Integrand size = 24, antiderivative size = 105 \[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{2 a x^3}+\frac {(3 b c-4 a d) \sqrt {a x^2+b x^3}}{4 a^2 x^2}-\frac {b (3 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{4 a^{5/2}} \] Output:
-1/2*c*(b*x^3+a*x^2)^(1/2)/a/x^3+1/4*(-4*a*d+3*b*c)*(b*x^3+a*x^2)^(1/2)/a^ 2/x^2-1/4*b*(-4*a*d+3*b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(5/2)
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\frac {-\sqrt {a} (a+b x) (-3 b c x+2 a (c+2 d x))-b (3 b c-4 a d) x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{5/2} x \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(c + d*x)/(x^2*Sqrt[a*x^2 + b*x^3]),x]
Output:
(-(Sqrt[a]*(a + b*x)*(-3*b*c*x + 2*a*(c + 2*d*x))) - b*(3*b*c - 4*a*d)*x^2 *Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(5/2)*x*Sqrt[x^2*(a + b*x)])
Time = 0.44 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1944, 1931, 1914, 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.
\(\displaystyle \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle -\frac {(3 b c-4 a d) \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {c \sqrt {a x^2+b x^3}}{2 a x^3}\) |
\(\Big \downarrow \) 1931 |
\(\displaystyle -\frac {(3 b c-4 a d) \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {c \sqrt {a x^2+b x^3}}{2 a x^3}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle -\frac {(3 b c-4 a d) \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {c \sqrt {a x^2+b x^3}}{2 a x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right ) (3 b c-4 a d)}{4 a}-\frac {c \sqrt {a x^2+b x^3}}{2 a x^3}\) |
Input:
Int[(c + d*x)/(x^2*Sqrt[a*x^2 + b*x^3]),x]
Output:
-1/2*(c*Sqrt[a*x^2 + b*x^3])/(a*x^3) - ((3*b*c - 4*a*d)*(-(Sqrt[a*x^2 + b* x^3]/(a*x^2)) + (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/a^(3/2)))/(4* a)
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_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]
Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol ] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))) I nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[ m + j*p + 1, 0]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 0.46 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.41
method | result | size |
pseudoelliptic | \(-\frac {c \sqrt {b x +a}}{a x}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \left (a d -\frac {b c}{2}\right )}{a^{\frac {3}{2}}}\) | \(43\) |
risch | \(-\frac {\left (b x +a \right ) \left (4 a d x -3 c b x +2 a c \right )}{4 a^{2} x \sqrt {x^{2} \left (b x +a \right )}}+\frac {\left (4 a d -3 b c \right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{4 a^{\frac {5}{2}} \sqrt {x^{2} \left (b x +a \right )}}\) | \(87\) |
default | \(-\frac {\sqrt {b x +a}\, \left (4 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {7}{2}} d -3 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}} b c -4 \sqrt {b x +a}\, a^{\frac {9}{2}} d +5 \sqrt {b x +a}\, a^{\frac {7}{2}} b c -4 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{3} b^{2} d \,x^{2}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{3} c \,x^{2}\right )}{4 x b \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {9}{2}}}\) | \(135\) |
Input:
int((d*x+c)/x^2/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-c/a*(b*x+a)^(1/2)/x-2*arctanh((b*x+a)^(1/2)/a^(1/2))*(a*d-1/2*b*c)/a^(3/2 )
Time = 0.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.88 \[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\left [-\frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (2 \, a^{2} c - {\left (3 \, a b c - 4 \, a^{2} d\right )} x\right )}}{8 \, a^{3} x^{3}}, \frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) - \sqrt {b x^{3} + a x^{2}} {\left (2 \, a^{2} c - {\left (3 \, a b c - 4 \, a^{2} d\right )} x\right )}}{4 \, a^{3} x^{3}}\right ] \] Input:
integrate((d*x+c)/x^2/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
[-1/8*((3*b^2*c - 4*a*b*d)*sqrt(a)*x^3*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*sqrt(b*x^3 + a*x^2)*(2*a^2*c - (3*a*b*c - 4*a^2* d)*x))/(a^3*x^3), 1/4*((3*b^2*c - 4*a*b*d)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) - sqrt(b*x^3 + a*x^2)*(2*a^2*c - (3*a*b*c - 4*a^2*d)*x))/(a^3*x^3)]
\[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\int \frac {c + d x}{x^{2} \sqrt {x^{2} \left (a + b x\right )}}\, dx \] Input:
integrate((d*x+c)/x**2/(b*x**3+a*x**2)**(1/2),x)
Output:
Integral((c + d*x)/(x**2*sqrt(x**2*(a + b*x))), x)
\[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\int { \frac {d x + c}{\sqrt {b x^{3} + a x^{2}} x^{2}} \,d x } \] Input:
integrate((d*x+c)/x^2/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate((d*x + c)/(sqrt(b*x^3 + a*x^2)*x^2), x)
Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.10 \[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\frac {\frac {{\left (3 \, b^{3} c - 4 \, a b^{2} d\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} c - 5 \, \sqrt {b x + a} a b^{3} c - 4 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{2} d + 4 \, \sqrt {b x + a} a^{2} b^{2} d}{a^{2} b^{2} x^{2}}}{4 \, b \mathrm {sgn}\left (x\right )} \] Input:
integrate((d*x+c)/x^2/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
1/4*((3*b^3*c - 4*a*b^2*d)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^(3/2)*b^3*c - 5*sqrt(b*x + a)*a*b^3*c - 4*(b*x + a)^(3/2)*a* b^2*d + 4*sqrt(b*x + a)*a^2*b^2*d)/(a^2*b^2*x^2))/(b*sgn(x))
Timed out. \[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\int \frac {c+d\,x}{x^2\,\sqrt {b\,x^3+a\,x^2}} \,d x \] Input:
int((c + d*x)/(x^2*(a*x^2 + b*x^3)^(1/2)),x)
Output:
int((c + d*x)/(x^2*(a*x^2 + b*x^3)^(1/2)), x)
Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.26 \[ \int \frac {c+d x}{x^2 \sqrt {a x^2+b x^3}} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2} c -8 \sqrt {b x +a}\, a^{2} d x +6 \sqrt {b x +a}\, a b c x -4 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b d \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} c \,x^{2}+4 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b d \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} c \,x^{2}}{8 a^{3} x^{2}} \] Input:
int((d*x+c)/x^2/(b*x^3+a*x^2)^(1/2),x)
Output:
( - 4*sqrt(a + b*x)*a**2*c - 8*sqrt(a + b*x)*a**2*d*x + 6*sqrt(a + b*x)*a* b*c*x - 4*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b*d*x**2 + 3*sqrt(a)*log( sqrt(a + b*x) - sqrt(a))*b**2*c*x**2 + 4*sqrt(a)*log(sqrt(a + b*x) + sqrt( a))*a*b*d*x**2 - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**2*c*x**2)/(8*a* *3*x**2)