Integrand size = 24, antiderivative size = 83 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{x^2}+\frac {2 d \sqrt {a x^2+b x^3}}{x}-\frac {(b c+2 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{\sqrt {a}} \] Output:
-c*(b*x^3+a*x^2)^(1/2)/x^2+2*d*(b*x^3+a*x^2)^(1/2)/x-(2*a*d+b*c)*arctanh(( b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(1/2)
Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=\frac {\sqrt {a} (a+b x) (-c+2 d x)-(b c+2 a d) x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*Sqrt[a*x^2 + b*x^3])/x^3,x]
Output:
(Sqrt[a]*(a + b*x)*(-c + 2*d*x) - (b*c + 2*a*d)*x*Sqrt[a + b*x]*ArcTanh[Sq rt[a + b*x]/Sqrt[a]])/(Sqrt[a]*Sqrt[x^2*(a + b*x)])
Time = 0.43 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1944, 1927, 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.
\(\displaystyle \int \frac {\sqrt {a x^2+b x^3} (c+d x)}{x^3} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle \frac {(2 a d+b c) \int \frac {\sqrt {b x^3+a x^2}}{x^2}dx}{2 a}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{a x^4}\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle \frac {(2 a d+b c) \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )}{2 a}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{a x^4}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle \frac {(2 a d+b c) \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )}{2 a}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{a x^4}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )\right ) (2 a d+b c)}{2 a}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{a x^4}\) |
Input:
Int[((c + d*x)*Sqrt[a*x^2 + b*x^3])/x^3,x]
Output:
-((c*(a*x^2 + b*x^3)^(3/2))/(a*x^4)) + ((b*c + 2*a*d)*((2*Sqrt[a*x^2 + b*x ^3])/x - 2*Sqrt[a]*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]]))/(2*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*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* (n - j)*(p/(c^j*(m + n*p + 1))) Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) , x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[p] && LtQ[0, j, n] && (Int egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*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 = 1.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77
method | result | size |
pseudoelliptic | \(-\frac {b \,x^{2} \left (a d -\frac {b c}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {\left (\left (4 d x +2 c \right ) a^{\frac {3}{2}}+\sqrt {a}\, b c x \right ) \sqrt {b x +a}}{4}}{a^{\frac {3}{2}} x^{2}}\) | \(64\) |
risch | \(-\frac {c \sqrt {x^{2} \left (b x +a \right )}}{x^{2}}+\frac {\left (2 \sqrt {b x +a}\, d -\frac {\left (2 a d +b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{x \sqrt {b x +a}}\) | \(77\) |
default | \(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (-2 \sqrt {b x +a}\, d x \sqrt {a}+2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a d x +\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b c x +c \sqrt {b x +a}\, \sqrt {a}\right )}{x^{2} \sqrt {b x +a}\, \sqrt {a}}\) | \(89\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/a^(3/2)*(b*x^2*(a*d-1/4*b*c)*arctanh((b*x+a)^(1/2)/a^(1/2))+1/4*((4*d*x +2*c)*a^(3/2)+a^(1/2)*b*c*x)*(b*x+a)^(1/2))/x^2
Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.96 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=\left [\frac {{\left (b c + 2 \, a d\right )} \sqrt {a} x^{2} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (2 \, a d x - a c\right )}}{2 \, a x^{2}}, \frac {{\left (b c + 2 \, a d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (2 \, a d x - a c\right )}}{a x^{2}}\right ] \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^3,x, algorithm="fricas")
Output:
[1/2*((b*c + 2*a*d)*sqrt(a)*x^2*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*d*x - a*c))/(a*x^2), ((b*c + 2 *a*d)*sqrt(-a)*x^2*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + sq rt(b*x^3 + a*x^2)*(2*a*d*x - a*c))/(a*x^2)]
\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x\right )} \left (c + d x\right )}{x^{3}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(1/2)/x**3,x)
Output:
Integral(sqrt(x**2*(a + b*x))*(c + d*x)/x**3, x)
\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=\int { \frac {\sqrt {b x^{3} + a x^{2}} {\left (d x + c\right )}}{x^{3}} \,d x } \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^3,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^3 + a*x^2)*(d*x + c)/x^3, x)
Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx={\left (\frac {2 \, \sqrt {b x + a} d \mathrm {sgn}\left (x\right )}{b} + \frac {{\left (b c \mathrm {sgn}\left (x\right ) + 2 \, a d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} b} - \frac {\sqrt {b x + a} c \mathrm {sgn}\left (x\right )}{b x}\right )} b \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^3,x, algorithm="giac")
Output:
(2*sqrt(b*x + a)*d*sgn(x)/b + (b*c*sgn(x) + 2*a*d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*b) - sqrt(b*x + a)*c*sgn(x)/(b*x))*b
Timed out. \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=\int \frac {\sqrt {b\,x^3+a\,x^2}\,\left (c+d\,x\right )}{x^3} \,d x \] Input:
int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^3,x)
Output:
int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^3, x)
Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^3} \, dx=\frac {-2 \sqrt {b x +a}\, a c +4 \sqrt {b x +a}\, a d x +2 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a d x +\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b c x -2 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a d x -\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b c x}{2 a x} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^3,x)
Output:
( - 2*sqrt(a + b*x)*a*c + 4*sqrt(a + b*x)*a*d*x + 2*sqrt(a)*log(sqrt(a + b *x) - sqrt(a))*a*d*x + sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b*c*x - 2*sqrt (a)*log(sqrt(a + b*x) + sqrt(a))*a*d*x - sqrt(a)*log(sqrt(a + b*x) + sqrt( a))*b*c*x)/(2*a*x)