Integrand size = 24, antiderivative size = 148 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=-\frac {2 (B d-A e) (a+b x)^{3/2}}{e (b d-a e) \sqrt {d+e x}}+\frac {(3 b B d-2 A b e-a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^2 (b d-a e)}-\frac {(3 b B d-2 A b e-a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{5/2}} \]
-(-2*A*b*e-B*a*e+3*B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1 /2))/e^(5/2)/b^(1/2)-2*(-A*e+B*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)/(e*x+d)^(1/2) +(-2*A*b*e-B*a*e+3*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/e^2/(-a*e+b*d)
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {a+b x} (3 B d-2 A e+B e x)}{e^2 \sqrt {d+e x}}+\frac {(-3 b B d+2 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{5/2}} \]
(Sqrt[a + b*x]*(3*B*d - 2*A*e + B*e*x))/(e^2*Sqrt[d + e*x]) + ((-3*b*B*d + 2*A*b*e + a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])] )/(Sqrt[b]*e^(5/2))
Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 60, 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.
\(\displaystyle \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(-a B e-2 A b e+3 b B d) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}}dx}{e (b d-a e)}-\frac {2 (a+b x)^{3/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(-a B e-2 A b e+3 b B d) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{2 e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{3/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {(-a B e-2 A b e+3 b B d) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{e}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{3/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(-a B e-2 A b e+3 b B d) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}}\right )}{e (b d-a e)}-\frac {2 (a+b x)^{3/2} (B d-A e)}{e \sqrt {d+e x} (b d-a e)}\) |
(-2*(B*d - A*e)*(a + b*x)^(3/2))/(e*(b*d - a*e)*Sqrt[d + e*x]) + ((3*b*B*d - 2*A*b*e - a*B*e)*((Sqrt[a + b*x]*Sqrt[d + e*x])/e - ((b*d - a*e)*ArcTan h[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(Sqrt[b]*e^(3/2))))/(e *(b*d - a*e))
3.23.4.3.1 Defintions of rubi rules used
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[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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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. \(385\) vs. \(2(128)=256\).
Time = 3.48 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.61
method | result | size |
default | \(\frac {\sqrt {b x +a}\, \left (2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b \,e^{2} x +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,e^{2} x -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b d e x +2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b d e +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a d e -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b \,d^{2}+2 B e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-4 A e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+6 B d \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right )}{2 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, e^{2} \sqrt {e x +d}}\) | \(386\) |
1/2*(b*x+a)^(1/2)*(2*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/ 2)+a*e+b*d)/(b*e)^(1/2))*b*e^2*x+B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/ 2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*e^2*x-3*B*ln(1/2*(2*b*e*x+2*((b*x+a )*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b*d*e*x+2*A*ln(1/2*(2*b *e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b*d*e+B*l n(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2)) *a*d*e-3*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/ (b*e)^(1/2))*b*d^2+2*B*e*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-4*A*e*((b*x +a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*B*d*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/ (b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/e^2/(e*x+d)^(1/2)
Time = 0.45 (sec) , antiderivative size = 358, normalized size of antiderivative = 2.42 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\left [-\frac {{\left (3 \, B b d^{2} - {\left (B a + 2 \, A b\right )} d e + {\left (3 \, B b d e - {\left (B a + 2 \, A b\right )} e^{2}\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (B b e^{2} x + 3 \, B b d e - 2 \, A b e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{4 \, {\left (b e^{4} x + b d e^{3}\right )}}, \frac {{\left (3 \, B b d^{2} - {\left (B a + 2 \, A b\right )} d e + {\left (3 \, B b d e - {\left (B a + 2 \, A b\right )} e^{2}\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \, {\left (B b e^{2} x + 3 \, B b d e - 2 \, A b e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b e^{4} x + b d e^{3}\right )}}\right ] \]
[-1/4*((3*B*b*d^2 - (B*a + 2*A*b)*d*e + (3*B*b*d*e - (B*a + 2*A*b)*e^2)*x) *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) - 4*(B*b*e^2*x + 3*B*b*d*e - 2*A*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/( b*e^4*x + b*d*e^3), 1/2*((3*B*b*d^2 - (B*a + 2*A*b)*d*e + (3*B*b*d*e - (B* a + 2*A*b)*e^2)*x)*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 )) + 2*(B*b*e^2*x + 3*B*b*d*e - 2*A*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b *e^4*x + b*d*e^3)]
\[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {a + b x}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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*(a*e-b*d)>0)', see `assume?` f or more de
Time = 0.34 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} B {\left | b \right |}}{b e} + \frac {3 \, B b^{2} d e {\left | b \right |} - B a b e^{2} {\left | b \right |} - 2 \, A b^{2} e^{2} {\left | b \right |}}{b^{2} e^{3}}\right )}}{\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} + \frac {{\left (3 \, B b d {\left | b \right |} - B a e {\left | b \right |} - 2 \, A b e {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b e^{2}} \]
sqrt(b*x + a)*((b*x + a)*B*abs(b)/(b*e) + (3*B*b^2*d*e*abs(b) - B*a*b*e^2* abs(b) - 2*A*b^2*e^2*abs(b))/(b^2*e^3))/sqrt(b^2*d + (b*x + a)*b*e - a*b*e ) + (3*B*b*d*abs(b) - B*a*e*abs(b) - 2*A*b*e*abs(b))*log(abs(-sqrt(b*e)*sq rt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e^2)
Timed out. \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,\sqrt {a+b\,x}}{{\left (d+e\,x\right )}^{3/2}} \,d x \]