt(-e^2*f + d*e*g))/((e^4*f^3 - 3*d*e^3*f^2*g + 3*d^2*e^2*f*g^2 - d^3*e*g^3)*sqrt(-e^2*f + d*e*g)) + 2*(c*f^2 + a*g^2)/((e^3*f^3 - 3*d*e^2 *f^2*g + 3*d^2*e*f*g^2 - d^3*g^3)*sqrt(g*x + f)) + 1/4*(8*(g*x + f)^(3/2)* c*d*e^2*f*g - 8*sqrt(g*x + f)*c*d*e^2*f^2*g - (g*x + f)^(3/2)*c*d^2*e*g^2 + 7*(g*x + f)^(3/2)*a*e^3*g^2 + 7*sqrt(g*x + f)*c*d^2*e*f*g^2 - 9*sqrt(g*x + f)*a*e^3*f*g^2 + sqrt(g*x + f)*c*d^3*g^3 + 9*sqrt(g*x + f)*a*d*e^2*g^3) /((e^4*f^3 - 3*d*e^3*f^2*g + 3*d^2*e^2*f*g^2 - d^3*e*g^3)*((g*x + f)*e - e *f + d*g)^2)
Time = 12.68 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.45 \[ \int \frac {a+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {f+g\,x}\,\left (-d^3\,e\,g^3+3\,d^2\,e^2\,f\,g^2-3\,d\,e^3\,f^2\,g+e^4\,f^3\right )}{\sqrt {e}\,{\left (d\,g-e\,f\right )}^{7/2}}\right )\,\left (-c\,d^2\,g^2+8\,c\,d\,e\,f\,g+8\,c\,e^2\,f^2+15\,a\,e^2\,g^2\right )}{4\,e^{3/2}\,{\left (d\,g-e\,f\right )}^{7/2}}-\frac {\frac {2\,\left (c\,f^2+a\,g^2\right )}{d\,g-e\,f}+\frac {{\left (f+g\,x\right )}^2\,\left (-c\,d^2\,g^2+8\,c\,d\,e\,f\,g+8\,c\,e^2\,f^2+15\,a\,e^2\,g^2\right )}{4\,{\left (d\,g-e\,f\right )}^3}+\frac {\left (f+g\,x\right )\,\left (c\,d^2\,g^2+8\,c\,d\,e\,f\,g+16\,c\,e^2\,f^2+25\,a\,e^2\,g^2\right )}{4\,e\,{\left (d\,g-e\,f\right )}^2}}{e^2\,{\left (f+g\,x\right )}^{5/2}-{\left (f+g\,x\right )}^{3/2}\,\left (2\,e^2\,f-2\,d\,e\,g\right )+\sqrt {f+g\,x}\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )} \]
(atan(((f + g*x)^(1/2)*(e^4*f^3 - d^3*e*g^3 + 3*d^2*e^2*f*g^2 - 3*d*e^3*f^ 2*g))/(e^(1/2)*(d*g - e*f)^(7/2)))*(15*a*e^2*g^2 - c*d^2*g^2 + 8*c*e^2*f^2 + 8*c*d*e*f*g))/(4*e^(3/2)*(d*g - e*f)^(7/2)) - ((2*(a*g^2 + c*f^2))/(d*g - e*f) + ((f + g*x)^2*(15*a*e^2*g^2 - c*d^2*g^2 + 8*c*e^2*f^2 + 8*c*d*e*f *g))/(4*(d*g - e*f)^3) + ((f + g*x)*(25*a*e^2*g^2 + c*d^2*g^2 + 16*c*e^2*f ^2 + 8*c*d*e*f*g))/(4*e*(d*g - e*f)^2))/(e^2*(f + g*x)^(5/2) - (f + g*x)^( 3/2)*(2*e^2*f - 2*d*e*g) + (f + g*x)^(1/2)*(d^2*g^2 + e^2*f^2 - 2*d*e*f*g) )