Integrand size = 20, antiderivative size = 58 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=\frac {2 b \sqrt {c+d x}}{d}-\frac {a \sqrt {c+d x}}{c x}+\frac {a d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \] Output:
2*b*(d*x+c)^(1/2)/d-a*(d*x+c)^(1/2)/c/x+a*d*arctanh((d*x+c)^(1/2)/c^(1/2)) /c^(3/2)
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=\frac {(-a d+2 b c x) \sqrt {c+d x}}{c d x}+\frac {a d \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \] Input:
Integrate[(a + b*x^2)/(x^2*Sqrt[c + d*x]),x]
Output:
((-(a*d) + 2*b*c*x)*Sqrt[c + d*x])/(c*d*x) + (a*d*ArcTanh[Sqrt[c + d*x]/Sq rt[c]])/c^(3/2)
Time = 0.35 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {517, 1471, 25, 299, 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.
\(\displaystyle \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 517 |
\(\displaystyle \frac {2 \int \frac {b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2}{d^2 x^2}d\sqrt {c+d x}}{d}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {2 \left (-\frac {\int \frac {2 b c^2-2 b (c+d x) c+a d^2}{d x}d\sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x}}{2 c x}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \left (\frac {\int -\frac {2 b c^2-2 b (c+d x) c+a d^2}{d x}d\sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x}}{2 c x}\right )}{d}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {2 \left (\frac {a d^2 \int -\frac {1}{d x}d\sqrt {c+d x}+2 b c \sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x}}{2 c x}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \left (\frac {\frac {a d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}+2 b c \sqrt {c+d x}}{2 c}-\frac {a d \sqrt {c+d x}}{2 c x}\right )}{d}\) |
Input:
Int[(a + b*x^2)/(x^2*Sqrt[c + d*x]),x]
Output:
(2*(-1/2*(a*d*Sqrt[c + d*x])/(c*x) + (2*b*c*Sqrt[c + d*x] + (a*d^2*ArcTanh [Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c])/(2*c)))/d
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)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {\sqrt {d x +c}\, \left (-2 c b x +a d \right )}{d c x}+\frac {a d \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{c^{\frac {3}{2}}}\) | \(47\) |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right ) a c \,d^{2} x +2 \sqrt {d x +c}\, c^{\frac {3}{2}} \left (c b x -\frac {a d}{2}\right )}{d \,c^{\frac {5}{2}} x}\) | \(52\) |
derivativedivides | \(\frac {2 b \sqrt {d x +c}+2 a \,d^{2} \left (-\frac {\sqrt {d x +c}}{2 c d x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\right )}{d}\) | \(57\) |
default | \(\frac {2 b \sqrt {d x +c}+2 a \,d^{2} \left (-\frac {\sqrt {d x +c}}{2 c d x}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\right )}{d}\) | \(57\) |
Input:
int((b*x^2+a)/x^2/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-(d*x+c)^(1/2)*(-2*b*c*x+a*d)/d/c/x+a*d*arctanh((d*x+c)^(1/2)/c^(1/2))/c^( 3/2)
Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.19 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=\left [\frac {a \sqrt {c} d^{2} x \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (2 \, b c^{2} x - a c d\right )} \sqrt {d x + c}}{2 \, c^{2} d x}, -\frac {a \sqrt {-c} d^{2} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) - {\left (2 \, b c^{2} x - a c d\right )} \sqrt {d x + c}}{c^{2} d x}\right ] \] Input:
integrate((b*x^2+a)/x^2/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
[1/2*(a*sqrt(c)*d^2*x*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(2* b*c^2*x - a*c*d)*sqrt(d*x + c))/(c^2*d*x), -(a*sqrt(-c)*d^2*x*arctan(sqrt( -c)/sqrt(d*x + c)) - (2*b*c^2*x - a*c*d)*sqrt(d*x + c))/(c^2*d*x)]
Time = 6.88 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=- \frac {a \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c \sqrt {x}} + \frac {a d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {3}{2}}} + b \left (\begin {cases} \frac {2 \sqrt {c + d x}}{d} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x**2+a)/x**2/(d*x+c)**(1/2),x)
Output:
-a*sqrt(d)*sqrt(c/(d*x) + 1)/(c*sqrt(x)) + a*d*asinh(sqrt(c)/(sqrt(d)*sqrt (x)))/c**(3/2) + b*Piecewise((2*sqrt(c + d*x)/d, Ne(d, 0)), (x/sqrt(c), Tr ue))
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.29 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=-\frac {1}{2} \, d {\left (\frac {2 \, \sqrt {d x + c} a}{{\left (d x + c\right )} c - c^{2}} + \frac {a \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}}} - \frac {4 \, \sqrt {d x + c} b}{d^{2}}\right )} \] Input:
integrate((b*x^2+a)/x^2/(d*x+c)^(1/2),x, algorithm="maxima")
Output:
-1/2*d*(2*sqrt(d*x + c)*a/((d*x + c)*c - c^2) + a*log((sqrt(d*x + c) - sqr t(c))/(sqrt(d*x + c) + sqrt(c)))/c^(3/2) - 4*sqrt(d*x + c)*b/d^2)
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=-d {\left (\frac {a \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c} - \frac {2 \, \sqrt {d x + c} b}{d^{2}} + \frac {\sqrt {d x + c} a}{c d x}\right )} \] Input:
integrate((b*x^2+a)/x^2/(d*x+c)^(1/2),x, algorithm="giac")
Output:
-d*(a*arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c) - 2*sqrt(d*x + c)*b/d^2 + sqrt(d*x + c)*a/(c*d*x))
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=\frac {2\,b\,\sqrt {c+d\,x}}{d}-\frac {a\,\sqrt {c+d\,x}}{c\,x}+\frac {a\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )}{c^{3/2}} \] Input:
int((a + b*x^2)/(x^2*(c + d*x)^(1/2)),x)
Output:
(2*b*(c + d*x)^(1/2))/d - (a*(c + d*x)^(1/2))/(c*x) + (a*d*atanh((c + d*x) ^(1/2)/c^(1/2)))/c^(3/2)
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.29 \[ \int \frac {a+b x^2}{x^2 \sqrt {c+d x}} \, dx=\frac {-2 \sqrt {d x +c}\, a c d +4 \sqrt {d x +c}\, b \,c^{2} x -\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,d^{2} x +\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,d^{2} x}{2 c^{2} d x} \] Input:
int((b*x^2+a)/x^2/(d*x+c)^(1/2),x)
Output:
( - 2*sqrt(c + d*x)*a*c*d + 4*sqrt(c + d*x)*b*c**2*x - sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a*d**2*x + sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*d**2*x )/(2*c**2*d*x)