Integrand size = 24, antiderivative size = 164 \[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\frac {5 (b c-a d)^2 \sqrt {a+b x^2} \sqrt {c+d x^2}}{16 d^3}-\frac {5 (b c-a d) \left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{24 d^2}+\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{6 d}-\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{16 \sqrt {b} d^{7/2}} \]
-5/16*(-a*d+b*c)^3*arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d*x^2+c)^(1/2) )/d^(7/2)/b^(1/2)-5/24*(-a*d+b*c)*(b*x^2+a)^(3/2)*(d*x^2+c)^(1/2)/d^2+1/6* (b*x^2+a)^(5/2)*(d*x^2+c)^(1/2)/d+5/16*(-a*d+b*c)^2*(b*x^2+a)^(1/2)*(d*x^2 +c)^(1/2)/d^3
Time = 1.59 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.84 \[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (33 a^2 d^2+2 a b d \left (-20 c+13 d x^2\right )+b^2 \left (15 c^2-10 c d x^2+8 d^2 x^4\right )\right )}{48 d^3}-\frac {5 (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{16 \sqrt {b} d^{7/2}} \]
(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x^2) + b^2*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4)))/(48*d^3) - (5*(b*c - a*d)^3*ArcT anh[(Sqrt[b]*Sqrt[c + d*x^2])/(Sqrt[d]*Sqrt[a + b*x^2])])/(16*Sqrt[b]*d^(7 /2))
Time = 0.24 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {353, 60, 60, 60, 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.
| \(\displaystyle \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^{5/2}}{\sqrt {d x^2+c}}dx^2\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{3 d}-\frac {5 (b c-a d) \int \frac {\left (b x^2+a\right )^{3/2}}{\sqrt {d x^2+c}}dx^2}{6 d}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{3 d}-\frac {5 (b c-a d) \left (\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {b x^2+a}}{\sqrt {d x^2+c}}dx^2}{4 d}\right )}{6 d}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{3 d}-\frac {5 (b c-a d) \left (\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {b x^2+a} \sqrt {d x^2+c}}dx^2}{2 d}\right )}{4 d}\right )}{6 d}\right )\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{3 d}-\frac {5 (b c-a d) \left (\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{d}-\frac {(b c-a d) \int \frac {1}{b-d x^4}d\frac {\sqrt {b x^2+a}}{\sqrt {d x^2+c}}}{d}\right )}{4 d}\right )}{6 d}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}}{3 d}-\frac {5 (b c-a d) \left (\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right )\) |
(((a + b*x^2)^(5/2)*Sqrt[c + d*x^2])/(3*d) - (5*(b*c - a*d)*(((a + b*x^2)^ (3/2)*Sqrt[c + d*x^2])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b*x^2]*Sqrt[c + d *x^2])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d*x^2])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(6*d))/2
3.10.56.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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_)^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_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
Time = 3.17 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.26
| method | result | size |
| risch | \(\frac {\left (8 b^{2} d^{2} x^{4}+26 x^{2} a b \,d^{2}-10 x^{2} b^{2} c d +33 a^{2} d^{2}-40 a b c d +15 b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{48 d^{3}}+\frac {5 \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{32 d^{3} \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) | \(206\) |
| default | \(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (16 b^{2} d^{2} x^{4} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+52 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, a b \,d^{2} x^{2}-20 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, b^{2} c d \,x^{2}+15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}-45 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}+45 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d -15 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+66 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, a^{2} d^{2}-80 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, a b c d +30 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\, b^{2} c^{2}\right )}{96 d^{3} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}}\) | \(452\) |
| elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {5 a^{3} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{32 \sqrt {b d}}+\frac {b^{2} x^{4} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{6 d}+\frac {13 b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, x^{2} a}{24 d}-\frac {5 b^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, x^{2} c}{24 d^{2}}+\frac {11 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a^{2}}{16 d}-\frac {5 b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a c}{6 d^{2}}+\frac {5 b^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c^{2}}{16 d^{3}}-\frac {15 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{2} c}{32 d \sqrt {b d}}+\frac {15 b^{2} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a \,c^{2}}{32 d^{2} \sqrt {b d}}-\frac {5 b^{3} \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{3}}{32 d^{3} \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) | \(478\) |
1/48*(8*b^2*d^2*x^4+26*a*b*d^2*x^2-10*b^2*c*d*x^2+33*a^2*d^2-40*a*b*c*d+15 *b^2*c^2)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/d^3+5/32*(a^3*d^3-3*a^2*b*c*d^2+ 3*a*b^2*c^2*d-b^3*c^3)/d^3*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 = 0.30 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.68 \[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\left [-\frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{192 \, b d^{4}}, \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{4} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \, {\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{96 \, b d^{4}}\right ] \]
[-1/192*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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*(8*b^3*d^3*x^4 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5* b^3*c*d^2 - 13*a*b^2*d^3)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^4), 1 /96*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*arc tan(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*(8*b^3*d^3*x^4 + 1 5*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^ 3)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(b*d^4)]
\[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\int \frac {x \left (a + b x^{2}\right )^{\frac {5}{2}}}{\sqrt {c + d x^{2}}}\, dx \]
Exception generated. \[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.32 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.28 \[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (2 \, {\left (b x^{2} + a\right )} {\left (\frac {4 \, {\left (b x^{2} + a\right )}}{b d} - \frac {5 \, {\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \frac {15 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d^{3}}\right )} b}{48 \, {\left | b \right |}} \]
1/48*(sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)*(2*(b*x^2 + a) *(4*(b*x^2 + a)/(b*d) - 5*(b*c*d^3 - a*d^4)/(b*d^5)) + 15*(b^2*c^2*d^2 - 2 *a*b*c*d^3 + a^2*d^4)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d ^2 - a^3*d^3)*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)*d^3))*b/abs(b)
Timed out. \[ \int \frac {x \left (a+b x^2\right )^{5/2}}{\sqrt {c+d x^2}} \, dx=\int \frac {x\,{\left (b\,x^2+a\right )}^{5/2}}{\sqrt {d\,x^2+c}} \,d x \]