Integrand size = 19, antiderivative size = 106 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {16 b^3 \left (b x^2+c x^4\right )^{5/2}}{1155 c^4 x^5}+\frac {8 b^2 \left (b x^2+c x^4\right )^{5/2}}{231 c^3 x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{33 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c} \] Output:
-16/1155*b^3*(c*x^4+b*x^2)^(5/2)/c^4/x^5+8/231*b^2*(c*x^4+b*x^2)^(5/2)/c^3 /x^3-2/33*b*(c*x^4+b*x^2)^(5/2)/c^2/x+1/11*x*(c*x^4+b*x^2)^(5/2)/c
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.54 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (-16 b^3+40 b^2 c x^2-70 b c^2 x^4+105 c^3 x^6\right )}{1155 c^4 x^5} \] Input:
Integrate[x^4*(b*x^2 + c*x^4)^(3/2),x]
Output:
((x^2*(b + c*x^2))^(5/2)*(-16*b^3 + 40*b^2*c*x^2 - 70*b*c^2*x^4 + 105*c^3* x^6))/(1155*c^4*x^5)
Time = 0.43 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1421, 1421, 1399, 1420}
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.
\(\displaystyle \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \int x^2 \left (c x^4+b x^2\right )^{3/2}dx}{11 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \int \left (c x^4+b x^2\right )^{3/2}dx}{9 c}\right )}{11 c}\) |
\(\Big \downarrow \) 1399 |
\(\displaystyle \frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^2}dx}{7 c}\right )}{9 c}\right )}{11 c}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}\right )}{9 c}\right )}{11 c}\) |
Input:
Int[x^4*(b*x^2 + c*x^4)^(3/2),x]
Output:
(x*(b*x^2 + c*x^4)^(5/2))/(11*c) - (6*b*((b*x^2 + c*x^4)^(5/2)/(9*c*x) - ( 4*b*((-2*b*(b*x^2 + c*x^4)^(5/2))/(35*c^2*x^5) + (b*x^2 + c*x^4)^(5/2)/(7* c*x^3)))/(9*c)))/(11*c)
Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(b*x^2 + c*x^4)^( p + 1)/(c*(4*p + 1)*x^3), x] - Simp[b*((2*p - 1)/(c*(4*p + 1))) Int[(b*x^ 2 + c*x^4)^p/x^2, x], x] /; FreeQ[{b, c, p}, x] && IGtQ[p - 1/2, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(2*c*(p + 1))), x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 2*p - 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && IGtQ[Simplify[(m + 2*p - 1)/2], 0] && NeQ[m + 4*p + 1, 0]
Time = 0.67 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-105 c^{3} x^{6}+70 b \,c^{2} x^{4}-40 b^{2} c \,x^{2}+16 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 c^{4} x^{3}}\) | \(61\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-105 c^{3} x^{6}+70 b \,c^{2} x^{4}-40 b^{2} c \,x^{2}+16 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 c^{4} x^{3}}\) | \(61\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-105 c^{3} x^{6}+70 b \,c^{2} x^{4}-40 b^{2} c \,x^{2}+16 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 c^{4} x^{3}}\) | \(61\) |
trager | \(-\frac {\left (-105 c^{5} x^{10}-140 c^{4} x^{8} b -5 b^{2} c^{3} x^{6}+6 c^{2} x^{4} b^{3}-8 x^{2} c \,b^{4}+16 b^{5}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{1155 c^{4} x}\) | \(76\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-105 c^{5} x^{10}-140 c^{4} x^{8} b -5 b^{2} c^{3} x^{6}+6 c^{2} x^{4} b^{3}-8 x^{2} c \,b^{4}+16 b^{5}\right )}{1155 x \,c^{4}}\) | \(76\) |
Input:
int(x^4*(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/1155*(c*x^2+b)*(-105*c^3*x^6+70*b*c^2*x^4-40*b^2*c*x^2+16*b^3)*(c*x^4+b *x^2)^(3/2)/c^4/x^3
Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{1155 \, c^{4} x} \] Input:
integrate(x^4*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
1/1155*(105*c^5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b ^4*c*x^2 - 16*b^5)*sqrt(c*x^4 + b*x^2)/(c^4*x)
\[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^{4} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}\, dx \] Input:
integrate(x**4*(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(x**4*(x**2*(b + c*x**2))**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{2} + b}}{1155 \, c^{4}} \] Input:
integrate(x^4*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
1/1155*(105*c^5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b ^4*c*x^2 - 16*b^5)*sqrt(c*x^2 + b)/c^4
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {16 \, b^{\frac {11}{2}} \mathrm {sgn}\left (x\right )}{1155 \, c^{4}} + \frac {105 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 385 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} b \mathrm {sgn}\left (x\right ) + 495 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} b^{2} \mathrm {sgn}\left (x\right ) - 231 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b^{3} \mathrm {sgn}\left (x\right )}{1155 \, c^{4}} \] Input:
integrate(x^4*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
16/1155*b^(11/2)*sgn(x)/c^4 + 1/1155*(105*(c*x^2 + b)^(11/2)*sgn(x) - 385* (c*x^2 + b)^(9/2)*b*sgn(x) + 495*(c*x^2 + b)^(7/2)*b^2*sgn(x) - 231*(c*x^2 + b)^(5/2)*b^3*sgn(x))/c^4
Time = 17.80 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {{\left (c\,x^2+b\right )}^2\,\sqrt {c\,x^4+b\,x^2}\,\left (16\,b^3-40\,b^2\,c\,x^2+70\,b\,c^2\,x^4-105\,c^3\,x^6\right )}{1155\,c^4\,x} \] Input:
int(x^4*(b*x^2 + c*x^4)^(3/2),x)
Output:
-((b + c*x^2)^2*(b*x^2 + c*x^4)^(1/2)*(16*b^3 - 105*c^3*x^6 - 40*b^2*c*x^2 + 70*b*c^2*x^4))/(1155*c^4*x)
Time = 0.18 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63 \[ \int x^4 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {c \,x^{2}+b}\, \left (105 c^{5} x^{10}+140 b \,c^{4} x^{8}+5 b^{2} c^{3} x^{6}-6 b^{3} c^{2} x^{4}+8 b^{4} c \,x^{2}-16 b^{5}\right )}{1155 c^{4}} \] Input:
int(x^4*(c*x^4+b*x^2)^(3/2),x)
Output:
(sqrt(b + c*x**2)*( - 16*b**5 + 8*b**4*c*x**2 - 6*b**3*c**2*x**4 + 5*b**2* c**3*x**6 + 140*b*c**4*x**8 + 105*c**5*x**10))/(1155*c**4)