Integrand size = 24, antiderivative size = 244 \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\frac {\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt {a x+b x^3+c x^5}}{1280 c^3 \sqrt {x}}-\frac {x^{3/2} \left (b \left (5 b^2-4 a c\right )+4 c \left (5 b^2-16 a c\right ) x^2\right ) \sqrt {a x+b x^3+c x^5}}{640 c^2}+\frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 b \left (b^2-4 a c\right )^2 \sqrt {x} \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{512 c^{7/2} \sqrt {a x+b x^3+c x^5}} \] Output:
1/1280*(128*a^2*c^2-100*a*b^2*c+15*b^4)*(c*x^5+b*x^3+a*x)^(1/2)/c^3/x^(1/2 )-1/640*x^(3/2)*(b*(-4*a*c+5*b^2)+4*c*(-16*a*c+5*b^2)*x^2)*(c*x^5+b*x^3+a* x)^(1/2)/c^2+1/80*x^(1/2)*(8*c*x^2+3*b)*(c*x^5+b*x^3+a*x)^(3/2)/c-3/512*b* (-4*a*c+b^2)^2*x^(1/2)*(c*x^4+b*x^2+a)^(1/2)*arctanh(1/2*(2*c*x^2+b)/c^(1/ 2)/(c*x^4+b*x^2+a)^(1/2))/c^(7/2)/(c*x^5+b*x^3+a*x)^(1/2)
Time = 0.36 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.74 \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\frac {\sqrt {x} \sqrt {a+b x^2+c x^4} \left (2 \sqrt {c} \sqrt {a+b x^2+c x^4} \left (15 b^4-10 b^3 c x^2+128 c^2 \left (a+c x^4\right )^2+4 b^2 c \left (-25 a+2 c x^4\right )+8 b c^2 x^2 \left (7 a+22 c x^4\right )\right )+15 b \left (b^2-4 a c\right )^2 \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right )}{2560 c^{7/2} \sqrt {x \left (a+b x^2+c x^4\right )}} \] Input:
Integrate[x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2),x]
Output:
(Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]*(15*b^ 4 - 10*b^3*c*x^2 + 128*c^2*(a + c*x^4)^2 + 4*b^2*c*(-25*a + 2*c*x^4) + 8*b *c^2*x^2*(7*a + 22*c*x^4)) + 15*b*(b^2 - 4*a*c)^2*Log[b + 2*c*x^2 - 2*Sqrt [c]*Sqrt[a + b*x^2 + c*x^4]]))/(2560*c^(7/2)*Sqrt[x*(a + b*x^2 + c*x^4)])
Time = 0.56 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {1966, 25, 1992, 25, 1996, 27, 1961, 1432, 1092, 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.
\(\displaystyle \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1966 |
\(\displaystyle \frac {3 \int -\sqrt {x} \left (\left (5 b^2-16 a c\right ) x^2+2 a b\right ) \sqrt {c x^5+b x^3+a x}dx}{80 c}+\frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \int \sqrt {x} \left (\left (5 b^2-16 a c\right ) x^2+2 a b\right ) \sqrt {c x^5+b x^3+a x}dx}{80 c}\) |
\(\Big \downarrow \) 1992 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {\int -\frac {x^{3/2} \left (\left (15 b^4-100 a c b^2+128 a^2 c^2\right ) x^2+2 a b \left (5 b^2-28 a c\right )\right )}{\sqrt {c x^5+b x^3+a x}}dx}{24 c}+\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\int \frac {x^{3/2} \left (\left (15 b^4-100 a c b^2+128 a^2 c^2\right ) x^2+2 a b \left (5 b^2-28 a c\right )\right )}{\sqrt {c x^5+b x^3+a x}}dx}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 1996 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x+b x^3+c x^5}}{2 c \sqrt {x}}-\frac {\int \frac {15 b \left (b^2-4 a c\right )^2 x^{3/2}}{\sqrt {c x^5+b x^3+a x}}dx}{2 c}}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x+b x^3+c x^5}}{2 c \sqrt {x}}-\frac {15 b \left (b^2-4 a c\right )^2 \int \frac {x^{3/2}}{\sqrt {c x^5+b x^3+a x}}dx}{2 c}}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 1961 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x+b x^3+c x^5}}{2 c \sqrt {x}}-\frac {15 b \sqrt {x} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4} \int \frac {x}{\sqrt {c x^4+b x^2+a}}dx}{2 c \sqrt {a x+b x^3+c x^5}}}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 1432 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x+b x^3+c x^5}}{2 c \sqrt {x}}-\frac {15 b \sqrt {x} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4} \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx^2}{4 c \sqrt {a x+b x^3+c x^5}}}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x+b x^3+c x^5}}{2 c \sqrt {x}}-\frac {15 b \sqrt {x} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4} \int \frac {1}{4 c-x^4}d\frac {2 c x^2+b}{\sqrt {c x^4+b x^2+a}}}{2 c \sqrt {a x+b x^3+c x^5}}}{24 c}\right )}{80 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {x} \left (3 b+8 c x^2\right ) \left (a x+b x^3+c x^5\right )^{3/2}}{80 c}-\frac {3 \left (\frac {x^{3/2} \left (4 c x^2 \left (5 b^2-16 a c\right )+b \left (5 b^2-4 a c\right )\right ) \sqrt {a x+b x^3+c x^5}}{24 c}-\frac {\frac {\left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt {a x+b x^3+c x^5}}{2 c \sqrt {x}}-\frac {15 b \sqrt {x} \left (b^2-4 a c\right )^2 \sqrt {a+b x^2+c x^4} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{3/2} \sqrt {a x+b x^3+c x^5}}}{24 c}\right )}{80 c}\) |
Input:
Int[x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2),x]
Output:
(Sqrt[x]*(3*b + 8*c*x^2)*(a*x + b*x^3 + c*x^5)^(3/2))/(80*c) - (3*((x^(3/2 )*(b*(5*b^2 - 4*a*c) + 4*c*(5*b^2 - 16*a*c)*x^2)*Sqrt[a*x + b*x^3 + c*x^5] )/(24*c) - (((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a*x + b*x^3 + c*x^5 ])/(2*c*Sqrt[x]) - (15*b*(b^2 - 4*a*c)^2*Sqrt[x]*Sqrt[a + b*x^2 + c*x^4]*A rcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*c^(3/2)*Sqrt [a*x + b*x^3 + c*x^5]))/(24*c)))/(80*c)
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)^(-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_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x]
Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)] , x_Symbol] :> Simp[x^(q/2)*(Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]/Sqrt[a *x^q + b*x^n + c*x^(2*n - q)]) Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) + c*x ^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] && EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q, 1]))
Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ ), x_Symbol] :> Simp[x^(m - n + q + 1)*(b*(n - q)*p + c*(m + p*q + (n - q)* (2*p - 1) + 1)*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^p/(c*(m + p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1))), x] + Simp[(n - q)*(p/(c*(m + p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1))) Int[x^(m - (n - 2* q))*Simp[(-a)*b*(m + p*q - n + q + 1) + (2*a*c*(m + p*q + (n - q)*(2*p - 1) + 1) - b^2*(m + p*q + (n - q)*(p - 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] & & PosQ[n - q] && !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[ p, 0] && RationalQ[m, q] && GtQ[m + p*q + 1, n - q] && NeQ[m + p*(2*n - q) + 1, 0] && NeQ[m + p*q + (n - q)*(2*p - 1) + 1, 0]
Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ .)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[x^(m + 1)*(b*B*(n - q)*p + A*c*(m + p*q + (n - q)*(2*p + 1) + 1) + B*c*(m + p*q + 2*(n - q)*p + 1)*x^( n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^p/(c*(m + p*(2*n - q) + 1)*(m + p* q + (n - q)*(2*p + 1) + 1))), x] + Simp[(n - q)*(p/(c*(m + p*(2*n - q) + 1) *(m + p*q + (n - q)*(2*p + 1) + 1))) Int[x^(m + q)*Simp[2*a*A*c*(m + p*q + (n - q)*(2*p + 1) + 1) - a*b*B*(m + p*q + 1) + (2*a*B*c*(m + p*q + 2*(n - q)*p + 1) + A*b*c*(m + p*q + (n - q)*(2*p + 1) + 1) - b^2*B*(m + p*q + (n - q)*p + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] && !Integ erQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && GtQ[m + p*q, -(n - q) - 1] && NeQ[m + p*(2*n - q) + 1, 0] && NeQ[m + p* q + (n - q)*(2*p + 1) + 1, 0]
Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ .)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[B*x^(m - n + 1)*((a*x^q + b *x^n + c*x^(2*n - q))^(p + 1)/(c*(m + p*q + (n - q)*(2*p + 1) + 1))), x] - Simp[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1)) Int[x^(m - n + q)*Simp[a*B*( m + p*q - n + q + 1) + (b*B*(m + p*q + (n - q)*p + 1) - A*c*(m + p*q + (n - q)*(2*p + 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] && !Inte gerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]
Time = 0.13 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {\left (128 c^{4} x^{8}+176 b \,x^{6} c^{3}+256 a \,x^{4} c^{3}+8 b^{2} x^{4} c^{2}+56 a b \,c^{2} x^{2}-10 b^{3} c \,x^{2}+128 a^{2} c^{2}-100 a \,b^{2} c +15 b^{4}\right ) \left (c \,x^{4}+b \,x^{2}+a \right ) \sqrt {x}}{1280 c^{3} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}-\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) b \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {x}}{512 c^{\frac {7}{2}} \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\) | \(202\) |
default | \(-\frac {\sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}\, \left (-256 c^{\frac {9}{2}} x^{8} \sqrt {c \,x^{4}+b \,x^{2}+a}-352 b \,c^{\frac {7}{2}} x^{6} \sqrt {c \,x^{4}+b \,x^{2}+a}-512 a \,c^{\frac {7}{2}} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}-16 b^{2} c^{\frac {5}{2}} x^{4} \sqrt {c \,x^{4}+b \,x^{2}+a}-112 a b \,c^{\frac {5}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+20 b^{3} c^{\frac {3}{2}} x^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}+240 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a^{2} b \,c^{2}-120 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) a \,b^{3} c +15 \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{2 \sqrt {c}}\right ) b^{5}-256 a^{2} c^{\frac {5}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}+200 a \,b^{2} c^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}-30 b^{4} \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2560 c^{\frac {7}{2}} \sqrt {x}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\) | \(369\) |
Input:
int(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/1280*(128*c^4*x^8+176*b*c^3*x^6+256*a*c^3*x^4+8*b^2*c^2*x^4+56*a*b*c^2*x ^2-10*b^3*c*x^2+128*a^2*c^2-100*a*b^2*c+15*b^4)/c^3*(c*x^4+b*x^2+a)*x^(1/2 )/(x*(c*x^4+b*x^2+a))^(1/2)-3/512*(16*a^2*c^2-8*a*b^2*c+b^4)/c^(7/2)*b*ln( (1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))*(c*x^4+b*x^2+a)^(1/2)*x^(1/2) /(x*(c*x^4+b*x^2+a))^(1/2)
Time = 0.09 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.62 \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{5} + 8 \, b c x^{3} - 4 \, \sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {c} \sqrt {x} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{5120 \, c^{4} x}, \frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (2 \, c x^{2} + b\right )} \sqrt {-c} \sqrt {x}}{2 \, {\left (c^{2} x^{5} + b c x^{3} + a c x\right )}}\right ) + 2 \, {\left (128 \, c^{5} x^{8} + 176 \, b c^{4} x^{6} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{4} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x^{2}\right )} \sqrt {c x^{5} + b x^{3} + a x} \sqrt {x}}{2560 \, c^{4} x}\right ] \] Input:
integrate(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="fricas")
Output:
[1/5120*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*sqrt(c)*x*log(-(8*c^2*x^5 + 8 *b*c*x^3 - 4*sqrt(c*x^5 + b*x^3 + a*x)*(2*c*x^2 + b)*sqrt(c)*sqrt(x) + (b^ 2 + 4*a*c)*x)/x) + 4*(128*c^5*x^8 + 176*b*c^4*x^6 + 15*b^4*c - 100*a*b^2*c ^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^4 - 2*(5*b^3*c^2 - 28*a*b*c^3) *x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x))/(c^4*x), 1/2560*(15*(b^5 - 8*a*b^ 3*c + 16*a^2*b*c^2)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^5 + b*x^3 + a*x)*(2*c*x ^2 + b)*sqrt(-c)*sqrt(x)/(c^2*x^5 + b*c*x^3 + a*c*x)) + 2*(128*c^5*x^8 + 1 76*b*c^4*x^6 + 15*b^4*c - 100*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a* c^4)*x^4 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x^2)*sqrt(c*x^5 + b*x^3 + a*x)*sqrt( x))/(c^4*x)]
Timed out. \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\text {Timed out} \] Input:
integrate(x**(3/2)*(c*x**5+b*x**3+a*x)**(3/2),x)
Output:
Timed out
\[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\int { {\left (c x^{5} + b x^{3} + a x\right )}^{\frac {3}{2}} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="maxima")
Output:
integrate((c*x^5 + b*x^3 + a*x)^(3/2)*x^(3/2), x)
Time = 0.21 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.67 \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\frac {1}{96} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {3 \, b^{2} - 8 \, a c}{c^{2}}\right )} - \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}}\right )} a + \frac {1}{768} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {5 \, b^{2} c - 12 \, a c^{2}}{c^{3}}\right )} x^{2} + \frac {15 \, b^{3} - 52 \, a b c}{c^{3}}\right )} + \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {7}{2}}}\right )} b + \frac {1}{7680} \, {\left (2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x^{2} + \frac {b}{c}\right )} x^{2} - \frac {7 \, b^{2} c^{2} - 16 \, a c^{3}}{c^{4}}\right )} x^{2} + \frac {35 \, b^{3} c - 116 \, a b c^{2}}{c^{4}}\right )} x^{2} - \frac {105 \, b^{4} - 460 \, a b^{2} c + 256 \, a^{2} c^{2}}{c^{4}}\right )} - \frac {15 \, {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {9}{2}}}\right )} c \] Input:
integrate(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x, algorithm="giac")
Output:
1/96*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*x^2 + b/c)*x^2 - (3*b^2 - 8*a*c)/c^2 ) - 3*(b^3 - 4*a*b*c)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sq rt(c) + b))/c^(5/2))*a + 1/768*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*x^2 + b /c)*x^2 - (5*b^2*c - 12*a*c^2)/c^3)*x^2 + (15*b^3 - 52*a*b*c)/c^3) + 3*(5* b^4 - 24*a*b^2*c + 16*a^2*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(7/2))*b + 1/7680*(2*sqrt(c*x^4 + b*x^2 + a)*(2*(4* (6*(8*x^2 + b/c)*x^2 - (7*b^2*c^2 - 16*a*c^3)/c^4)*x^2 + (35*b^3*c - 116*a *b*c^2)/c^4)*x^2 - (105*b^4 - 460*a*b^2*c + 256*a^2*c^2)/c^4) - 15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) + b))/c^(9/2))*c
Timed out. \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx=\int x^{3/2}\,{\left (c\,x^5+b\,x^3+a\,x\right )}^{3/2} \,d x \] Input:
int(x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2),x)
Output:
int(x^(3/2)*(a*x + b*x^3 + c*x^5)^(3/2), x)
Time = 0.26 (sec) , antiderivative size = 3844, normalized size of antiderivative = 15.75 \[ \int x^{3/2} \left (a x+b x^3+c x^5\right )^{3/2} \, dx =\text {Too large to display} \] Input:
int(x^(3/2)*(c*x^5+b*x^3+a*x)^(3/2),x)
Output:
( - 7680*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**4*b*c**4 - 15360*sqrt(c)* sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2 *c*x**2)/sqrt(4*a*c - b**2))*a**3*b**3*c**3 - 92160*sqrt(c)*sqrt(a + b*x** 2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt( 4*a*c - b**2))*a**3*b**2*c**4*x**2 - 92160*sqrt(c)*sqrt(a + b*x**2 + c*x** 4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b **2))*a**3*b*c**5*x**4 + 6720*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqr t(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b* *5*c**2 + 7680*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b *x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b**4*c**3*x**2 - 115200*sqrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b**3*c**4*x**4 - 245760*s qrt(c)*sqrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*a**2*b**2*c**5*x**6 - 122880*sqrt(c)*s qrt(a + b*x**2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2* c*x**2)/sqrt(4*a*c - b**2))*a**2*b*c**6*x**8 + 13440*sqrt(c)*sqrt(a + b*x* *2 + c*x**4)*log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt (4*a*c - b**2))*a*b**6*c**2*x**2 + 74880*sqrt(c)*sqrt(a + b*x**2 + c*x**4) *log((2*sqrt(c)*sqrt(a + b*x**2 + c*x**4) + b + 2*c*x**2)/sqrt(4*a*c - ...