Integrand size = 29, antiderivative size = 229 \[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=-\frac {(a e-(c d-a f) x) (g+h x)^3}{a c \sqrt {a+c x^2}}-\frac {(3 c d-4 a f) h (g+h x)^2 \sqrt {a+c x^2}}{3 a c^2}-\frac {h \left (4 \left (3 c^2 d g^2+4 a^2 f h^2-a c \left (7 f g^2+3 h (3 e g+d h)\right )\right )+c h (6 c d g-11 a f g-9 a e h) x\right ) \sqrt {a+c x^2}}{6 a c^3}-\frac {\left (3 a h^2 (3 f g+e h)-2 c g \left (f g^2+3 h (e g+d h)\right )\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}} \]
-1/2*(3*a*h^2*(e*h+3*f*g)-2*c*g*(f*g^2+3*h*(d*h+e*g)))*arctanh(x*c^(1/2)/( c*x^2+a)^(1/2))/c^(5/2)-(a*e-(-a*f+c*d)*x)*(h*x+g)^3/a/c/(c*x^2+a)^(1/2)-1 /3*(-4*a*f+3*c*d)*h*(h*x+g)^2*(c*x^2+a)^(1/2)/a/c^2-1/6*h*(12*c^2*d*g^2+16 *a^2*f*h^2-4*a*c*(7*f*g^2+3*h*(d*h+3*e*g))+c*h*(-9*a*e*h-11*a*f*g+6*c*d*g) *x)*(c*x^2+a)^(1/2)/a/c^3
Time = 0.87 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.12 \[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {-16 a^3 f h^3+6 c^3 d g^3 x+a c^2 \left (6 d h \left (-3 g^2-3 g h x+h^2 x^2\right )-3 e \left (2 g^3+6 g^2 h x-6 g h^2 x^2-h^3 x^3\right )+f x \left (-6 g^3+18 g^2 h x+9 g h^2 x^2+2 h^3 x^3\right )\right )+a^2 c h \left (f \left (36 g^2+27 g h x-8 h^2 x^2\right )+3 h (4 d h+3 e (4 g+h x))\right )+3 a \sqrt {c} \left (3 a h^2 (3 f g+e h)-2 c \left (f g^3+3 g h (e g+d h)\right )\right ) \sqrt {a+c x^2} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{6 a c^3 \sqrt {a+c x^2}} \]
(-16*a^3*f*h^3 + 6*c^3*d*g^3*x + a*c^2*(6*d*h*(-3*g^2 - 3*g*h*x + h^2*x^2) - 3*e*(2*g^3 + 6*g^2*h*x - 6*g*h^2*x^2 - h^3*x^3) + f*x*(-6*g^3 + 18*g^2* h*x + 9*g*h^2*x^2 + 2*h^3*x^3)) + a^2*c*h*(f*(36*g^2 + 27*g*h*x - 8*h^2*x^ 2) + 3*h*(4*d*h + 3*e*(4*g + h*x))) + 3*a*Sqrt[c]*(3*a*h^2*(3*f*g + e*h) - 2*c*(f*g^3 + 3*g*h*(e*g + d*h)))*Sqrt[a + c*x^2]*Log[-(Sqrt[c]*x) + Sqrt[ a + c*x^2]])/(6*a*c^3*Sqrt[a + c*x^2])
Time = 0.50 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2176, 25, 687, 676, 224, 219}
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.
| \(\displaystyle \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle -\frac {\int -\frac {(g+h x)^2 (a (f g+3 e h)-(3 c d-4 a f) h x)}{\sqrt {c x^2+a}}dx}{a c}-\frac {(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(g+h x)^2 (a (f g+3 e h)-(3 c d-4 a f) h x)}{\sqrt {c x^2+a}}dx}{a c}-\frac {(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 687 |
| \(\displaystyle \frac {\frac {\int \frac {(g+h x) \left (a \left (2 (3 c d-4 a f) h^2+3 c g (f g+3 e h)\right )-c h (6 c d g-11 a f g-9 a e h) x\right )}{\sqrt {c x^2+a}}dx}{3 c}-\frac {h \sqrt {a+c x^2} (g+h x)^2 (3 c d-4 a f)}{3 c}}{a c}-\frac {(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\frac {\frac {3}{2} a \left (-3 a h^2 (e h+3 f g)+6 c g h (d h+e g)+2 c f g^3\right ) \int \frac {1}{\sqrt {c x^2+a}}dx-\frac {2 h \sqrt {a+c x^2} \left (4 a^2 f h^2-a c \left (3 h (d h+3 e g)+7 f g^2\right )+3 c^2 d g^2\right )}{c}-\frac {1}{2} h^2 x \sqrt {a+c x^2} (-9 a e h-11 a f g+6 c d g)}{3 c}-\frac {h \sqrt {a+c x^2} (g+h x)^2 (3 c d-4 a f)}{3 c}}{a c}-\frac {(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {3}{2} a \left (-3 a h^2 (e h+3 f g)+6 c g h (d h+e g)+2 c f g^3\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}-\frac {2 h \sqrt {a+c x^2} \left (4 a^2 f h^2-a c \left (3 h (d h+3 e g)+7 f g^2\right )+3 c^2 d g^2\right )}{c}-\frac {1}{2} h^2 x \sqrt {a+c x^2} (-9 a e h-11 a f g+6 c d g)}{3 c}-\frac {h \sqrt {a+c x^2} (g+h x)^2 (3 c d-4 a f)}{3 c}}{a c}-\frac {(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt {a+c x^2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {-\frac {2 h \sqrt {a+c x^2} \left (4 a^2 f h^2-a c \left (3 h (d h+3 e g)+7 f g^2\right )+3 c^2 d g^2\right )}{c}+\frac {3 a \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (-3 a h^2 (e h+3 f g)+6 c g h (d h+e g)+2 c f g^3\right )}{2 \sqrt {c}}-\frac {1}{2} h^2 x \sqrt {a+c x^2} (-9 a e h-11 a f g+6 c d g)}{3 c}-\frac {h \sqrt {a+c x^2} (g+h x)^2 (3 c d-4 a f)}{3 c}}{a c}-\frac {(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt {a+c x^2}}\) |
-(((a*e - (c*d - a*f)*x)*(g + h*x)^3)/(a*c*Sqrt[a + c*x^2])) + (-1/3*((3*c *d - 4*a*f)*h*(g + h*x)^2*Sqrt[a + c*x^2])/c + ((-2*h*(3*c^2*d*g^2 + 4*a^2 *f*h^2 - a*c*(7*f*g^2 + 3*h*(3*e*g + d*h)))*Sqrt[a + c*x^2])/c - (h^2*(6*c *d*g - 11*a*f*g - 9*a*e*h)*x*Sqrt[a + c*x^2])/2 + (3*a*(2*c*f*g^3 + 6*c*g* h*(e*g + d*h) - 3*a*h^2*(3*f*g + e*h))*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2] ])/(2*Sqrt[c]))/(3*c))/(a*c)
3.2.8.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2)) ), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*Simp [c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x ] /; FreeQ[{a, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && Eq Q[f, 0])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 , x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x ] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && !(IGtQ[m, 0] && R ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 0.85 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.24
| method | result | size |
| risch | \(-\frac {h \left (-2 f \,h^{2} c \,x^{2}-3 c e \,h^{2} x -9 c f g h x +10 a f \,h^{2}-6 c d \,h^{2}-18 c e g h -18 c f \,g^{2}\right ) \sqrt {c \,x^{2}+a}}{6 c^{3}}-\frac {\frac {a e \,h^{3} x}{\sqrt {c \,x^{2}+a}}-\frac {2 c^{2} d \,g^{3} x}{a \sqrt {c \,x^{2}+a}}+\frac {3 a f g \,h^{2} x}{\sqrt {c \,x^{2}+a}}+\left (3 a c e \,h^{3}+9 a c f g \,h^{2}-6 c^{2} d g \,h^{2}-6 c^{2} e \,g^{2} h -2 c^{2} f \,g^{3}\right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )-\frac {-2 a^{2} f \,h^{3}+2 a c d \,h^{3}+6 a c e g \,h^{2}+6 a c f \,g^{2} h -6 c^{2} d \,g^{2} h -2 c^{2} e \,g^{3}}{c \sqrt {c \,x^{2}+a}}}{2 c^{2}}\) | \(285\) |
| default | \(\frac {d \,g^{3} x}{a \sqrt {c \,x^{2}+a}}+f \,h^{3} \left (\frac {x^{4}}{3 c \sqrt {c \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+a}}+\frac {2 a}{c^{2} \sqrt {c \,x^{2}+a}}\right )}{3 c}\right )+\left (e \,h^{3}+3 f g \,h^{2}\right ) \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )-\frac {3 d \,g^{2} h +e \,g^{3}}{c \sqrt {c \,x^{2}+a}}+\left (d \,h^{3}+3 e g \,h^{2}+3 f \,g^{2} h \right ) \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+a}}+\frac {2 a}{c^{2} \sqrt {c \,x^{2}+a}}\right )+\left (3 d g \,h^{2}+3 e \,g^{2} h +f \,g^{3}\right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+a}}+\frac {\ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{c^{\frac {3}{2}}}\right )\) | \(292\) |
-1/6*h*(-2*c*f*h^2*x^2-3*c*e*h^2*x-9*c*f*g*h*x+10*a*f*h^2-6*c*d*h^2-18*c*e *g*h-18*c*f*g^2)/c^3*(c*x^2+a)^(1/2)-1/2/c^2*(a*e*h^3*x/(c*x^2+a)^(1/2)-2* c^2*d*g^3*x/a/(c*x^2+a)^(1/2)+3*a*f*g*h^2*x/(c*x^2+a)^(1/2)+(3*a*c*e*h^3+9 *a*c*f*g*h^2-6*c^2*d*g*h^2-6*c^2*e*g^2*h-2*c^2*f*g^3)*(-x/c/(c*x^2+a)^(1/2 )+1/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2)))-(-2*a^2*f*h^3+2*a*c*d*h^3+6*a*c *e*g*h^2+6*a*c*f*g^2*h-6*c^2*d*g^2*h-2*c^2*e*g^3)/c/(c*x^2+a)^(1/2))
Time = 0.32 (sec) , antiderivative size = 758, normalized size of antiderivative = 3.31 \[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (2 \, a^{2} c f g^{3} + 6 \, a^{2} c e g^{2} h - 3 \, a^{3} e h^{3} + 3 \, {\left (2 \, a^{2} c d - 3 \, a^{3} f\right )} g h^{2} + {\left (2 \, a c^{2} f g^{3} + 6 \, a c^{2} e g^{2} h - 3 \, a^{2} c e h^{3} + 3 \, {\left (2 \, a c^{2} d - 3 \, a^{2} c f\right )} g h^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) - 2 \, {\left (2 \, a c^{2} f h^{3} x^{4} - 6 \, a c^{2} e g^{3} + 36 \, a^{2} c e g h^{2} - 18 \, {\left (a c^{2} d - 2 \, a^{2} c f\right )} g^{2} h + 4 \, {\left (3 \, a^{2} c d - 4 \, a^{3} f\right )} h^{3} + 3 \, {\left (3 \, a c^{2} f g h^{2} + a c^{2} e h^{3}\right )} x^{3} + 2 \, {\left (9 \, a c^{2} f g^{2} h + 9 \, a c^{2} e g h^{2} + {\left (3 \, a c^{2} d - 4 \, a^{2} c f\right )} h^{3}\right )} x^{2} - 3 \, {\left (6 \, a c^{2} e g^{2} h - 3 \, a^{2} c e h^{3} - 2 \, {\left (c^{3} d - a c^{2} f\right )} g^{3} + 3 \, {\left (2 \, a c^{2} d - 3 \, a^{2} c f\right )} g h^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{12 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}}, -\frac {3 \, {\left (2 \, a^{2} c f g^{3} + 6 \, a^{2} c e g^{2} h - 3 \, a^{3} e h^{3} + 3 \, {\left (2 \, a^{2} c d - 3 \, a^{3} f\right )} g h^{2} + {\left (2 \, a c^{2} f g^{3} + 6 \, a c^{2} e g^{2} h - 3 \, a^{2} c e h^{3} + 3 \, {\left (2 \, a c^{2} d - 3 \, a^{2} c f\right )} g h^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (2 \, a c^{2} f h^{3} x^{4} - 6 \, a c^{2} e g^{3} + 36 \, a^{2} c e g h^{2} - 18 \, {\left (a c^{2} d - 2 \, a^{2} c f\right )} g^{2} h + 4 \, {\left (3 \, a^{2} c d - 4 \, a^{3} f\right )} h^{3} + 3 \, {\left (3 \, a c^{2} f g h^{2} + a c^{2} e h^{3}\right )} x^{3} + 2 \, {\left (9 \, a c^{2} f g^{2} h + 9 \, a c^{2} e g h^{2} + {\left (3 \, a c^{2} d - 4 \, a^{2} c f\right )} h^{3}\right )} x^{2} - 3 \, {\left (6 \, a c^{2} e g^{2} h - 3 \, a^{2} c e h^{3} - 2 \, {\left (c^{3} d - a c^{2} f\right )} g^{3} + 3 \, {\left (2 \, a c^{2} d - 3 \, a^{2} c f\right )} g h^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{6 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}}\right ] \]
[-1/12*(3*(2*a^2*c*f*g^3 + 6*a^2*c*e*g^2*h - 3*a^3*e*h^3 + 3*(2*a^2*c*d - 3*a^3*f)*g*h^2 + (2*a*c^2*f*g^3 + 6*a*c^2*e*g^2*h - 3*a^2*c*e*h^3 + 3*(2*a *c^2*d - 3*a^2*c*f)*g*h^2)*x^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*s qrt(c)*x - a) - 2*(2*a*c^2*f*h^3*x^4 - 6*a*c^2*e*g^3 + 36*a^2*c*e*g*h^2 - 18*(a*c^2*d - 2*a^2*c*f)*g^2*h + 4*(3*a^2*c*d - 4*a^3*f)*h^3 + 3*(3*a*c^2* f*g*h^2 + a*c^2*e*h^3)*x^3 + 2*(9*a*c^2*f*g^2*h + 9*a*c^2*e*g*h^2 + (3*a*c ^2*d - 4*a^2*c*f)*h^3)*x^2 - 3*(6*a*c^2*e*g^2*h - 3*a^2*c*e*h^3 - 2*(c^3*d - a*c^2*f)*g^3 + 3*(2*a*c^2*d - 3*a^2*c*f)*g*h^2)*x)*sqrt(c*x^2 + a))/(a* c^4*x^2 + a^2*c^3), -1/6*(3*(2*a^2*c*f*g^3 + 6*a^2*c*e*g^2*h - 3*a^3*e*h^3 + 3*(2*a^2*c*d - 3*a^3*f)*g*h^2 + (2*a*c^2*f*g^3 + 6*a*c^2*e*g^2*h - 3*a^ 2*c*e*h^3 + 3*(2*a*c^2*d - 3*a^2*c*f)*g*h^2)*x^2)*sqrt(-c)*arctan(sqrt(-c) *x/sqrt(c*x^2 + a)) - (2*a*c^2*f*h^3*x^4 - 6*a*c^2*e*g^3 + 36*a^2*c*e*g*h^ 2 - 18*(a*c^2*d - 2*a^2*c*f)*g^2*h + 4*(3*a^2*c*d - 4*a^3*f)*h^3 + 3*(3*a* c^2*f*g*h^2 + a*c^2*e*h^3)*x^3 + 2*(9*a*c^2*f*g^2*h + 9*a*c^2*e*g*h^2 + (3 *a*c^2*d - 4*a^2*c*f)*h^3)*x^2 - 3*(6*a*c^2*e*g^2*h - 3*a^2*c*e*h^3 - 2*(c ^3*d - a*c^2*f)*g^3 + 3*(2*a*c^2*d - 3*a^2*c*f)*g*h^2)*x)*sqrt(c*x^2 + a)) /(a*c^4*x^2 + a^2*c^3)]
\[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=\int \frac {\left (g + h x\right )^{3} \left (d + e x + f x^{2}\right )}{\left (a + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.51 \[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {f h^{3} x^{4}}{3 \, \sqrt {c x^{2} + a} c} - \frac {4 \, a f h^{3} x^{2}}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {d g^{3} x}{\sqrt {c x^{2} + a} a} - \frac {e g^{3}}{\sqrt {c x^{2} + a} c} - \frac {3 \, d g^{2} h}{\sqrt {c x^{2} + a} c} - \frac {8 \, a^{2} f h^{3}}{3 \, \sqrt {c x^{2} + a} c^{3}} + \frac {{\left (3 \, f g h^{2} + e h^{3}\right )} x^{3}}{2 \, \sqrt {c x^{2} + a} c} + \frac {{\left (3 \, f g^{2} h + 3 \, e g h^{2} + d h^{3}\right )} x^{2}}{\sqrt {c x^{2} + a} c} + \frac {3 \, {\left (3 \, f g h^{2} + e h^{3}\right )} a x}{2 \, \sqrt {c x^{2} + a} c^{2}} - \frac {{\left (f g^{3} + 3 \, e g^{2} h + 3 \, d g h^{2}\right )} x}{\sqrt {c x^{2} + a} c} - \frac {3 \, {\left (3 \, f g h^{2} + e h^{3}\right )} a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {5}{2}}} + \frac {{\left (f g^{3} + 3 \, e g^{2} h + 3 \, d g h^{2}\right )} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, f g^{2} h + 3 \, e g h^{2} + d h^{3}\right )} a}{\sqrt {c x^{2} + a} c^{2}} \]
1/3*f*h^3*x^4/(sqrt(c*x^2 + a)*c) - 4/3*a*f*h^3*x^2/(sqrt(c*x^2 + a)*c^2) + d*g^3*x/(sqrt(c*x^2 + a)*a) - e*g^3/(sqrt(c*x^2 + a)*c) - 3*d*g^2*h/(sqr t(c*x^2 + a)*c) - 8/3*a^2*f*h^3/(sqrt(c*x^2 + a)*c^3) + 1/2*(3*f*g*h^2 + e *h^3)*x^3/(sqrt(c*x^2 + a)*c) + (3*f*g^2*h + 3*e*g*h^2 + d*h^3)*x^2/(sqrt( c*x^2 + a)*c) + 3/2*(3*f*g*h^2 + e*h^3)*a*x/(sqrt(c*x^2 + a)*c^2) - (f*g^3 + 3*e*g^2*h + 3*d*g*h^2)*x/(sqrt(c*x^2 + a)*c) - 3/2*(3*f*g*h^2 + e*h^3)* a*arcsinh(c*x/sqrt(a*c))/c^(5/2) + (f*g^3 + 3*e*g^2*h + 3*d*g*h^2)*arcsinh (c*x/sqrt(a*c))/c^(3/2) + 2*(3*f*g^2*h + 3*e*g*h^2 + d*h^3)*a/(sqrt(c*x^2 + a)*c^2)
Time = 0.29 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.45 \[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=\frac {{\left ({\left ({\left (\frac {2 \, f h^{3} x}{c} + \frac {3 \, {\left (3 \, a c^{4} f g h^{2} + a c^{4} e h^{3}\right )}}{a c^{5}}\right )} x + \frac {2 \, {\left (9 \, a c^{4} f g^{2} h + 9 \, a c^{4} e g h^{2} + 3 \, a c^{4} d h^{3} - 4 \, a^{2} c^{3} f h^{3}\right )}}{a c^{5}}\right )} x + \frac {3 \, {\left (2 \, c^{5} d g^{3} - 2 \, a c^{4} f g^{3} - 6 \, a c^{4} e g^{2} h - 6 \, a c^{4} d g h^{2} + 9 \, a^{2} c^{3} f g h^{2} + 3 \, a^{2} c^{3} e h^{3}\right )}}{a c^{5}}\right )} x - \frac {2 \, {\left (3 \, a c^{4} e g^{3} + 9 \, a c^{4} d g^{2} h - 18 \, a^{2} c^{3} f g^{2} h - 18 \, a^{2} c^{3} e g h^{2} - 6 \, a^{2} c^{3} d h^{3} + 8 \, a^{3} c^{2} f h^{3}\right )}}{a c^{5}}}{6 \, \sqrt {c x^{2} + a}} - \frac {{\left (2 \, c f g^{3} + 6 \, c e g^{2} h + 6 \, c d g h^{2} - 9 \, a f g h^{2} - 3 \, a e h^{3}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c^{\frac {5}{2}}} \]
1/6*((((2*f*h^3*x/c + 3*(3*a*c^4*f*g*h^2 + a*c^4*e*h^3)/(a*c^5))*x + 2*(9* a*c^4*f*g^2*h + 9*a*c^4*e*g*h^2 + 3*a*c^4*d*h^3 - 4*a^2*c^3*f*h^3)/(a*c^5) )*x + 3*(2*c^5*d*g^3 - 2*a*c^4*f*g^3 - 6*a*c^4*e*g^2*h - 6*a*c^4*d*g*h^2 + 9*a^2*c^3*f*g*h^2 + 3*a^2*c^3*e*h^3)/(a*c^5))*x - 2*(3*a*c^4*e*g^3 + 9*a* c^4*d*g^2*h - 18*a^2*c^3*f*g^2*h - 18*a^2*c^3*e*g*h^2 - 6*a^2*c^3*d*h^3 + 8*a^3*c^2*f*h^3)/(a*c^5))/sqrt(c*x^2 + a) - 1/2*(2*c*f*g^3 + 6*c*e*g^2*h + 6*c*d*g*h^2 - 9*a*f*g*h^2 - 3*a*e*h^3)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2)
Timed out. \[ \int \frac {(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (g+h\,x\right )}^3\,\left (f\,x^2+e\,x+d\right )}{{\left (c\,x^2+a\right )}^{3/2}} \,d x \]