Integrand size = 19, antiderivative size = 216 \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\frac {5 a^2 d \left (8 c d^2-3 a e^2\right ) x \sqrt {a+c x^2}}{128 c}+\frac {5 a d \left (8 c d^2-3 a e^2\right ) x \left (a+c x^2\right )^{3/2}}{192 c}+\frac {d \left (8 c d^2-3 a e^2\right ) x \left (a+c x^2\right )^{5/2}}{48 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (16 \left (10 c d^2-a e^2\right )+77 c d e x\right ) \left (a+c x^2\right )^{7/2}}{504 c^2}+\frac {5 a^3 d \left (8 c d^2-3 a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{3/2}} \]
5/192*a*d*(-3*a*e^2+8*c*d^2)*x*(c*x^2+a)^(3/2)/c+1/48*d*(-3*a*e^2+8*c*d^2) *x*(c*x^2+a)^(5/2)/c+1/9*e*(e*x+d)^2*(c*x^2+a)^(7/2)/c+1/504*e*(77*c*d*e*x -16*a*e^2+160*c*d^2)*(c*x^2+a)^(7/2)/c^2+5/128*a^3*d*(-3*a*e^2+8*c*d^2)*ar ctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)+5/128*a^2*d*(-3*a*e^2+8*c*d^2)*x* (c*x^2+a)^(1/2)/c
Time = 0.80 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00 \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\frac {\sqrt {a+c x^2} \left (-256 a^4 e^3+a^3 c e \left (3456 d^2+945 d e x+128 e^2 x^2\right )+16 c^4 x^5 \left (84 d^3+216 d^2 e x+189 d e^2 x^2+56 e^3 x^3\right )+8 a c^3 x^3 \left (546 d^3+1296 d^2 e x+1071 d e^2 x^2+304 e^3 x^3\right )+6 a^2 c^2 x \left (924 d^3+1728 d^2 e x+1239 d e^2 x^2+320 e^3 x^3\right )\right )+315 a^3 \sqrt {c} d \left (-8 c d^2+3 a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{8064 c^2} \]
(Sqrt[a + c*x^2]*(-256*a^4*e^3 + a^3*c*e*(3456*d^2 + 945*d*e*x + 128*e^2*x ^2) + 16*c^4*x^5*(84*d^3 + 216*d^2*e*x + 189*d*e^2*x^2 + 56*e^3*x^3) + 8*a *c^3*x^3*(546*d^3 + 1296*d^2*e*x + 1071*d*e^2*x^2 + 304*e^3*x^3) + 6*a^2*c ^2*x*(924*d^3 + 1728*d^2*e*x + 1239*d*e^2*x^2 + 320*e^3*x^3)) + 315*a^3*Sq rt[c]*d*(-8*c*d^2 + 3*a*e^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(8064*c^ 2)
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {497, 676, 211, 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.
| \(\displaystyle \int \left (a+c x^2\right )^{5/2} (d+e x)^3 \, dx\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle \frac {\int (d+e x) \left (9 c d^2+11 c e x d-2 a e^2\right ) \left (c x^2+a\right )^{5/2}dx}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\frac {9}{8} d \left (8 c d^2-3 a e^2\right ) \int \left (c x^2+a\right )^{5/2}dx+\frac {2 e \left (a+c x^2\right )^{7/2} \left (10 c d^2-a e^2\right )}{7 c}+\frac {11}{8} d e^2 x \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {9}{8} d \left (8 c d^2-3 a e^2\right ) \left (\frac {5}{6} a \int \left (c x^2+a\right )^{3/2}dx+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {2 e \left (a+c x^2\right )^{7/2} \left (10 c d^2-a e^2\right )}{7 c}+\frac {11}{8} d e^2 x \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {9}{8} d \left (8 c d^2-3 a e^2\right ) \left (\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {c x^2+a}dx+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {2 e \left (a+c x^2\right )^{7/2} \left (10 c d^2-a e^2\right )}{7 c}+\frac {11}{8} d e^2 x \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {9}{8} d \left (8 c d^2-3 a e^2\right ) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {c x^2+a}}dx+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {2 e \left (a+c x^2\right )^{7/2} \left (10 c d^2-a e^2\right )}{7 c}+\frac {11}{8} d e^2 x \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {9}{8} d \left (8 c d^2-3 a e^2\right ) \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right )+\frac {2 e \left (a+c x^2\right )^{7/2} \left (10 c d^2-a e^2\right )}{7 c}+\frac {11}{8} d e^2 x \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {9}{8} d \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+c x^2\right )^{5/2}\right ) \left (8 c d^2-3 a e^2\right )+\frac {2 e \left (a+c x^2\right )^{7/2} \left (10 c d^2-a e^2\right )}{7 c}+\frac {11}{8} d e^2 x \left (a+c x^2\right )^{7/2}}{9 c}+\frac {e \left (a+c x^2\right )^{7/2} (d+e x)^2}{9 c}\) |
(e*(d + e*x)^2*(a + c*x^2)^(7/2))/(9*c) + ((2*e*(10*c*d^2 - a*e^2)*(a + c* x^2)^(7/2))/(7*c) + (11*d*e^2*x*(a + c*x^2)^(7/2))/8 + (9*d*(8*c*d^2 - 3*a *e^2)*((x*(a + c*x^2)^(5/2))/6 + (5*a*((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*Sqrt[c])) )/4))/6))/8)/(9*c)
3.6.46.3.1 Defintions of rubi rules used
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[1/(b *(n + 2*p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^p*Simp[b*c^2*(n + 2*p + 1) - a*d^2*(n - 1) + 2*b*c*d*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, n , p}, x] && If[RationalQ[n], GtQ[n, 1], SumSimplerQ[n, -2]] && NeQ[n + 2*p + 1, 0] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
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]
Time = 2.46 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.03
| method | result | size |
| default | \(d^{3} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6}\right )+e^{3} \left (\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {7}{2}}}{9 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {7}{2}}}{63 c^{2}}\right )+3 d \,e^{2} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {7}{2}}}{8 c}-\frac {a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6}\right )}{8 c}\right )+\frac {3 d^{2} e \left (c \,x^{2}+a \right )^{\frac {7}{2}}}{7 c}\) | \(223\) |
| risch | \(-\frac {\left (-896 e^{3} c^{4} x^{8}-3024 d \,e^{2} c^{4} x^{7}-2432 e^{3} c^{3} a \,x^{6}-3456 d^{2} e \,c^{4} x^{6}-8568 d \,e^{2} c^{3} a \,x^{5}-1344 d^{3} c^{4} x^{5}-1920 a^{2} c^{2} e^{3} x^{4}-10368 a \,c^{3} d^{2} e \,x^{4}-7434 a^{2} c^{2} d \,e^{2} x^{3}-4368 d^{3} c^{3} a \,x^{3}-128 e^{3} c \,a^{3} x^{2}-10368 d^{2} e \,a^{2} c^{2} x^{2}-945 d \,e^{2} c \,a^{3} x -5544 d^{3} a^{2} c^{2} x +256 e^{3} a^{4}-3456 d^{2} e c \,a^{3}\right ) \sqrt {c \,x^{2}+a}}{8064 c^{2}}-\frac {5 a^{3} d \left (3 e^{2} a -8 c \,d^{2}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{128 c^{\frac {3}{2}}}\) | \(248\) |
d^3*(1/6*x*(c*x^2+a)^(5/2)+5/6*a*(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(c^(1/2)*x+(c*x^2+a)^(1/2)))))+e^3*(1/9*x^2*(c* x^2+a)^(7/2)/c-2/63*a/c^2*(c*x^2+a)^(7/2))+3*d*e^2*(1/8*x*(c*x^2+a)^(7/2)/ c-1/8*a/c*(1/6*x*(c*x^2+a)^(5/2)+5/6*a*(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(c^(1/2)*x+(c*x^2+a)^(1/2))))))+3/7*d^2*e *(c*x^2+a)^(7/2)/c
Time = 0.29 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.40 \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\left [\frac {315 \, {\left (8 \, a^{3} c d^{3} - 3 \, a^{4} d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (896 \, c^{4} e^{3} x^{8} + 3024 \, c^{4} d e^{2} x^{7} + 3456 \, a^{3} c d^{2} e - 256 \, a^{4} e^{3} + 128 \, {\left (27 \, c^{4} d^{2} e + 19 \, a c^{3} e^{3}\right )} x^{6} + 168 \, {\left (8 \, c^{4} d^{3} + 51 \, a c^{3} d e^{2}\right )} x^{5} + 384 \, {\left (27 \, a c^{3} d^{2} e + 5 \, a^{2} c^{2} e^{3}\right )} x^{4} + 42 \, {\left (104 \, a c^{3} d^{3} + 177 \, a^{2} c^{2} d e^{2}\right )} x^{3} + 128 \, {\left (81 \, a^{2} c^{2} d^{2} e + a^{3} c e^{3}\right )} x^{2} + 63 \, {\left (88 \, a^{2} c^{2} d^{3} + 15 \, a^{3} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{16128 \, c^{2}}, -\frac {315 \, {\left (8 \, a^{3} c d^{3} - 3 \, a^{4} d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (896 \, c^{4} e^{3} x^{8} + 3024 \, c^{4} d e^{2} x^{7} + 3456 \, a^{3} c d^{2} e - 256 \, a^{4} e^{3} + 128 \, {\left (27 \, c^{4} d^{2} e + 19 \, a c^{3} e^{3}\right )} x^{6} + 168 \, {\left (8 \, c^{4} d^{3} + 51 \, a c^{3} d e^{2}\right )} x^{5} + 384 \, {\left (27 \, a c^{3} d^{2} e + 5 \, a^{2} c^{2} e^{3}\right )} x^{4} + 42 \, {\left (104 \, a c^{3} d^{3} + 177 \, a^{2} c^{2} d e^{2}\right )} x^{3} + 128 \, {\left (81 \, a^{2} c^{2} d^{2} e + a^{3} c e^{3}\right )} x^{2} + 63 \, {\left (88 \, a^{2} c^{2} d^{3} + 15 \, a^{3} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{8064 \, c^{2}}\right ] \]
[1/16128*(315*(8*a^3*c*d^3 - 3*a^4*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c* x^2 + a)*sqrt(c)*x - a) + 2*(896*c^4*e^3*x^8 + 3024*c^4*d*e^2*x^7 + 3456*a ^3*c*d^2*e - 256*a^4*e^3 + 128*(27*c^4*d^2*e + 19*a*c^3*e^3)*x^6 + 168*(8* c^4*d^3 + 51*a*c^3*d*e^2)*x^5 + 384*(27*a*c^3*d^2*e + 5*a^2*c^2*e^3)*x^4 + 42*(104*a*c^3*d^3 + 177*a^2*c^2*d*e^2)*x^3 + 128*(81*a^2*c^2*d^2*e + a^3* c*e^3)*x^2 + 63*(88*a^2*c^2*d^3 + 15*a^3*c*d*e^2)*x)*sqrt(c*x^2 + a))/c^2, -1/8064*(315*(8*a^3*c*d^3 - 3*a^4*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt( c*x^2 + a)) - (896*c^4*e^3*x^8 + 3024*c^4*d*e^2*x^7 + 3456*a^3*c*d^2*e - 2 56*a^4*e^3 + 128*(27*c^4*d^2*e + 19*a*c^3*e^3)*x^6 + 168*(8*c^4*d^3 + 51*a *c^3*d*e^2)*x^5 + 384*(27*a*c^3*d^2*e + 5*a^2*c^2*e^3)*x^4 + 42*(104*a*c^3 *d^3 + 177*a^2*c^2*d*e^2)*x^3 + 128*(81*a^2*c^2*d^2*e + a^3*c*e^3)*x^2 + 6 3*(88*a^2*c^2*d^3 + 15*a^3*c*d*e^2)*x)*sqrt(c*x^2 + a))/c^2]
Leaf count of result is larger than twice the leaf count of optimal. 617 vs. \(2 (206) = 412\).
Time = 0.65 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.86 \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\begin {cases} \sqrt {a + c x^{2}} \cdot \left (\frac {3 c^{2} d e^{2} x^{7}}{8} + \frac {c^{2} e^{3} x^{8}}{9} + \frac {x^{6} \cdot \left (\frac {19 a c^{2} e^{3}}{9} + 3 c^{3} d^{2} e\right )}{7 c} + \frac {x^{5} \cdot \left (\frac {51 a c^{2} d e^{2}}{8} + c^{3} d^{3}\right )}{6 c} + \frac {x^{4} \cdot \left (3 a^{2} c e^{3} + 9 a c^{2} d^{2} e - \frac {6 a \left (\frac {19 a c^{2} e^{3}}{9} + 3 c^{3} d^{2} e\right )}{7 c}\right )}{5 c} + \frac {x^{3} \cdot \left (9 a^{2} c d e^{2} + 3 a c^{2} d^{3} - \frac {5 a \left (\frac {51 a c^{2} d e^{2}}{8} + c^{3} d^{3}\right )}{6 c}\right )}{4 c} + \frac {x^{2} \left (a^{3} e^{3} + 9 a^{2} c d^{2} e - \frac {4 a \left (3 a^{2} c e^{3} + 9 a c^{2} d^{2} e - \frac {6 a \left (\frac {19 a c^{2} e^{3}}{9} + 3 c^{3} d^{2} e\right )}{7 c}\right )}{5 c}\right )}{3 c} + \frac {x \left (3 a^{3} d e^{2} + 3 a^{2} c d^{3} - \frac {3 a \left (9 a^{2} c d e^{2} + 3 a c^{2} d^{3} - \frac {5 a \left (\frac {51 a c^{2} d e^{2}}{8} + c^{3} d^{3}\right )}{6 c}\right )}{4 c}\right )}{2 c} + \frac {3 a^{3} d^{2} e - \frac {2 a \left (a^{3} e^{3} + 9 a^{2} c d^{2} e - \frac {4 a \left (3 a^{2} c e^{3} + 9 a c^{2} d^{2} e - \frac {6 a \left (\frac {19 a c^{2} e^{3}}{9} + 3 c^{3} d^{2} e\right )}{7 c}\right )}{5 c}\right )}{3 c}}{c}\right ) + \left (a^{3} d^{3} - \frac {a \left (3 a^{3} d e^{2} + 3 a^{2} c d^{3} - \frac {3 a \left (9 a^{2} c d e^{2} + 3 a c^{2} d^{3} - \frac {5 a \left (\frac {51 a c^{2} d e^{2}}{8} + c^{3} d^{3}\right )}{6 c}\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 {5}{2}} \left (\begin {cases} d^{3} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{4}}{4 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + c*x**2)*(3*c**2*d*e**2*x**7/8 + c**2*e**3*x**8/9 + x** 6*(19*a*c**2*e**3/9 + 3*c**3*d**2*e)/(7*c) + x**5*(51*a*c**2*d*e**2/8 + c* *3*d**3)/(6*c) + x**4*(3*a**2*c*e**3 + 9*a*c**2*d**2*e - 6*a*(19*a*c**2*e* *3/9 + 3*c**3*d**2*e)/(7*c))/(5*c) + x**3*(9*a**2*c*d*e**2 + 3*a*c**2*d**3 - 5*a*(51*a*c**2*d*e**2/8 + c**3*d**3)/(6*c))/(4*c) + x**2*(a**3*e**3 + 9 *a**2*c*d**2*e - 4*a*(3*a**2*c*e**3 + 9*a*c**2*d**2*e - 6*a*(19*a*c**2*e** 3/9 + 3*c**3*d**2*e)/(7*c))/(5*c))/(3*c) + x*(3*a**3*d*e**2 + 3*a**2*c*d** 3 - 3*a*(9*a**2*c*d*e**2 + 3*a*c**2*d**3 - 5*a*(51*a*c**2*d*e**2/8 + c**3* d**3)/(6*c))/(4*c))/(2*c) + (3*a**3*d**2*e - 2*a*(a**3*e**3 + 9*a**2*c*d** 2*e - 4*a*(3*a**2*c*e**3 + 9*a*c**2*d**2*e - 6*a*(19*a*c**2*e**3/9 + 3*c** 3*d**2*e)/(7*c))/(5*c))/(3*c))/c) + (a**3*d**3 - a*(3*a**3*d*e**2 + 3*a**2 *c*d**3 - 3*a*(9*a**2*c*d*e**2 + 3*a*c**2*d**3 - 5*a*(51*a*c**2*d*e**2/8 + c**3*d**3)/(6*c))/(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**(5/2)*Piecewise((d**3*x, Eq(e, 0)), ((d + e*x)**4/(4*e), True)), True))
Time = 0.21 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.06 \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} e^{3} x^{2}}{9 \, c} + \frac {1}{6} \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{3} x + \frac {5}{24} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a d^{3} x + \frac {5}{16} \, \sqrt {c x^{2} + a} a^{2} d^{3} x + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} d e^{2} x}{8 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} a d e^{2} x}{16 \, c} - \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2} d e^{2} x}{64 \, c} - \frac {15 \, \sqrt {c x^{2} + a} a^{3} d e^{2} x}{128 \, c} + \frac {5 \, a^{3} d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {c}} - \frac {15 \, a^{4} d e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{128 \, c^{\frac {3}{2}}} + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} d^{2} e}{7 \, c} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {7}{2}} a e^{3}}{63 \, c^{2}} \]
1/9*(c*x^2 + a)^(7/2)*e^3*x^2/c + 1/6*(c*x^2 + a)^(5/2)*d^3*x + 5/24*(c*x^ 2 + a)^(3/2)*a*d^3*x + 5/16*sqrt(c*x^2 + a)*a^2*d^3*x + 3/8*(c*x^2 + a)^(7 /2)*d*e^2*x/c - 1/16*(c*x^2 + a)^(5/2)*a*d*e^2*x/c - 5/64*(c*x^2 + a)^(3/2 )*a^2*d*e^2*x/c - 15/128*sqrt(c*x^2 + a)*a^3*d*e^2*x/c + 5/16*a^3*d^3*arcs inh(c*x/sqrt(a*c))/sqrt(c) - 15/128*a^4*d*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/ 2) + 3/7*(c*x^2 + a)^(7/2)*d^2*e/c - 2/63*(c*x^2 + a)^(7/2)*a*e^3/c^2
Time = 0.30 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.32 \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\frac {1}{8064} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left ({\left (2 \, {\left (7 \, {\left (8 \, c^{2} e^{3} x + 27 \, c^{2} d e^{2}\right )} x + \frac {8 \, {\left (27 \, c^{9} d^{2} e + 19 \, a c^{8} e^{3}\right )}}{c^{7}}\right )} x + \frac {21 \, {\left (8 \, c^{9} d^{3} + 51 \, a c^{8} d e^{2}\right )}}{c^{7}}\right )} x + \frac {48 \, {\left (27 \, a c^{8} d^{2} e + 5 \, a^{2} c^{7} e^{3}\right )}}{c^{7}}\right )} x + \frac {21 \, {\left (104 \, a c^{8} d^{3} + 177 \, a^{2} c^{7} d e^{2}\right )}}{c^{7}}\right )} x + \frac {64 \, {\left (81 \, a^{2} c^{7} d^{2} e + a^{3} c^{6} e^{3}\right )}}{c^{7}}\right )} x + \frac {63 \, {\left (88 \, a^{2} c^{7} d^{3} + 15 \, a^{3} c^{6} d e^{2}\right )}}{c^{7}}\right )} x + \frac {128 \, {\left (27 \, a^{3} c^{6} d^{2} e - 2 \, a^{4} c^{5} e^{3}\right )}}{c^{7}}\right )} - \frac {5 \, {\left (8 \, a^{3} c d^{3} - 3 \, a^{4} d e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{128 \, c^{\frac {3}{2}}} \]
1/8064*sqrt(c*x^2 + a)*((2*((4*((2*(7*(8*c^2*e^3*x + 27*c^2*d*e^2)*x + 8*( 27*c^9*d^2*e + 19*a*c^8*e^3)/c^7)*x + 21*(8*c^9*d^3 + 51*a*c^8*d*e^2)/c^7) *x + 48*(27*a*c^8*d^2*e + 5*a^2*c^7*e^3)/c^7)*x + 21*(104*a*c^8*d^3 + 177* a^2*c^7*d*e^2)/c^7)*x + 64*(81*a^2*c^7*d^2*e + a^3*c^6*e^3)/c^7)*x + 63*(8 8*a^2*c^7*d^3 + 15*a^3*c^6*d*e^2)/c^7)*x + 128*(27*a^3*c^6*d^2*e - 2*a^4*c ^5*e^3)/c^7) - 5/128*(8*a^3*c*d^3 - 3*a^4*d*e^2)*log(abs(-sqrt(c)*x + sqrt (c*x^2 + a)))/c^(3/2)
Timed out. \[ \int (d+e x)^3 \left (a+c x^2\right )^{5/2} \, dx=\int {\left (c\,x^2+a\right )}^{5/2}\,{\left (d+e\,x\right )}^3 \,d x \]