Integrand size = 26, antiderivative size = 138 \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\frac {a \sqrt {c+d x^2}}{5 b (b c-a d) \left (a+b x^2\right )^{5/2}}-\frac {(5 b c-a d) \sqrt {c+d x^2}}{15 b (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac {2 d (5 b c-a d) \sqrt {c+d x^2}}{15 b (b c-a d)^3 \sqrt {a+b x^2}} \] Output:
1/5*a*(d*x^2+c)^(1/2)/b/(-a*d+b*c)/(b*x^2+a)^(5/2)-1/15*(-a*d+5*b*c)*(d*x^ 2+c)^(1/2)/b/(-a*d+b*c)^2/(b*x^2+a)^(3/2)+2/15*d*(-a*d+5*b*c)*(d*x^2+c)^(1 /2)/b/(-a*d+b*c)^3/(b*x^2+a)^(1/2)
Time = 2.01 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.66 \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {c+d x^2} \left (-5 b^2 c x^2 \left (c-2 d x^2\right )-5 a^2 d \left (-2 c+d x^2\right )-2 a b \left (c^2-13 c d x^2+d^2 x^4\right )\right )}{15 (b c-a d)^3 \left (a+b x^2\right )^{5/2}} \] Input:
Integrate[x^3/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]
Output:
(Sqrt[c + d*x^2]*(-5*b^2*c*x^2*(c - 2*d*x^2) - 5*a^2*d*(-2*c + d*x^2) - 2* a*b*(c^2 - 13*c*d*x^2 + d^2*x^4)))/(15*(b*c - a*d)^3*(a + b*x^2)^(5/2))
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {354, 87, 55, 48}
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.
\(\displaystyle \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (b x^2+a\right )^{7/2} \sqrt {d x^2+c}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (\frac {(5 b c-a d) \int \frac {1}{\left (b x^2+a\right )^{5/2} \sqrt {d x^2+c}}dx^2}{5 b (b c-a d)}+\frac {2 a \sqrt {c+d x^2}}{5 b \left (a+b x^2\right )^{5/2} (b c-a d)}\right )\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{2} \left (\frac {(5 b c-a d) \left (-\frac {2 d \int \frac {1}{\left (b x^2+a\right )^{3/2} \sqrt {d x^2+c}}dx^2}{3 (b c-a d)}-\frac {2 \sqrt {c+d x^2}}{3 \left (a+b x^2\right )^{3/2} (b c-a d)}\right )}{5 b (b c-a d)}+\frac {2 a \sqrt {c+d x^2}}{5 b \left (a+b x^2\right )^{5/2} (b c-a d)}\right )\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {1}{2} \left (\frac {2 a \sqrt {c+d x^2}}{5 b \left (a+b x^2\right )^{5/2} (b c-a d)}+\frac {(5 b c-a d) \left (\frac {4 d \sqrt {c+d x^2}}{3 \sqrt {a+b x^2} (b c-a d)^2}-\frac {2 \sqrt {c+d x^2}}{3 \left (a+b x^2\right )^{3/2} (b c-a d)}\right )}{5 b (b c-a d)}\right )\) |
Input:
Int[x^3/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]
Output:
((2*a*Sqrt[c + d*x^2])/(5*b*(b*c - a*d)*(a + b*x^2)^(5/2)) + ((5*b*c - a*d )*((-2*Sqrt[c + d*x^2])/(3*(b*c - a*d)*(a + b*x^2)^(3/2)) + (4*d*Sqrt[c + d*x^2])/(3*(b*c - a*d)^2*Sqrt[a + b*x^2])))/(5*b*(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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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[(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 = 0.56 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {\sqrt {x^{2} d +c}\, \left (-2 a b \,d^{2} x^{4}+10 b^{2} c d \,x^{4}-5 a^{2} d^{2} x^{2}+26 a b c d \,x^{2}-5 b^{2} c^{2} x^{2}+10 a^{2} c d -2 b \,c^{2} a \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (a d -b c \right ) \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\) | \(120\) |
gosper | \(-\frac {\sqrt {x^{2} d +c}\, \left (-2 a b \,d^{2} x^{4}+10 b^{2} c d \,x^{4}-5 a^{2} d^{2} x^{2}+26 a b c d \,x^{2}-5 b^{2} c^{2} x^{2}+10 a^{2} c d -2 b \,c^{2} a \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(125\) |
orering | \(-\frac {\sqrt {x^{2} d +c}\, \left (-2 a b \,d^{2} x^{4}+10 b^{2} c d \,x^{4}-5 a^{2} d^{2} x^{2}+26 a b c d \,x^{2}-5 b^{2} c^{2} x^{2}+10 a^{2} c d -2 b \,c^{2} a \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(125\) |
elliptic | \(-\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {x^{2} d +c}\, \left (-2 a b \,d^{2} x^{4}+10 b^{2} c d \,x^{4}-5 a^{2} d^{2} x^{2}+26 a b c d \,x^{2}-5 b^{2} c^{2} x^{2}+10 a^{2} c d -2 b \,c^{2} a \right )}{15 \sqrt {b \,x^{2}+a}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}\, \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\) | \(186\) |
Input:
int(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(d*x^2+c)^(1/2)*(-2*a*b*d^2*x^4+10*b^2*c*d*x^4-5*a^2*d^2*x^2+26*a*b* c*d*x^2-5*b^2*c^2*x^2+10*a^2*c*d-2*a*b*c^2)/(b*x^2+a)^(5/2)/(a*d-b*c)/(a^2 *d^2-2*a*b*c*d+b^2*c^2)
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (120) = 240\).
Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.95 \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\frac {{\left (2 \, {\left (5 \, b^{2} c d - a b d^{2}\right )} x^{4} - 2 \, a b c^{2} + 10 \, a^{2} c d - {\left (5 \, b^{2} c^{2} - 26 \, a b c d + 5 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{15 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{6} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{4} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x^{2}\right )}} \] Input:
integrate(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")
Output:
1/15*(2*(5*b^2*c*d - a*b*d^2)*x^4 - 2*a*b*c^2 + 10*a^2*c*d - (5*b^2*c^2 - 26*a*b*c*d + 5*a^2*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3 *a^2*b^4*c*d^2 - a^3*b^3*d^3)*x^6 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3 *b^3*c*d^2 - a^4*b^2*d^3)*x^4 + 3*(a^2*b^4*c^3 - 3*a^3*b^3*c^2*d + 3*a^4*b ^2*c*d^2 - a^5*b*d^3)*x^2)
\[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\int \frac {x^{3}}{\left (a + b x^{2}\right )^{\frac {7}{2}} \sqrt {c + d x^{2}}}\, dx \] Input:
integrate(x**3/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)
Output:
Integral(x**3/((a + b*x**2)**(7/2)*sqrt(c + d*x**2)), x)
Exception generated. \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),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
Leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (120) = 240\).
Time = 0.18 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.42 \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\frac {4 \, {\left (5 \, \sqrt {b d} b^{8} c^{3} d - 11 \, \sqrt {b d} a b^{7} c^{2} d^{2} + 7 \, \sqrt {b d} a^{2} b^{6} c d^{3} - \sqrt {b d} a^{3} b^{5} d^{4} - 25 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{6} c^{2} d + 30 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b^{5} c d^{2} - 5 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b^{4} d^{3} + 35 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} b^{4} c d + 5 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a b^{3} d^{2} - 15 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} b^{2} d\right )}}{15 \, {\left (b^{2} c - a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}^{5} b {\left | b \right |}} \] Input:
integrate(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")
Output:
4/15*(5*sqrt(b*d)*b^8*c^3*d - 11*sqrt(b*d)*a*b^7*c^2*d^2 + 7*sqrt(b*d)*a^2 *b^6*c*d^3 - sqrt(b*d)*a^3*b^5*d^4 - 25*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b* d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^6*c^2*d + 30*sqrt(b*d)*(sq rt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b^5*c *d^2 - 5*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b *d - a*b*d))^2*a^2*b^4*d^3 + 35*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqr t(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*b^4*c*d + 5*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a*b^3*d^2 - 15*sqr t(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)) ^6*b^2*d)/((b^2*c - a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x ^2 + a)*b*d - a*b*d))^2)^5*b*abs(b))
Time = 2.14 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.64 \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {b\,x^2+a}\,\left (\frac {x^2\,\left (5\,a^2\,c\,d^2+24\,a\,b\,c^2\,d-5\,b^2\,c^3\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x^4\,\left (-5\,a^2\,d^3+24\,a\,b\,c\,d^2+5\,b^2\,c^2\,d\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}-\frac {2\,d^2\,x^6\,\left (a\,d-5\,b\,c\right )}{15\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {2\,a\,c^2\,\left (5\,a\,d-b\,c\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}\right )}{x^6\,\sqrt {d\,x^2+c}+\frac {a^3\,\sqrt {d\,x^2+c}}{b^3}+\frac {3\,a\,x^4\,\sqrt {d\,x^2+c}}{b}+\frac {3\,a^2\,x^2\,\sqrt {d\,x^2+c}}{b^2}} \] Input:
int(x^3/((a + b*x^2)^(7/2)*(c + d*x^2)^(1/2)),x)
Output:
-((a + b*x^2)^(1/2)*((x^2*(5*a^2*c*d^2 - 5*b^2*c^3 + 24*a*b*c^2*d))/(15*b^ 3*(a*d - b*c)^3) + (x^4*(5*b^2*c^2*d - 5*a^2*d^3 + 24*a*b*c*d^2))/(15*b^3* (a*d - b*c)^3) - (2*d^2*x^6*(a*d - 5*b*c))/(15*b^2*(a*d - b*c)^3) + (2*a*c ^2*(5*a*d - b*c))/(15*b^3*(a*d - b*c)^3)))/(x^6*(c + d*x^2)^(1/2) + (a^3*( c + d*x^2)^(1/2))/b^3 + (3*a*x^4*(c + d*x^2)^(1/2))/b + (3*a^2*x^2*(c + d* x^2)^(1/2))/b^2)
Time = 0.17 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.12 \[ \int \frac {x^3}{\left (a+b x^2\right )^{7/2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, \left (2 a b \,d^{2} x^{4}-10 b^{2} c d \,x^{4}+5 a^{2} d^{2} x^{2}-26 a b c d \,x^{2}+5 b^{2} c^{2} x^{2}-10 a^{2} c d +2 a b \,c^{2}\right )}{15 a^{3} b^{3} d^{3} x^{6}-45 a^{2} b^{4} c \,d^{2} x^{6}+45 a \,b^{5} c^{2} d \,x^{6}-15 b^{6} c^{3} x^{6}+45 a^{4} b^{2} d^{3} x^{4}-135 a^{3} b^{3} c \,d^{2} x^{4}+135 a^{2} b^{4} c^{2} d \,x^{4}-45 a \,b^{5} c^{3} x^{4}+45 a^{5} b \,d^{3} x^{2}-135 a^{4} b^{2} c \,d^{2} x^{2}+135 a^{3} b^{3} c^{2} d \,x^{2}-45 a^{2} b^{4} c^{3} x^{2}+15 a^{6} d^{3}-45 a^{5} b c \,d^{2}+45 a^{4} b^{2} c^{2} d -15 a^{3} b^{3} c^{3}} \] Input:
int(x^3/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)
Output:
(sqrt(c + d*x**2)*sqrt(a + b*x**2)*( - 10*a**2*c*d + 5*a**2*d**2*x**2 + 2* a*b*c**2 - 26*a*b*c*d*x**2 + 2*a*b*d**2*x**4 + 5*b**2*c**2*x**2 - 10*b**2* c*d*x**4))/(15*(a**6*d**3 - 3*a**5*b*c*d**2 + 3*a**5*b*d**3*x**2 + 3*a**4* b**2*c**2*d - 9*a**4*b**2*c*d**2*x**2 + 3*a**4*b**2*d**3*x**4 - a**3*b**3* c**3 + 9*a**3*b**3*c**2*d*x**2 - 9*a**3*b**3*c*d**2*x**4 + a**3*b**3*d**3* x**6 - 3*a**2*b**4*c**3*x**2 + 9*a**2*b**4*c**2*d*x**4 - 3*a**2*b**4*c*d** 2*x**6 - 3*a*b**5*c**3*x**4 + 3*a*b**5*c**2*d*x**6 - b**6*c**3*x**6))