Integrand size = 24, antiderivative size = 103 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {c}{3 d (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac {a}{(b c-a d)^2 \sqrt {c+d x^2}}+\frac {a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}} \] Output:
-1/3*c/d/(-a*d+b*c)/(d*x^2+c)^(3/2)-a/(-a*d+b*c)^2/(d*x^2+c)^(1/2)+a*b^(1/ 2)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/(-a*d+b*c)^(5/2)
Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {-b c^2-a d \left (2 c+3 d x^2\right )}{3 d (b c-a d)^2 \left (c+d x^2\right )^{3/2}}-\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}} \] Input:
Integrate[x^3/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
Output:
(-(b*c^2) - a*d*(2*c + 3*d*x^2))/(3*d*(b*c - a*d)^2*(c + d*x^2)^(3/2)) - ( a*Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(5/2)
Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {354, 87, 61, 73, 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.
\(\displaystyle \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (-\frac {a \int \frac {1}{\left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx^2}{b c-a d}-\frac {2 c}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {a \left (\frac {b \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{b c-a d}+\frac {2}{\sqrt {c+d x^2} (b c-a d)}\right )}{b c-a d}-\frac {2 c}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (-\frac {a \left (\frac {2 b \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{d (b c-a d)}+\frac {2}{\sqrt {c+d x^2} (b c-a d)}\right )}{b c-a d}-\frac {2 c}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {a \left (\frac {2}{\sqrt {c+d x^2} (b c-a d)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )}{b c-a d}-\frac {2 c}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\right )\) |
Input:
Int[x^3/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
Output:
((-2*c)/(3*d*(b*c - a*d)*(c + d*x^2)^(3/2)) - (a*(2/((b*c - a*d)*Sqrt[c + d*x^2]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/( b*c - a*d)^(3/2)))/(b*c - a*d))/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.75 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(-\frac {a b \arctan \left (\frac {\sqrt {x^{2} d +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {a}{\left (a d -b c \right )^{2} \sqrt {x^{2} d +c}}+\frac {c}{3 d \left (a d -b c \right ) \left (x^{2} d +c \right )^{\frac {3}{2}}}\) | \(99\) |
default | \(\text {Expression too large to display}\) | \(1409\) |
Input:
int(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
-a*b/(a*d-b*c)^2/((a*d-b*c)*b)^(1/2)*arctan((d*x^2+c)^(1/2)*b/((a*d-b*c)*b )^(1/2))-a/(a*d-b*c)^2/(d*x^2+c)^(1/2)+1/3/d*c/(a*d-b*c)/(d*x^2+c)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (87) = 174\).
Time = 0.12 (sec) , antiderivative size = 535, normalized size of antiderivative = 5.19 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a d^{3} x^{4} + 2 \, a c d^{2} x^{2} + a c^{2} d\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b^{2} c^{2} - 3 \, a b c d + a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b}{b c - a d}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (3 \, a d^{2} x^{2} + b c^{2} + 2 \, a c d\right )} \sqrt {d x^{2} + c}}{12 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}, -\frac {3 \, {\left (a d^{3} x^{4} + 2 \, a c d^{2} x^{2} + a c^{2} d\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b}{b c - a d}}}{2 \, {\left (b d x^{2} + b c\right )}}\right ) + 2 \, {\left (3 \, a d^{2} x^{2} + b c^{2} + 2 \, a c d\right )} \sqrt {d x^{2} + c}}{6 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{4} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{2}\right )}}\right ] \] Input:
integrate(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
Output:
[1/12*(3*(a*d^3*x^4 + 2*a*c*d^2*x^2 + a*c^2*d)*sqrt(b/(b*c - a*d))*log((b^ 2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b*d^2)*x^ 2 + 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x ^2 + c)*sqrt(b/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*(3*a*d^2*x^2 + b*c^2 + 2*a*c*d)*sqrt(d*x^2 + c))/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2* d^3 + (b^2*c^2*d^3 - 2*a*b*c*d^4 + a^2*d^5)*x^4 + 2*(b^2*c^3*d^2 - 2*a*b*c ^2*d^3 + a^2*c*d^4)*x^2), -1/6*(3*(a*d^3*x^4 + 2*a*c*d^2*x^2 + a*c^2*d)*sq rt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt (-b/(b*c - a*d))/(b*d*x^2 + b*c)) + 2*(3*a*d^2*x^2 + b*c^2 + 2*a*c*d)*sqrt (d*x^2 + c))/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3 + (b^2*c^2*d^3 - 2*a *b*c*d^4 + a^2*d^5)*x^4 + 2*(b^2*c^3*d^2 - 2*a*b*c^2*d^3 + a^2*c*d^4)*x^2) ]
\[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{3}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**3/(b*x**2+a)/(d*x**2+c)**(5/2),x)
Output:
Integral(x**3/((a + b*x**2)*(c + d*x**2)**(5/2)), x)
Exception generated. \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
Output:
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 = 0.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.23 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {\frac {3 \, a b d \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {b c^{2} + 3 \, {\left (d x^{2} + c\right )} a d - a c d}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}}}{3 \, d} \] Input:
integrate(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")
Output:
-1/3*(3*a*b*d*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((b^2*c^2 - 2 *a*b*c*d + a^2*d^2)*sqrt(-b^2*c + a*b*d)) + (b*c^2 + 3*(d*x^2 + c)*a*d - a *c*d)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x^2 + c)^(3/2)))/d
Time = 1.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.07 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {\frac {c}{3\,\left (a\,d-b\,c\right )}-\frac {a\,d\,\left (d\,x^2+c\right )}{{\left (a\,d-b\,c\right )}^2}}{d\,{\left (d\,x^2+c\right )}^{3/2}}-\frac {a\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^{5/2}}\right )}{{\left (a\,d-b\,c\right )}^{5/2}} \] Input:
int(x^3/((a + b*x^2)*(c + d*x^2)^(5/2)),x)
Output:
(c/(3*(a*d - b*c)) - (a*d*(c + d*x^2))/(a*d - b*c)^2)/(d*(c + d*x^2)^(3/2) ) - (a*b^(1/2)*atan((b^(1/2)*(c + d*x^2)^(1/2)*(a^2*d^2 + b^2*c^2 - 2*a*b* c*d))/(a*d - b*c)^(5/2)))/(a*d - b*c)^(5/2)
Time = 0.21 (sec) , antiderivative size = 488, normalized size of antiderivative = 4.74 \[ \int \frac {x^3}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b x +b c +b d \,x^{2}}{\sqrt {b}\, \sqrt {d \,x^{2}+c}\, \sqrt {a d -b c}+\sqrt {d}\, \sqrt {b}\, \sqrt {a d -b c}\, x}\right ) a \,c^{2} d -6 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b x +b c +b d \,x^{2}}{\sqrt {b}\, \sqrt {d \,x^{2}+c}\, \sqrt {a d -b c}+\sqrt {d}\, \sqrt {b}\, \sqrt {a d -b c}\, x}\right ) a c \,d^{2} x^{2}-3 \sqrt {b}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b x +b c +b d \,x^{2}}{\sqrt {b}\, \sqrt {d \,x^{2}+c}\, \sqrt {a d -b c}+\sqrt {d}\, \sqrt {b}\, \sqrt {a d -b c}\, x}\right ) a \,d^{3} x^{4}-2 \sqrt {d \,x^{2}+c}\, a^{2} c \,d^{2}-3 \sqrt {d \,x^{2}+c}\, a^{2} d^{3} x^{2}+\sqrt {d \,x^{2}+c}\, a b \,c^{2} d +3 \sqrt {d \,x^{2}+c}\, a b c \,d^{2} x^{2}+\sqrt {d \,x^{2}+c}\, b^{2} c^{3}}{3 d \left (a^{3} d^{5} x^{4}-3 a^{2} b c \,d^{4} x^{4}+3 a \,b^{2} c^{2} d^{3} x^{4}-b^{3} c^{3} d^{2} x^{4}+2 a^{3} c \,d^{4} x^{2}-6 a^{2} b \,c^{2} d^{3} x^{2}+6 a \,b^{2} c^{3} d^{2} x^{2}-2 b^{3} c^{4} d \,x^{2}+a^{3} c^{2} d^{3}-3 a^{2} b \,c^{3} d^{2}+3 a \,b^{2} c^{4} d -b^{3} c^{5}\right )} \] Input:
int(x^3/(b*x^2+a)/(d*x^2+c)^(5/2),x)
Output:
( - 3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b *d*x**2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt( a*d - b*c)*x))*a*c**2*d - 6*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x**2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d - b*c)*x))*a*c*d**2*x**2 - 3*sqrt(b)*sqrt(a*d - b*c)*atan((sqrt(d)*sqrt(c + d*x**2)*b*x + b*c + b*d*x**2)/(sqrt(b)*sqrt(c + d*x**2)*sqrt(a*d - b*c) + sqrt(d)*sqrt(b)*sqrt(a*d - b*c)*x))*a*d**3*x** 4 - 2*sqrt(c + d*x**2)*a**2*c*d**2 - 3*sqrt(c + d*x**2)*a**2*d**3*x**2 + s qrt(c + d*x**2)*a*b*c**2*d + 3*sqrt(c + d*x**2)*a*b*c*d**2*x**2 + sqrt(c + d*x**2)*b**2*c**3)/(3*d*(a**3*c**2*d**3 + 2*a**3*c*d**4*x**2 + a**3*d**5* x**4 - 3*a**2*b*c**3*d**2 - 6*a**2*b*c**2*d**3*x**2 - 3*a**2*b*c*d**4*x**4 + 3*a*b**2*c**4*d + 6*a*b**2*c**3*d**2*x**2 + 3*a*b**2*c**2*d**3*x**4 - b **3*c**5 - 2*b**3*c**4*d*x**2 - b**3*c**3*d**2*x**4))