Integrand size = 37, antiderivative size = 222 \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {63 e^2 \left (c d^2-a e^2\right )^2 \sqrt {d+e x}}{4 c^5 d^5}+\frac {21 e^2 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{4 c^4 d^4}+\frac {63 e^2 (d+e x)^{5/2}}{20 c^3 d^3}-\frac {9 e (d+e x)^{7/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}-\frac {63 e^2 \left (c d^2-a e^2\right )^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{11/2} d^{11/2}} \] Output:
63/4*e^2*(-a*e^2+c*d^2)^2*(e*x+d)^(1/2)/c^5/d^5+21/4*e^2*(-a*e^2+c*d^2)*(e *x+d)^(3/2)/c^4/d^4+63/20*e^2*(e*x+d)^(5/2)/c^3/d^3-9/4*e*(e*x+d)^(7/2)/c^ 2/d^2/(c*d*x+a*e)-1/2*(e*x+d)^(9/2)/c/d/(c*d*x+a*e)^2-63/4*e^2*(-a*e^2+c*d ^2)^(5/2)*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/c^(1 1/2)/d^(11/2)
Time = 0.84 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.13 \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (-315 a^4 e^8+105 a^3 c d e^6 (7 d-5 e x)-21 a^2 c^2 d^2 e^4 \left (23 d^2-59 d e x+8 e^2 x^2\right )+3 a c^3 d^3 e^2 \left (15 d^3-277 d^2 e x+136 d e^2 x^2+8 e^3 x^3\right )+c^4 d^4 \left (10 d^4+85 d^3 e x-288 d^2 e^2 x^2-56 d e^3 x^3-8 e^4 x^4\right )\right )}{20 c^5 d^5 (a e+c d x)^2}-\frac {63 e^2 \left (-c d^2+a e^2\right )^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 c^{11/2} d^{11/2}} \] Input:
Integrate[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
Output:
-1/20*(Sqrt[d + e*x]*(-315*a^4*e^8 + 105*a^3*c*d*e^6*(7*d - 5*e*x) - 21*a^ 2*c^2*d^2*e^4*(23*d^2 - 59*d*e*x + 8*e^2*x^2) + 3*a*c^3*d^3*e^2*(15*d^3 - 277*d^2*e*x + 136*d*e^2*x^2 + 8*e^3*x^3) + c^4*d^4*(10*d^4 + 85*d^3*e*x - 288*d^2*e^2*x^2 - 56*d*e^3*x^3 - 8*e^4*x^4)))/(c^5*d^5*(a*e + c*d*x)^2) - (63*e^2*(-(c*d^2) + a*e^2)^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sq rt[-(c*d^2) + a*e^2]])/(4*c^(11/2)*d^(11/2))
Time = 0.52 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1121, 51, 51, 60, 60, 60, 73, 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.
| \(\displaystyle \int \frac {(d+e x)^{15/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1121 |
| \(\displaystyle \int \frac {(d+e x)^{9/2}}{(a e+c d x)^3}dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {9 e \int \frac {(d+e x)^{7/2}}{(a e+c d x)^2}dx}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \int \frac {(d+e x)^{5/2}}{a e+c d x}dx}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {(d+e x)^{3/2}}{a e+c d x}dx}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {\sqrt {d+e x}}{a e+c d x}dx}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{d}+\frac {2 \sqrt {d+e x}}{c d}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{d e}+\frac {2 \sqrt {d+e x}}{c d}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {9 e \left (\frac {7 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \sqrt {d+e x}}{c d}-\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} d^{3/2} \sqrt {c d^2-a e^2}}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\right )}{d}+\frac {2 (d+e x)^{5/2}}{5 c d}\right )}{2 c d}-\frac {(d+e x)^{7/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{9/2}}{2 c d (a e+c d x)^2}\) |
Input:
Int[(d + e*x)^(15/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
Output:
-1/2*(d + e*x)^(9/2)/(c*d*(a*e + c*d*x)^2) + (9*e*(-((d + e*x)^(7/2)/(c*d* (a*e + c*d*x))) + (7*e*((2*(d + e*x)^(5/2))/(5*c*d) + ((d^2 - (a*e^2)/c)*( (2*(d + e*x)^(3/2))/(3*c*d) + ((d^2 - (a*e^2)/c)*((2*Sqrt[d + e*x])/(c*d) - (2*(d^2 - (a*e^2)/c)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(Sqrt[c]*d^(3/2)*Sqrt[c*d^2 - a*e^2])))/d))/d))/(2*c*d)))/(4*c* d)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
Time = 16.65 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.06
| method | result | size |
| risch | \(\frac {2 e^{2} \left (x^{2} c^{2} d^{2} e^{2}-5 x a c d \,e^{3}+7 x \,c^{2} d^{3} e +30 a^{2} e^{4}-65 a c \,d^{2} e^{2}+36 c^{2} d^{4}\right ) \sqrt {e x +d}}{5 d^{5} c^{5}}-\frac {\left (2 e^{6} a^{3}-6 d^{2} e^{4} a^{2} c +6 d^{4} e^{2} a \,c^{2}-2 d^{6} c^{3}\right ) e^{2} \left (\frac {-\frac {17 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (-\frac {15 a \,e^{2}}{8}+\frac {15 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {63 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{d^{5} c^{5}}\) | \(235\) |
| pseudoelliptic | \(-\frac {63 \left (e^{2} \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )-\sqrt {e x +d}\, \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (-\frac {2 \left (-\frac {4}{5} e^{4} x^{4}-\frac {28}{5} d \,e^{3} x^{3}-\frac {144}{5} d^{2} e^{2} x^{2}+\frac {17}{2} d^{3} e x +d^{4}\right ) d^{4} c^{4}}{63}-\frac {e^{2} \left (\frac {8}{15} e^{3} x^{3}+\frac {136}{15} d \,e^{2} x^{2}-\frac {277}{15} d^{2} e x +d^{3}\right ) a \,d^{3} c^{3}}{7}+\frac {23 e^{4} \left (\frac {8}{23} e^{2} x^{2}-\frac {59}{23} d e x +d^{2}\right ) a^{2} d^{2} c^{2}}{15}-\frac {7 e^{6} \left (-\frac {5 e x}{7}+d \right ) a^{3} d c}{3}+a^{4} e^{8}\right )\right )}{4 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, d^{5} c^{5} \left (c d x +a e \right )^{2}}\) | \(261\) |
| derivativedivides | \(2 e^{2} \left (\frac {\frac {c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}+c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} e^{4} \sqrt {e x +d}-12 a c \,d^{2} e^{2} \sqrt {e x +d}+6 c^{2} d^{4} \sqrt {e x +d}}{c^{5} d^{5}}-\frac {\frac {\left (-\frac {17}{8} d \,e^{6} c \,a^{3}+\frac {51}{8} d^{3} e^{4} a^{2} c^{2}-\frac {51}{8} d^{5} e^{2} a \,c^{3}+\frac {17}{8} d^{7} c^{4}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {15}{8} a^{4} e^{8}+\frac {15}{2} a^{3} c \,d^{2} e^{6}-\frac {45}{4} a^{2} c^{2} d^{4} e^{4}+\frac {15}{2} a \,c^{3} d^{6} e^{2}-\frac {15}{8} c^{4} d^{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {63 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{d^{5} c^{5}}\right )\) | \(342\) |
| default | \(2 e^{2} \left (\frac {\frac {c^{2} d^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-a c d \,e^{2} \left (e x +d \right )^{\frac {3}{2}}+c^{2} d^{3} \left (e x +d \right )^{\frac {3}{2}}+6 a^{2} e^{4} \sqrt {e x +d}-12 a c \,d^{2} e^{2} \sqrt {e x +d}+6 c^{2} d^{4} \sqrt {e x +d}}{c^{5} d^{5}}-\frac {\frac {\left (-\frac {17}{8} d \,e^{6} c \,a^{3}+\frac {51}{8} d^{3} e^{4} a^{2} c^{2}-\frac {51}{8} d^{5} e^{2} a \,c^{3}+\frac {17}{8} d^{7} c^{4}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {15}{8} a^{4} e^{8}+\frac {15}{2} a^{3} c \,d^{2} e^{6}-\frac {45}{4} a^{2} c^{2} d^{4} e^{4}+\frac {15}{2} a \,c^{3} d^{6} e^{2}-\frac {15}{8} c^{4} d^{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {63 \left (e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} a \,c^{2}-d^{6} c^{3}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{d^{5} c^{5}}\right )\) | \(342\) |
Input:
int((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^3,x,method=_RETURNVER BOSE)
Output:
2/5*e^2*(c^2*d^2*e^2*x^2-5*a*c*d*e^3*x+7*c^2*d^3*e*x+30*a^2*e^4-65*a*c*d^2 *e^2+36*c^2*d^4)*(e*x+d)^(1/2)/d^5/c^5-1/d^5/c^5*(2*a^3*e^6-6*a^2*c*d^2*e^ 4+6*a*c^2*d^4*e^2-2*c^3*d^6)*e^2*((-17/8*c*d*(e*x+d)^(3/2)+(-15/8*a*e^2+15 /8*c*d^2)*(e*x+d)^(1/2))/(c*d*(e*x+d)+a*e^2-c*d^2)^2+63/8/(c*d*(a*e^2-c*d^ 2))^(1/2)*arctan(c*d*(e*x+d)^(1/2)/(c*d*(a*e^2-c*d^2))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (186) = 372\).
Time = 0.15 (sec) , antiderivative size = 858, normalized size of antiderivative = 3.86 \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx =\text {Too large to display} \] Input:
integrate((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= "fricas")
Output:
[1/40*(315*(a^2*c^2*d^4*e^4 - 2*a^3*c*d^2*e^6 + a^4*e^8 + (c^4*d^6*e^2 - 2 *a*c^3*d^4*e^4 + a^2*c^2*d^2*e^6)*x^2 + 2*(a*c^3*d^5*e^3 - 2*a^2*c^2*d^3*e ^5 + a^3*c*d*e^7)*x)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(8*c^4*d^4*e^4*x^4 - 10*c^4*d^8 - 45*a*c^3*d^6*e^2 + 483*a^2*c^2*d^4*e^4 - 735*a^3*c*d^2*e^6 + 315*a^4*e^8 + 8*(7*c^4*d^5*e^3 - 3*a*c^3*d^3*e^5)*x ^3 + 24*(12*c^4*d^6*e^2 - 17*a*c^3*d^4*e^4 + 7*a^2*c^2*d^2*e^6)*x^2 - (85* c^4*d^7*e - 831*a*c^3*d^5*e^3 + 1239*a^2*c^2*d^3*e^5 - 525*a^3*c*d*e^7)*x) *sqrt(e*x + d))/(c^7*d^7*x^2 + 2*a*c^6*d^6*e*x + a^2*c^5*d^5*e^2), -1/20*( 315*(a^2*c^2*d^4*e^4 - 2*a^3*c*d^2*e^6 + a^4*e^8 + (c^4*d^6*e^2 - 2*a*c^3* d^4*e^4 + a^2*c^2*d^2*e^6)*x^2 + 2*(a*c^3*d^5*e^3 - 2*a^2*c^2*d^3*e^5 + a^ 3*c*d*e^7)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c*d*sqrt( -(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (8*c^4*d^4*e^4*x^4 - 10*c^4*d^8 - 45*a*c^3*d^6*e^2 + 483*a^2*c^2*d^4*e^4 - 735*a^3*c*d^2*e^6 + 315*a^4*e^ 8 + 8*(7*c^4*d^5*e^3 - 3*a*c^3*d^3*e^5)*x^3 + 24*(12*c^4*d^6*e^2 - 17*a*c^ 3*d^4*e^4 + 7*a^2*c^2*d^2*e^6)*x^2 - (85*c^4*d^7*e - 831*a*c^3*d^5*e^3 + 1 239*a^2*c^2*d^3*e^5 - 525*a^3*c*d*e^7)*x)*sqrt(e*x + d))/(c^7*d^7*x^2 + 2* a*c^6*d^6*e*x + a^2*c^5*d^5*e^2)]
Timed out. \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x+d)**(15/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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(a*e^2-c*d^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (186) = 372\).
Time = 0.19 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.88 \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {63 \, {\left (c^{3} d^{6} e^{2} - 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} - a^{3} e^{8}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} c^{5} d^{5}} - \frac {17 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{4} d^{7} e^{2} - 15 \, \sqrt {e x + d} c^{4} d^{8} e^{2} - 51 \, {\left (e x + d\right )}^{\frac {3}{2}} a c^{3} d^{5} e^{4} + 60 \, \sqrt {e x + d} a c^{3} d^{6} e^{4} + 51 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} c^{2} d^{3} e^{6} - 90 \, \sqrt {e x + d} a^{2} c^{2} d^{4} e^{6} - 17 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{3} c d e^{8} + 60 \, \sqrt {e x + d} a^{3} c d^{2} e^{8} - 15 \, \sqrt {e x + d} a^{4} e^{10}}{4 \, {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{5} d^{5}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {5}{2}} c^{12} d^{12} e^{2} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{12} d^{13} e^{2} + 30 \, \sqrt {e x + d} c^{12} d^{14} e^{2} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a c^{11} d^{11} e^{4} - 60 \, \sqrt {e x + d} a c^{11} d^{12} e^{4} + 30 \, \sqrt {e x + d} a^{2} c^{10} d^{10} e^{6}\right )}}{5 \, c^{15} d^{15}} \] Input:
integrate((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= "giac")
Output:
63/4*(c^3*d^6*e^2 - 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 - a^3*e^8)*arctan(sq rt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/(sqrt(-c^2*d^3 + a*c*d*e^2)*c^ 5*d^5) - 1/4*(17*(e*x + d)^(3/2)*c^4*d^7*e^2 - 15*sqrt(e*x + d)*c^4*d^8*e^ 2 - 51*(e*x + d)^(3/2)*a*c^3*d^5*e^4 + 60*sqrt(e*x + d)*a*c^3*d^6*e^4 + 51 *(e*x + d)^(3/2)*a^2*c^2*d^3*e^6 - 90*sqrt(e*x + d)*a^2*c^2*d^4*e^6 - 17*( e*x + d)^(3/2)*a^3*c*d*e^8 + 60*sqrt(e*x + d)*a^3*c*d^2*e^8 - 15*sqrt(e*x + d)*a^4*e^10)/(((e*x + d)*c*d - c*d^2 + a*e^2)^2*c^5*d^5) + 2/5*((e*x + d )^(5/2)*c^12*d^12*e^2 + 5*(e*x + d)^(3/2)*c^12*d^13*e^2 + 30*sqrt(e*x + d) *c^12*d^14*e^2 - 5*(e*x + d)^(3/2)*a*c^11*d^11*e^4 - 60*sqrt(e*x + d)*a*c^ 11*d^12*e^4 + 30*sqrt(e*x + d)*a^2*c^10*d^10*e^6)/(c^15*d^15)
Time = 5.46 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.94 \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {\sqrt {d+e\,x}\,\left (\frac {15\,a^4\,e^{10}}{4}-15\,a^3\,c\,d^2\,e^8+\frac {45\,a^2\,c^2\,d^4\,e^6}{2}-15\,a\,c^3\,d^6\,e^4+\frac {15\,c^4\,d^8\,e^2}{4}\right )-{\left (d+e\,x\right )}^{3/2}\,\left (-\frac {17\,a^3\,c\,d\,e^8}{4}+\frac {51\,a^2\,c^2\,d^3\,e^6}{4}-\frac {51\,a\,c^3\,d^5\,e^4}{4}+\frac {17\,c^4\,d^7\,e^2}{4}\right )}{c^7\,d^9-\left (2\,c^7\,d^8-2\,a\,c^6\,d^6\,e^2\right )\,\left (d+e\,x\right )+c^7\,d^7\,{\left (d+e\,x\right )}^2-2\,a\,c^6\,d^7\,e^2+a^2\,c^5\,d^5\,e^4}+\left (\frac {2\,e^2\,{\left (3\,c^3\,d^4-3\,a\,c^2\,d^2\,e^2\right )}^2}{c^9\,d^9}-\frac {6\,e^2\,{\left (a\,e^2-c\,d^2\right )}^2}{c^5\,d^5}\right )\,\sqrt {d+e\,x}+\frac {2\,e^2\,{\left (d+e\,x\right )}^{5/2}}{5\,c^3\,d^3}-\frac {63\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,{\left (a\,e^2-c\,d^2\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^8-3\,a^2\,c\,d^2\,e^6+3\,a\,c^2\,d^4\,e^4-c^3\,d^6\,e^2}\right )\,{\left (a\,e^2-c\,d^2\right )}^{5/2}}{4\,c^{11/2}\,d^{11/2}}+\frac {2\,e^2\,\left (3\,c^3\,d^4-3\,a\,c^2\,d^2\,e^2\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,c^6\,d^6} \] Input:
int((d + e*x)^(15/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)
Output:
((d + e*x)^(1/2)*((15*a^4*e^10)/4 + (15*c^4*d^8*e^2)/4 - 15*a*c^3*d^6*e^4 - 15*a^3*c*d^2*e^8 + (45*a^2*c^2*d^4*e^6)/2) - (d + e*x)^(3/2)*((17*c^4*d^ 7*e^2)/4 - (51*a*c^3*d^5*e^4)/4 + (51*a^2*c^2*d^3*e^6)/4 - (17*a^3*c*d*e^8 )/4))/(c^7*d^9 - (2*c^7*d^8 - 2*a*c^6*d^6*e^2)*(d + e*x) + c^7*d^7*(d + e* x)^2 - 2*a*c^6*d^7*e^2 + a^2*c^5*d^5*e^4) + ((2*e^2*(3*c^3*d^4 - 3*a*c^2*d ^2*e^2)^2)/(c^9*d^9) - (6*e^2*(a*e^2 - c*d^2)^2)/(c^5*d^5))*(d + e*x)^(1/2 ) + (2*e^2*(d + e*x)^(5/2))/(5*c^3*d^3) - (63*e^2*atan((c^(1/2)*d^(1/2)*e^ 2*(a*e^2 - c*d^2)^(5/2)*(d + e*x)^(1/2))/(a^3*e^8 - c^3*d^6*e^2 + 3*a*c^2* d^4*e^4 - 3*a^2*c*d^2*e^6))*(a*e^2 - c*d^2)^(5/2))/(4*c^(11/2)*d^(11/2)) + (2*e^2*(3*c^3*d^4 - 3*a*c^2*d^2*e^2)*(d + e*x)^(3/2))/(3*c^6*d^6)
Time = 0.26 (sec) , antiderivative size = 895, normalized size of antiderivative = 4.03 \[ \int \frac {(d+e x)^{15/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx =\text {Too large to display} \] Input:
int((e*x+d)^(15/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)
Output:
( - 315*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sq rt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**4*e**8 + 630*sqrt(d)*sqrt(c)*sqrt (a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**3*c*d**2*e**6 - 630*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*ata n((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**3*c*d*e* *7*x - 315*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/ (sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**2*c**2*d**4*e**4 + 1260*sqrt(d )*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)* sqrt(a*e**2 - c*d**2)))*a**2*c**2*d**3*e**5*x - 315*sqrt(d)*sqrt(c)*sqrt(a *e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c* d**2)))*a**2*c**2*d**2*e**6*x**2 - 630*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d** 2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a*c** 3*d**5*e**3*x + 630*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e *x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a*c**3*d**4*e**4*x**2 - 315*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d )*sqrt(c)*sqrt(a*e**2 - c*d**2)))*c**4*d**6*e**2*x**2 + 315*sqrt(d + e*x)* a**4*c*d*e**8 - 735*sqrt(d + e*x)*a**3*c**2*d**3*e**6 + 525*sqrt(d + e*x)* a**3*c**2*d**2*e**7*x + 483*sqrt(d + e*x)*a**2*c**3*d**5*e**4 - 1239*sqrt( d + e*x)*a**2*c**3*d**4*e**5*x + 168*sqrt(d + e*x)*a**2*c**3*d**3*e**6*x** 2 - 45*sqrt(d + e*x)*a*c**4*d**7*e**2 + 831*sqrt(d + e*x)*a*c**4*d**6*e...