Integrand size = 26, antiderivative size = 141 \[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=-\frac {3 (b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b^2 d^2}+\frac {x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{4 b d}-\frac {\left (4 a b c d-3 (b c+a d)^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{5/2} d^{5/2}} \]
-1/8*(4*a*b*c*d-3*(a*d+b*c)^2)*arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d* x^2+c)^(1/2))/b^(5/2)/d^(5/2)-3/8*(a*d+b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2 )/b^2/d^2+1/4*x^2*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/d
Time = 1.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-3 b c-3 a d+2 b d x^2\right )}{8 b^2 d^2}+\frac {\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{5/2} d^{5/2}} \]
(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-3*b*c - 3*a*d + 2*b*d*x^2))/(8*b^2*d^2) + ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/ (Sqrt[b]*Sqrt[c + d*x^2])])/(8*b^(5/2)*d^(5/2))
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {354, 101, 27, 90, 66, 221}
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.
| \(\displaystyle \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {x^4}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx^2\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {3 (b c+a d) x^2+2 a c}{2 \sqrt {b x^2+a} \sqrt {d x^2+c}}dx^2}{2 b d}+\frac {x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {\int \frac {3 (b c+a d) x^2+2 a c}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx^2}{4 b d}\right )\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \int \frac {1}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx^2}{2 b d}+\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{b d}}{4 b d}\right )\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \int \frac {1}{b-d x^4}d\frac {\sqrt {b x^2+a}}{\sqrt {d x^2+c}}}{b d}+\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{b d}}{4 b d}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {\frac {\left (4 a b c d-3 (a d+b c)^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{b^{3/2} d^{3/2}}+\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (a d+b c)}{b d}}{4 b d}\right )\) |
((x^2*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*b*d) - ((3*(b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(b*d) + ((4*a*b*c*d - 3*(b*c + a*d)^2)*ArcTanh[(Sq rt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(b^(3/2)*d^(3/2)))/(4*b *d))/2
3.10.69.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
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] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 3.18 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.14
| method | result | size |
| risch | \(-\frac {\left (-2 b d \,x^{2}+3 a d +3 b c \right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{8 b^{2} d^{2}}+\frac {\left (3 a^{2} d^{2}+2 a b c d +3 b^{2} c^{2}\right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{16 b^{2} d^{2} \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) | \(161\) |
| elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{4 b d}-\frac {3 \left (a d +b c \right ) \left (\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{b d}-\frac {\left (a d +b c \right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{2 b d \sqrt {b d}}\right )}{8 b d}-\frac {a c \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 b d \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) | \(245\) |
| default | \(\frac {\left (4 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b d \,x^{2}+3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{2}+2 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d +3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}-6 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a d -6 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b c \right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{16 \sqrt {b d}\, d^{2} b^{2} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}\) | \(291\) |
-1/8*(-2*b*d*x^2+3*a*d+3*b*c)/b^2/d^2*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)+1/16 *(3*a^2*d^2+2*a*b*c*d+3*b^2*c^2)/b^2/d^2*ln((1/2*a*d+1/2*b*c+b*d*x^2)/(b*d )^(1/2)+(b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2))/(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c ))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)
Time = 0.36 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.38 \[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\left [\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) + 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, b^{3} d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, b^{3} d^{3}}\right ] \]
[1/32*((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*sqrt(b*d)*log(8*b^2*d^2*x^4 + b ^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)) + 4*(2*b^2*d^2*x^2 - 3*b^2*c*d - 3*a*b*d^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^3*d^3), -1/16* ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b* c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)) - 2*(2*b^2*d^2*x^2 - 3*b^2*c*d - 3*a*b*d^2)*s qrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b^3*d^3)]
\[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int \frac {x^{5}}{\sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \]
Exception generated. \[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.30 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.13 \[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (\frac {2 \, {\left (b x^{2} + a\right )}}{b^{3} d} - \frac {3 \, b^{6} c d + 5 \, a b^{5} d^{2}}{b^{8} d^{3}}\right )} - \frac {{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{2} d^{2}}\right )} b}{8 \, {\left | b \right |}} \]
1/8*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a)/ (b^3*d) - (3*b^6*c*d + 5*a*b^5*d^2)/(b^8*d^3)) - (3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b *d - a*b*d)))/(sqrt(b*d)*b^2*d^2))*b/abs(b)
Time = 28.28 (sec) , antiderivative size = 550, normalized size of antiderivative = 3.90 \[ \int \frac {x^5}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}\right )\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d+3\,b^2\,c^2\right )}{4\,b^{5/2}\,d^{5/2}}-\frac {\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (\frac {3\,a^2\,b\,d^2}{4}+\frac {a\,b^2\,c\,d}{2}+\frac {3\,b^3\,c^2}{4}\right )}{d^6\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}-\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (\frac {11\,a^2\,d^2}{4}+\frac {25\,a\,b\,c\,d}{2}+\frac {11\,b^2\,c^2}{4}\right )}{d^5\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^7\,\left (\frac {3\,a^2\,d^2}{4}+\frac {a\,b\,c\,d}{2}+\frac {3\,b^2\,c^2}{4}\right )}{b^2\,d^3\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^7}-\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^5\,\left (\frac {11\,a^2\,d^2}{4}+\frac {25\,a\,b\,c\,d}{2}+\frac {11\,b^2\,c^2}{4}\right )}{b\,d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^5}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4\,\left (16\,a\,d+16\,b\,c\right )}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^8}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^8}+\frac {b^4}{d^4}-\frac {4\,b^3\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^3\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}-\frac {4\,b\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^6}{d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^6}} \]
(atanh((d^(1/2)*((a + b*x^2)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x^2)^(1/2) - c^(1/2))))*(3*a^2*d^2 + 3*b^2*c^2 + 2*a*b*c*d))/(4*b^(5/2)*d^(5/2)) - ( (((a + b*x^2)^(1/2) - a^(1/2))*((3*b^3*c^2)/4 + (3*a^2*b*d^2)/4 + (a*b^2*c *d)/2))/(d^6*((c + d*x^2)^(1/2) - c^(1/2))) - (((a + b*x^2)^(1/2) - a^(1/2 ))^3*((11*a^2*d^2)/4 + (11*b^2*c^2)/4 + (25*a*b*c*d)/2))/(d^5*((c + d*x^2) ^(1/2) - c^(1/2))^3) + (((a + b*x^2)^(1/2) - a^(1/2))^7*((3*a^2*d^2)/4 + ( 3*b^2*c^2)/4 + (a*b*c*d)/2))/(b^2*d^3*((c + d*x^2)^(1/2) - c^(1/2))^7) - ( ((a + b*x^2)^(1/2) - a^(1/2))^5*((11*a^2*d^2)/4 + (11*b^2*c^2)/4 + (25*a*b *c*d)/2))/(b*d^4*((c + d*x^2)^(1/2) - c^(1/2))^5) + (a^(1/2)*c^(1/2)*((a + b*x^2)^(1/2) - a^(1/2))^4*(16*a*d + 16*b*c))/(d^4*((c + d*x^2)^(1/2) - c^ (1/2))^4))/(((a + b*x^2)^(1/2) - a^(1/2))^8/((c + d*x^2)^(1/2) - c^(1/2))^ 8 + b^4/d^4 - (4*b^3*((a + b*x^2)^(1/2) - a^(1/2))^2)/(d^3*((c + d*x^2)^(1 /2) - c^(1/2))^2) + (6*b^2*((a + b*x^2)^(1/2) - a^(1/2))^4)/(d^2*((c + d*x ^2)^(1/2) - c^(1/2))^4) - (4*b*((a + b*x^2)^(1/2) - a^(1/2))^6)/(d*((c + d *x^2)^(1/2) - c^(1/2))^6))