Integrand size = 31, antiderivative size = 118 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {2 c^2 \left (4 b c^2+5 a d^2\right ) \sqrt {-c+d x} \sqrt {c+d x}}{15 d^6}+\frac {\left (4 b c^2+5 a d^2\right ) x^2 \sqrt {-c+d x} \sqrt {c+d x}}{15 d^4}+\frac {b x^4 \sqrt {-c+d x} \sqrt {c+d x}}{5 d^2} \] Output:
2/15*c^2*(5*a*d^2+4*b*c^2)*(d*x-c)^(1/2)*(d*x+c)^(1/2)/d^6+1/15*(5*a*d^2+4 *b*c^2)*x^2*(d*x-c)^(1/2)*(d*x+c)^(1/2)/d^4+1/5*b*x^4*(d*x-c)^(1/2)*(d*x+c )^(1/2)/d^2
Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.63 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {\sqrt {-c+d x} \sqrt {c+d x} \left (5 a d^2 \left (2 c^2+d^2 x^2\right )+b \left (8 c^4+4 c^2 d^2 x^2+3 d^4 x^4\right )\right )}{15 d^6} \] Input:
Integrate[(x^3*(a + b*x^2))/(Sqrt[-c + d*x]*Sqrt[c + d*x]),x]
Output:
(Sqrt[-c + d*x]*Sqrt[c + d*x]*(5*a*d^2*(2*c^2 + d^2*x^2) + b*(8*c^4 + 4*c^ 2*d^2*x^2 + 3*d^4*x^4)))/(15*d^6)
Time = 0.39 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {960, 111, 27, 83}
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.
\(\displaystyle \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {d x-c} \sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 960 |
\(\displaystyle \frac {1}{5} \left (5 a+\frac {4 b c^2}{d^2}\right ) \int \frac {x^3}{\sqrt {d x-c} \sqrt {c+d x}}dx+\frac {b x^4 \sqrt {d x-c} \sqrt {c+d x}}{5 d^2}\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {1}{5} \left (5 a+\frac {4 b c^2}{d^2}\right ) \left (\frac {\int \frac {2 c^2 x}{\sqrt {d x-c} \sqrt {c+d x}}dx}{3 d^2}+\frac {x^2 \sqrt {d x-c} \sqrt {c+d x}}{3 d^2}\right )+\frac {b x^4 \sqrt {d x-c} \sqrt {c+d x}}{5 d^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (5 a+\frac {4 b c^2}{d^2}\right ) \left (\frac {2 c^2 \int \frac {x}{\sqrt {d x-c} \sqrt {c+d x}}dx}{3 d^2}+\frac {x^2 \sqrt {d x-c} \sqrt {c+d x}}{3 d^2}\right )+\frac {b x^4 \sqrt {d x-c} \sqrt {c+d x}}{5 d^2}\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {1}{5} \left (\frac {2 c^2 \sqrt {d x-c} \sqrt {c+d x}}{3 d^4}+\frac {x^2 \sqrt {d x-c} \sqrt {c+d x}}{3 d^2}\right ) \left (5 a+\frac {4 b c^2}{d^2}\right )+\frac {b x^4 \sqrt {d x-c} \sqrt {c+d x}}{5 d^2}\) |
Input:
Int[(x^3*(a + b*x^2))/(Sqrt[-c + d*x]*Sqrt[c + d*x]),x]
Output:
(b*x^4*Sqrt[-c + d*x]*Sqrt[c + d*x])/(5*d^2) + ((5*a + (4*b*c^2)/d^2)*((2* c^2*Sqrt[-c + d*x]*Sqrt[c + d*x])/(3*d^4) + (x^2*Sqrt[-c + d*x]*Sqrt[c + d *x])/(3*d^2)))/5
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) *(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^( m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(b1*b2*e*(m + n *(p + 1) + 1))), x] - Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/ (b1*b2*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n /2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (3 b \,x^{4} d^{4}+5 a \,d^{4} x^{2}+4 b \,c^{2} d^{2} x^{2}+10 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{15 d^{6}}\) | \(68\) |
default | \(\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (3 b \,x^{4} d^{4}+5 a \,d^{4} x^{2}+4 b \,c^{2} d^{2} x^{2}+10 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{15 d^{6}}\) | \(68\) |
risch | \(-\frac {\sqrt {d x +c}\, \left (-d x +c \right ) \left (3 b \,x^{4} d^{4}+5 a \,d^{4} x^{2}+4 b \,c^{2} d^{2} x^{2}+10 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{15 d^{6} \sqrt {d x -c}}\) | \(74\) |
orering | \(-\frac {\sqrt {d x +c}\, \left (-d x +c \right ) \left (3 b \,x^{4} d^{4}+5 a \,d^{4} x^{2}+4 b \,c^{2} d^{2} x^{2}+10 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{15 d^{6} \sqrt {d x -c}}\) | \(74\) |
Input:
int(x^3*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/15/d^6*(d*x-c)^(1/2)*(d*x+c)^(1/2)*(3*b*d^4*x^4+5*a*d^4*x^2+4*b*c^2*d^2* x^2+10*a*c^2*d^2+8*b*c^4)
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.56 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {{\left (3 \, b d^{4} x^{4} + 8 \, b c^{4} + 10 \, a c^{2} d^{2} + {\left (4 \, b c^{2} d^{2} + 5 \, a d^{4}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{15 \, d^{6}} \] Input:
integrate(x^3*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
1/15*(3*b*d^4*x^4 + 8*b*c^4 + 10*a*c^2*d^2 + (4*b*c^2*d^2 + 5*a*d^4)*x^2)* sqrt(d*x + c)*sqrt(d*x - c)/d^6
Result contains complex when optimal does not.
Time = 10.11 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.03 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {a c^{3} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} & -1, -1, - \frac {1}{2}, 1 \\- \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} + \frac {i a c^{3} {G_{6, 6}^{2, 6}\left (\begin {matrix} -2, - \frac {7}{4}, - \frac {3}{2}, - \frac {5}{4}, -1, 1 & \\- \frac {7}{4}, - \frac {5}{4} & -2, - \frac {3}{2}, - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{4}} + \frac {b c^{5} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {9}{4}, - \frac {7}{4} & -2, -2, - \frac {3}{2}, 1 \\- \frac {5}{2}, - \frac {9}{4}, -2, - \frac {7}{4}, - \frac {3}{2}, 0 & \end {matrix} \middle | {\frac {c^{2}}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{6}} + \frac {i b c^{5} {G_{6, 6}^{2, 6}\left (\begin {matrix} -3, - \frac {11}{4}, - \frac {5}{2}, - \frac {9}{4}, -2, 1 & \\- \frac {11}{4}, - \frac {9}{4} & -3, - \frac {5}{2}, - \frac {5}{2}, 0 \end {matrix} \middle | {\frac {c^{2} e^{2 i \pi }}{d^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} d^{6}} \] Input:
integrate(x**3*(b*x**2+a)/(d*x-c)**(1/2)/(d*x+c)**(1/2),x)
Output:
a*c**3*meijerg(((-5/4, -3/4), (-1, -1, -1/2, 1)), ((-3/2, -5/4, -1, -3/4, -1/2, 0), ()), c**2/(d**2*x**2))/(4*pi**(3/2)*d**4) + I*a*c**3*meijerg(((- 2, -7/4, -3/2, -5/4, -1, 1), ()), ((-7/4, -5/4), (-2, -3/2, -3/2, 0)), c** 2*exp_polar(2*I*pi)/(d**2*x**2))/(4*pi**(3/2)*d**4) + b*c**5*meijerg(((-9/ 4, -7/4), (-2, -2, -3/2, 1)), ((-5/2, -9/4, -2, -7/4, -3/2, 0), ()), c**2/ (d**2*x**2))/(4*pi**(3/2)*d**6) + I*b*c**5*meijerg(((-3, -11/4, -5/2, -9/4 , -2, 1), ()), ((-11/4, -9/4), (-3, -5/2, -5/2, 0)), c**2*exp_polar(2*I*pi )/(d**2*x**2))/(4*pi**(3/2)*d**6)
Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {\sqrt {d^{2} x^{2} - c^{2}} b x^{4}}{5 \, d^{2}} + \frac {4 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{2} x^{2}}{15 \, d^{4}} + \frac {\sqrt {d^{2} x^{2} - c^{2}} a x^{2}}{3 \, d^{2}} + \frac {8 \, \sqrt {d^{2} x^{2} - c^{2}} b c^{4}}{15 \, d^{6}} + \frac {2 \, \sqrt {d^{2} x^{2} - c^{2}} a c^{2}}{3 \, d^{4}} \] Input:
integrate(x^3*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="maxima")
Output:
1/5*sqrt(d^2*x^2 - c^2)*b*x^4/d^2 + 4/15*sqrt(d^2*x^2 - c^2)*b*c^2*x^2/d^4 + 1/3*sqrt(d^2*x^2 - c^2)*a*x^2/d^2 + 8/15*sqrt(d^2*x^2 - c^2)*b*c^4/d^6 + 2/3*sqrt(d^2*x^2 - c^2)*a*c^2/d^4
Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {{\left ({\left ({\left (d x + c\right )} {\left (3 \, {\left (d x + c\right )} {\left (\frac {{\left (d x + c\right )} b}{d^{5}} - \frac {4 \, b c}{d^{5}}\right )} + \frac {22 \, b c^{2} d^{25} + 5 \, a d^{27}}{d^{30}}\right )} - \frac {10 \, {\left (2 \, b c^{3} d^{25} + a c d^{27}\right )}}{d^{30}}\right )} {\left (d x + c\right )} + \frac {15 \, {\left (b c^{4} d^{25} + a c^{2} d^{27}\right )}}{d^{30}}\right )} \sqrt {d x + c} \sqrt {d x - c}}{15 \, d} \] Input:
integrate(x^3*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")
Output:
1/15*(((d*x + c)*(3*(d*x + c)*((d*x + c)*b/d^5 - 4*b*c/d^5) + (22*b*c^2*d^ 25 + 5*a*d^27)/d^30) - 10*(2*b*c^3*d^25 + a*c*d^27)/d^30)*(d*x + c) + 15*( b*c^4*d^25 + a*c^2*d^27)/d^30)*sqrt(d*x + c)*sqrt(d*x - c)/d
Time = 5.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {\sqrt {d\,x-c}\,\left (\frac {8\,b\,c^5+10\,a\,c^3\,d^2}{15\,d^6}+\frac {x^3\,\left (4\,b\,c^2\,d^3+5\,a\,d^5\right )}{15\,d^6}+\frac {x\,\left (8\,b\,c^4\,d+10\,a\,c^2\,d^3\right )}{15\,d^6}+\frac {b\,x^5}{5\,d}+\frac {x^2\,\left (4\,b\,c^3\,d^2+5\,a\,c\,d^4\right )}{15\,d^6}+\frac {b\,c\,x^4}{5\,d^2}\right )}{\sqrt {c+d\,x}} \] Input:
int((x^3*(a + b*x^2))/((c + d*x)^(1/2)*(d*x - c)^(1/2)),x)
Output:
((d*x - c)^(1/2)*((8*b*c^5 + 10*a*c^3*d^2)/(15*d^6) + (x^3*(5*a*d^5 + 4*b* c^2*d^3))/(15*d^6) + (x*(10*a*c^2*d^3 + 8*b*c^4*d))/(15*d^6) + (b*x^5)/(5* d) + (x^2*(4*b*c^3*d^2 + 5*a*c*d^4))/(15*d^6) + (b*c*x^4)/(5*d^2)))/(c + d *x)^(1/2)
Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.55 \[ \int \frac {x^3 \left (a+b x^2\right )}{\sqrt {-c+d x} \sqrt {c+d x}} \, dx=\frac {\sqrt {d x +c}\, \sqrt {d x -c}\, \left (3 b \,d^{4} x^{4}+5 a \,d^{4} x^{2}+4 b \,c^{2} d^{2} x^{2}+10 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{15 d^{6}} \] Input:
int(x^3*(b*x^2+a)/(d*x-c)^(1/2)/(d*x+c)^(1/2),x)
Output:
(sqrt(c + d*x)*sqrt( - c + d*x)*(10*a*c**2*d**2 + 5*a*d**4*x**2 + 8*b*c**4 + 4*b*c**2*d**2*x**2 + 3*b*d**4*x**4))/(15*d**6)