Integrand size = 24, antiderivative size = 81 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\frac {2 c \sqrt {a x^2+b x^3}}{x}+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b x^3}-2 \sqrt {a} c \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right ) \] Output:
2*c*(b*x^3+a*x^2)^(1/2)/x+2/3*d*(b*x^3+a*x^2)^(3/2)/b/x^3-2*a^(1/2)*c*arct anh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)
Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\frac {2 x \left ((a+b x) (3 b c+a d+b d x)-3 \sqrt {a} b c \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{3 b \sqrt {x^2 (a+b x)}} \] Input:
Integrate[((c + d*x)*Sqrt[a*x^2 + b*x^3])/x^2,x]
Output:
(2*x*((a + b*x)*(3*b*c + a*d + b*d*x) - 3*Sqrt[a]*b*c*Sqrt[a + b*x]*ArcTan h[Sqrt[a + b*x]/Sqrt[a]]))/(3*b*Sqrt[x^2*(a + b*x)])
Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1945, 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.
\(\displaystyle \int \frac {\sqrt {a x^2+b x^3} (c+d x)}{x^2} \, dx\) |
\(\Big \downarrow \) 1945 |
\(\displaystyle c \int \frac {\sqrt {b x^3+a x^2}}{x^2}dx+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b x^3}\) |
\(\Big \downarrow \) 1927 |
\(\displaystyle c \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b x^3}\) |
\(\Big \downarrow \) 1914 |
\(\displaystyle c \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle c \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )\right )+\frac {2 d \left (a x^2+b x^3\right )^{3/2}}{3 b x^3}\) |
Input:
Int[((c + d*x)*Sqrt[a*x^2 + b*x^3])/x^2,x]
Output:
(2*d*(a*x^2 + b*x^3)^(3/2))/(3*b*x^3) + c*((2*Sqrt[a*x^2 + b*x^3])/x - 2*S qrt[a]*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])
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[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 1.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.60
method | result | size |
pseudoelliptic | \(-\frac {\left (2 a d +b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) x +\sqrt {b x +a}\, \sqrt {a}\, \left (-2 d x +c \right )}{\sqrt {a}\, x}\) | \(49\) |
default | \(\frac {2 \sqrt {b \,x^{3}+a \,x^{2}}\, \left (\left (b x +a \right )^{\frac {3}{2}} d -3 \sqrt {a}\, b c \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+3 \sqrt {b x +a}\, b c \right )}{3 x \sqrt {b x +a}\, b}\) | \(69\) |
Input:
int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/a^(1/2)*((2*a*d+b*c)*arctanh((b*x+a)^(1/2)/a^(1/2))*x+(b*x+a)^(1/2)*a^( 1/2)*(-2*d*x+c))/x
Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.90 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\left [\frac {3 \, \sqrt {a} b c x \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (b d x + 3 \, b c + a d\right )}}{3 \, b x}, \frac {2 \, {\left (3 \, \sqrt {-a} b c x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (b d x + 3 \, b c + a d\right )}\right )}}{3 \, b x}\right ] \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^2,x, algorithm="fricas")
Output:
[1/3*(3*sqrt(a)*b*c*x*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)*(b*d*x + 3*b*c + a*d))/(b*x), 2/3*(3*sqrt(-a) *b*c*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + sqrt(b*x^3 + a *x^2)*(b*d*x + 3*b*c + a*d))/(b*x)]
\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x\right )} \left (c + d x\right )}{x^{2}}\, dx \] Input:
integrate((d*x+c)*(b*x**3+a*x**2)**(1/2)/x**2,x)
Output:
Integral(sqrt(x**2*(a + b*x))*(c + d*x)/x**2, x)
\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\int { \frac {\sqrt {b x^{3} + a x^{2}} {\left (d x + c\right )}}{x^{2}} \,d x } \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^2,x, algorithm="maxima")
Output:
integrate(sqrt(b*x^3 + a*x^2)*(d*x + c)/x^2, x)
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\frac {2 \, a c \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} - \frac {2 \, {\left (3 \, a b c \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 3 \, \sqrt {-a} \sqrt {a} b c + \sqrt {-a} a^{\frac {3}{2}} d\right )} \mathrm {sgn}\left (x\right )}{3 \, \sqrt {-a} b} + \frac {2 \, {\left (3 \, \sqrt {b x + a} b^{3} c \mathrm {sgn}\left (x\right ) + {\left (b x + a\right )}^{\frac {3}{2}} b^{2} d \mathrm {sgn}\left (x\right )\right )}}{3 \, b^{3}} \] Input:
integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^2,x, algorithm="giac")
Output:
2*a*c*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/sqrt(-a) - 2/3*(3*a*b*c*arctan (sqrt(a)/sqrt(-a)) + 3*sqrt(-a)*sqrt(a)*b*c + sqrt(-a)*a^(3/2)*d)*sgn(x)/( sqrt(-a)*b) + 2/3*(3*sqrt(b*x + a)*b^3*c*sgn(x) + (b*x + a)^(3/2)*b^2*d*sg n(x))/b^3
Timed out. \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\int \frac {\sqrt {b\,x^3+a\,x^2}\,\left (c+d\,x\right )}{x^2} \,d x \] Input:
int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^2,x)
Output:
int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^2, x)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^2} \, dx=\frac {2 \sqrt {b x +a}\, a d +6 \sqrt {b x +a}\, b c +2 \sqrt {b x +a}\, b d x +3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b c -3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b c}{3 b} \] Input:
int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^2,x)
Output:
(2*sqrt(a + b*x)*a*d + 6*sqrt(a + b*x)*b*c + 2*sqrt(a + b*x)*b*d*x + 3*sqr t(a)*log(sqrt(a + b*x) - sqrt(a))*b*c - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt (a))*b*c)/(3*b)