Integrand size = 28, antiderivative size = 77 \[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 c e \sqrt {a x^2+b x^3}}{a (e x)^{3/2}}+\frac {2 d \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a x^2+b x^3}}\right )}{\sqrt {b} \sqrt {e}} \] Output:
-2*c*e*(b*x^3+a*x^2)^(1/2)/a/(e*x)^(3/2)+2*d*arctanh(b^(1/2)*(e*x)^(3/2)/e ^(3/2)/(b*x^3+a*x^2)^(1/2))/b^(1/2)/e^(1/2)
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \left (\sqrt {b} c x (a+b x)+a d x^{3/2} \sqrt {a+b x} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{a \sqrt {b} \sqrt {e x} \sqrt {x^2 (a+b x)}} \] Input:
Integrate[(c + d*x)/(Sqrt[e*x]*Sqrt[a*x^2 + b*x^3]),x]
Output:
(-2*(Sqrt[b]*c*x*(a + b*x) + a*d*x^(3/2)*Sqrt[a + b*x]*Log[-(Sqrt[b]*Sqrt[ x]) + Sqrt[a + b*x]]))/(a*Sqrt[b]*Sqrt[e*x]*Sqrt[x^2*(a + b*x)])
Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1944, 1937, 1935, 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.
\(\displaystyle \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx\) |
\(\Big \downarrow \) 1944 |
\(\displaystyle \frac {d \int \frac {\sqrt {e x}}{\sqrt {b x^3+a x^2}}dx}{e}-\frac {2 c e \sqrt {a x^2+b x^3}}{a (e x)^{3/2}}\) |
\(\Big \downarrow \) 1937 |
\(\displaystyle \frac {d \sqrt {e x} \int \frac {\sqrt {x}}{\sqrt {b x^3+a x^2}}dx}{e \sqrt {x}}-\frac {2 c e \sqrt {a x^2+b x^3}}{a (e x)^{3/2}}\) |
\(\Big \downarrow \) 1935 |
\(\displaystyle \frac {2 d \sqrt {e x} \int \frac {1}{1-\frac {b x^3}{b x^3+a x^2}}d\frac {x^{3/2}}{\sqrt {b x^3+a x^2}}}{e \sqrt {x}}-\frac {2 c e \sqrt {a x^2+b x^3}}{a (e x)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 d \sqrt {e x} \text {arctanh}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {b} e \sqrt {x}}-\frac {2 c e \sqrt {a x^2+b x^3}}{a (e x)^{3/2}}\) |
Input:
Int[(c + d*x)/(Sqrt[e*x]*Sqrt[a*x^2 + b*x^3]),x]
Output:
(-2*c*e*Sqrt[a*x^2 + b*x^3])/(a*(e*x)^(3/2)) + (2*d*Sqrt[e*x]*ArcTanh[(Sqr t[b]*x^(3/2))/Sqrt[a*x^2 + b*x^3]])/(Sqrt[b]*e*Sqrt[x])
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[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp [-2/(n - j) Subst[Int[1/(1 - a*x^2), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]
Int[((c_)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPart[m]) Int[x^m*(a*x^j + b *x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && IntegerQ[p + 1/2] && NeQ[n, j] && EqQ[Simplify[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 = 0.39 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.22
method | result | size |
risch | \(-\frac {2 c \left (b x +a \right ) x}{a \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}+\frac {d \ln \left (\frac {\frac {1}{2} a e +b e x}{\sqrt {b e}}+\sqrt {b e \,x^{2}+a e x}\right ) x \sqrt {e x \left (b x +a \right )}}{\sqrt {b e}\, \sqrt {x^{2} \left (b x +a \right )}\, \sqrt {e x}}\) | \(94\) |
default | \(-\frac {x \left (b x +a \right ) \left (-\ln \left (\frac {2 b e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}+a e}{2 \sqrt {b e}}\right ) a d e x +2 \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}\, c \right )}{\sqrt {b \,x^{3}+a \,x^{2}}\, a \sqrt {e x}\, \sqrt {e x \left (b x +a \right )}\, \sqrt {b e}}\) | \(104\) |
Input:
int((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-2*c/a*(b*x+a)/(x^2*(b*x+a))^(1/2)*x/(e*x)^(1/2)+d*ln((1/2*a*e+b*e*x)/(b*e )^(1/2)+(b*e*x^2+a*e*x)^(1/2))/(b*e)^(1/2)/(x^2*(b*x+a))^(1/2)*x*(e*x*(b*x +a))^(1/2)/(e*x)^(1/2)
Time = 0.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.30 \[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=\left [\frac {\sqrt {b e} a d x^{2} \log \left (\frac {2 \, b e x^{2} + a e x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b e} \sqrt {e x}}{x}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {e x} b c}{a b e x^{2}}, -\frac {2 \, {\left (\sqrt {-b e} a d x^{2} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b e} \sqrt {e x}}{b e x^{2} + a e x}\right ) + \sqrt {b x^{3} + a x^{2}} \sqrt {e x} b c\right )}}{a b e x^{2}}\right ] \] Input:
integrate((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
[(sqrt(b*e)*a*d*x^2*log((2*b*e*x^2 + a*e*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(b* e)*sqrt(e*x))/x) - 2*sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*c)/(a*b*e*x^2), -2*(s qrt(-b*e)*a*d*x^2*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-b*e)*sqrt(e*x)/(b*e*x^2 + a*e*x)) + sqrt(b*x^3 + a*x^2)*sqrt(e*x)*b*c)/(a*b*e*x^2)]
\[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=\int \frac {c + d x}{\sqrt {e x} \sqrt {x^{2} \left (a + b x\right )}}\, dx \] Input:
integrate((d*x+c)/(e*x)**(1/2)/(b*x**3+a*x**2)**(1/2),x)
Output:
Integral((c + d*x)/(sqrt(e*x)*sqrt(x**2*(a + b*x))), x)
\[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=\int { \frac {d x + c}{\sqrt {b x^{3} + a x^{2}} \sqrt {e x}} \,d x } \] Input:
integrate((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate((d*x + c)/(sqrt(b*x^3 + a*x^2)*sqrt(e*x)), x)
Time = 0.21 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10 \[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=-\frac {2 \, b^{2} {\left (\frac {d \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {{\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b} + \frac {\sqrt {b x + a} c}{\sqrt {{\left (b x + a\right )} b e - a b e} a}\right )}}{{\left | b \right |} \mathrm {sgn}\left (x\right )} \] Input:
integrate((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
-2*b^2*(d*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt((b*x + a)*b*e - a*b*e))) /(sqrt(b*e)*b) + sqrt(b*x + a)*c/(sqrt((b*x + a)*b*e - a*b*e)*a))/(abs(b)* sgn(x))
Timed out. \[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=\int \frac {c+d\,x}{\sqrt {e\,x}\,\sqrt {b\,x^3+a\,x^2}} \,d x \] Input:
int((c + d*x)/((e*x)^(1/2)*(a*x^2 + b*x^3)^(1/2)),x)
Output:
int((c + d*x)/((e*x)^(1/2)*(a*x^2 + b*x^3)^(1/2)), x)
Time = 0.21 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int \frac {c+d x}{\sqrt {e x} \sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {e}\, \left (-\sqrt {x}\, \sqrt {b x +a}\, b c +\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a d x -\sqrt {b}\, b c x \right )}{a b e x} \] Input:
int((d*x+c)/(e*x)^(1/2)/(b*x^3+a*x^2)^(1/2),x)
Output:
(2*sqrt(e)*( - sqrt(x)*sqrt(a + b*x)*b*c + sqrt(b)*log((sqrt(a + b*x) + sq rt(x)*sqrt(b))/sqrt(a))*a*d*x - sqrt(b)*b*c*x))/(a*b*e*x)