Integrand size = 24, antiderivative size = 111 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {b (b c+4 a d) x \sqrt {c+d x^2}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}-\frac {2 a b \left (c+d x^2\right )^{3/2}}{c x}+\frac {b (b c+4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}} \] Output:
1/2*b*(4*a*d+b*c)*x*(d*x^2+c)^(1/2)/c-1/3*a^2*(d*x^2+c)^(3/2)/c/x^3-2*a*b* (d*x^2+c)^(3/2)/c/x+1/2*b*(4*a*d+b*c)*arctanh(d^(1/2)*x/(d*x^2+c)^(1/2))/d ^(1/2)
Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {1}{6} \left (\frac {\sqrt {c+d x^2} \left (-12 a b c x^2+3 b^2 c x^4-2 a^2 \left (c+d x^2\right )\right )}{c x^3}-\frac {3 b (b c+4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{\sqrt {d}}\right ) \] Input:
Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^4,x]
Output:
((Sqrt[c + d*x^2]*(-12*a*b*c*x^2 + 3*b^2*c*x^4 - 2*a^2*(c + d*x^2)))/(c*x^ 3) - (3*b*(b*c + 4*a*d)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/Sqrt[d])/6
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {365, 27, 359, 211, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx\) |
\(\Big \downarrow \) 365 |
\(\displaystyle \frac {\int \frac {3 b c \left (b x^2+2 a\right ) \sqrt {d x^2+c}}{x^2}dx}{3 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b \int \frac {\left (b x^2+2 a\right ) \sqrt {d x^2+c}}{x^2}dx-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle b \left (\frac {(4 a d+b c) \int \sqrt {d x^2+c}dx}{c}-\frac {2 a \left (c+d x^2\right )^{3/2}}{c x}\right )-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle b \left (\frac {(4 a d+b c) \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{3/2}}{c x}\right )-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle b \left (\frac {(4 a d+b c) \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{3/2}}{c x}\right )-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle b \left (\frac {(4 a d+b c) \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )}{c}-\frac {2 a \left (c+d x^2\right )^{3/2}}{c x}\right )-\frac {a^2 \left (c+d x^2\right )^{3/2}}{3 c x^3}\) |
Input:
Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x^4,x]
Output:
-1/3*(a^2*(c + d*x^2)^(3/2))/(c*x^3) + b*((-2*a*(c + d*x^2)^(3/2))/(c*x) + ((b*c + 4*a*d)*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d *x^2]])/(2*Sqrt[d])))/c)
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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Time = 0.51 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.74
method | result | size |
risch | \(-\frac {\sqrt {x^{2} d +c}\, \left (-3 b^{2} c \,x^{4}+2 a^{2} d \,x^{2}+12 a b c \,x^{2}+2 a^{2} c \right )}{6 x^{3} c}+\frac {\left (4 a d +b c \right ) b \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\) | \(82\) |
pseudoelliptic | \(-\frac {-6 c \left (a d +\frac {b c}{4}\right ) b \,x^{3} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )+\left (d^{\frac {3}{2}} a^{2} x^{2}+c \sqrt {d}\, \left (-\frac {3}{2} b^{2} x^{4}+6 a b \,x^{2}+a^{2}\right )\right ) \sqrt {x^{2} d +c}}{3 \sqrt {d}\, x^{3} c}\) | \(90\) |
default | \(b^{2} \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )-\frac {a^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}{3 c \,x^{3}}+2 a b \left (-\frac {\left (x^{2} d +c \right )^{\frac {3}{2}}}{c x}+\frac {2 d \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )}{c}\right )\) | \(124\) |
Input:
int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/6*(d*x^2+c)^(1/2)*(-3*b^2*c*x^4+2*a^2*d*x^2+12*a*b*c*x^2+2*a^2*c)/x^3/c +1/2*(4*a*d+b*c)*b*ln(d^(1/2)*x+(d*x^2+c)^(1/2))/d^(1/2)
Time = 0.10 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {d} x^{3} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (3 \, b^{2} c d x^{4} - 2 \, a^{2} c d - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{12 \, c d x^{3}}, -\frac {3 \, {\left (b^{2} c^{2} + 4 \, a b c d\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (3 \, b^{2} c d x^{4} - 2 \, a^{2} c d - 2 \, {\left (6 \, a b c d + a^{2} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{6 \, c d x^{3}}\right ] \] Input:
integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x, algorithm="fricas")
Output:
[1/12*(3*(b^2*c^2 + 4*a*b*c*d)*sqrt(d)*x^3*log(-2*d*x^2 - 2*sqrt(d*x^2 + c )*sqrt(d)*x - c) + 2*(3*b^2*c*d*x^4 - 2*a^2*c*d - 2*(6*a*b*c*d + a^2*d^2)* x^2)*sqrt(d*x^2 + c))/(c*d*x^3), -1/6*(3*(b^2*c^2 + 4*a*b*c*d)*sqrt(-d)*x^ 3*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (3*b^2*c*d*x^4 - 2*a^2*c*d - 2*(6*a *b*c*d + a^2*d^2)*x^2)*sqrt(d*x^2 + c))/(c*d*x^3)]
Time = 1.40 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.71 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {a^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c} - \frac {2 a b \sqrt {c}}{x \sqrt {1 + \frac {d x^{2}}{c}}} + 2 a b \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )} - \frac {2 a b d x}{\sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} + b^{2} \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c + d x^{2}}}{2} & \text {for}\: d \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x**4,x)
Output:
-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - a**2*d**(3/2)*sqrt(c/(d*x**2 ) + 1)/(3*c) - 2*a*b*sqrt(c)/(x*sqrt(1 + d*x**2/c)) + 2*a*b*sqrt(d)*asinh( sqrt(d)*x/sqrt(c)) - 2*a*b*d*x/(sqrt(c)*sqrt(1 + d*x**2/c)) + b**2*Piecewi se((c*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0) ), (x*log(x)/sqrt(d*x**2), True))/2 + x*sqrt(c + d*x**2)/2, Ne(d, 0)), (sq rt(c)*x, True))
Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} x + \frac {b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} + 2 \, a b \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {2 \, \sqrt {d x^{2} + c} a b}{x} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{3 \, c x^{3}} \] Input:
integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x, algorithm="maxima")
Output:
1/2*sqrt(d*x^2 + c)*b^2*x + 1/2*b^2*c*arcsinh(d*x/sqrt(c*d))/sqrt(d) + 2*a *b*sqrt(d)*arcsinh(d*x/sqrt(c*d)) - 2*sqrt(d*x^2 + c)*a*b/x - 1/3*(d*x^2 + c)^(3/2)*a^2/(c*x^3)
Time = 0.15 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.65 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} x - \frac {{\left (b^{2} c + 4 \, a b d\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, \sqrt {d}} + \frac {2 \, {\left (6 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c \sqrt {d} + 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} d^{\frac {3}{2}} - 12 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{2} \sqrt {d} + 6 \, a b c^{3} \sqrt {d} + a^{2} c^{2} d^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{3}} \] Input:
integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x, algorithm="giac")
Output:
1/2*sqrt(d*x^2 + c)*b^2*x - 1/4*(b^2*c + 4*a*b*d)*log((sqrt(d)*x - sqrt(d* x^2 + c))^2)/sqrt(d) + 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c*sqrt(d ) + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*d^(3/2) - 12*(sqrt(d)*x - sqrt(d *x^2 + c))^2*a*b*c^2*sqrt(d) + 6*a*b*c^3*sqrt(d) + a^2*c^2*d^(3/2))/((sqrt (d)*x - sqrt(d*x^2 + c))^2 - c)^3
Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c}}{x^4} \,d x \] Input:
int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^4,x)
Output:
int(((a + b*x^2)^2*(c + d*x^2)^(1/2))/x^4, x)
Time = 0.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.61 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^4} \, dx=\frac {-4 \sqrt {d \,x^{2}+c}\, a^{2} c d -4 \sqrt {d \,x^{2}+c}\, a^{2} d^{2} x^{2}-24 \sqrt {d \,x^{2}+c}\, a b c d \,x^{2}+6 \sqrt {d \,x^{2}+c}\, b^{2} c d \,x^{4}+24 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c d \,x^{3}+6 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{2} x^{3}-4 \sqrt {d}\, a^{2} d^{2} x^{3}+8 \sqrt {d}\, a b c d \,x^{3}+\sqrt {d}\, b^{2} c^{2} x^{3}}{12 c d \,x^{3}} \] Input:
int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x^4,x)
Output:
( - 4*sqrt(c + d*x**2)*a**2*c*d - 4*sqrt(c + d*x**2)*a**2*d**2*x**2 - 24*s qrt(c + d*x**2)*a*b*c*d*x**2 + 6*sqrt(c + d*x**2)*b**2*c*d*x**4 + 24*sqrt( d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a*b*c*d*x**3 + 6*sqrt(d)*lo g((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*b**2*c**2*x**3 - 4*sqrt(d)*a**2* d**2*x**3 + 8*sqrt(d)*a*b*c*d*x**3 + sqrt(d)*b**2*c**2*x**3)/(12*c*d*x**3)