Integrand size = 27, antiderivative size = 213 \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\frac {a (6 c d g-a f g-a e h) x \sqrt {a+c x^2}}{16 c}+\frac {(6 c d g-a (f g+e h)) x \left (a+c x^2\right )^{3/2}}{24 c}+\frac {f (g+h x)^2 \left (a+c x^2\right )^{5/2}}{7 c h}-\frac {\left (6 \left (2 a f h^2+c \left (5 f g^2-7 h (e g+d h)\right )\right )+5 c h (5 f g-7 e h) x\right ) \left (a+c x^2\right )^{5/2}}{210 c^2 h}+\frac {a^2 (6 c d g-a f g-a e h) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}} \]
1/24*(6*c*d*g-a*(e*h+f*g))*x*(c*x^2+a)^(3/2)/c+1/7*f*(h*x+g)^2*(c*x^2+a)^( 5/2)/c/h-1/210*(12*a*f*h^2+6*c*(5*f*g^2-7*h*(d*h+e*g))+5*c*h*(-7*e*h+5*f*g )*x)*(c*x^2+a)^(5/2)/c^2/h+1/16*a^2*(-a*e*h-a*f*g+6*c*d*g)*arctanh(x*c^(1/ 2)/(c*x^2+a)^(1/2))/c^(3/2)+1/16*a*(-a*e*h-a*f*g+6*c*d*g)*x*(c*x^2+a)^(1/2 )/c
Time = 0.72 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.92 \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {a+c x^2} \left (-96 a^3 f h+3 a^2 c (112 d h+7 e (16 g+5 h x)+f x (35 g+16 h x))+4 c^3 x^3 (21 d (5 g+4 h x)+2 x (7 e (6 g+5 h x)+5 f x (7 g+6 h x)))+2 a c^2 x (21 d (25 g+16 h x)+x (7 e (48 g+35 h x)+f x (245 g+192 h x)))\right )+105 a^2 \sqrt {c} (-6 c d g+a f g+a e h) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{1680 c^2} \]
(Sqrt[a + c*x^2]*(-96*a^3*f*h + 3*a^2*c*(112*d*h + 7*e*(16*g + 5*h*x) + f* x*(35*g + 16*h*x)) + 4*c^3*x^3*(21*d*(5*g + 4*h*x) + 2*x*(7*e*(6*g + 5*h*x ) + 5*f*x*(7*g + 6*h*x))) + 2*a*c^2*x*(21*d*(25*g + 16*h*x) + x*(7*e*(48*g + 35*h*x) + f*x*(245*g + 192*h*x)))) + 105*a^2*Sqrt[c]*(-6*c*d*g + a*f*g + a*e*h)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(1680*c^2)
Time = 0.39 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2185, 27, 676, 211, 211, 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.
| \(\displaystyle \int \left (a+c x^2\right )^{3/2} (g+h x) \left (d+e x+f x^2\right ) \, dx\) |
| \(\Big \downarrow \) 2185 |
| \(\displaystyle \frac {\int h (g+h x) ((7 c d-2 a f) h-c (5 f g-7 e h) x) \left (c x^2+a\right )^{3/2}dx}{7 c h^2}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (g+h x) ((7 c d-2 a f) h-c (5 f g-7 e h) x) \left (c x^2+a\right )^{3/2}dx}{7 c h}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\frac {7}{6} h (-a e h-a f g+6 c d g) \int \left (c x^2+a\right )^{3/2}dx-\frac {\left (a+c x^2\right )^{5/2} \left (2 a f h^2-7 c h (d h+e g)+5 c f g^2\right )}{5 c}-\frac {1}{6} h x \left (a+c x^2\right )^{5/2} (5 f g-7 e h)}{7 c h}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {7}{6} h (-a e h-a f g+6 c d g) \left (\frac {3}{4} a \int \sqrt {c x^2+a}dx+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )-\frac {\left (a+c x^2\right )^{5/2} \left (2 a f h^2-7 c h (d h+e g)+5 c f g^2\right )}{5 c}-\frac {1}{6} h x \left (a+c x^2\right )^{5/2} (5 f g-7 e h)}{7 c h}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {7}{6} h (-a e h-a f g+6 c d g) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {c x^2+a}}dx+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )-\frac {\left (a+c x^2\right )^{5/2} \left (2 a f h^2-7 c h (d h+e g)+5 c f g^2\right )}{5 c}-\frac {1}{6} h x \left (a+c x^2\right )^{5/2} (5 f g-7 e h)}{7 c h}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {7}{6} h (-a e h-a f g+6 c d g) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )-\frac {\left (a+c x^2\right )^{5/2} \left (2 a f h^2-7 c h (d h+e g)+5 c f g^2\right )}{5 c}-\frac {1}{6} h x \left (a+c x^2\right )^{5/2} (5 f g-7 e h)}{7 c h}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {7}{6} h \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right ) (-a e h-a f g+6 c d g)-\frac {\left (a+c x^2\right )^{5/2} \left (2 a f h^2-7 c h (d h+e g)+5 c f g^2\right )}{5 c}-\frac {1}{6} h x \left (a+c x^2\right )^{5/2} (5 f g-7 e h)}{7 c h}+\frac {f \left (a+c x^2\right )^{5/2} (g+h x)^2}{7 c h}\) |
(f*(g + h*x)^2*(a + c*x^2)^(5/2))/(7*c*h) + (-1/5*((5*c*f*g^2 + 2*a*f*h^2 - 7*c*h*(e*g + d*h))*(a + c*x^2)^(5/2))/c - (h*(5*f*g - 7*e*h)*x*(a + c*x^ 2)^(5/2))/6 + (7*h*(6*c*d*g - a*f*g - a*e*h)*((x*(a + c*x^2)^(3/2))/4 + (3 *a*((x*Sqrt[a + c*x^2])/2 + (a*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*Sq rt[c])))/4))/6)/(7*c*h)
3.1.90.3.1 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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x )^(q - 2)*(a*e^2*(m + q - 1) - b*d^2*(m + q + 2*p + 1) - 2*b*d*e*(m + q + p )*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d , e, m, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && !(EqQ[d, 0] && True) && !(IGtQ[m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 0.53 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.91
| method | result | size |
| default | \(d g \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )+h f \left (\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{7 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{35 c^{2}}\right )+\left (e h +f g \right ) \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6 c}-\frac {a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6 c}\right )+\frac {\left (d h +e g \right ) \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5 c}\) | \(194\) |
| risch | \(-\frac {\left (-240 c^{3} f h \,x^{6}-280 c^{3} e h \,x^{5}-280 c^{3} f g \,x^{5}-384 a \,c^{2} f h \,x^{4}-336 c^{3} d h \,x^{4}-336 c^{3} e g \,x^{4}-490 a \,c^{2} e h \,x^{3}-490 a \,c^{2} f g \,x^{3}-420 c^{3} d g \,x^{3}-48 a^{2} c f h \,x^{2}-672 a \,c^{2} d h \,x^{2}-672 a \,c^{2} e g \,x^{2}-105 a^{2} c e h x -105 a^{2} c f g x -1050 a \,c^{2} d g x +96 a^{3} f h -336 a^{2} c d h -336 a^{2} c e g \right ) \sqrt {c \,x^{2}+a}}{1680 c^{2}}-\frac {a^{2} \left (a e h +a f g -6 c d g \right ) \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}\) | \(231\) |
d*g*(1/4*x*(c*x^2+a)^(3/2)+3/4*a*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(x *c^(1/2)+(c*x^2+a)^(1/2))))+h*f*(1/7*x^2*(c*x^2+a)^(5/2)/c-2/35*a/c^2*(c*x ^2+a)^(5/2))+(e*h+f*g)*(1/6*x*(c*x^2+a)^(5/2)/c-1/6*a/c*(1/4*x*(c*x^2+a)^( 3/2)+3/4*a*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/ 2)))))+1/5*(d*h+e*g)*(c*x^2+a)^(5/2)/c
Time = 0.29 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.24 \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\left [\frac {105 \, {\left (a^{3} e h - {\left (6 \, a^{2} c d - a^{3} f\right )} g\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (240 \, c^{3} f h x^{6} + 280 \, {\left (c^{3} f g + c^{3} e h\right )} x^{5} + 336 \, a^{2} c e g + 48 \, {\left (7 \, c^{3} e g + {\left (7 \, c^{3} d + 8 \, a c^{2} f\right )} h\right )} x^{4} + 70 \, {\left (7 \, a c^{2} e h + {\left (6 \, c^{3} d + 7 \, a c^{2} f\right )} g\right )} x^{3} + 48 \, {\left (14 \, a c^{2} e g + {\left (14 \, a c^{2} d + a^{2} c f\right )} h\right )} x^{2} + 48 \, {\left (7 \, a^{2} c d - 2 \, a^{3} f\right )} h + 105 \, {\left (a^{2} c e h + {\left (10 \, a c^{2} d + a^{2} c f\right )} g\right )} x\right )} \sqrt {c x^{2} + a}}{3360 \, c^{2}}, \frac {105 \, {\left (a^{3} e h - {\left (6 \, a^{2} c d - a^{3} f\right )} g\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (240 \, c^{3} f h x^{6} + 280 \, {\left (c^{3} f g + c^{3} e h\right )} x^{5} + 336 \, a^{2} c e g + 48 \, {\left (7 \, c^{3} e g + {\left (7 \, c^{3} d + 8 \, a c^{2} f\right )} h\right )} x^{4} + 70 \, {\left (7 \, a c^{2} e h + {\left (6 \, c^{3} d + 7 \, a c^{2} f\right )} g\right )} x^{3} + 48 \, {\left (14 \, a c^{2} e g + {\left (14 \, a c^{2} d + a^{2} c f\right )} h\right )} x^{2} + 48 \, {\left (7 \, a^{2} c d - 2 \, a^{3} f\right )} h + 105 \, {\left (a^{2} c e h + {\left (10 \, a c^{2} d + a^{2} c f\right )} g\right )} x\right )} \sqrt {c x^{2} + a}}{1680 \, c^{2}}\right ] \]
[1/3360*(105*(a^3*e*h - (6*a^2*c*d - a^3*f)*g)*sqrt(c)*log(-2*c*x^2 + 2*sq rt(c*x^2 + a)*sqrt(c)*x - a) + 2*(240*c^3*f*h*x^6 + 280*(c^3*f*g + c^3*e*h )*x^5 + 336*a^2*c*e*g + 48*(7*c^3*e*g + (7*c^3*d + 8*a*c^2*f)*h)*x^4 + 70* (7*a*c^2*e*h + (6*c^3*d + 7*a*c^2*f)*g)*x^3 + 48*(14*a*c^2*e*g + (14*a*c^2 *d + a^2*c*f)*h)*x^2 + 48*(7*a^2*c*d - 2*a^3*f)*h + 105*(a^2*c*e*h + (10*a *c^2*d + a^2*c*f)*g)*x)*sqrt(c*x^2 + a))/c^2, 1/1680*(105*(a^3*e*h - (6*a^ 2*c*d - a^3*f)*g)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (240*c^3*f *h*x^6 + 280*(c^3*f*g + c^3*e*h)*x^5 + 336*a^2*c*e*g + 48*(7*c^3*e*g + (7* c^3*d + 8*a*c^2*f)*h)*x^4 + 70*(7*a*c^2*e*h + (6*c^3*d + 7*a*c^2*f)*g)*x^3 + 48*(14*a*c^2*e*g + (14*a*c^2*d + a^2*c*f)*h)*x^2 + 48*(7*a^2*c*d - 2*a^ 3*f)*h + 105*(a^2*c*e*h + (10*a*c^2*d + a^2*c*f)*g)*x)*sqrt(c*x^2 + a))/c^ 2]
Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (196) = 392\).
Time = 0.59 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.38 \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\begin {cases} \sqrt {a + c x^{2}} \left (\frac {c f h x^{6}}{7} + \frac {x^{5} \left (c^{2} e h + c^{2} f g\right )}{6 c} + \frac {x^{4} \cdot \left (\frac {8 a c f h}{7} + c^{2} d h + c^{2} e g\right )}{5 c} + \frac {x^{3} \cdot \left (2 a c e h + 2 a c f g - \frac {5 a \left (c^{2} e h + c^{2} f g\right )}{6 c} + c^{2} d g\right )}{4 c} + \frac {x^{2} \left (a^{2} f h + 2 a c d h + 2 a c e g - \frac {4 a \left (\frac {8 a c f h}{7} + c^{2} d h + c^{2} e g\right )}{5 c}\right )}{3 c} + \frac {x \left (a^{2} e h + a^{2} f g + 2 a c d g - \frac {3 a \left (2 a c e h + 2 a c f g - \frac {5 a \left (c^{2} e h + c^{2} f g\right )}{6 c} + c^{2} d g\right )}{4 c}\right )}{2 c} + \frac {a^{2} d h + a^{2} e g - \frac {2 a \left (a^{2} f h + 2 a c d h + 2 a c e g - \frac {4 a \left (\frac {8 a c f h}{7} + c^{2} d h + c^{2} e g\right )}{5 c}\right )}{3 c}}{c}\right ) + \left (a^{2} d g - \frac {a \left (a^{2} e h + a^{2} f g + 2 a c d g - \frac {3 a \left (2 a c e h + 2 a c f g - \frac {5 a \left (c^{2} e h + c^{2} f g\right )}{6 c} + c^{2} d g\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\a^{\frac {3}{2}} \left (d g x + \frac {f h x^{4}}{4} + \frac {x^{3} \left (e h + f g\right )}{3} + \frac {x^{2} \left (d h + e g\right )}{2}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + c*x**2)*(c*f*h*x**6/7 + x**5*(c**2*e*h + c**2*f*g)/(6* c) + x**4*(8*a*c*f*h/7 + c**2*d*h + c**2*e*g)/(5*c) + x**3*(2*a*c*e*h + 2* a*c*f*g - 5*a*(c**2*e*h + c**2*f*g)/(6*c) + c**2*d*g)/(4*c) + x**2*(a**2*f *h + 2*a*c*d*h + 2*a*c*e*g - 4*a*(8*a*c*f*h/7 + c**2*d*h + c**2*e*g)/(5*c) )/(3*c) + x*(a**2*e*h + a**2*f*g + 2*a*c*d*g - 3*a*(2*a*c*e*h + 2*a*c*f*g - 5*a*(c**2*e*h + c**2*f*g)/(6*c) + c**2*d*g)/(4*c))/(2*c) + (a**2*d*h + a **2*e*g - 2*a*(a**2*f*h + 2*a*c*d*h + 2*a*c*e*g - 4*a*(8*a*c*f*h/7 + c**2* d*h + c**2*e*g)/(5*c))/(3*c))/c) + (a**2*d*g - a*(a**2*e*h + a**2*f*g + 2* a*c*d*g - 3*a*(2*a*c*e*h + 2*a*c*f*g - 5*a*(c**2*e*h + c**2*f*g)/(6*c) + c **2*d*g)/(4*c))/(2*c))*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) + 2*c*x)/ sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2), True)), Ne(c, 0)), (a**(3/2)*( d*g*x + f*h*x**4/4 + x**3*(e*h + f*g)/3 + x**2*(d*h + e*g)/2), True))
Time = 0.19 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.99 \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} f h x^{2}}{7 \, c} + \frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d g x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d g x + \frac {3 \, a^{2} d g \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e g}{5 \, c} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} d h}{5 \, c} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a f h}{35 \, c^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} {\left (f g + e h\right )} x}{6 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (f g + e h\right )} a x}{24 \, c} - \frac {\sqrt {c x^{2} + a} {\left (f g + e h\right )} a^{2} x}{16 \, c} - \frac {{\left (f g + e h\right )} a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {3}{2}}} \]
1/7*(c*x^2 + a)^(5/2)*f*h*x^2/c + 1/4*(c*x^2 + a)^(3/2)*d*g*x + 3/8*sqrt(c *x^2 + a)*a*d*g*x + 3/8*a^2*d*g*arcsinh(c*x/sqrt(a*c))/sqrt(c) + 1/5*(c*x^ 2 + a)^(5/2)*e*g/c + 1/5*(c*x^2 + a)^(5/2)*d*h/c - 2/35*(c*x^2 + a)^(5/2)* a*f*h/c^2 + 1/6*(c*x^2 + a)^(5/2)*(f*g + e*h)*x/c - 1/24*(c*x^2 + a)^(3/2) *(f*g + e*h)*a*x/c - 1/16*sqrt(c*x^2 + a)*(f*g + e*h)*a^2*x/c - 1/16*(f*g + e*h)*a^3*arcsinh(c*x/sqrt(a*c))/c^(3/2)
Time = 0.30 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.21 \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\frac {1}{1680} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, c f h x + \frac {7 \, {\left (c^{6} f g + c^{6} e h\right )}}{c^{5}}\right )} x + \frac {6 \, {\left (7 \, c^{6} e g + 7 \, c^{6} d h + 8 \, a c^{5} f h\right )}}{c^{5}}\right )} x + \frac {35 \, {\left (6 \, c^{6} d g + 7 \, a c^{5} f g + 7 \, a c^{5} e h\right )}}{c^{5}}\right )} x + \frac {24 \, {\left (14 \, a c^{5} e g + 14 \, a c^{5} d h + a^{2} c^{4} f h\right )}}{c^{5}}\right )} x + \frac {105 \, {\left (10 \, a c^{5} d g + a^{2} c^{4} f g + a^{2} c^{4} e h\right )}}{c^{5}}\right )} x + \frac {48 \, {\left (7 \, a^{2} c^{4} e g + 7 \, a^{2} c^{4} d h - 2 \, a^{3} c^{3} f h\right )}}{c^{5}}\right )} - \frac {{\left (6 \, a^{2} c d g - a^{3} f g - a^{3} e h\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} \]
1/1680*sqrt(c*x^2 + a)*((2*((4*(5*(6*c*f*h*x + 7*(c^6*f*g + c^6*e*h)/c^5)* x + 6*(7*c^6*e*g + 7*c^6*d*h + 8*a*c^5*f*h)/c^5)*x + 35*(6*c^6*d*g + 7*a*c ^5*f*g + 7*a*c^5*e*h)/c^5)*x + 24*(14*a*c^5*e*g + 14*a*c^5*d*h + a^2*c^4*f *h)/c^5)*x + 105*(10*a*c^5*d*g + a^2*c^4*f*g + a^2*c^4*e*h)/c^5)*x + 48*(7 *a^2*c^4*e*g + 7*a^2*c^4*d*h - 2*a^3*c^3*f*h)/c^5) - 1/16*(6*a^2*c*d*g - a ^3*f*g - a^3*e*h)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)
Timed out. \[ \int (g+h x) \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right ) \, dx=\int \left (g+h\,x\right )\,{\left (c\,x^2+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right ) \,d x \]