Integrand size = 20, antiderivative size = 120 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}+\frac {3 B c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \]
3/2*B*c*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+(B*x+A)/a/x^3/(c*x^2+a)^( 1/2)-4/3*A*(c*x^2+a)^(1/2)/a^2/x^3-3/2*B*(c*x^2+a)^(1/2)/a^2/x^2+8/3*A*c*( c*x^2+a)^(1/2)/a^3/x
Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=\frac {16 A c^2 x^4+a c x^2 (8 A-9 B x)-a^2 (2 A+3 B x)}{6 a^3 x^3 \sqrt {a+c x^2}}-\frac {3 B c \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}} \]
(16*A*c^2*x^4 + a*c*x^2*(8*A - 9*B*x) - a^2*(2*A + 3*B*x))/(6*a^3*x^3*Sqrt [a + c*x^2]) - (3*B*c*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/a^(5 /2)
Time = 0.45 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {532, 25, 2338, 25, 2338, 27, 534, 243, 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.
\(\displaystyle \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 532 |
\(\displaystyle -\frac {\int -\frac {-\frac {B c x^3}{a}-\frac {A c x^2}{a}+B x+A}{x^4 \sqrt {c x^2+a}}dx}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {-\frac {B c x^3}{a}-\frac {A c x^2}{a}+B x+A}{x^4 \sqrt {c x^2+a}}dx}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {-\frac {\int -\frac {-3 B c x^2-5 A c x+3 a B}{x^3 \sqrt {c x^2+a}}dx}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {-3 B c x^2-5 A c x+3 a B}{x^3 \sqrt {c x^2+a}}dx}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle \frac {\frac {-\frac {\int \frac {a c (10 A+9 B x)}{x^2 \sqrt {c x^2+a}}dx}{2 a}-\frac {3 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {-\frac {1}{2} c \int \frac {10 A+9 B x}{x^2 \sqrt {c x^2+a}}dx-\frac {3 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 534 |
\(\displaystyle \frac {\frac {-\frac {1}{2} c \left (9 B \int \frac {1}{x \sqrt {c x^2+a}}dx-\frac {10 A \sqrt {a+c x^2}}{a x}\right )-\frac {3 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\frac {-\frac {1}{2} c \left (\frac {9}{2} B \int \frac {1}{x^2 \sqrt {c x^2+a}}dx^2-\frac {10 A \sqrt {a+c x^2}}{a x}\right )-\frac {3 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {-\frac {1}{2} c \left (\frac {9 B \int \frac {1}{\frac {x^4}{c}-\frac {a}{c}}d\sqrt {c x^2+a}}{c}-\frac {10 A \sqrt {a+c x^2}}{a x}\right )-\frac {3 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {-\frac {1}{2} c \left (-\frac {10 A \sqrt {a+c x^2}}{a x}-\frac {9 B \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\right )-\frac {3 B \sqrt {a+c x^2}}{2 x^2}}{3 a}-\frac {A \sqrt {a+c x^2}}{3 a x^3}}{a}-\frac {c (a B-A c x)}{a^3 \sqrt {a+c x^2}}\) |
-((c*(a*B - A*c*x))/(a^3*Sqrt[a + c*x^2])) + (-1/3*(A*Sqrt[a + c*x^2])/(a* x^3) + ((-3*B*Sqrt[a + c*x^2])/(2*x^2) - (c*((-10*A*Sqrt[a + c*x^2])/(a*x) - (9*B*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/Sqrt[a]))/2)/(3*a))/a
3.4.73.3.1 Defintions of rubi rules used
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_))^(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_)^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_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[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[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {\sqrt {c \,x^{2}+a}\, \left (-10 A c \,x^{2}+3 a B x +2 a A \right )}{6 a^{3} x^{3}}+\frac {c^{2} A x}{a^{3} \sqrt {c \,x^{2}+a}}-\frac {c B}{a^{2} \sqrt {c \,x^{2}+a}}+\frac {3 c B \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{2 a^{\frac {5}{2}}}\) | \(100\) |
default | \(B \left (-\frac {1}{2 a \,x^{2} \sqrt {c \,x^{2}+a}}-\frac {3 c \left (\frac {1}{a \sqrt {c \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )+A \left (-\frac {1}{3 a \,x^{3} \sqrt {c \,x^{2}+a}}-\frac {4 c \left (-\frac {1}{a x \sqrt {c \,x^{2}+a}}-\frac {2 c x}{a^{2} \sqrt {c \,x^{2}+a}}\right )}{3 a}\right )\) | \(130\) |
-1/6*(c*x^2+a)^(1/2)*(-10*A*c*x^2+3*B*a*x+2*A*a)/a^3/x^3+c^2/a^3*A*x/(c*x^ 2+a)^(1/2)-c/a^2*B/(c*x^2+a)^(1/2)+3/2*c/a^(5/2)*B*ln((2*a+2*a^(1/2)*(c*x^ 2+a)^(1/2))/x)
Time = 0.33 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.93 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=\left [\frac {9 \, {\left (B c^{2} x^{5} + B a c x^{3}\right )} \sqrt {a} \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (16 \, A c^{2} x^{4} - 9 \, B a c x^{3} + 8 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, {\left (a^{3} c x^{5} + a^{4} x^{3}\right )}}, -\frac {9 \, {\left (B c^{2} x^{5} + B a c x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (16 \, A c^{2} x^{4} - 9 \, B a c x^{3} + 8 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (a^{3} c x^{5} + a^{4} x^{3}\right )}}\right ] \]
[1/12*(9*(B*c^2*x^5 + B*a*c*x^3)*sqrt(a)*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*s qrt(a) + 2*a)/x^2) + 2*(16*A*c^2*x^4 - 9*B*a*c*x^3 + 8*A*a*c*x^2 - 3*B*a^2 *x - 2*A*a^2)*sqrt(c*x^2 + a))/(a^3*c*x^5 + a^4*x^3), -1/6*(9*(B*c^2*x^5 + B*a*c*x^3)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (16*A*c^2*x^4 - 9* B*a*c*x^3 + 8*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/(a^3*c*x^5 + a^4*x^3)]
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (114) = 228\).
Time = 11.79 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.59 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=A \left (- \frac {a^{3} c^{\frac {9}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}} + \frac {3 a^{2} c^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}} + \frac {12 a c^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}} + \frac {8 c^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}}\right ) + B \left (- \frac {1}{2 a \sqrt {c} x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 \sqrt {c}}{2 a^{2} x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {3 c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2 a^{\frac {5}{2}}}\right ) \]
A*(-a**3*c**(9/2)*sqrt(a/(c*x**2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5*x** 4 + 3*a**3*c**6*x**6) + 3*a**2*c**(11/2)*x**2*sqrt(a/(c*x**2) + 1)/(3*a**5 *c**4*x**2 + 6*a**4*c**5*x**4 + 3*a**3*c**6*x**6) + 12*a*c**(13/2)*x**4*sq rt(a/(c*x**2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5*x**4 + 3*a**3*c**6*x**6 ) + 8*c**(15/2)*x**6*sqrt(a/(c*x**2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5* x**4 + 3*a**3*c**6*x**6)) + B*(-1/(2*a*sqrt(c)*x**3*sqrt(a/(c*x**2) + 1)) - 3*sqrt(c)/(2*a**2*x*sqrt(a/(c*x**2) + 1)) + 3*c*asinh(sqrt(a)/(sqrt(c)*x ))/(2*a**(5/2)))
Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=\frac {8 \, A c^{2} x}{3 \, \sqrt {c x^{2} + a} a^{3}} + \frac {3 \, B c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {3 \, B c}{2 \, \sqrt {c x^{2} + a} a^{2}} + \frac {4 \, A c}{3 \, \sqrt {c x^{2} + a} a^{2} x} - \frac {B}{2 \, \sqrt {c x^{2} + a} a x^{2}} - \frac {A}{3 \, \sqrt {c x^{2} + a} a x^{3}} \]
8/3*A*c^2*x/(sqrt(c*x^2 + a)*a^3) + 3/2*B*c*arcsinh(a/(sqrt(a*c)*abs(x)))/ a^(5/2) - 3/2*B*c/(sqrt(c*x^2 + a)*a^2) + 4/3*A*c/(sqrt(c*x^2 + a)*a^2*x) - 1/2*B/(sqrt(c*x^2 + a)*a*x^2) - 1/3*A/(sqrt(c*x^2 + a)*a*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (98) = 196\).
Time = 0.29 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=\frac {\frac {A c^{2} x}{a^{3}} - \frac {B c}{a^{2}}}{\sqrt {c x^{2} + a}} - \frac {3 \, B c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B c - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A c^{\frac {3}{2}} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{2} c - 10 \, A a^{2} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3} a^{2}} \]
(A*c^2*x/a^3 - B*c/a^2)/sqrt(c*x^2 + a) - 3*B*c*arctan(-(sqrt(c)*x - sqrt( c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a) )^5*B*c - 6*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*c^(3/2) + 24*(sqrt(c)*x - sq rt(c*x^2 + a))^2*A*a*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^2*c - 1 0*A*a^2*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3*a^2)
Time = 10.65 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx=\frac {3\,B\,c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}}-\frac {B}{2\,a\,x^2\,\sqrt {c\,x^2+a}}-\frac {3\,B\,c}{2\,a^2\,\sqrt {c\,x^2+a}}+\frac {A\,\left (-a^2+4\,a\,c\,x^2+8\,c^2\,x^4\right )}{3\,a^3\,x^3\,\sqrt {c\,x^2+a}} \]