Integrand size = 20, antiderivative size = 137 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=-\frac {5 a c (A-B x) \sqrt {a+c x^2}}{2 x}-\frac {5 (a B-A c x) \left (a+c x^2\right )^{3/2}}{6 x^2}-\frac {(A-B x) \left (a+c x^2\right )^{5/2}}{3 x^3}+\frac {5}{2} a A c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\frac {5}{2} a^{3/2} B c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right ) \]
-5/6*(-A*c*x+B*a)*(c*x^2+a)^(3/2)/x^2-1/3*(-B*x+A)*(c*x^2+a)^(5/2)/x^3+5/2 *a*A*c^(3/2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))-5/2*a^(3/2)*B*c*arctanh((c *x^2+a)^(1/2)/a^(1/2))-5/2*a*c*(-B*x+A)*(c*x^2+a)^(1/2)/x
Time = 0.52 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=\frac {\sqrt {a+c x^2} \left (14 a c x^2 (-A+B x)+c^2 x^4 (3 A+2 B x)-a^2 (2 A+3 B x)\right )}{6 x^3}+5 a^{3/2} B c \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )-\frac {5}{2} a A c^{3/2} \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right ) \]
(Sqrt[a + c*x^2]*(14*a*c*x^2*(-A + B*x) + c^2*x^4*(3*A + 2*B*x) - a^2*(2*A + 3*B*x)))/(6*x^3) + 5*a^(3/2)*B*c*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/ Sqrt[a]] - (5*a*A*c^(3/2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/2
Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {537, 25, 536, 535, 27, 538, 224, 219, 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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^{5/2} (A+B x)}{x^4} \, dx\) |
| \(\Big \downarrow \) 537 |
| \(\displaystyle -\frac {5}{6} c \int -\frac {(2 A+3 B x) \left (c x^2+a\right )^{3/2}}{x^2}dx-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5}{6} c \int \frac {(2 A+3 B x) \left (c x^2+a\right )^{3/2}}{x^2}dx-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 536 |
| \(\displaystyle \frac {5}{6} c \left (\int \frac {(3 a B+6 A c x) \sqrt {c x^2+a}}{x}dx-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 535 |
| \(\displaystyle \frac {5}{6} c \left (\frac {1}{2} a \int \frac {6 (a B+A c x)}{x \sqrt {c x^2+a}}dx-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{6} c \left (3 a \int \frac {a B+A c x}{x \sqrt {c x^2+a}}dx-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 538 |
| \(\displaystyle \frac {5}{6} c \left (3 a \left (A c \int \frac {1}{\sqrt {c x^2+a}}dx+a B \int \frac {1}{x \sqrt {c x^2+a}}dx\right )-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {5}{6} c \left (3 a \left (A c \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}+a B \int \frac {1}{x \sqrt {c x^2+a}}dx\right )-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {5}{6} c \left (3 a \left (a B \int \frac {1}{x \sqrt {c x^2+a}}dx+A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )\right )-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {5}{6} c \left (3 a \left (\frac {1}{2} a B \int \frac {1}{x^2 \sqrt {c x^2+a}}dx^2+A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )\right )-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {5}{6} c \left (3 a \left (\frac {a B \int \frac {1}{\frac {x^4}{c}-\frac {a}{c}}d\sqrt {c x^2+a}}{c}+A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )\right )-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {5}{6} c \left (3 a \left (A \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-\sqrt {a} B \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )\right )-\frac {\left (a+c x^2\right )^{3/2} (2 A-B x)}{x}+3 \sqrt {a+c x^2} (a B+A c x)\right )-\frac {\left (a+c x^2\right )^{5/2} (2 A+3 B x)}{6 x^3}\) |
-1/6*((2*A + 3*B*x)*(a + c*x^2)^(5/2))/x^3 + (5*c*(3*(a*B + A*c*x)*Sqrt[a + c*x^2] - ((2*A - B*x)*(a + c*x^2)^(3/2))/x + 3*a*(A*Sqrt[c]*ArcTanh[(Sqr t[c]*x)/Sqrt[a + c*x^2]] - Sqrt[a]*B*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])))/6
3.4.47.3.1 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[(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[((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[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_.)*((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[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p + 1) Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_)^2, x_Symbol] :> S imp[(-(2*c*p - d*x))*((a + b*x^2)^p/(2*p*x)), x] + Int[(a*d + 2*b*c*p*x)*(( a + b*x^2)^(p - 1)/x), x] /; FreeQ[{a, b, c, d}, x] && GtQ[p, 0] && Integer Q[2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), x] - Simp[2*b*(p/((m + 1)*(m + 2))) Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) *x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && GtQ[p, 0] && !ILtQ[m + 2*p + 3, 0] && IntegerQ[2*p]
Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp [c Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d Int[1/Sqrt[a + b*x^2], x] , x] /; FreeQ[{a, b, c, d}, x]
Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99
| method | result | size |
| risch | \(-\frac {a \sqrt {c \,x^{2}+a}\, \left (14 A c \,x^{2}+3 a B x +2 a A \right )}{6 x^{3}}+\frac {5 c^{\frac {3}{2}} A a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2}+\frac {B \,c^{2} x^{2} \sqrt {c \,x^{2}+a}}{3}+\frac {7 B c a \sqrt {c \,x^{2}+a}}{3}+\frac {A \,c^{2} x \sqrt {c \,x^{2}+a}}{2}-\frac {5 c B \,a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{2}\) | \(135\) |
| default | \(B \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 c \left (\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {c \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )+A \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 c \left (-\frac {\left (c \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 c \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )\) | \(211\) |
-1/6*a*(c*x^2+a)^(1/2)*(14*A*c*x^2+3*B*a*x+2*A*a)/x^3+5/2*c^(3/2)*A*a*ln(x *c^(1/2)+(c*x^2+a)^(1/2))+1/3*B*c^2*x^2*(c*x^2+a)^(1/2)+7/3*B*c*a*(c*x^2+a )^(1/2)+1/2*A*c^2*x*(c*x^2+a)^(1/2)-5/2*c*B*a^(3/2)*ln((2*a+2*a^(1/2)*(c*x ^2+a)^(1/2))/x)
Time = 0.32 (sec) , antiderivative size = 529, normalized size of antiderivative = 3.86 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=\left [\frac {15 \, A a c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 15 \, B a^{\frac {3}{2}} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, -\frac {30 \, A a \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 15 \, B a^{\frac {3}{2}} c x^{3} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, \frac {30 \, B \sqrt {-a} a c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + 15 \, A a c^{\frac {3}{2}} x^{3} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, x^{3}}, -\frac {15 \, A a \sqrt {-c} c x^{3} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - 15 \, B \sqrt {-a} a c x^{3} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 14 \, B a c x^{3} - 14 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, x^{3}}\right ] \]
[1/12*(15*A*a*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 15*B*a^(3/2)*c*x^3*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, -1/12*(30*A*a*sqrt(-c)*c*x^3*arctan(sqrt(-c )*x/sqrt(c*x^2 + a)) - 15*B*a^(3/2)*c*x^3*log(-(c*x^2 - 2*sqrt(c*x^2 + a)* sqrt(a) + 2*a)/x^2) - 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a*c*x^3 - 14*A*a *c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, 1/12*(30*B*sqrt(-a)*a* c*x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + 15*A*a*c^(3/2)*x^3*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B*a* c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3, -1/6*(15 *A*a*sqrt(-c)*c*x^3*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 15*B*sqrt(-a)*a*c *x^3*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (2*B*c^2*x^5 + 3*A*c^2*x^4 + 14*B* a*c*x^3 - 14*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/x^3]
Time = 3.57 (sec) , antiderivative size = 338, normalized size of antiderivative = 2.47 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=- \frac {2 A a^{\frac {3}{2}} c}{x \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {2 A \sqrt {a} c^{2} x}{\sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A a^{2} \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{3 x^{2}} - \frac {A a c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3} + 2 A a c^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )} + A c^{2} \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + c x^{2}}}{2} & \text {for}\: c \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) - \frac {5 B a^{\frac {3}{2}} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2} - \frac {B a^{2} \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} + \frac {2 B a^{2} \sqrt {c}}{x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {2 B a c^{\frac {3}{2}} x}{\sqrt {\frac {a}{c x^{2}} + 1}} + B c^{2} \left (\begin {cases} \frac {a \sqrt {a + c x^{2}}}{3 c} + \frac {x^{2} \sqrt {a + c x^{2}}}{3} & \text {for}\: c \neq 0 \\\frac {\sqrt {a} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
-2*A*a**(3/2)*c/(x*sqrt(1 + c*x**2/a)) - 2*A*sqrt(a)*c**2*x/sqrt(1 + c*x** 2/a) - A*a**2*sqrt(c)*sqrt(a/(c*x**2) + 1)/(3*x**2) - A*a*c**(3/2)*sqrt(a/ (c*x**2) + 1)/3 + 2*A*a*c**(3/2)*asinh(sqrt(c)*x/sqrt(a)) + A*c**2*Piecewi se((a*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) + 2*c*x)/sqrt(c), Ne(a, 0) ), (x*log(x)/sqrt(c*x**2), True))/2 + x*sqrt(a + c*x**2)/2, Ne(c, 0)), (sq rt(a)*x, True)) - 5*B*a**(3/2)*c*asinh(sqrt(a)/(sqrt(c)*x))/2 - B*a**2*sqr t(c)*sqrt(a/(c*x**2) + 1)/(2*x) + 2*B*a**2*sqrt(c)/(x*sqrt(a/(c*x**2) + 1) ) + 2*B*a*c**(3/2)*x/sqrt(a/(c*x**2) + 1) + B*c**2*Piecewise((a*sqrt(a + c *x**2)/(3*c) + x**2*sqrt(a + c*x**2)/3, Ne(c, 0)), (sqrt(a)*x**2/2, True))
Time = 0.24 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=\frac {5}{2} \, \sqrt {c x^{2} + a} A c^{2} x + \frac {5 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A c^{2} x}{3 \, a} + \frac {5}{2} \, A a c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right ) - \frac {5}{2} \, B a^{\frac {3}{2}} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} B c + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c}{2 \, a} + \frac {5}{2} \, \sqrt {c x^{2} + a} B a c - \frac {4 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} A c}{3 \, a x} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} B}{2 \, a x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {7}{2}} A}{3 \, a x^{3}} \]
5/2*sqrt(c*x^2 + a)*A*c^2*x + 5/3*(c*x^2 + a)^(3/2)*A*c^2*x/a + 5/2*A*a*c^ (3/2)*arcsinh(c*x/sqrt(a*c)) - 5/2*B*a^(3/2)*c*arcsinh(a/(sqrt(a*c)*abs(x) )) + 5/6*(c*x^2 + a)^(3/2)*B*c + 1/2*(c*x^2 + a)^(5/2)*B*c/a + 5/2*sqrt(c* x^2 + a)*B*a*c - 4/3*(c*x^2 + a)^(5/2)*A*c/(a*x) - 1/2*(c*x^2 + a)^(7/2)*B /(a*x^2) - 1/3*(c*x^2 + a)^(7/2)*A/(a*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (111) = 222\).
Time = 0.30 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=\frac {5 \, B a^{2} c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5}{2} \, A a c^{\frac {3}{2}} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right ) + \frac {1}{6} \, {\left (14 \, B a c + {\left (2 \, B c^{2} x + 3 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + a} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B a^{2} c + 18 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {3}{2}} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a^{3} c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c + 14 \, A a^{4} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3}} \]
5*B*a^2*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - 5/2*A *a*c^(3/2)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a))) + 1/6*(14*B*a*c + (2*B*c ^2*x + 3*A*c^2)*x)*sqrt(c*x^2 + a) + 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^ 5*B*a^2*c + 18*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(3/2) - 24*(sqrt(c) *x - sqrt(c*x^2 + a))^2*A*a^3*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B* a^4*c + 14*A*a^4*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^3
Timed out. \[ \int \frac {(A+B x) \left (a+c x^2\right )^{5/2}}{x^4} \, dx=\int \frac {{\left (c\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right )}{x^4} \,d x \]