Integrand size = 24, antiderivative size = 83 \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}-\frac {(b B d-2 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{3/2} e^{3/2}} \] Output:
B*(b*x+a)^(1/2)*(e*x+d)^(1/2)/b/e-(-2*A*b*e+B*a*e+B*b*d)*arctanh(e^(1/2)*( b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2))/b^(3/2)/e^(3/2)
Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}-\frac {(b B d-2 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{3/2} e^{3/2}} \] Input:
Integrate[(A + B*x)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]
Output:
(B*Sqrt[a + b*x]*Sqrt[d + e*x])/(b*e) - ((b*B*d - 2*A*b*e + a*B*e)*ArcTanh [(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])])/(b^(3/2)*e^(3/2))
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {90, 66, 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.
\(\displaystyle \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(2 A b e-B (a e+b d)) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{2 b e}+\frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {(2 A b e-B (a e+b d)) \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{b e}+\frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(2 A b e-B (a e+b d)) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{3/2} e^{3/2}}+\frac {B \sqrt {a+b x} \sqrt {d+e x}}{b e}\) |
Input:
Int[(A + B*x)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]
Output:
(B*Sqrt[a + b*x]*Sqrt[d + e*x])/(b*e) + ((2*A*b*e - B*(b*d + a*e))*ArcTanh [(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(3/2)*e^(3/2))
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 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]
Leaf count of result is larger than twice the leaf count of optimal. \(197\) vs. \(2(67)=134\).
Time = 0.29 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.39
method | result | size |
default | \(\frac {\left (2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b e -B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a e -B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b d +2 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\right ) \sqrt {b x +a}\, \sqrt {e x +d}}{2 e \sqrt {b e}\, b \sqrt {\left (e x +d \right ) \left (b x +a \right )}}\) | \(198\) |
Input:
int((B*x+A)/(b*x+a)^(1/2)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*(2*A*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b *e)^(1/2))*b*e-B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e +d*b)/(b*e)^(1/2))*a*e-B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^( 1/2)+a*e+d*b)/(b*e)^(1/2))*b*d+2*B*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2))*(b *x+a)^(1/2)*(e*x+d)^(1/2)/e/(b*e)^(1/2)/b/((e*x+d)*(b*x+a))^(1/2)
Time = 0.12 (sec) , antiderivative size = 251, normalized size of antiderivative = 3.02 \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\left [\frac {4 \, \sqrt {b x + a} \sqrt {e x + d} B b e - {\left (B b d + {\left (B a - 2 \, A b\right )} e\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right )}{4 \, b^{2} e^{2}}, \frac {2 \, \sqrt {b x + a} \sqrt {e x + d} B b e + {\left (B b d + {\left (B a - 2 \, A b\right )} e\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right )}{2 \, b^{2} e^{2}}\right ] \] Input:
integrate((B*x+A)/(b*x+a)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")
Output:
[1/4*(4*sqrt(b*x + a)*sqrt(e*x + d)*B*b*e - (B*b*d + (B*a - 2*A*b)*e)*sqrt (b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x))/ (b^2*e^2), 1/2*(2*sqrt(b*x + a)*sqrt(e*x + d)*B*b*e + (B*b*d + (B*a - 2*A* b)*e)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a) *sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x)))/(b^2*e^2) ]
\[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\int \frac {A + B x}{\sqrt {a + b x} \sqrt {d + e x}}\, dx \] Input:
integrate((B*x+A)/(b*x+a)**(1/2)/(e*x+d)**(1/2),x)
Output:
Integral((A + B*x)/(sqrt(a + b*x)*sqrt(d + e*x)), x)
Exception generated. \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)/(b*x+a)^(1/2)/(e*x+d)^(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(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\frac {b {\left (\frac {{\left (B b d + B a e - 2 \, A b e\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b e} + \frac {\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} B}{b^{2} e}\right )}}{{\left | b \right |}} \] Input:
integrate((B*x+A)/(b*x+a)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")
Output:
b*((B*b*d + B*a*e - 2*A*b*e)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*B/(b^2*e))/abs(b)
Time = 3.85 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.75 \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\frac {\frac {\left (2\,B\,a\,e+2\,B\,b\,d\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{e^3\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}+\frac {\left (2\,B\,a\,e+2\,B\,b\,d\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{b\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}-\frac {8\,B\,\sqrt {a}\,\sqrt {d}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}+\frac {b^2}{e^2}-\frac {2\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}}-\frac {4\,A\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {-b\,e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {-b\,e}}-\frac {2\,B\,\mathrm {atanh}\left (\frac {\sqrt {e}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )\,\left (a\,e+b\,d\right )}{b^{3/2}\,e^{3/2}} \] Input:
int((A + B*x)/((a + b*x)^(1/2)*(d + e*x)^(1/2)),x)
Output:
(((2*B*a*e + 2*B*b*d)*((a + b*x)^(1/2) - a^(1/2)))/(e^3*((d + e*x)^(1/2) - d^(1/2))) + ((2*B*a*e + 2*B*b*d)*((a + b*x)^(1/2) - a^(1/2))^3)/(b*e^2*(( d + e*x)^(1/2) - d^(1/2))^3) - (8*B*a^(1/2)*d^(1/2)*((a + b*x)^(1/2) - a^( 1/2))^2)/(e^2*((d + e*x)^(1/2) - d^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2)) ^4/((d + e*x)^(1/2) - d^(1/2))^4 + b^2/e^2 - (2*b*((a + b*x)^(1/2) - a^(1/ 2))^2)/(e*((d + e*x)^(1/2) - d^(1/2))^2)) - (4*A*atan((b*((d + e*x)^(1/2) - d^(1/2)))/((-b*e)^(1/2)*((a + b*x)^(1/2) - a^(1/2)))))/(-b*e)^(1/2) - (2 *B*atanh((e^(1/2)*((a + b*x)^(1/2) - a^(1/2)))/(b^(1/2)*((d + e*x)^(1/2) - d^(1/2))))*(a*e + b*d))/(b^(3/2)*e^(3/2))
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x}{\sqrt {a+b x} \sqrt {d+e x}} \, dx=\frac {\sqrt {e x +d}\, \sqrt {b x +a}\, b e +\sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a e -\sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b d}{b \,e^{2}} \] Input:
int((B*x+A)/(b*x+a)^(1/2)/(e*x+d)^(1/2),x)
Output:
(sqrt(d + e*x)*sqrt(a + b*x)*b*e + sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b *x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a*e - sqrt(e)*sqrt(b)*log((s qrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b*d)/(b*e** 2)