Integrand size = 24, antiderivative size = 111 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=-\frac {3 b c \sqrt {e x} \sqrt {c+d x}}{4 d^2 e}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}+\frac {\left (3 b c^2+8 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{4 d^{5/2} \sqrt {e}} \] Output:
-3/4*b*c*(e*x)^(1/2)*(d*x+c)^(1/2)/d^2/e+1/2*b*(e*x)^(3/2)*(d*x+c)^(1/2)/d /e^2+1/4*(8*a*d^2+3*b*c^2)*arctanh(d^(1/2)*(e*x)^(1/2)/e^(1/2)/(d*x+c)^(1/ 2))/d^(5/2)/e^(1/2)
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.84 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {b \sqrt {d} x \sqrt {c+d x} (-3 c+2 d x)+2 \left (3 b c^2+8 a d^2\right ) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{-\sqrt {c}+\sqrt {c+d x}}\right )}{4 d^{5/2} \sqrt {e x}} \] Input:
Integrate[(a + b*x^2)/(Sqrt[e*x]*Sqrt[c + d*x]),x]
Output:
(b*Sqrt[d]*x*Sqrt[c + d*x]*(-3*c + 2*d*x) + 2*(3*b*c^2 + 8*a*d^2)*Sqrt[x]* ArcTanh[(Sqrt[d]*Sqrt[x])/(-Sqrt[c] + Sqrt[c + d*x])])/(4*d^(5/2)*Sqrt[e*x ])
Time = 0.36 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {521, 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.
\(\displaystyle \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 521 |
\(\displaystyle \frac {\int \frac {e^2 (4 a d-3 b c x)}{2 \sqrt {e x} \sqrt {c+d x}}dx}{2 d e^2}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 a d-3 b c x}{\sqrt {e x} \sqrt {c+d x}}dx}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {\frac {\left (8 a d^2+3 b c^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}-\frac {3 b c \sqrt {e x} \sqrt {c+d x}}{d e}}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {\frac {\left (8 a d^2+3 b c^2\right ) \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}-\frac {3 b c \sqrt {e x} \sqrt {c+d x}}{d e}}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\left (8 a d^2+3 b c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2} \sqrt {e}}-\frac {3 b c \sqrt {e x} \sqrt {c+d x}}{d e}}{4 d}+\frac {b (e x)^{3/2} \sqrt {c+d x}}{2 d e^2}\) |
Input:
Int[(a + b*x^2)/(Sqrt[e*x]*Sqrt[c + d*x]),x]
Output:
(b*(e*x)^(3/2)*Sqrt[c + d*x])/(2*d*e^2) + ((-3*b*c*Sqrt[e*x]*Sqrt[c + d*x] )/(d*e) + ((3*b*c^2 + 8*a*d^2)*ArcTanh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[c + d*x])])/(d^(3/2)*Sqrt[e]))/(4*d)
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] :> Simp[b^p*(e*x)^(m + 2*p)*((c + d*x)^(n + 1)/(d*e^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(d*e^(2*p)*(m + n + 2*p + 1)) Int[(e*x)^m*(c + d*x)^n*ExpandToSum[d*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x^2)^p - b^p*(e*x)^ (2*p)) - b^p*(e*c)*(m + 2*p)*(e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && !IntegerQ[m] && !I ntegerQ[n]
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {b \left (-2 d x +3 c \right ) x \sqrt {d x +c}}{4 d^{2} \sqrt {e x}}+\frac {\left (8 a \,d^{2}+3 b \,c^{2}\right ) \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{8 d^{2} \sqrt {d e}\, \sqrt {e x}\, \sqrt {d x +c}}\) | \(104\) |
default | \(\frac {\sqrt {d x +c}\, x \left (8 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) a \,d^{2} e +3 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{2} e +4 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b d x -6 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b c \right )}{8 \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, d^{2} \sqrt {d e}}\) | \(156\) |
Input:
int((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4*b*(-2*d*x+3*c)/d^2*x*(d*x+c)^(1/2)/(e*x)^(1/2)+1/8*(8*a*d^2+3*b*c^2)/ d^2*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 = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.54 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\left [\frac {{\left (3 \, b c^{2} + 8 \, a d^{2}\right )} \sqrt {d e} \log \left (2 \, d e x + c e + 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (2 \, b d^{2} x - 3 \, b c d\right )} \sqrt {d x + c} \sqrt {e x}}{8 \, d^{3} e}, -\frac {{\left (3 \, b c^{2} + 8 \, a d^{2}\right )} \sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) - {\left (2 \, b d^{2} x - 3 \, b c d\right )} \sqrt {d x + c} \sqrt {e x}}{4 \, d^{3} e}\right ] \] Input:
integrate((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
[1/8*((3*b*c^2 + 8*a*d^2)*sqrt(d*e)*log(2*d*e*x + c*e + 2*sqrt(d*e)*sqrt(d *x + c)*sqrt(e*x)) + 2*(2*b*d^2*x - 3*b*c*d)*sqrt(d*x + c)*sqrt(e*x))/(d^3 *e), -1/4*((3*b*c^2 + 8*a*d^2)*sqrt(-d*e)*arctan(sqrt(-d*e)*sqrt(d*x + c)* sqrt(e*x)/(d*e*x + c*e)) - (2*b*d^2*x - 3*b*c*d)*sqrt(d*x + c)*sqrt(e*x))/ (d^3*e)]
Time = 3.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.42 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {2 a \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{\sqrt {d} \sqrt {e}} - \frac {3 b c^{\frac {3}{2}} \sqrt {x}}{4 d^{2} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} - \frac {b \sqrt {c} x^{\frac {3}{2}}}{4 d \sqrt {e} \sqrt {1 + \frac {d x}{c}}} + \frac {3 b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{4 d^{\frac {5}{2}} \sqrt {e}} + \frac {b x^{\frac {5}{2}}}{2 \sqrt {c} \sqrt {e} \sqrt {1 + \frac {d x}{c}}} \] Input:
integrate((b*x**2+a)/(e*x)**(1/2)/(d*x+c)**(1/2),x)
Output:
2*a*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(sqrt(d)*sqrt(e)) - 3*b*c**(3/2)*sqrt(x )/(4*d**2*sqrt(e)*sqrt(1 + d*x/c)) - b*sqrt(c)*x**(3/2)/(4*d*sqrt(e)*sqrt( 1 + d*x/c)) + 3*b*c**2*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(4*d**(5/2)*sqrt(e)) + b*x**(5/2)/(2*sqrt(c)*sqrt(e)*sqrt(1 + d*x/c))
Exception generated. \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(1/2),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 = 0.15 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c} {\left (\frac {2 \, {\left (d x + c\right )} b}{d^{3} e} - \frac {5 \, b c}{d^{3} e}\right )} - \frac {{\left (3 \, b c^{2} + 8 \, a d^{2}\right )} \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{2}}\right )} d}{4 \, {\left | d \right |}} \] Input:
integrate((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")
Output:
1/4*(sqrt((d*x + c)*d*e - c*d*e)*sqrt(d*x + c)*(2*(d*x + c)*b/(d^3*e) - 5* b*c/(d^3*e)) - (3*b*c^2 + 8*a*d^2)*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt ((d*x + c)*d*e - c*d*e)))/(sqrt(d*e)*d^2))*d/abs(d)
Time = 12.42 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.73 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {3\,b\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {d}\,\sqrt {e\,x}}{\sqrt {e}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )}{2\,d^{5/2}\,\sqrt {e}}-\frac {4\,a\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-d\,e}\,\sqrt {e\,x}}\right )}{\sqrt {-d\,e}}-\frac {\frac {3\,b\,c^2\,d\,{\left (e\,x\right )}^{7/2}}{2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}-\frac {11\,b\,c^2\,e\,{\left (e\,x\right )}^{5/2}}{2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {3\,b\,c^2\,e^3\,\sqrt {e\,x}}{2\,d^2\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {11\,b\,c^2\,e^2\,{\left (e\,x\right )}^{3/2}}{2\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}}{e^4-\frac {4\,d\,e^4\,x}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {6\,d^2\,e^4\,x^2}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {4\,d^3\,e^4\,x^3}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {d^4\,e^4\,x^4}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}} \] Input:
int((a + b*x^2)/((e*x)^(1/2)*(c + d*x)^(1/2)),x)
Output:
(3*b*c^2*atanh((d^(1/2)*(e*x)^(1/2))/(e^(1/2)*((c + d*x)^(1/2) - c^(1/2))) ))/(2*d^(5/2)*e^(1/2)) - (4*a*atan((e*((c + d*x)^(1/2) - c^(1/2)))/((-d*e) ^(1/2)*(e*x)^(1/2))))/(-d*e)^(1/2) - ((3*b*c^2*d*(e*x)^(7/2))/(2*((c + d*x )^(1/2) - c^(1/2))^7) - (11*b*c^2*e*(e*x)^(5/2))/(2*((c + d*x)^(1/2) - c^( 1/2))^5) + (3*b*c^2*e^3*(e*x)^(1/2))/(2*d^2*((c + d*x)^(1/2) - c^(1/2))) - (11*b*c^2*e^2*(e*x)^(3/2))/(2*d*((c + d*x)^(1/2) - c^(1/2))^3))/(e^4 - (4 *d*e^4*x)/((c + d*x)^(1/2) - c^(1/2))^2 + (6*d^2*e^4*x^2)/((c + d*x)^(1/2) - c^(1/2))^4 - (4*d^3*e^4*x^3)/((c + d*x)^(1/2) - c^(1/2))^6 + (d^4*e^4*x ^4)/((c + d*x)^(1/2) - c^(1/2))^8)
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.82 \[ \int \frac {a+b x^2}{\sqrt {e x} \sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \left (-3 \sqrt {x}\, \sqrt {d x +c}\, b c d +2 \sqrt {x}\, \sqrt {d x +c}\, b \,d^{2} x +8 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) a \,d^{2}+3 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{2}\right )}{4 d^{3} e} \] Input:
int((b*x^2+a)/(e*x)^(1/2)/(d*x+c)^(1/2),x)
Output:
(sqrt(e)*( - 3*sqrt(x)*sqrt(c + d*x)*b*c*d + 2*sqrt(x)*sqrt(c + d*x)*b*d** 2*x + 8*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*a*d**2 + 3* sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b*c**2))/(4*d**3*e)