Integrand size = 19, antiderivative size = 255 \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\frac {3 a \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \sqrt {a+c x^2}}{128 c^2}+\frac {\left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) x \left (a+c x^2\right )^{3/2}}{64 c^2}+\frac {11 d e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{56 c}+\frac {e (d+e x)^3 \left (a+c x^2\right )^{5/2}}{8 c}+\frac {e \left (4 d \left (67 c d^2-32 a e^2\right )+5 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (a+c x^2\right )^{5/2}}{560 c^2}+\frac {3 a^2 \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{128 c^{5/2}} \]
1/64*(a^2*e^4-16*a*c*d^2*e^2+16*c^2*d^4)*x*(c*x^2+a)^(3/2)/c^2+11/56*d*e*( e*x+d)^2*(c*x^2+a)^(5/2)/c+1/8*e*(e*x+d)^3*(c*x^2+a)^(5/2)/c+1/560*e*(4*d* (-32*a*e^2+67*c*d^2)+5*e*(-7*a*e^2+26*c*d^2)*x)*(c*x^2+a)^(5/2)/c^2+3/128* a^2*(a^2*e^4-16*a*c*d^2*e^2+16*c^2*d^4)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2)) /c^(5/2)+3/128*a*(a^2*e^4-16*a*c*d^2*e^2+16*c^2*d^4)*x*(c*x^2+a)^(1/2)/c^2
Time = 0.76 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.90 \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\frac {\sqrt {c} \sqrt {a+c x^2} \left (-a^3 e^3 (1024 d+105 e x)+2 a^2 c e \left (1792 d^3+840 d^2 e x+256 d e^2 x^2+35 e^3 x^3\right )+16 c^3 x^3 \left (70 d^4+224 d^3 e x+280 d^2 e^2 x^2+160 d e^3 x^3+35 e^4 x^4\right )+8 a c^2 x \left (350 d^4+896 d^3 e x+980 d^2 e^2 x^2+512 d e^3 x^3+105 e^4 x^4\right )\right )-105 a^2 \left (16 c^2 d^4-16 a c d^2 e^2+a^2 e^4\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{4480 c^{5/2}} \]
(Sqrt[c]*Sqrt[a + c*x^2]*(-(a^3*e^3*(1024*d + 105*e*x)) + 2*a^2*c*e*(1792* d^3 + 840*d^2*e*x + 256*d*e^2*x^2 + 35*e^3*x^3) + 16*c^3*x^3*(70*d^4 + 224 *d^3*e*x + 280*d^2*e^2*x^2 + 160*d*e^3*x^3 + 35*e^4*x^4) + 8*a*c^2*x*(350* d^4 + 896*d^3*e*x + 980*d^2*e^2*x^2 + 512*d*e^3*x^3 + 105*e^4*x^4)) - 105* a^2*(16*c^2*d^4 - 16*a*c*d^2*e^2 + a^2*e^4)*Log[-(Sqrt[c]*x) + Sqrt[a + c* x^2]])/(4480*c^(5/2))
Time = 0.38 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {497, 687, 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.
| \(\displaystyle \int \left (a+c x^2\right )^{3/2} (d+e x)^4 \, dx\) |
| \(\Big \downarrow \) 497 |
| \(\displaystyle \frac {\int (d+e x)^2 \left (8 c d^2+11 c e x d-3 a e^2\right ) \left (c x^2+a\right )^{3/2}dx}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 687 |
| \(\displaystyle \frac {\frac {\int c (d+e x) \left (d \left (56 c d^2-43 a e^2\right )+3 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (c x^2+a\right )^{3/2}dx}{7 c}+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{7} \int (d+e x) \left (d \left (56 c d^2-43 a e^2\right )+3 e \left (26 c d^2-7 a e^2\right ) x\right ) \left (c x^2+a\right )^{3/2}dx+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 676 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {7 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \int \left (c x^2+a\right )^{3/2}dx}{2 c}+\frac {2 d e \left (a+c x^2\right )^{5/2} \left (67 c d^2-32 a e^2\right )}{5 c}+\frac {e^2 x \left (a+c x^2\right )^{5/2} \left (26 c d^2-7 a e^2\right )}{2 c}\right )+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {7 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \left (\frac {3}{4} a \int \sqrt {c x^2+a}dx+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )}{2 c}+\frac {2 d e \left (a+c x^2\right )^{5/2} \left (67 c d^2-32 a e^2\right )}{5 c}+\frac {e^2 x \left (a+c x^2\right )^{5/2} \left (26 c d^2-7 a e^2\right )}{2 c}\right )+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {7 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {c x^2+a}}dx+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )}{2 c}+\frac {2 d e \left (a+c x^2\right )^{5/2} \left (67 c d^2-32 a e^2\right )}{5 c}+\frac {e^2 x \left (a+c x^2\right )^{5/2} \left (26 c d^2-7 a e^2\right )}{2 c}\right )+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {7 \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right )}{2 c}+\frac {2 d e \left (a+c x^2\right )^{5/2} \left (67 c d^2-32 a e^2\right )}{5 c}+\frac {e^2 x \left (a+c x^2\right )^{5/2} \left (26 c d^2-7 a e^2\right )}{2 c}\right )+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{7} \left (\frac {7 \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c}}+\frac {1}{2} x \sqrt {a+c x^2}\right )+\frac {1}{4} x \left (a+c x^2\right )^{3/2}\right ) \left (a^2 e^4-16 a c d^2 e^2+16 c^2 d^4\right )}{2 c}+\frac {2 d e \left (a+c x^2\right )^{5/2} \left (67 c d^2-32 a e^2\right )}{5 c}+\frac {e^2 x \left (a+c x^2\right )^{5/2} \left (26 c d^2-7 a e^2\right )}{2 c}\right )+\frac {11}{7} d e \left (a+c x^2\right )^{5/2} (d+e x)^2}{8 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^3}{8 c}\) |
(e*(d + e*x)^3*(a + c*x^2)^(5/2))/(8*c) + ((11*d*e*(d + e*x)^2*(a + c*x^2) ^(5/2))/7 + ((2*d*e*(67*c*d^2 - 32*a*e^2)*(a + c*x^2)^(5/2))/(5*c) + (e^2* (26*c*d^2 - 7*a*e^2)*x*(a + c*x^2)^(5/2))/(2*c) + (7*(16*c^2*d^4 - 16*a*c* d^2*e^2 + a^2*e^4)*((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))/(2*c))/7)/(8* c)
3.6.34.3.1 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[((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]
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])
Time = 2.21 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(-\frac {\left (-560 e^{4} c^{3} x^{7}-2560 d \,e^{3} c^{3} x^{6}-840 e^{4} c^{2} a \,x^{5}-4480 d^{2} e^{2} c^{3} x^{5}-4096 d \,e^{3} c^{2} a \,x^{4}-3584 d^{3} e \,c^{3} x^{4}-70 e^{4} a^{2} c \,x^{3}-7840 d^{2} e^{2} c^{2} a \,x^{3}-1120 d^{4} c^{3} x^{3}-512 a^{2} c d \,e^{3} x^{2}-7168 d^{3} e \,c^{2} a \,x^{2}+105 e^{4} a^{3} x -1680 d^{2} e^{2} a^{2} c x -2800 a \,c^{2} d^{4} x +1024 a^{3} d \,e^{3}-3584 a^{2} c \,d^{3} e \right ) \sqrt {c \,x^{2}+a}}{4480 c^{2}}+\frac {3 a^{2} \left (a^{2} e^{4}-16 a c \,d^{2} e^{2}+16 c^{2} d^{4}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{128 c^{\frac {5}{2}}}\) | \(256\) |
| default | \(d^{4} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )+e^{4} \left (\frac {x^{3} \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{8 c}-\frac {3 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6 c}-\frac {a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6 c}\right )}{8 c}\right )+4 d \,e^{3} \left (\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{7 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{35 c^{2}}\right )+6 d^{2} e^{2} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6 c}-\frac {a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6 c}\right )+\frac {4 d^{3} e \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5 c}\) | \(296\) |
-1/4480/c^2*(-560*c^3*e^4*x^7-2560*c^3*d*e^3*x^6-840*a*c^2*e^4*x^5-4480*c^ 3*d^2*e^2*x^5-4096*a*c^2*d*e^3*x^4-3584*c^3*d^3*e*x^4-70*a^2*c*e^4*x^3-784 0*a*c^2*d^2*e^2*x^3-1120*c^3*d^4*x^3-512*a^2*c*d*e^3*x^2-7168*a*c^2*d^3*e* x^2+105*a^3*e^4*x-1680*a^2*c*d^2*e^2*x-2800*a*c^2*d^4*x+1024*a^3*d*e^3-358 4*a^2*c*d^3*e)*(c*x^2+a)^(1/2)+3/128*a^2*(a^2*e^4-16*a*c*d^2*e^2+16*c^2*d^ 4)/c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))
Time = 0.32 (sec) , antiderivative size = 544, normalized size of antiderivative = 2.13 \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\left [\frac {105 \, {\left (16 \, a^{2} c^{2} d^{4} - 16 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (560 \, c^{4} e^{4} x^{7} + 2560 \, c^{4} d e^{3} x^{6} + 3584 \, a^{2} c^{2} d^{3} e - 1024 \, a^{3} c d e^{3} + 280 \, {\left (16 \, c^{4} d^{2} e^{2} + 3 \, a c^{3} e^{4}\right )} x^{5} + 512 \, {\left (7 \, c^{4} d^{3} e + 8 \, a c^{3} d e^{3}\right )} x^{4} + 70 \, {\left (16 \, c^{4} d^{4} + 112 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{3} + 512 \, {\left (14 \, a c^{3} d^{3} e + a^{2} c^{2} d e^{3}\right )} x^{2} + 35 \, {\left (80 \, a c^{3} d^{4} + 48 \, a^{2} c^{2} d^{2} e^{2} - 3 \, a^{3} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{8960 \, c^{3}}, -\frac {105 \, {\left (16 \, a^{2} c^{2} d^{4} - 16 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (560 \, c^{4} e^{4} x^{7} + 2560 \, c^{4} d e^{3} x^{6} + 3584 \, a^{2} c^{2} d^{3} e - 1024 \, a^{3} c d e^{3} + 280 \, {\left (16 \, c^{4} d^{2} e^{2} + 3 \, a c^{3} e^{4}\right )} x^{5} + 512 \, {\left (7 \, c^{4} d^{3} e + 8 \, a c^{3} d e^{3}\right )} x^{4} + 70 \, {\left (16 \, c^{4} d^{4} + 112 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{3} + 512 \, {\left (14 \, a c^{3} d^{3} e + a^{2} c^{2} d e^{3}\right )} x^{2} + 35 \, {\left (80 \, a c^{3} d^{4} + 48 \, a^{2} c^{2} d^{2} e^{2} - 3 \, a^{3} c e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{4480 \, c^{3}}\right ] \]
[1/8960*(105*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(c)*log(-2* c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(560*c^4*e^4*x^7 + 2560*c^4*d *e^3*x^6 + 3584*a^2*c^2*d^3*e - 1024*a^3*c*d*e^3 + 280*(16*c^4*d^2*e^2 + 3 *a*c^3*e^4)*x^5 + 512*(7*c^4*d^3*e + 8*a*c^3*d*e^3)*x^4 + 70*(16*c^4*d^4 + 112*a*c^3*d^2*e^2 + a^2*c^2*e^4)*x^3 + 512*(14*a*c^3*d^3*e + a^2*c^2*d*e^ 3)*x^2 + 35*(80*a*c^3*d^4 + 48*a^2*c^2*d^2*e^2 - 3*a^3*c*e^4)*x)*sqrt(c*x^ 2 + a))/c^3, -1/4480*(105*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*sq rt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (560*c^4*e^4*x^7 + 2560*c^4*d* e^3*x^6 + 3584*a^2*c^2*d^3*e - 1024*a^3*c*d*e^3 + 280*(16*c^4*d^2*e^2 + 3* a*c^3*e^4)*x^5 + 512*(7*c^4*d^3*e + 8*a*c^3*d*e^3)*x^4 + 70*(16*c^4*d^4 + 112*a*c^3*d^2*e^2 + a^2*c^2*e^4)*x^3 + 512*(14*a*c^3*d^3*e + a^2*c^2*d*e^3 )*x^2 + 35*(80*a*c^3*d^4 + 48*a^2*c^2*d^2*e^2 - 3*a^3*c*e^4)*x)*sqrt(c*x^2 + a))/c^3]
Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (248) = 496\).
Time = 0.62 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.03 \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a + c x^{2}} \cdot \left (\frac {4 c d e^{3} x^{6}}{7} + \frac {c e^{4} x^{7}}{8} + \frac {x^{5} \cdot \left (\frac {9 a c e^{4}}{8} + 6 c^{2} d^{2} e^{2}\right )}{6 c} + \frac {x^{4} \cdot \left (\frac {32 a c d e^{3}}{7} + 4 c^{2} d^{3} e\right )}{5 c} + \frac {x^{3} \left (a^{2} e^{4} + 12 a c d^{2} e^{2} - \frac {5 a \left (\frac {9 a c e^{4}}{8} + 6 c^{2} d^{2} e^{2}\right )}{6 c} + c^{2} d^{4}\right )}{4 c} + \frac {x^{2} \cdot \left (4 a^{2} d e^{3} + 8 a c d^{3} e - \frac {4 a \left (\frac {32 a c d e^{3}}{7} + 4 c^{2} d^{3} e\right )}{5 c}\right )}{3 c} + \frac {x \left (6 a^{2} d^{2} e^{2} + 2 a c d^{4} - \frac {3 a \left (a^{2} e^{4} + 12 a c d^{2} e^{2} - \frac {5 a \left (\frac {9 a c e^{4}}{8} + 6 c^{2} d^{2} e^{2}\right )}{6 c} + c^{2} d^{4}\right )}{4 c}\right )}{2 c} + \frac {4 a^{2} d^{3} e - \frac {2 a \left (4 a^{2} d e^{3} + 8 a c d^{3} e - \frac {4 a \left (\frac {32 a c d e^{3}}{7} + 4 c^{2} d^{3} e\right )}{5 c}\right )}{3 c}}{c}\right ) + \left (a^{2} d^{4} - \frac {a \left (6 a^{2} d^{2} e^{2} + 2 a c d^{4} - \frac {3 a \left (a^{2} e^{4} + 12 a c d^{2} e^{2} - \frac {5 a \left (\frac {9 a c e^{4}}{8} + 6 c^{2} d^{2} e^{2}\right )}{6 c} + c^{2} d^{4}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\a^{\frac {3}{2}} \left (\begin {cases} d^{4} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{5}}{5 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + c*x**2)*(4*c*d*e**3*x**6/7 + c*e**4*x**7/8 + x**5*(9*a *c*e**4/8 + 6*c**2*d**2*e**2)/(6*c) + x**4*(32*a*c*d*e**3/7 + 4*c**2*d**3* e)/(5*c) + x**3*(a**2*e**4 + 12*a*c*d**2*e**2 - 5*a*(9*a*c*e**4/8 + 6*c**2 *d**2*e**2)/(6*c) + c**2*d**4)/(4*c) + x**2*(4*a**2*d*e**3 + 8*a*c*d**3*e - 4*a*(32*a*c*d*e**3/7 + 4*c**2*d**3*e)/(5*c))/(3*c) + x*(6*a**2*d**2*e**2 + 2*a*c*d**4 - 3*a*(a**2*e**4 + 12*a*c*d**2*e**2 - 5*a*(9*a*c*e**4/8 + 6* c**2*d**2*e**2)/(6*c) + c**2*d**4)/(4*c))/(2*c) + (4*a**2*d**3*e - 2*a*(4* a**2*d*e**3 + 8*a*c*d**3*e - 4*a*(32*a*c*d*e**3/7 + 4*c**2*d**3*e)/(5*c))/ (3*c))/c) + (a**2*d**4 - a*(6*a**2*d**2*e**2 + 2*a*c*d**4 - 3*a*(a**2*e**4 + 12*a*c*d**2*e**2 - 5*a*(9*a*c*e**4/8 + 6*c**2*d**2*e**2)/(6*c) + c**2*d **4)/(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)*Piece wise((d**4*x, Eq(e, 0)), ((d + e*x)**5/(5*e), True)), True))
Time = 0.20 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.18 \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e^{4} x^{3}}{8 \, c} + \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d e^{3} x^{2}}{7 \, c} + \frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{4} x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d^{4} x + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{2} e^{2} x}{c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a d^{2} e^{2} x}{4 \, c} - \frac {3 \, \sqrt {c x^{2} + a} a^{2} d^{2} e^{2} x}{8 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} a e^{4} x}{16 \, c^{2}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2} e^{4} x}{64 \, c^{2}} + \frac {3 \, \sqrt {c x^{2} + a} a^{3} e^{4} x}{128 \, c^{2}} + \frac {3 \, a^{2} d^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} - \frac {3 \, a^{3} d^{2} e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {3}{2}}} + \frac {3 \, a^{4} e^{4} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{128 \, c^{\frac {5}{2}}} + \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{3} e}{5 \, c} - \frac {8 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a d e^{3}}{35 \, c^{2}} \]
1/8*(c*x^2 + a)^(5/2)*e^4*x^3/c + 4/7*(c*x^2 + a)^(5/2)*d*e^3*x^2/c + 1/4* (c*x^2 + a)^(3/2)*d^4*x + 3/8*sqrt(c*x^2 + a)*a*d^4*x + (c*x^2 + a)^(5/2)* d^2*e^2*x/c - 1/4*(c*x^2 + a)^(3/2)*a*d^2*e^2*x/c - 3/8*sqrt(c*x^2 + a)*a^ 2*d^2*e^2*x/c - 1/16*(c*x^2 + a)^(5/2)*a*e^4*x/c^2 + 1/64*(c*x^2 + a)^(3/2 )*a^2*e^4*x/c^2 + 3/128*sqrt(c*x^2 + a)*a^3*e^4*x/c^2 + 3/8*a^2*d^4*arcsin h(c*x/sqrt(a*c))/sqrt(c) - 3/8*a^3*d^2*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) + 3/128*a^4*e^4*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 4/5*(c*x^2 + a)^(5/2)*d^3 *e/c - 8/35*(c*x^2 + a)^(5/2)*a*d*e^3/c^2
Time = 0.28 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.13 \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\frac {1}{4480} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (2 \, {\left (7 \, c e^{4} x + 32 \, c d e^{3}\right )} x + \frac {7 \, {\left (16 \, c^{7} d^{2} e^{2} + 3 \, a c^{6} e^{4}\right )}}{c^{6}}\right )} x + \frac {64 \, {\left (7 \, c^{7} d^{3} e + 8 \, a c^{6} d e^{3}\right )}}{c^{6}}\right )} x + \frac {35 \, {\left (16 \, c^{7} d^{4} + 112 \, a c^{6} d^{2} e^{2} + a^{2} c^{5} e^{4}\right )}}{c^{6}}\right )} x + \frac {256 \, {\left (14 \, a c^{6} d^{3} e + a^{2} c^{5} d e^{3}\right )}}{c^{6}}\right )} x + \frac {35 \, {\left (80 \, a c^{6} d^{4} + 48 \, a^{2} c^{5} d^{2} e^{2} - 3 \, a^{3} c^{4} e^{4}\right )}}{c^{6}}\right )} x + \frac {512 \, {\left (7 \, a^{2} c^{5} d^{3} e - 2 \, a^{3} c^{4} d e^{3}\right )}}{c^{6}}\right )} - \frac {3 \, {\left (16 \, a^{2} c^{2} d^{4} - 16 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{128 \, c^{\frac {5}{2}}} \]
1/4480*sqrt(c*x^2 + a)*((2*((4*(5*(2*(7*c*e^4*x + 32*c*d*e^3)*x + 7*(16*c^ 7*d^2*e^2 + 3*a*c^6*e^4)/c^6)*x + 64*(7*c^7*d^3*e + 8*a*c^6*d*e^3)/c^6)*x + 35*(16*c^7*d^4 + 112*a*c^6*d^2*e^2 + a^2*c^5*e^4)/c^6)*x + 256*(14*a*c^6 *d^3*e + a^2*c^5*d*e^3)/c^6)*x + 35*(80*a*c^6*d^4 + 48*a^2*c^5*d^2*e^2 - 3 *a^3*c^4*e^4)/c^6)*x + 512*(7*a^2*c^5*d^3*e - 2*a^3*c^4*d*e^3)/c^6) - 3/12 8*(16*a^2*c^2*d^4 - 16*a^3*c*d^2*e^2 + a^4*e^4)*log(abs(-sqrt(c)*x + sqrt( c*x^2 + a)))/c^(5/2)
Timed out. \[ \int (d+e x)^4 \left (a+c x^2\right )^{3/2} \, dx=\int {\left (c\,x^2+a\right )}^{3/2}\,{\left (d+e\,x\right )}^4 \,d x \]