Integrand size = 24, antiderivative size = 105 \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {2 \left (b c^2+a d^2\right ) \sqrt {e x}}{c d^2 e \sqrt {c+d x}}+\frac {b \sqrt {e x} \sqrt {c+d x}}{d^2 e}-\frac {3 b c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{5/2} \sqrt {e}} \] Output:
2*(a*d^2+b*c^2)*(e*x)^(1/2)/c/d^2/e/(d*x+c)^(1/2)+b*(e*x)^(1/2)*(d*x+c)^(1 /2)/d^2/e-3*b*c*arctanh(d^(1/2)*(e*x)^(1/2)/e^(1/2)/(d*x+c)^(1/2))/d^(5/2) /e^(1/2)
Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.83 \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {x \left (3 b c^2+2 a d^2+b c d x\right )}{c d^2 \sqrt {e x} \sqrt {c+d x}}+\frac {3 b c \sqrt {x} \log \left (-\sqrt {d} \sqrt {x}+\sqrt {c+d x}\right )}{d^{5/2} \sqrt {e x}} \] Input:
Integrate[(a + b*x^2)/(Sqrt[e*x]*(c + d*x)^(3/2)),x]
Output:
(x*(3*b*c^2 + 2*a*d^2 + b*c*d*x))/(c*d^2*Sqrt[e*x]*Sqrt[c + d*x]) + (3*b*c *Sqrt[x]*Log[-(Sqrt[d]*Sqrt[x]) + Sqrt[c + d*x]])/(d^(5/2)*Sqrt[e*x])
Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {519, 27, 90, 65, 221}
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.
\(\displaystyle \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 519 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {2 \int \frac {b c (c-d x)}{2 d^2 \sqrt {e x} \sqrt {c+d x}}dx}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {b \int \frac {c-d x}{\sqrt {e x} \sqrt {c+d x}}dx}{d^2}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {b \left (\frac {3}{2} c \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx-\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{d^2}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {b \left (3 c \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}-\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{d^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {e x} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {b \left (\frac {3 c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}-\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{d^2}\) |
Input:
Int[(a + b*x^2)/(Sqrt[e*x]*(c + d*x)^(3/2)),x]
Output:
(2*(a + (b*c^2)/d^2)*Sqrt[e*x])/(c*e*Sqrt[c + d*x]) - (b*(-((Sqrt[e*x]*Sqr t[c + d*x])/e) + (3*c*ArcTanh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[c + d*x])] )/(Sqrt[d]*Sqrt[e])))/d^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, c + d*x, x], R = PolynomialRemainder[(a + b*x^2)^p, c + d*x, x]}, Simp[(-R)*(e*x)^(m + 1)*( (c + d*x)^(n + 1)/(c*e*(n + 1))), x] + Simp[1/(c*(n + 1)) Int[(e*x)^m*(c + d*x)^(n + 1)*ExpandToSum[c*(n + 1)*Qx + R*(m + n + 2), x], x], x]] /; Fre eQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0] && LtQ[n, -1] && !IntegerQ[m]
Time = 0.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.39
method | result | size |
risch | \(\frac {b x \sqrt {d x +c}}{d^{2} \sqrt {e x}}-\frac {\left (\frac {2 \left (-2 a \,d^{2}-2 b \,c^{2}\right ) \sqrt {d e \left (x +\frac {c}{d}\right )^{2}-c e \left (x +\frac {c}{d}\right )}}{d c e \left (x +\frac {c}{d}\right )}+\frac {3 b c \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}\right ) \sqrt {\left (d x +c \right ) e x}}{2 d^{2} \sqrt {e x}\, \sqrt {d x +c}}\) | \(146\) |
default | \(-\frac {\left (3 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{2} d e x +3 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{3} e -2 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b c d x -4 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, a \,d^{2}-6 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b \,c^{2}\right ) x}{2 c \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {e x}\, d^{2} \sqrt {d x +c}}\) | \(185\) |
Input:
int((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
b/d^2*x*(d*x+c)^(1/2)/(e*x)^(1/2)-1/2/d^2*(2*(-2*a*d^2-2*b*c^2)/d/c/e/(x+c /d)*(d*e*(x+c/d)^2-c*e*(x+c/d))^(1/2)+3*b*c*ln((1/2*c*e+d*e*x)/(d*e)^(1/2) +(d*e*x^2+c*e*x)^(1/2))/(d*e)^(1/2))*((d*x+c)*e*x)^(1/2)/(e*x)^(1/2)/(d*x+ c)^(1/2)
Time = 0.12 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.01 \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (b c^{2} d x + b c^{3}\right )} \sqrt {d e} \log \left (2 \, d e x + c e - 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (b c d^{2} x + 3 \, b c^{2} d + 2 \, a d^{3}\right )} \sqrt {d x + c} \sqrt {e x}}{2 \, {\left (c d^{4} e x + c^{2} d^{3} e\right )}}, \frac {3 \, {\left (b c^{2} d x + b c^{3}\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) + {\left (b c d^{2} x + 3 \, b c^{2} d + 2 \, a d^{3}\right )} \sqrt {d x + c} \sqrt {e x}}{c d^{4} e x + c^{2} d^{3} e}\right ] \] Input:
integrate((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(3/2),x, algorithm="fricas")
Output:
[1/2*(3*(b*c^2*d*x + b*c^3)*sqrt(d*e)*log(2*d*e*x + c*e - 2*sqrt(d*e)*sqrt (d*x + c)*sqrt(e*x)) + 2*(b*c*d^2*x + 3*b*c^2*d + 2*a*d^3)*sqrt(d*x + c)*s qrt(e*x))/(c*d^4*e*x + c^2*d^3*e), (3*(b*c^2*d*x + b*c^3)*sqrt(-d*e)*arcta n(sqrt(-d*e)*sqrt(d*x + c)*sqrt(e*x)/(d*e*x + c*e)) + (b*c*d^2*x + 3*b*c^2 *d + 2*a*d^3)*sqrt(d*x + c)*sqrt(e*x))/(c*d^4*e*x + c^2*d^3*e)]
Time = 4.53 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.09 \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {2 a}{c \sqrt {d} \sqrt {e} \sqrt {\frac {c}{d x} + 1}} + b \left (\frac {3 \sqrt {c} \sqrt {x}}{d^{2} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} - \frac {3 c \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{d^{\frac {5}{2}} \sqrt {e}} + \frac {x^{\frac {3}{2}}}{\sqrt {c} d \sqrt {e} \sqrt {1 + \frac {d x}{c}}}\right ) \] Input:
integrate((b*x**2+a)/(e*x)**(1/2)/(d*x+c)**(3/2),x)
Output:
2*a/(c*sqrt(d)*sqrt(e)*sqrt(c/(d*x) + 1)) + b*(3*sqrt(c)*sqrt(x)/(d**2*sqr t(e)*sqrt(1 + d*x/c)) - 3*c*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(d**(5/2)*sqrt( e)) + x**(3/2)/(sqrt(c)*d*sqrt(e)*sqrt(1 + d*x/c)))
Exception generated. \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(3/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.16 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.43 \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {3 \, b c \log \left ({\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )}{2 \, \sqrt {d e} d {\left | d \right |}} + \frac {4 \, {\left (b c^{2} e + a d^{2} e\right )}}{{\left (c d e + {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )} \sqrt {d e} {\left | d \right |}} + \frac {\sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c} b {\left | d \right |}}{d^{4} e} \] Input:
integrate((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(3/2),x, algorithm="giac")
Output:
3/2*b*c*log((sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)/(sq rt(d*e)*d*abs(d)) + 4*(b*c^2*e + a*d^2*e)/((c*d*e + (sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)*sqrt(d*e)*abs(d)) + sqrt((d*x + c)*d* e - c*d*e)*sqrt(d*x + c)*b*abs(d)/(d^4*e)
Timed out. \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\int \frac {b\,x^2+a}{\sqrt {e\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((a + b*x^2)/((e*x)^(1/2)*(c + d*x)^(3/2)),x)
Output:
int((a + b*x^2)/((e*x)^(1/2)*(c + d*x)^(3/2)), x)
Time = 0.19 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04 \[ \int \frac {a+b x^2}{\sqrt {e x} (c+d x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (-12 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{2}+8 \sqrt {d}\, \sqrt {d x +c}\, a \,d^{2}+9 \sqrt {d}\, \sqrt {d x +c}\, b \,c^{2}+8 \sqrt {x}\, a \,d^{3}+12 \sqrt {x}\, b \,c^{2} d +4 \sqrt {x}\, b c \,d^{2} x \right )}{4 \sqrt {d x +c}\, c \,d^{3} e} \] Input:
int((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(3/2),x)
Output:
(sqrt(e)*( - 12*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d) )/sqrt(c))*b*c**2 + 8*sqrt(d)*sqrt(c + d*x)*a*d**2 + 9*sqrt(d)*sqrt(c + d* x)*b*c**2 + 8*sqrt(x)*a*d**3 + 12*sqrt(x)*b*c**2*d + 4*sqrt(x)*b*c*d**2*x) )/(4*sqrt(c + d*x)*c*d**3*e)