Integrand size = 18, antiderivative size = 93 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}+\frac {d x}{35 a b \left (a+b x^2\right )^{5/2}}+\frac {4 d x}{105 a^2 b \left (a+b x^2\right )^{3/2}}+\frac {8 d x}{105 a^3 b \sqrt {a+b x^2}} \] Output:
-1/7*(d*x+c)/b/(b*x^2+a)^(7/2)+1/35*d*x/a/b/(b*x^2+a)^(5/2)+4/105*d*x/a^2/ b/(b*x^2+a)^(3/2)+8/105*d*x/a^3/b/(b*x^2+a)^(1/2)
Time = 0.55 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-15 a^3 c+35 a^2 b d x^3+28 a b^2 d x^5+8 b^3 d x^7}{105 a^3 b \left (a+b x^2\right )^{7/2}} \] Input:
Integrate[(x*(c + d*x))/(a + b*x^2)^(9/2),x]
Output:
(-15*a^3*c + 35*a^2*b*d*x^3 + 28*a*b^2*d*x^5 + 8*b^3*d*x^7)/(105*a^3*b*(a + b*x^2)^(7/2))
Time = 0.32 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {530, 25, 27, 209, 209, 208}
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.
\(\displaystyle \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx\) |
\(\Big \downarrow \) 530 |
\(\displaystyle -\frac {\int -\frac {a d}{b \left (b x^2+a\right )^{7/2}}dx}{7 a}-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a d}{b \left (b x^2+a\right )^{7/2}}dx}{7 a}-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \int \frac {1}{\left (b x^2+a\right )^{7/2}}dx}{7 b}-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {d \left (\frac {4 \int \frac {1}{\left (b x^2+a\right )^{5/2}}dx}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 b}-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {d \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 b}-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {d \left (\frac {4 \left (\frac {2 x}{3 a^2 \sqrt {a+b x^2}}+\frac {x}{3 a \left (a+b x^2\right )^{3/2}}\right )}{5 a}+\frac {x}{5 a \left (a+b x^2\right )^{5/2}}\right )}{7 b}-\frac {c+d x}{7 b \left (a+b x^2\right )^{7/2}}\) |
Input:
Int[(x*(c + d*x))/(a + b*x^2)^(9/2),x]
Output:
-1/7*(c + d*x)/(b*(a + b*x^2)^(7/2)) + (d*(x/(5*a*(a + b*x^2)^(5/2)) + (4* (x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^2])))/(5*a)))/(7*b)
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)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 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]
Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {-8 b^{3} d \,x^{7}-28 a \,b^{2} d \,x^{5}-35 a^{2} b d \,x^{3}+15 c \,a^{3}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3} b}\) | \(54\) |
trager | \(-\frac {-8 b^{3} d \,x^{7}-28 a \,b^{2} d \,x^{5}-35 a^{2} b d \,x^{3}+15 c \,a^{3}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3} b}\) | \(54\) |
orering | \(-\frac {-8 b^{3} d \,x^{7}-28 a \,b^{2} d \,x^{5}-35 a^{2} b d \,x^{3}+15 c \,a^{3}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{3} b}\) | \(54\) |
default | \(-\frac {c}{7 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+d \left (-\frac {x}{6 b \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {a \left (\frac {x}{7 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}+\frac {\frac {6 x}{35 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}\right )}{7 a}}{a}\right )}{6 b}\right )\) | \(114\) |
Input:
int(x*(d*x+c)/(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/105*(-8*b^3*d*x^7-28*a*b^2*d*x^5-35*a^2*b*d*x^3+15*a^3*c)/(b*x^2+a)^(7/ 2)/a^3/b
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (8 \, b^{3} d x^{7} + 28 \, a b^{2} d x^{5} + 35 \, a^{2} b d x^{3} - 15 \, a^{3} c\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{3} b^{5} x^{8} + 4 \, a^{4} b^{4} x^{6} + 6 \, a^{5} b^{3} x^{4} + 4 \, a^{6} b^{2} x^{2} + a^{7} b\right )}} \] Input:
integrate(x*(d*x+c)/(b*x^2+a)^(9/2),x, algorithm="fricas")
Output:
1/105*(8*b^3*d*x^7 + 28*a*b^2*d*x^5 + 35*a^2*b*d*x^3 - 15*a^3*c)*sqrt(b*x^ 2 + a)/(a^3*b^5*x^8 + 4*a^4*b^4*x^6 + 6*a^5*b^3*x^4 + 4*a^6*b^2*x^2 + a^7* b)
Time = 21.45 (sec) , antiderivative size = 612, normalized size of antiderivative = 6.58 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=c \left (\begin {cases} - \frac {1}{7 a^{3} b \sqrt {a + b x^{2}} + 21 a^{2} b^{2} x^{2} \sqrt {a + b x^{2}} + 21 a b^{3} x^{4} \sqrt {a + b x^{2}} + 7 b^{4} x^{6} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {9}{2}}} & \text {otherwise} \end {cases}\right ) + d \left (\frac {35 a^{5} x^{3}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {63 a^{4} b x^{5}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {36 a^{3} b^{2} x^{7}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {8 a^{2} b^{3} x^{9}}{105 a^{\frac {19}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {17}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 630 a^{\frac {15}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 420 a^{\frac {13}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}} + 105 a^{\frac {11}{2}} b^{4} x^{8} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:
integrate(x*(d*x+c)/(b*x**2+a)**(9/2),x)
Output:
c*Piecewise((-1/(7*a**3*b*sqrt(a + b*x**2) + 21*a**2*b**2*x**2*sqrt(a + b* x**2) + 21*a*b**3*x**4*sqrt(a + b*x**2) + 7*b**4*x**6*sqrt(a + b*x**2)), N e(b, 0)), (x**2/(2*a**(9/2)), True)) + d*(35*a**5*x**3/(105*a**(19/2)*sqrt (1 + b*x**2/a) + 420*a**(17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b **2*x**4*sqrt(1 + b*x**2/a) + 420*a**(13/2)*b**3*x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x**2/a)) + 63*a**4*b*x**5/(105*a**(19/ 2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**( 15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a**(13/2)*b**3*x**6*sqrt(1 + b*x* *2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x**2/a)) + 36*a**3*b**2*x**7/(1 05*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x**2*sqrt(1 + b*x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a**(13/2)*b**3*x**6*sqr t(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x**2/a)) + 8*a**2*b** 3*x**9/(105*a**(19/2)*sqrt(1 + b*x**2/a) + 420*a**(17/2)*b*x**2*sqrt(1 + b *x**2/a) + 630*a**(15/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 420*a**(13/2)*b**3 *x**6*sqrt(1 + b*x**2/a) + 105*a**(11/2)*b**4*x**8*sqrt(1 + b*x**2/a)))
Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=-\frac {d x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, d x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, d x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {d x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {c}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} \] Input:
integrate(x*(d*x+c)/(b*x^2+a)^(9/2),x, algorithm="maxima")
Output:
-1/7*d*x/((b*x^2 + a)^(7/2)*b) + 8/105*d*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105 *d*x/((b*x^2 + a)^(3/2)*a^2*b) + 1/35*d*x/((b*x^2 + a)^(5/2)*a*b) - 1/7*c/ ((b*x^2 + a)^(7/2)*b)
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.58 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {{\left (4 \, {\left (\frac {2 \, b^{2} d x^{2}}{a^{3}} + \frac {7 \, b d}{a^{2}}\right )} x^{2} + \frac {35 \, d}{a}\right )} x^{3} - \frac {15 \, c}{b}}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \] Input:
integrate(x*(d*x+c)/(b*x^2+a)^(9/2),x, algorithm="giac")
Output:
1/105*((4*(2*b^2*d*x^2/a^3 + 7*b*d/a^2)*x^2 + 35*d/a)*x^3 - 15*c/b)/(b*x^2 + a)^(7/2)
Time = 8.58 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {8\,d\,x}{105\,a^3\,b\,\sqrt {b\,x^2+a}}-\frac {\frac {c}{7\,b}+\frac {d\,x}{7\,b}}{{\left (b\,x^2+a\right )}^{7/2}}+\frac {4\,d\,x}{105\,a^2\,b\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {d\,x}{35\,a\,b\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:
int((x*(c + d*x))/(a + b*x^2)^(9/2),x)
Output:
(8*d*x)/(105*a^3*b*(a + b*x^2)^(1/2)) - (c/(7*b) + (d*x)/(7*b))/(a + b*x^2 )^(7/2) + (4*d*x)/(105*a^2*b*(a + b*x^2)^(3/2)) + (d*x)/(35*a*b*(a + b*x^2 )^(5/2))
Time = 0.21 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.91 \[ \int \frac {x (c+d x)}{\left (a+b x^2\right )^{9/2}} \, dx=\frac {-15 \sqrt {b \,x^{2}+a}\, a^{3} b c +35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d \,x^{3}+28 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{5}+8 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{7}-8 \sqrt {b}\, a^{4} d -32 \sqrt {b}\, a^{3} b d \,x^{2}-48 \sqrt {b}\, a^{2} b^{2} d \,x^{4}-32 \sqrt {b}\, a \,b^{3} d \,x^{6}-8 \sqrt {b}\, b^{4} d \,x^{8}}{105 a^{3} b^{2} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(x*(d*x+c)/(b*x^2+a)^(9/2),x)
Output:
( - 15*sqrt(a + b*x**2)*a**3*b*c + 35*sqrt(a + b*x**2)*a**2*b**2*d*x**3 + 28*sqrt(a + b*x**2)*a*b**3*d*x**5 + 8*sqrt(a + b*x**2)*b**4*d*x**7 - 8*sqr t(b)*a**4*d - 32*sqrt(b)*a**3*b*d*x**2 - 48*sqrt(b)*a**2*b**2*d*x**4 - 32* sqrt(b)*a*b**3*d*x**6 - 8*sqrt(b)*b**4*d*x**8)/(105*a**3*b**2*(a**4 + 4*a* *3*b*x**2 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b**4*x**8))