Integrand size = 31, antiderivative size = 109 \[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {c^2 \left (b c^2+a d^2\right ) (-c+d x)^{3/2} (c+d x)^{3/2}}{3 d^6}+\frac {\left (2 b c^2+a d^2\right ) (-c+d x)^{5/2} (c+d x)^{5/2}}{5 d^6}+\frac {b (-c+d x)^{7/2} (c+d x)^{7/2}}{7 d^6} \]
1/3*c^2*(a*d^2+b*c^2)*(d*x-c)^(3/2)*(d*x+c)^(3/2)/d^6+1/5*(a*d^2+2*b*c^2)* (d*x-c)^(5/2)*(d*x+c)^(5/2)/d^6+1/7*b*(d*x-c)^(7/2)*(d*x+c)^(7/2)/d^6
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {(-c+d x)^{3/2} (c+d x)^{3/2} \left (7 a d^2 \left (2 c^2+3 d^2 x^2\right )+b \left (8 c^4+12 c^2 d^2 x^2+15 d^4 x^4\right )\right )}{105 d^6} \]
((-c + d*x)^(3/2)*(c + d*x)^(3/2)*(7*a*d^2*(2*c^2 + 3*d^2*x^2) + b*(8*c^4 + 12*c^2*d^2*x^2 + 15*d^4*x^4)))/(105*d^6)
Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int x^3 \left (a+b x^2\right ) \sqrt {d x-c} \sqrt {c+d x} \, dx\) |
\(\Big \downarrow \) 960 |
\(\displaystyle \frac {1}{7} \left (7 a+\frac {4 b c^2}{d^2}\right ) \int x^3 \sqrt {d x-c} \sqrt {c+d x}dx+\frac {b x^4 (d x-c)^{3/2} (c+d x)^{3/2}}{7 d^2}\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {1}{7} \left (7 a+\frac {4 b c^2}{d^2}\right ) \left (\frac {\int 2 c^2 x \sqrt {d x-c} \sqrt {c+d x}dx}{5 d^2}+\frac {x^2 (d x-c)^{3/2} (c+d x)^{3/2}}{5 d^2}\right )+\frac {b x^4 (d x-c)^{3/2} (c+d x)^{3/2}}{7 d^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (7 a+\frac {4 b c^2}{d^2}\right ) \left (\frac {2 c^2 \int x \sqrt {d x-c} \sqrt {c+d x}dx}{5 d^2}+\frac {x^2 (d x-c)^{3/2} (c+d x)^{3/2}}{5 d^2}\right )+\frac {b x^4 (d x-c)^{3/2} (c+d x)^{3/2}}{7 d^2}\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {1}{7} \left (\frac {2 c^2 (d x-c)^{3/2} (c+d x)^{3/2}}{15 d^4}+\frac {x^2 (d x-c)^{3/2} (c+d x)^{3/2}}{5 d^2}\right ) \left (7 a+\frac {4 b c^2}{d^2}\right )+\frac {b x^4 (d x-c)^{3/2} (c+d x)^{3/2}}{7 d^2}\) |
(b*x^4*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(7*d^2) + ((7*a + (4*b*c^2)/d^2)* ((2*c^2*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(15*d^4) + (x^2*(-c + d*x)^(3/2) *(c + d*x)^(3/2))/(5*d^2)))/7
3.4.38.3.1 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_))*((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 = 4.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(\frac {\left (d x -c \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}} \left (15 b \,d^{4} x^{4}+21 a \,d^{4} x^{2}+12 b \,c^{2} d^{2} x^{2}+14 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{105 d^{6}}\) | \(68\) |
default | \(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (-d^{2} x^{2}+c^{2}\right ) \left (15 b \,d^{4} x^{4}+21 a \,d^{4} x^{2}+12 b \,c^{2} d^{2} x^{2}+14 a \,c^{2} d^{2}+8 b \,c^{4}\right )}{105 d^{6}}\) | \(80\) |
risch | \(\frac {\sqrt {d x +c}\, \left (-15 b \,x^{6} d^{6}-21 a \,d^{6} x^{4}+3 b \,c^{2} d^{4} x^{4}+7 a \,c^{2} d^{4} x^{2}+4 b \,c^{4} d^{2} x^{2}+14 a \,c^{4} d^{2}+8 b \,c^{6}\right ) \left (-d x +c \right )}{105 \sqrt {d x -c}\, d^{6}}\) | \(98\) |
1/105/d^6*(d*x-c)^(3/2)*(d*x+c)^(3/2)*(15*b*d^4*x^4+21*a*d^4*x^2+12*b*c^2* d^2*x^2+14*a*c^2*d^2+8*b*c^4)
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83 \[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {{\left (15 \, b d^{6} x^{6} - 8 \, b c^{6} - 14 \, a c^{4} d^{2} - 3 \, {\left (b c^{2} d^{4} - 7 \, a d^{6}\right )} x^{4} - {\left (4 \, b c^{4} d^{2} + 7 \, a c^{2} d^{4}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{105 \, d^{6}} \]
1/105*(15*b*d^6*x^6 - 8*b*c^6 - 14*a*c^4*d^2 - 3*(b*c^2*d^4 - 7*a*d^6)*x^4 - (4*b*c^4*d^2 + 7*a*c^2*d^4)*x^2)*sqrt(d*x + c)*sqrt(d*x - c)/d^6
\[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\int x^{3} \left (a + b x^{2}\right ) \sqrt {- c + d x} \sqrt {c + d x}\, dx \]
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14 \[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b x^{4}}{7 \, d^{2}} + \frac {4 \, {\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b c^{2} x^{2}}{35 \, d^{4}} + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a x^{2}}{5 \, d^{2}} + \frac {8 \, {\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} b c^{4}}{105 \, d^{6}} + \frac {2 \, {\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a c^{2}}{15 \, d^{4}} \]
1/7*(d^2*x^2 - c^2)^(3/2)*b*x^4/d^2 + 4/35*(d^2*x^2 - c^2)^(3/2)*b*c^2*x^2 /d^4 + 1/5*(d^2*x^2 - c^2)^(3/2)*a*x^2/d^2 + 8/105*(d^2*x^2 - c^2)^(3/2)*b *c^4/d^6 + 2/15*(d^2*x^2 - c^2)^(3/2)*a*c^2/d^4
Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (91) = 182\).
Time = 0.38 (sec) , antiderivative size = 495, normalized size of antiderivative = 4.54 \[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {70 \, {\left ({\left ({\left (d x + c\right )} {\left (2 \, {\left (d x + c\right )} {\left (\frac {3 \, {\left (d x + c\right )}}{d^{3}} - \frac {13 \, c}{d^{3}}\right )} + \frac {43 \, c^{2}}{d^{3}}\right )} - \frac {39 \, c^{3}}{d^{3}}\right )} \sqrt {d x + c} \sqrt {d x - c} - \frac {18 \, c^{4} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{3}}\right )} a c + 7 \, {\left ({\left ({\left (2 \, {\left ({\left (d x + c\right )} {\left (4 \, {\left (d x + c\right )} {\left (\frac {5 \, {\left (d x + c\right )}}{d^{5}} - \frac {31 \, c}{d^{5}}\right )} + \frac {321 \, c^{2}}{d^{5}}\right )} - \frac {451 \, c^{3}}{d^{5}}\right )} {\left (d x + c\right )} + \frac {745 \, c^{4}}{d^{5}}\right )} {\left (d x + c\right )} - \frac {405 \, c^{5}}{d^{5}}\right )} \sqrt {d x + c} \sqrt {d x - c} - \frac {150 \, c^{6} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{5}}\right )} b c + 14 \, {\left ({\left ({\left (2 \, {\left (d x + c\right )} {\left (3 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )}}{d^{4}} - \frac {21 \, c}{d^{4}}\right )} + \frac {133 \, c^{2}}{d^{4}}\right )} - \frac {295 \, c^{3}}{d^{4}}\right )} {\left (d x + c\right )} + \frac {195 \, c^{4}}{d^{4}}\right )} \sqrt {d x + c} \sqrt {d x - c} + \frac {90 \, c^{5} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{4}}\right )} a d + {\left ({\left ({\left (2 \, {\left ({\left (4 \, {\left (d x + c\right )} {\left (5 \, {\left (d x + c\right )} {\left (\frac {6 \, {\left (d x + c\right )}}{d^{6}} - \frac {43 \, c}{d^{6}}\right )} + \frac {661 \, c^{2}}{d^{6}}\right )} - \frac {4551 \, c^{3}}{d^{6}}\right )} {\left (d x + c\right )} + \frac {4781 \, c^{4}}{d^{6}}\right )} {\left (d x + c\right )} - \frac {6335 \, c^{5}}{d^{6}}\right )} {\left (d x + c\right )} + \frac {2835 \, c^{6}}{d^{6}}\right )} \sqrt {d x + c} \sqrt {d x - c} + \frac {1050 \, c^{7} \log \left ({\left | -\sqrt {d x + c} + \sqrt {d x - c} \right |}\right )}{d^{6}}\right )} b d}{1680 \, d} \]
1/1680*(70*(((d*x + c)*(2*(d*x + c)*(3*(d*x + c)/d^3 - 13*c/d^3) + 43*c^2/ d^3) - 39*c^3/d^3)*sqrt(d*x + c)*sqrt(d*x - c) - 18*c^4*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^3)*a*c + 7*(((2*((d*x + c)*(4*(d*x + c)*(5*(d*x + c)/d^5 - 31*c/d^5) + 321*c^2/d^5) - 451*c^3/d^5)*(d*x + c) + 745*c^4/d^5) *(d*x + c) - 405*c^5/d^5)*sqrt(d*x + c)*sqrt(d*x - c) - 150*c^6*log(abs(-s qrt(d*x + c) + sqrt(d*x - c)))/d^5)*b*c + 14*(((2*(d*x + c)*(3*(d*x + c)*( 4*(d*x + c)/d^4 - 21*c/d^4) + 133*c^2/d^4) - 295*c^3/d^4)*(d*x + c) + 195* c^4/d^4)*sqrt(d*x + c)*sqrt(d*x - c) + 90*c^5*log(abs(-sqrt(d*x + c) + sqr t(d*x - c)))/d^4)*a*d + (((2*((4*(d*x + c)*(5*(d*x + c)*(6*(d*x + c)/d^6 - 43*c/d^6) + 661*c^2/d^6) - 4551*c^3/d^6)*(d*x + c) + 4781*c^4/d^6)*(d*x + c) - 6335*c^5/d^6)*(d*x + c) + 2835*c^6/d^6)*sqrt(d*x + c)*sqrt(d*x - c) + 1050*c^7*log(abs(-sqrt(d*x + c) + sqrt(d*x - c)))/d^6)*b*d)/d
Time = 5.79 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.08 \[ \int x^3 \sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right ) \, dx=-\sqrt {d\,x-c}\,\left (\frac {\left (8\,b\,c^6+14\,a\,c^4\,d^2\right )\,\sqrt {c+d\,x}}{105\,d^6}-\frac {b\,x^6\,\sqrt {c+d\,x}}{7}+\frac {x^2\,\left (4\,b\,c^4\,d^2+7\,a\,c^2\,d^4\right )\,\sqrt {c+d\,x}}{105\,d^6}-\frac {x^4\,\left (21\,a\,d^6-3\,b\,c^2\,d^4\right )\,\sqrt {c+d\,x}}{105\,d^6}\right ) \]