Integrand size = 20, antiderivative size = 77 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {a (A+B x)}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {3 A+4 B x}{3 b^2 \sqrt {a+b x^2}}+\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:
1/3*a*(B*x+A)/b^2/(b*x^2+a)^(3/2)-1/3*(4*B*x+3*A)/b^2/(b*x^2+a)^(1/2)+B*ar ctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-2 a A-3 a B x-3 A b x^2-4 b B x^3}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{5/2}} \] Input:
Integrate[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]
Output:
(-2*a*A - 3*a*B*x - 3*A*b*x^2 - 4*b*B*x^3)/(3*b^2*(a + b*x^2)^(3/2)) - (B* Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/b^(5/2)
Time = 0.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {530, 2345, 27, 224, 219}
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.
\(\displaystyle \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle \frac {a (A+B x)}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {\frac {B a^2}{b^2}-\frac {3 B x^2 a}{b}-\frac {3 A x a}{b}}{\left (b x^2+a\right )^{3/2}}dx}{3 a}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {a (A+B x)}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {\frac {a (3 A+4 B x)}{b^2 \sqrt {a+b x^2}}-\frac {\int \frac {3 a^2 B}{b^2 \sqrt {b x^2+a}}dx}{a}}{3 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a (A+B x)}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {\frac {a (3 A+4 B x)}{b^2 \sqrt {a+b x^2}}-\frac {3 a B \int \frac {1}{\sqrt {b x^2+a}}dx}{b^2}}{3 a}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {a (A+B x)}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {\frac {a (3 A+4 B x)}{b^2 \sqrt {a+b x^2}}-\frac {3 a B \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{b^2}}{3 a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a (A+B x)}{3 b^2 \left (a+b x^2\right )^{3/2}}-\frac {\frac {a (3 A+4 B x)}{b^2 \sqrt {a+b x^2}}-\frac {3 a B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}}}{3 a}\) |
Input:
Int[(x^3*(A + B*x))/(a + b*x^2)^(5/2),x]
Output:
(a*(A + B*x))/(3*b^2*(a + b*x^2)^(3/2)) - ((a*(3*A + 4*B*x))/(b^2*Sqrt[a + b*x^2]) - (3*a*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/b^(5/2))/(3*a)
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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x )*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 1] && IntegerQ[2*p]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.26
method | result | size |
default | \(A \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+B \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )\) | \(97\) |
Input:
int(x^3*(B*x+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
A*(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))+B*(-1/3*x^3/b/(b*x^2+ a)^(3/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2)) ))
Time = 0.11 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.10 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (B b^{2} x^{4} + 2 \, B a b x^{2} + B a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (4 \, B b^{2} x^{3} + 3 \, A b^{2} x^{2} + 3 \, B a b x + 2 \, A a b\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac {3 \, {\left (B b^{2} x^{4} + 2 \, B a b x^{2} + B a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (4 \, B b^{2} x^{3} + 3 \, A b^{2} x^{2} + 3 \, B a b x + 2 \, A a b\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:
integrate(x^3*(B*x+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
Output:
[1/6*(3*(B*b^2*x^4 + 2*B*a*b*x^2 + B*a^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b* x^2 + a)*sqrt(b)*x - a) - 2*(4*B*b^2*x^3 + 3*A*b^2*x^2 + 3*B*a*b*x + 2*A*a *b)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3), -1/3*(3*(B*b^2*x^4 + 2*B*a*b*x^2 + B*a^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (4*B *b^2*x^3 + 3*A*b^2*x^2 + 3*B*a*b*x + 2*A*a*b)*sqrt(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)]
Time = 5.86 (sec) , antiderivative size = 400, normalized size of antiderivative = 5.19 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=A \left (\begin {cases} - \frac {2 a}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} - \frac {3 b x^{2}}{3 a b^{2} \sqrt {a + b x^{2}} + 3 b^{3} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + B \left (\frac {3 a^{\frac {39}{2}} b^{11} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{\frac {37}{2}} b^{12} x^{2} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a^{19} b^{\frac {23}{2}} x}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {4 a^{18} b^{\frac {25}{2}} x^{3}}{3 a^{\frac {39}{2}} b^{\frac {27}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 3 a^{\frac {37}{2}} b^{\frac {29}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:
integrate(x**3*(B*x+A)/(b*x**2+a)**(5/2),x)
Output:
A*Piecewise((-2*a/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2 )) - 3*b*x**2/(3*a*b**2*sqrt(a + b*x**2) + 3*b**3*x**2*sqrt(a + b*x**2)), Ne(b, 0)), (x**4/(4*a**(5/2)), True)) + B*(3*a**(39/2)*b**11*sqrt(1 + b*x* *2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) + 3*a**(37/2)*b**12*x**2*s qrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**( 23/2)*x/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)* x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(25/2)*x**3/(3*a**(39/2)*b**(27/2)*s qrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)))
Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, B x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} - \frac {A x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {B x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {B \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, A a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \] Input:
integrate(x^3*(B*x+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
Output:
-1/3*B*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) - A*x ^2/((b*x^2 + a)^(3/2)*b) - 1/3*B*x/(sqrt(b*x^2 + a)*b^2) + B*arcsinh(b*x/s qrt(a*b))/b^(5/2) - 2/3*A*a/((b*x^2 + a)^(3/2)*b^2)
Time = 0.14 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {{\left ({\left (\frac {4 \, B x}{b} + \frac {3 \, A}{b}\right )} x + \frac {3 \, B a}{b^{2}}\right )} x + \frac {2 \, A a}{b^{2}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {B \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:
integrate(x^3*(B*x+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
Output:
-1/3*(((4*B*x/b + 3*A/b)*x + 3*B*a/b^2)*x + 2*A*a/b^2)/(b*x^2 + a)^(3/2) - B*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:
int((x^3*(A + B*x))/(a + b*x^2)^(5/2),x)
Output:
int((x^3*(A + B*x))/(a + b*x^2)^(5/2), x)
Time = 0.21 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.16 \[ \int \frac {x^3 (A+B x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a^{2}-3 \sqrt {b \,x^{2}+a}\, a b \,x^{2}-3 \sqrt {b \,x^{2}+a}\, a b x -4 \sqrt {b \,x^{2}+a}\, b^{2} x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} x^{4}}{3 b^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^3*(B*x+A)/(b*x^2+a)^(5/2),x)
Output:
( - 2*sqrt(a + b*x**2)*a**2 - 3*sqrt(a + b*x**2)*a*b*x**2 - 3*sqrt(a + b*x **2)*a*b*x - 4*sqrt(a + b*x**2)*b**2*x**3 + 3*sqrt(b)*log((sqrt(a + b*x**2 ) + sqrt(b)*x)/sqrt(a))*a**2 + 6*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x )/sqrt(a))*a*b*x**2 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a) )*b**2*x**4)/(3*b**2*(a**2 + 2*a*b*x**2 + b**2*x**4))