Integrand size = 27, antiderivative size = 209 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=-\frac {(a B-(A c-a C) x) (d+e x)^3}{4 a c \left (a+c x^2\right )^2}-\frac {(d+e x) \left (a e (3 A c d+5 a C d+3 a B e)-\left (3 A c^2 d^2-a \left (4 a C e^2-c d (C d+3 B e)\right )\right ) x\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 a e^2 (A c d+3 a C d+a B e)+c d^2 (3 A c d+a C d+3 a B e)\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{5/2}}+\frac {C e^3 \log \left (a+c x^2\right )}{2 c^3} \]
-1/4*(a*B-(A*c-C*a)*x)*(e*x+d)^3/a/c/(c*x^2+a)^2-1/8*(e*x+d)*(a*e*(3*A*c*d +3*B*a*e+5*C*a*d)-(3*A*c^2*d^2-a*(4*a*C*e^2-c*d*(3*B*e+C*d)))*x)/a^2/c^2/( c*x^2+a)+1/8*(3*A*c*d*(a*e^2+c*d^2)+a*(3*a*e^2*(B*e+3*C*d)+c*d^2*(3*B*e+C* d)))*arctan(x*c^(1/2)/a^(1/2))/a^(5/2)/c^(5/2)+1/2*C*e^3*ln(c*x^2+a)/c^3
Time = 0.16 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.34 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {-2 a^3 C e^3+2 A c^3 d^3 x-2 a c^2 d \left (C d^2 x+3 A e (d+e x)+B d (d+3 e x)\right )+2 a^2 c e (3 C d (d+e x)+e (3 B d+A e+B e x))}{a \left (a+c x^2\right )^2}+\frac {8 a^3 C e^3+3 A c^3 d^3 x+a c^2 d \left (C d^2+3 e (B d+A e)\right ) x-a^2 c e (3 C d (4 d+5 e x)+e (12 B d+4 A e+5 B e x))}{a^2 \left (a+c x^2\right )}+\frac {\sqrt {c} \left (3 A c d \left (c d^2+a e^2\right )+a \left (3 a e^2 (3 C d+B e)+c d^2 (C d+3 B e)\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{5/2}}+4 C e^3 \log \left (a+c x^2\right )}{8 c^3} \]
((-2*a^3*C*e^3 + 2*A*c^3*d^3*x - 2*a*c^2*d*(C*d^2*x + 3*A*e*(d + e*x) + B* d*(d + 3*e*x)) + 2*a^2*c*e*(3*C*d*(d + e*x) + e*(3*B*d + A*e + B*e*x)))/(a *(a + c*x^2)^2) + (8*a^3*C*e^3 + 3*A*c^3*d^3*x + a*c^2*d*(C*d^2 + 3*e*(B*d + A*e))*x - a^2*c*e*(3*C*d*(4*d + 5*e*x) + e*(12*B*d + 4*A*e + 5*B*e*x))) /(a^2*(a + c*x^2)) + (Sqrt[c]*(3*A*c*d*(c*d^2 + a*e^2) + a*(3*a*e^2*(3*C*d + B*e) + c*d^2*(C*d + 3*B*e)))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(5/2) + 4*C *e^3*Log[a + c*x^2])/(8*c^3)
Time = 0.43 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2176, 25, 684, 452, 218, 240}
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.
| \(\displaystyle \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 2176 |
| \(\displaystyle -\frac {\int -\frac {(d+e x)^2 (3 A c d+a C d+3 a B e+4 a C e x)}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {(d+e x)^3 (a B-x (A c-a C))}{4 a c \left (a+c x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(d+e x)^2 (3 A c d+a C d+3 a B e+4 a C e x)}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {(d+e x)^3 (a B-x (A c-a C))}{4 a c \left (a+c x^2\right )^2}\) |
| \(\Big \downarrow \) 684 |
| \(\displaystyle \frac {\frac {\int \frac {8 a^2 C x e^3+3 a (A c d+3 a C d+a B e) e^2+c d^2 (3 A c d+a C d+3 a B e)}{c x^2+a}dx}{2 a c}-\frac {(d+e x) \left (x \left (4 a^2 C e^2-c d (3 a B e+a C d+3 A c d)\right )+a e (3 a B e+5 a C d+3 A c d)\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^3 (a B-x (A c-a C))}{4 a c \left (a+c x^2\right )^2}\) |
| \(\Big \downarrow \) 452 |
| \(\displaystyle \frac {\frac {8 a^2 C e^3 \int \frac {x}{c x^2+a}dx+\left (c d^2 (3 a B e+a C d+3 A c d)+3 a e^2 (a B e+3 a C d+A c d)\right ) \int \frac {1}{c x^2+a}dx}{2 a c}-\frac {(d+e x) \left (x \left (4 a^2 C e^2-c d (3 a B e+a C d+3 A c d)\right )+a e (3 a B e+5 a C d+3 A c d)\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^3 (a B-x (A c-a C))}{4 a c \left (a+c x^2\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {8 a^2 C e^3 \int \frac {x}{c x^2+a}dx+\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (c d^2 (3 a B e+a C d+3 A c d)+3 a e^2 (a B e+3 a C d+A c d)\right )}{\sqrt {a} \sqrt {c}}}{2 a c}-\frac {(d+e x) \left (x \left (4 a^2 C e^2-c d (3 a B e+a C d+3 A c d)\right )+a e (3 a B e+5 a C d+3 A c d)\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^3 (a B-x (A c-a C))}{4 a c \left (a+c x^2\right )^2}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {\frac {\frac {4 a^2 C e^3 \log \left (a+c x^2\right )}{c}+\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (c d^2 (3 a B e+a C d+3 A c d)+3 a e^2 (a B e+3 a C d+A c d)\right )}{\sqrt {a} \sqrt {c}}}{2 a c}-\frac {(d+e x) \left (x \left (4 a^2 C e^2-c d (3 a B e+a C d+3 A c d)\right )+a e (3 a B e+5 a C d+3 A c d)\right )}{2 a c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^3 (a B-x (A c-a C))}{4 a c \left (a+c x^2\right )^2}\) |
-1/4*((a*B - (A*c - a*C)*x)*(d + e*x)^3)/(a*c*(a + c*x^2)^2) + (-1/2*((d + e*x)*(a*e*(3*A*c*d + 5*a*C*d + 3*a*B*e) + (4*a^2*C*e^2 - c*d*(3*A*c*d + a *C*d + 3*a*B*e))*x))/(a*c*(a + c*x^2)) + (((3*a*e^2*(A*c*d + 3*a*C*d + a*B *e) + c*d^2*(3*A*c*d + a*C*d + 3*a*B*e))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqr t[a]*Sqrt[c]) + (4*a^2*C*e^3*Log[a + c*x^2])/c)/(2*a*c))/(4*a*c)
3.1.57.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g ) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Simp[1/(2*a*c*(p + 1)) Int[ (d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^ 2*f*(2*p + 3) + e*(a*e*g*m - c*d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a , c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) || !ILtQ[m + 2*p + 3, 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 = 0.57 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.59
| method | result | size |
| default | \(\frac {\frac {\left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}-5 a^{2} B \,e^{3}+3 B a c \,d^{2} e -15 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x^{3}}{8 c \,a^{2}}-\frac {e \left (A c \,e^{2}+3 B c d e -2 a C \,e^{2}+3 C c \,d^{2}\right ) x^{2}}{2 c^{2}}-\frac {\left (3 A a c d \,e^{2}-5 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x}{8 a \,c^{2}}-\frac {A a c \,e^{3}+3 A \,c^{2} d^{2} e +3 B a c d \,e^{2}+B \,c^{2} d^{3}-3 C \,a^{2} e^{3}+3 C a c \,d^{2} e}{4 c^{3}}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\frac {4 C \,a^{2} e^{3} \ln \left (c \,x^{2}+a \right )}{c}+\frac {\left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{8 a^{2} c^{2}}\) | \(333\) |
| risch | \(\frac {\frac {\left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}-5 a^{2} B \,e^{3}+3 B a c \,d^{2} e -15 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x^{3}}{8 c \,a^{2}}-\frac {e \left (A c \,e^{2}+3 B c d e -2 a C \,e^{2}+3 C c \,d^{2}\right ) x^{2}}{2 c^{2}}-\frac {\left (3 A a c d \,e^{2}-5 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x}{8 a \,c^{2}}-\frac {A a c \,e^{3}+3 A \,c^{2} d^{2} e +3 B a c d \,e^{2}+B \,c^{2} d^{3}-3 C \,a^{2} e^{3}+3 C a c \,d^{2} e}{4 c^{3}}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\ln \left (3 A \,a^{2} c d \,e^{2}+3 A \,d^{3} a \,c^{2}+3 B \,e^{3} a^{3}+3 B \,a^{2} c \,d^{2} e +9 C \,a^{3} d \,e^{2}+C \,a^{2} c \,d^{3}-\sqrt {-a c \left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right )^{2}}\, x \right ) C \,e^{3}}{2 c^{3}}+\frac {\ln \left (3 A \,a^{2} c d \,e^{2}+3 A \,d^{3} a \,c^{2}+3 B \,e^{3} a^{3}+3 B \,a^{2} c \,d^{2} e +9 C \,a^{3} d \,e^{2}+C \,a^{2} c \,d^{3}-\sqrt {-a c \left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right )^{2}}\, x \right ) \sqrt {-a c \left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right )^{2}}}{16 a^{3} c^{3}}+\frac {\ln \left (3 A \,a^{2} c d \,e^{2}+3 A \,d^{3} a \,c^{2}+3 B \,e^{3} a^{3}+3 B \,a^{2} c \,d^{2} e +9 C \,a^{3} d \,e^{2}+C \,a^{2} c \,d^{3}+\sqrt {-a c \left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right )^{2}}\, x \right ) C \,e^{3}}{2 c^{3}}-\frac {\ln \left (3 A \,a^{2} c d \,e^{2}+3 A \,d^{3} a \,c^{2}+3 B \,e^{3} a^{3}+3 B \,a^{2} c \,d^{2} e +9 C \,a^{3} d \,e^{2}+C \,a^{2} c \,d^{3}+\sqrt {-a c \left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right )^{2}}\, x \right ) \sqrt {-a c \left (3 A a c d \,e^{2}+3 A \,d^{3} c^{2}+3 a^{2} B \,e^{3}+3 B a c \,d^{2} e +9 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right )^{2}}}{16 a^{3} c^{3}}\) | \(899\) |
(1/8*(3*A*a*c*d*e^2+3*A*c^2*d^3-5*B*a^2*e^3+3*B*a*c*d^2*e-15*C*a^2*d*e^2+C *a*c*d^3)/c/a^2*x^3-1/2*e*(A*c*e^2+3*B*c*d*e-2*C*a*e^2+3*C*c*d^2)/c^2*x^2- 1/8*(3*A*a*c*d*e^2-5*A*c^2*d^3+3*B*a^2*e^3+3*B*a*c*d^2*e+9*C*a^2*d*e^2+C*a *c*d^3)/a/c^2*x-1/4*(A*a*c*e^3+3*A*c^2*d^2*e+3*B*a*c*d*e^2+B*c^2*d^3-3*C*a ^2*e^3+3*C*a*c*d^2*e)/c^3)/(c*x^2+a)^2+1/8/a^2/c^2*(4*C*a^2*e^3/c*ln(c*x^2 +a)+(3*A*a*c*d*e^2+3*A*c^2*d^3+3*B*a^2*e^3+3*B*a*c*d^2*e+9*C*a^2*d*e^2+C*a *c*d^3)/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (198) = 396\).
Time = 0.48 (sec) , antiderivative size = 1138, normalized size of antiderivative = 5.44 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=\text {Too large to display} \]
[-1/16*(4*B*a^3*c^2*d^3 + 12*B*a^4*c*d*e^2 + 12*(C*a^4*c + A*a^3*c^2)*d^2* e - 4*(3*C*a^5 - A*a^4*c)*e^3 - 2*(3*B*a^2*c^3*d^2*e - 5*B*a^3*c^2*e^3 + ( C*a^2*c^3 + 3*A*a*c^4)*d^3 - 3*(5*C*a^3*c^2 - A*a^2*c^3)*d*e^2)*x^3 + 8*(3 *C*a^3*c^2*d^2*e + 3*B*a^3*c^2*d*e^2 - (2*C*a^4*c - A*a^3*c^2)*e^3)*x^2 + (3*B*a^3*c*d^2*e + 3*B*a^4*e^3 + (3*B*a*c^3*d^2*e + 3*B*a^2*c^2*e^3 + (C*a *c^3 + 3*A*c^4)*d^3 + 3*(3*C*a^2*c^2 + A*a*c^3)*d*e^2)*x^4 + (C*a^3*c + 3* A*a^2*c^2)*d^3 + 3*(3*C*a^4 + A*a^3*c)*d*e^2 + 2*(3*B*a^2*c^2*d^2*e + 3*B* a^3*c*e^3 + (C*a^2*c^2 + 3*A*a*c^3)*d^3 + 3*(3*C*a^3*c + A*a^2*c^2)*d*e^2) *x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(3*B*a^ 3*c^2*d^2*e + 3*B*a^4*c*e^3 + (C*a^3*c^2 - 5*A*a^2*c^3)*d^3 + 3*(3*C*a^4*c + A*a^3*c^2)*d*e^2)*x - 8*(C*a^3*c^2*e^3*x^4 + 2*C*a^4*c*e^3*x^2 + C*a^5* e^3)*log(c*x^2 + a))/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3), -1/8*(2*B*a^ 3*c^2*d^3 + 6*B*a^4*c*d*e^2 + 6*(C*a^4*c + A*a^3*c^2)*d^2*e - 2*(3*C*a^5 - A*a^4*c)*e^3 - (3*B*a^2*c^3*d^2*e - 5*B*a^3*c^2*e^3 + (C*a^2*c^3 + 3*A*a* c^4)*d^3 - 3*(5*C*a^3*c^2 - A*a^2*c^3)*d*e^2)*x^3 + 4*(3*C*a^3*c^2*d^2*e + 3*B*a^3*c^2*d*e^2 - (2*C*a^4*c - A*a^3*c^2)*e^3)*x^2 - (3*B*a^3*c*d^2*e + 3*B*a^4*e^3 + (3*B*a*c^3*d^2*e + 3*B*a^2*c^2*e^3 + (C*a*c^3 + 3*A*c^4)*d^ 3 + 3*(3*C*a^2*c^2 + A*a*c^3)*d*e^2)*x^4 + (C*a^3*c + 3*A*a^2*c^2)*d^3 + 3 *(3*C*a^4 + A*a^3*c)*d*e^2 + 2*(3*B*a^2*c^2*d^2*e + 3*B*a^3*c*e^3 + (C*a^2 *c^2 + 3*A*a*c^3)*d^3 + 3*(3*C*a^3*c + A*a^2*c^2)*d*e^2)*x^2)*sqrt(a*c)...
Timed out. \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.81 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=\frac {C e^{3} \log \left (c x^{2} + a\right )}{2 \, c^{3}} - \frac {2 \, B a^{2} c^{2} d^{3} + 6 \, B a^{3} c d e^{2} + 6 \, {\left (C a^{3} c + A a^{2} c^{2}\right )} d^{2} e - 2 \, {\left (3 \, C a^{4} - A a^{3} c\right )} e^{3} - {\left (3 \, B a c^{3} d^{2} e - 5 \, B a^{2} c^{2} e^{3} + {\left (C a c^{3} + 3 \, A c^{4}\right )} d^{3} - 3 \, {\left (5 \, C a^{2} c^{2} - A a c^{3}\right )} d e^{2}\right )} x^{3} + 4 \, {\left (3 \, C a^{2} c^{2} d^{2} e + 3 \, B a^{2} c^{2} d e^{2} - {\left (2 \, C a^{3} c - A a^{2} c^{2}\right )} e^{3}\right )} x^{2} + {\left (3 \, B a^{2} c^{2} d^{2} e + 3 \, B a^{3} c e^{3} + {\left (C a^{2} c^{2} - 5 \, A a c^{3}\right )} d^{3} + 3 \, {\left (3 \, C a^{3} c + A a^{2} c^{2}\right )} d e^{2}\right )} x}{8 \, {\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}} + \frac {{\left (3 \, B a c d^{2} e + 3 \, B a^{2} e^{3} + {\left (C a c + 3 \, A c^{2}\right )} d^{3} + 3 \, {\left (3 \, C a^{2} + A a c\right )} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{2}} \]
1/2*C*e^3*log(c*x^2 + a)/c^3 - 1/8*(2*B*a^2*c^2*d^3 + 6*B*a^3*c*d*e^2 + 6* (C*a^3*c + A*a^2*c^2)*d^2*e - 2*(3*C*a^4 - A*a^3*c)*e^3 - (3*B*a*c^3*d^2*e - 5*B*a^2*c^2*e^3 + (C*a*c^3 + 3*A*c^4)*d^3 - 3*(5*C*a^2*c^2 - A*a*c^3)*d *e^2)*x^3 + 4*(3*C*a^2*c^2*d^2*e + 3*B*a^2*c^2*d*e^2 - (2*C*a^3*c - A*a^2* c^2)*e^3)*x^2 + (3*B*a^2*c^2*d^2*e + 3*B*a^3*c*e^3 + (C*a^2*c^2 - 5*A*a*c^ 3)*d^3 + 3*(3*C*a^3*c + A*a^2*c^2)*d*e^2)*x)/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3) + 1/8*(3*B*a*c*d^2*e + 3*B*a^2*e^3 + (C*a*c + 3*A*c^2)*d^3 + 3* (3*C*a^2 + A*a*c)*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c^2)
Time = 0.27 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.71 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=\frac {C e^{3} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {{\left (C a c d^{3} + 3 \, A c^{2} d^{3} + 3 \, B a c d^{2} e + 9 \, C a^{2} d e^{2} + 3 \, A a c d e^{2} + 3 \, B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{2}} + \frac {{\left (C a c^{2} d^{3} + 3 \, A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 15 \, C a^{2} c d e^{2} + 3 \, A a c^{2} d e^{2} - 5 \, B a^{2} c e^{3}\right )} x^{3} - 4 \, {\left (3 \, C a^{2} c d^{2} e + 3 \, B a^{2} c d e^{2} - 2 \, C a^{3} e^{3} + A a^{2} c e^{3}\right )} x^{2} - {\left (C a^{2} c d^{3} - 5 \, A a c^{2} d^{3} + 3 \, B a^{2} c d^{2} e + 9 \, C a^{3} d e^{2} + 3 \, A a^{2} c d e^{2} + 3 \, B a^{3} e^{3}\right )} x - \frac {2 \, {\left (B a^{2} c^{2} d^{3} + 3 \, C a^{3} c d^{2} e + 3 \, A a^{2} c^{2} d^{2} e + 3 \, B a^{3} c d e^{2} - 3 \, C a^{4} e^{3} + A a^{3} c e^{3}\right )}}{c}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \]
1/2*C*e^3*log(c*x^2 + a)/c^3 + 1/8*(C*a*c*d^3 + 3*A*c^2*d^3 + 3*B*a*c*d^2* e + 9*C*a^2*d*e^2 + 3*A*a*c*d*e^2 + 3*B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sq rt(a*c)*a^2*c^2) + 1/8*((C*a*c^2*d^3 + 3*A*c^3*d^3 + 3*B*a*c^2*d^2*e - 15* C*a^2*c*d*e^2 + 3*A*a*c^2*d*e^2 - 5*B*a^2*c*e^3)*x^3 - 4*(3*C*a^2*c*d^2*e + 3*B*a^2*c*d*e^2 - 2*C*a^3*e^3 + A*a^2*c*e^3)*x^2 - (C*a^2*c*d^3 - 5*A*a* c^2*d^3 + 3*B*a^2*c*d^2*e + 9*C*a^3*d*e^2 + 3*A*a^2*c*d*e^2 + 3*B*a^3*e^3) *x - 2*(B*a^2*c^2*d^3 + 3*C*a^3*c*d^2*e + 3*A*a^2*c^2*d^2*e + 3*B*a^3*c*d* e^2 - 3*C*a^4*e^3 + A*a^3*c*e^3)/c)/((c*x^2 + a)^2*a^2*c^2)
Time = 1.76 (sec) , antiderivative size = 920, normalized size of antiderivative = 4.40 \[ \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^3} \, dx=\frac {5\,A\,d^3\,x}{8\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}-\frac {B\,d^3}{4\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {3\,C\,a^2\,e^3}{4\,\left (a^2\,c^3+2\,a\,c^4\,x^2+c^5\,x^4\right )}-\frac {3\,A\,d^2\,e}{4\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {C\,d^3\,x^3}{8\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}-\frac {C\,d^3\,x}{8\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {A\,a\,e^3}{4\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )}-\frac {A\,e^3\,x^2}{2\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {5\,B\,e^3\,x^3}{8\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {C\,e^3\,\ln \left (c\,x^2+a\right )}{2\,c^3}-\frac {3\,B\,a\,d\,e^2}{4\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )}-\frac {3\,C\,a\,d^2\,e}{4\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )}+\frac {3\,A\,c\,d^3\,x^3}{8\,\left (a^4+2\,a^3\,c\,x^2+a^2\,c^2\,x^4\right )}-\frac {3\,B\,a\,e^3\,x}{8\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )}-\frac {3\,B\,d\,e^2\,x^2}{2\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {3\,C\,d^2\,e\,x^2}{2\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {15\,C\,d\,e^2\,x^3}{8\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {C\,a\,e^3\,x^2}{a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4}+\frac {3\,A\,d^3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {c}}+\frac {3\,B\,e^3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,c^{5/2}}+\frac {C\,d^3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,c^{3/2}}+\frac {3\,A\,d\,e^2\,x^3}{8\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}+\frac {3\,B\,d^2\,e\,x^3}{8\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}-\frac {3\,A\,d\,e^2\,x}{8\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {3\,B\,d^2\,e\,x}{8\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {3\,A\,d\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,c^{3/2}}+\frac {3\,B\,d^2\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{3/2}\,c^{3/2}}+\frac {9\,C\,d\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,c^{5/2}}-\frac {9\,C\,a\,d\,e^2\,x}{8\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )} \]
(5*A*d^3*x)/(8*(a^3 + 2*a^2*c*x^2 + a*c^2*x^4)) - (B*d^3)/(4*(a^2*c + c^3* x^4 + 2*a*c^2*x^2)) + (3*C*a^2*e^3)/(4*(a^2*c^3 + c^5*x^4 + 2*a*c^4*x^2)) - (3*A*d^2*e)/(4*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) + (C*d^3*x^3)/(8*(a^3 + 2*a^2*c*x^2 + a*c^2*x^4)) - (C*d^3*x)/(8*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) - (A*a*e^3)/(4*(a^2*c^2 + c^4*x^4 + 2*a*c^3*x^2)) - (A*e^3*x^2)/(2*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) - (5*B*e^3*x^3)/(8*(a^2*c + c^3*x^4 + 2*a*c^2*x^ 2)) + (C*e^3*log(a + c*x^2))/(2*c^3) - (3*B*a*d*e^2)/(4*(a^2*c^2 + c^4*x^4 + 2*a*c^3*x^2)) - (3*C*a*d^2*e)/(4*(a^2*c^2 + c^4*x^4 + 2*a*c^3*x^2)) + ( 3*A*c*d^3*x^3)/(8*(a^4 + 2*a^3*c*x^2 + a^2*c^2*x^4)) - (3*B*a*e^3*x)/(8*(a ^2*c^2 + c^4*x^4 + 2*a*c^3*x^2)) - (3*B*d*e^2*x^2)/(2*(a^2*c + c^3*x^4 + 2 *a*c^2*x^2)) - (3*C*d^2*e*x^2)/(2*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) - (15*C *d*e^2*x^3)/(8*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) + (C*a*e^3*x^2)/(a^2*c^2 + c^4*x^4 + 2*a*c^3*x^2) + (3*A*d^3*atan((c^(1/2)*x)/a^(1/2)))/(8*a^(5/2)*c ^(1/2)) + (3*B*e^3*atan((c^(1/2)*x)/a^(1/2)))/(8*a^(1/2)*c^(5/2)) + (C*d^3 *atan((c^(1/2)*x)/a^(1/2)))/(8*a^(3/2)*c^(3/2)) + (3*A*d*e^2*x^3)/(8*(a^3 + 2*a^2*c*x^2 + a*c^2*x^4)) + (3*B*d^2*e*x^3)/(8*(a^3 + 2*a^2*c*x^2 + a*c^ 2*x^4)) - (3*A*d*e^2*x)/(8*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) - (3*B*d^2*e*x )/(8*(a^2*c + c^3*x^4 + 2*a*c^2*x^2)) + (3*A*d*e^2*atan((c^(1/2)*x)/a^(1/2 )))/(8*a^(3/2)*c^(3/2)) + (3*B*d^2*e*atan((c^(1/2)*x)/a^(1/2)))/(8*a^(3/2) *c^(3/2)) + (9*C*d*e^2*atan((c^(1/2)*x)/a^(1/2)))/(8*a^(1/2)*c^(5/2)) -...