Integrand size = 18, antiderivative size = 100 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\frac {d x}{b^3}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}+\frac {a (8 c+9 d x)}{8 b^3 \left (a+b x^2\right )}-\frac {15 \sqrt {a} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{7/2}}+\frac {c \log \left (a+b x^2\right )}{2 b^3} \] Output:
d*x/b^3-1/4*a^2*(d*x+c)/b^3/(b*x^2+a)^2+1/8*a*(9*d*x+8*c)/b^3/(b*x^2+a)-15 /8*a^(1/2)*d*arctan(b^(1/2)*x/a^(1/2))/b^(7/2)+1/2*c*ln(b*x^2+a)/b^3
Time = 0.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\frac {d x}{b^3}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}+\frac {a (8 c+9 d x)}{8 b^3 \left (a+b x^2\right )}-\frac {15 \sqrt {a} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{7/2}}+\frac {c \log \left (a+b x^2\right )}{2 b^3} \] Input:
Integrate[(x^5*(c + d*x))/(a + b*x^2)^3,x]
Output:
(d*x)/b^3 - (a^2*(c + d*x))/(4*b^3*(a + b*x^2)^2) + (a*(8*c + 9*d*x))/(8*b ^3*(a + b*x^2)) - (15*Sqrt[a]*d*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(7/2)) + (c*Log[a + b*x^2])/(2*b^3)
Time = 0.72 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {530, 25, 2345, 2341, 2009}
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.
\(\displaystyle \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle -\frac {\int -\frac {\frac {4 a d x^4}{b}+\frac {4 a c x^3}{b}-\frac {4 a^2 d x^2}{b^2}-\frac {4 a^2 c x}{b^2}+\frac {a^3 d}{b^3}}{\left (b x^2+a\right )^2}dx}{4 a}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\frac {4 a d x^4}{b}+\frac {4 a c x^3}{b}-\frac {4 a^2 d x^2}{b^2}-\frac {4 a^2 c x}{b^2}+\frac {a^3 d}{b^3}}{\left (b x^2+a\right )^2}dx}{4 a}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {\frac {a^2 (8 c+9 d x)}{2 b^3 \left (a+b x^2\right )}-\frac {\int \frac {\frac {7 d a^3}{b^3}-\frac {8 d x^2 a^2}{b^2}-\frac {8 c x a^2}{b^2}}{b x^2+a}dx}{2 a}}{4 a}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \frac {\frac {a^2 (8 c+9 d x)}{2 b^3 \left (a+b x^2\right )}-\frac {\int \left (\frac {15 a^3 d-8 a^2 b c x}{b^3 \left (b x^2+a\right )}-\frac {8 a^2 d}{b^3}\right )dx}{2 a}}{4 a}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2 (8 c+9 d x)}{2 b^3 \left (a+b x^2\right )}-\frac {\frac {15 a^{5/2} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}-\frac {4 a^2 c \log \left (a+b x^2\right )}{b^3}-\frac {8 a^2 d x}{b^3}}{2 a}}{4 a}-\frac {a^2 (c+d x)}{4 b^3 \left (a+b x^2\right )^2}\) |
Input:
Int[(x^5*(c + d*x))/(a + b*x^2)^3,x]
Output:
-1/4*(a^2*(c + d*x))/(b^3*(a + b*x^2)^2) + ((a^2*(8*c + 9*d*x))/(2*b^3*(a + b*x^2)) - ((-8*a^2*d*x)/b^3 + (15*a^(5/2)*d*ArcTan[(Sqrt[b]*x)/Sqrt[a]]) /b^(7/2) - (4*a^2*c*Log[a + b*x^2])/b^3)/(2*a))/(4*a)
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] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {d x}{b^{3}}-\frac {\frac {-\frac {9}{8} a b d \,x^{3}-a b c \,x^{2}-\frac {7}{8} a^{2} d x -\frac {3}{4} a^{2} c}{\left (b \,x^{2}+a \right )^{2}}-\frac {c \ln \left (b \,x^{2}+a \right )}{2}+\frac {15 a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{3}}\) | \(83\) |
risch | \(\frac {d x}{b^{3}}+\frac {\frac {9}{8} a b d \,x^{3}+a b c \,x^{2}+\frac {7}{8} a^{2} d x +\frac {3}{4} a^{2} c}{b^{3} \left (b \,x^{2}+a \right )^{2}}+\frac {15 \ln \left (-\sqrt {-a b}\, x -a \right ) \sqrt {-a b}\, d}{16 b^{4}}+\frac {\ln \left (-\sqrt {-a b}\, x -a \right ) c}{2 b^{3}}-\frac {15 \ln \left (\sqrt {-a b}\, x -a \right ) \sqrt {-a b}\, d}{16 b^{4}}+\frac {\ln \left (\sqrt {-a b}\, x -a \right ) c}{2 b^{3}}\) | \(140\) |
Input:
int(x^5*(d*x+c)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
Output:
d*x/b^3-1/b^3*((-9/8*a*b*d*x^3-a*b*c*x^2-7/8*a^2*d*x-3/4*a^2*c)/(b*x^2+a)^ 2-1/2*c*ln(b*x^2+a)+15/8*a*d/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
Time = 0.13 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.08 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, b^{2} d x^{5} + 50 \, a b d x^{3} + 16 \, a b c x^{2} + 30 \, a^{2} d x + 12 \, a^{2} c + 15 \, {\left (b^{2} d x^{4} + 2 \, a b d x^{2} + a^{2} d\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 8 \, {\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, \frac {8 \, b^{2} d x^{5} + 25 \, a b d x^{3} + 8 \, a b c x^{2} + 15 \, a^{2} d x + 6 \, a^{2} c - 15 \, {\left (b^{2} d x^{4} + 2 \, a b d x^{2} + a^{2} d\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 4 \, {\left (b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:
integrate(x^5*(d*x+c)/(b*x^2+a)^3,x, algorithm="fricas")
Output:
[1/16*(16*b^2*d*x^5 + 50*a*b*d*x^3 + 16*a*b*c*x^2 + 30*a^2*d*x + 12*a^2*c + 15*(b^2*d*x^4 + 2*a*b*d*x^2 + a^2*d)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt( -a/b) - a)/(b*x^2 + a)) + 8*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*log(b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3), 1/8*(8*b^2*d*x^5 + 25*a*b*d*x^3 + 8 *a*b*c*x^2 + 15*a^2*d*x + 6*a^2*c - 15*(b^2*d*x^4 + 2*a*b*d*x^2 + a^2*d)*s qrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 4*(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*log (b*x^2 + a))/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3)]
Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (94) = 188\).
Time = 0.82 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.01 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\left (\frac {c}{2 b^{3}} - \frac {15 d \sqrt {- a b^{7}}}{16 b^{7}}\right ) \log {\left (x + \frac {- 16 b^{3} \left (\frac {c}{2 b^{3}} - \frac {15 d \sqrt {- a b^{7}}}{16 b^{7}}\right ) + 8 c}{15 d} \right )} + \left (\frac {c}{2 b^{3}} + \frac {15 d \sqrt {- a b^{7}}}{16 b^{7}}\right ) \log {\left (x + \frac {- 16 b^{3} \left (\frac {c}{2 b^{3}} + \frac {15 d \sqrt {- a b^{7}}}{16 b^{7}}\right ) + 8 c}{15 d} \right )} + \frac {6 a^{2} c + 7 a^{2} d x + 8 a b c x^{2} + 9 a b d x^{3}}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} + \frac {d x}{b^{3}} \] Input:
integrate(x**5*(d*x+c)/(b*x**2+a)**3,x)
Output:
(c/(2*b**3) - 15*d*sqrt(-a*b**7)/(16*b**7))*log(x + (-16*b**3*(c/(2*b**3) - 15*d*sqrt(-a*b**7)/(16*b**7)) + 8*c)/(15*d)) + (c/(2*b**3) + 15*d*sqrt(- a*b**7)/(16*b**7))*log(x + (-16*b**3*(c/(2*b**3) + 15*d*sqrt(-a*b**7)/(16* b**7)) + 8*c)/(15*d)) + (6*a**2*c + 7*a**2*d*x + 8*a*b*c*x**2 + 9*a*b*d*x* *3)/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4) + d*x/b**3
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\frac {9 \, a b d x^{3} + 8 \, a b c x^{2} + 7 \, a^{2} d x + 6 \, a^{2} c}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} - \frac {15 \, a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {d x}{b^{3}} + \frac {c \log \left (b x^{2} + a\right )}{2 \, b^{3}} \] Input:
integrate(x^5*(d*x+c)/(b*x^2+a)^3,x, algorithm="maxima")
Output:
1/8*(9*a*b*d*x^3 + 8*a*b*c*x^2 + 7*a^2*d*x + 6*a^2*c)/(b^5*x^4 + 2*a*b^4*x ^2 + a^2*b^3) - 15/8*a*d*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + d*x/b^3 + 1/2*c*log(b*x^2 + a)/b^3
Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=-\frac {15 \, a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {d x}{b^{3}} + \frac {c \log \left (b x^{2} + a\right )}{2 \, b^{3}} + \frac {9 \, a b d x^{3} + 8 \, a b c x^{2} + 7 \, a^{2} d x + 6 \, a^{2} c}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \] Input:
integrate(x^5*(d*x+c)/(b*x^2+a)^3,x, algorithm="giac")
Output:
-15/8*a*d*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + d*x/b^3 + 1/2*c*log(b*x^ 2 + a)/b^3 + 1/8*(9*a*b*d*x^3 + 8*a*b*c*x^2 + 7*a^2*d*x + 6*a^2*c)/((b*x^2 + a)^2*b^3)
Time = 6.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {7\,d\,a^2\,x}{8}+\frac {3\,c\,a^2}{4}+\frac {9\,b\,d\,a\,x^3}{8}+b\,c\,a\,x^2}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {c\,\ln \left (b\,x^2+a\right )}{2\,b^3}+\frac {d\,x}{b^3}-\frac {15\,\sqrt {a}\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,b^{7/2}} \] Input:
int((x^5*(c + d*x))/(a + b*x^2)^3,x)
Output:
((3*a^2*c)/4 + (7*a^2*d*x)/8 + a*b*c*x^2 + (9*a*b*d*x^3)/8)/(a^2*b^3 + b^5 *x^4 + 2*a*b^4*x^2) + (c*log(a + b*x^2))/(2*b^3) + (d*x)/b^3 - (15*a^(1/2) *d*atan((b^(1/2)*x)/a^(1/2)))/(8*b^(7/2))
Time = 0.18 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.90 \[ \int \frac {x^5 (c+d x)}{\left (a+b x^2\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} d -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b d \,x^{2}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} d \,x^{4}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b c +8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} c \,x^{2}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} c \,x^{4}+2 a^{2} b c +15 a^{2} b d x +25 a \,b^{2} d \,x^{3}-4 b^{3} c \,x^{4}+8 b^{3} d \,x^{5}}{8 b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^5*(d*x+c)/(b*x^2+a)^3,x)
Output:
( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*d - 30*sqrt(b)*s qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*d*x**2 - 15*sqrt(b)*sqrt(a)*atan( (b*x)/(sqrt(b)*sqrt(a)))*b**2*d*x**4 + 4*log(a + b*x**2)*a**2*b*c + 8*log( a + b*x**2)*a*b**2*c*x**2 + 4*log(a + b*x**2)*b**3*c*x**4 + 2*a**2*b*c + 1 5*a**2*b*d*x + 25*a*b**2*d*x**3 - 4*b**3*c*x**4 + 8*b**3*d*x**5)/(8*b**4*( a**2 + 2*a*b*x**2 + b**2*x**4))