Integrand size = 22, antiderivative size = 98 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\frac {2 (A b-a B) \sqrt {d+e x}}{b^2}+\frac {2 B (d+e x)^{3/2}}{3 b e}-\frac {2 (A b-a B) \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2}} \] Output:
2*(A*b-B*a)*(e*x+d)^(1/2)/b^2+2/3*B*(e*x+d)^(3/2)/b/e-2*(A*b-B*a)*(-a*e+b* d)^(1/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)
Time = 0.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\frac {2 \sqrt {d+e x} (3 A b e-3 a B e+b B (d+e x))}{3 b^2 e}+\frac {2 (-A b+a B) \sqrt {-b d+a e} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{5/2}} \] Input:
Integrate[((A + B*x)*Sqrt[d + e*x])/(a + b*x),x]
Output:
(2*Sqrt[d + e*x]*(3*A*b*e - 3*a*B*e + b*B*(d + e*x)))/(3*b^2*e) + (2*(-(A* b) + a*B)*Sqrt[-(b*d) + a*e]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/b^(5/2)
Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {90, 60, 73, 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.
\(\displaystyle \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {(A b-a B) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(A b-a B) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(A b-a B) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 b e}\) |
Input:
Int[((A + B*x)*Sqrt[d + e*x])/(a + b*x),x]
Output:
(2*B*(d + e*x)^(3/2))/(3*b*e) + ((A*b - a*B)*((2*Sqrt[d + e*x])/b - (2*Sqr t[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/b
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {2 \left (e b B x +3 A b e -3 B a e +B b d \right ) \sqrt {e x +d}}{3 e \,b^{2}}-\frac {2 \left (A a b e -A \,b^{2} d -B \,a^{2} e +B a b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{b^{2} \sqrt {\left (a e -d b \right ) b}}\) | \(101\) |
pseudoelliptic | \(\frac {-2 e \left (a e -d b \right ) \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )+2 \left (\left (\left (\frac {B x}{3}+A \right ) e +\frac {B d}{3}\right ) b -B a e \right ) \sqrt {\left (a e -d b \right ) b}\, \sqrt {e x +d}}{e \,b^{2} \sqrt {\left (a e -d b \right ) b}}\) | \(104\) |
derivativedivides | \(\frac {\frac {2 \left (\frac {b B \left (e x +d \right )^{\frac {3}{2}}}{3}+A b e \sqrt {e x +d}-B a e \sqrt {e x +d}\right )}{b^{2}}-\frac {2 e \left (A a b e -A \,b^{2} d -B \,a^{2} e +B a b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{b^{2} \sqrt {\left (a e -d b \right ) b}}}{e}\) | \(111\) |
default | \(\frac {\frac {2 \left (\frac {b B \left (e x +d \right )^{\frac {3}{2}}}{3}+A b e \sqrt {e x +d}-B a e \sqrt {e x +d}\right )}{b^{2}}-\frac {2 e \left (A a b e -A \,b^{2} d -B \,a^{2} e +B a b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -d b \right ) b}}\right )}{b^{2} \sqrt {\left (a e -d b \right ) b}}}{e}\) | \(111\) |
Input:
int((B*x+A)*(e*x+d)^(1/2)/(b*x+a),x,method=_RETURNVERBOSE)
Output:
2/3*(B*b*e*x+3*A*b*e-3*B*a*e+B*b*d)*(e*x+d)^(1/2)/e/b^2-2*(A*a*b*e-A*b^2*d -B*a^2*e+B*a*b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d )*b)^(1/2))
Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.15 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\left [-\frac {3 \, {\left (B a - A b\right )} e \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (B b e x + B b d - 3 \, {\left (B a - A b\right )} e\right )} \sqrt {e x + d}}{3 \, b^{2} e}, \frac {2 \, {\left (3 \, {\left (B a - A b\right )} e \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (B b e x + B b d - 3 \, {\left (B a - A b\right )} e\right )} \sqrt {e x + d}\right )}}{3 \, b^{2} e}\right ] \] Input:
integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a),x, algorithm="fricas")
Output:
[-1/3*(3*(B*a - A*b)*e*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sq rt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) - 2*(B*b*e*x + B*b*d - 3*(B* a - A*b)*e)*sqrt(e*x + d))/(b^2*e), 2/3*(3*(B*a - A*b)*e*sqrt(-(b*d - a*e) /b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (B*b*e*x + B*b*d - 3*(B*a - A*b)*e)*sqrt(e*x + d))/(b^2*e)]
Time = 4.58 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.27 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\begin {cases} \frac {2 \left (\frac {B \left (d + e x\right )^{\frac {3}{2}}}{3 b} + \frac {\sqrt {d + e x} \left (A b e - B a e\right )}{b^{2}} + \frac {e \left (- A b + B a\right ) \left (a e - b d\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{3} \sqrt {\frac {a e - b d}{b}}}\right )}{e} & \text {for}\: e \neq 0 \\\sqrt {d} \left (\frac {B x}{b} - \frac {\left (- A b + B a\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((B*x+A)*(e*x+d)**(1/2)/(b*x+a),x)
Output:
Piecewise((2*(B*(d + e*x)**(3/2)/(3*b) + sqrt(d + e*x)*(A*b*e - B*a*e)/b** 2 + e*(-A*b + B*a)*(a*e - b*d)*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(b* *3*sqrt((a*e - b*d)/b)))/e, Ne(e, 0)), (sqrt(d)*(B*x/b - (-A*b + B*a)*Piec ewise((x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b), True))
Exception generated. \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a),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 = 0.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=-\frac {2 \, {\left (B a b d - A b^{2} d - B a^{2} e + A a b e\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} B b^{2} e^{2} - 3 \, \sqrt {e x + d} B a b e^{3} + 3 \, \sqrt {e x + d} A b^{2} e^{3}\right )}}{3 \, b^{3} e^{3}} \] Input:
integrate((B*x+A)*(e*x+d)^(1/2)/(b*x+a),x, algorithm="giac")
Output:
-2*(B*a*b*d - A*b^2*d - B*a^2*e + A*a*b*e)*arctan(sqrt(e*x + d)*b/sqrt(-b^ 2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^2) + 2/3*((e*x + d)^(3/2)*B*b^2*e^2 - 3*sqrt(e*x + d)*B*a*b*e^3 + 3*sqrt(e*x + d)*A*b^2*e^3)/(b^3*e^3)
Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\sqrt {d+e\,x}-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {b\,d-a\,e}}\right )\,\left (A\,b-B\,a\right )\,\sqrt {b\,d-a\,e}}{b^{5/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{3/2}}{3\,b\,e} \] Input:
int(((A + B*x)*(d + e*x)^(1/2))/(a + b*x),x)
Output:
((2*A*e - 2*B*d)/(b*e) - (2*B*(a*e^2 - b*d*e))/(b^2*e^2))*(d + e*x)^(1/2) - (2*atanh((b^(1/2)*(d + e*x)^(1/2))/(b*d - a*e)^(1/2))*(A*b - B*a)*(b*d - a*e)^(1/2))/b^(5/2) + (2*B*(d + e*x)^(3/2))/(3*b*e)
Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.16 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a+b x} \, dx=\frac {2 \sqrt {e x +d}\, \left (e x +d \right )}{3 e} \] Input:
int((B*x+A)*(e*x+d)^(1/2)/(b*x+a),x)
Output:
(2*sqrt(d + e*x)*(d + e*x))/(3*e)