Integrand size = 24, antiderivative size = 145 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\frac {(b B d+2 A b e-3 a B e) \sqrt {a+b x} \sqrt {d+e x}}{b^2 (b d-a e)}-\frac {2 (A b-a B) (d+e x)^{3/2}}{b (b d-a e) \sqrt {a+b x}}+\frac {(b B d+2 A b e-3 a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{5/2} \sqrt {e}} \]
(2*A*b*e-3*B*a*e+B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2 ))/b^(5/2)/e^(1/2)-2*(A*b-B*a)*(e*x+d)^(3/2)/b/(-a*e+b*d)/(b*x+a)^(1/2)+(2 *A*b*e-3*B*a*e+B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/b^2/(-a*e+b*d)
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\frac {(-2 A b+3 a B+b B x) \sqrt {d+e x}}{b^2 \sqrt {a+b x}}+\frac {(b B d+2 A b e-3 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{b^{5/2} \sqrt {e}} \]
((-2*A*b + 3*a*B + b*B*x)*Sqrt[d + e*x])/(b^2*Sqrt[a + b*x]) + ((b*B*d + 2 *A*b*e - 3*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a + b*x])] )/(b^(5/2)*Sqrt[e])
Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99, 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.
\(\displaystyle \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(-3 a B e+2 A b e+b B d) \int \frac {\sqrt {d+e x}}{\sqrt {a+b x}}dx}{b (b d-a e)}-\frac {2 (d+e x)^{3/2} (A b-a B)}{b \sqrt {a+b x} (b d-a e)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(-3 a B e+2 A b e+b B d) \left (\frac {(b d-a e) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {d+e x}}{b}\right )}{b (b d-a e)}-\frac {2 (d+e x)^{3/2} (A b-a B)}{b \sqrt {a+b x} (b d-a e)}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {(-3 a B e+2 A b e+b B d) \left (\frac {(b d-a e) \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{b}+\frac {\sqrt {a+b x} \sqrt {d+e x}}{b}\right )}{b (b d-a e)}-\frac {2 (d+e x)^{3/2} (A b-a B)}{b \sqrt {a+b x} (b d-a e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(-3 a B e+2 A b e+b B d) \left (\frac {(b d-a e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{b^{3/2} \sqrt {e}}+\frac {\sqrt {a+b x} \sqrt {d+e x}}{b}\right )}{b (b d-a e)}-\frac {2 (d+e x)^{3/2} (A b-a B)}{b \sqrt {a+b x} (b d-a e)}\) |
(-2*(A*b - a*B)*(d + e*x)^(3/2))/(b*(b*d - a*e)*Sqrt[a + b*x]) + ((b*B*d + 2*A*b*e - 3*a*B*e)*((Sqrt[a + b*x]*Sqrt[d + e*x])/b + ((b*d - a*e)*ArcTan h[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(3/2)*Sqrt[e])))/(b *(b*d - a*e))
3.23.47.3.1 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. \(2(125)=250\).
Time = 1.10 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.66
method | result | size |
default | \(\frac {\sqrt {e x +d}\, \left (2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} e x -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b e x +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d x +2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b e -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a^{2} e +B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b d +2 B b x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-4 A b \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+6 B a \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right )}{2 \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, b^{2} \sqrt {b x +a}}\) | \(386\) |
1/2*(e*x+d)^(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^2*e*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))*a*b*e*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))*b^2*d*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))*a*b*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 ))*a^2*e+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*b*d+2*B*b*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-4*A*b*((b*x +a)*(e*x+d))^(1/2)*(b*e)^(1/2)+6*B*a*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2))/ (b*e)^(1/2)/((b*x+a)*(e*x+d))^(1/2)/b^2/(b*x+a)^(1/2)
Time = 0.45 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.52 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\left [\frac {{\left (B a b d - {\left (3 \, B a^{2} - 2 \, A a b\right )} e + {\left (B b^{2} d - {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (B b^{2} e x + {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {e x + d}}{4 \, {\left (b^{4} e x + a b^{3} e\right )}}, -\frac {{\left (B a b d - {\left (3 \, B a^{2} - 2 \, A a b\right )} e + {\left (B b^{2} d - {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \, {\left (B b^{2} e x + {\left (3 \, B a b - 2 \, A b^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{4} e x + a b^{3} e\right )}}\right ] \]
[1/4*((B*a*b*d - (3*B*a^2 - 2*A*a*b)*e + (B*b^2*d - (3*B*a*b - 2*A*b^2)*e) *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^2*e*x + (3*B*a*b - 2*A*b^2)*e)*sqrt(b*x + a)*sqrt(e*x + d) )/(b^4*e*x + a*b^3*e), -1/2*((B*a*b*d - (3*B*a^2 - 2*A*a*b)*e + (B*b^2*d - (3*B*a*b - 2*A*b^2)*e)*x)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqr t(-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^2*e*x + (3*B*a*b - 2*A*b^2)*e)*sqrt(b*x + a)*sqrt(e*x + d))/(b^4*e*x + a*b^3*e)]
\[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {d + e x}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.37 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} B {\left | b \right |}}{b^{4}} - \frac {{\left (B b d {\left | b \right |} - 3 \, B a e {\left | b \right |} + 2 \, A b e {\left | b \right |}\right )} \log \left ({\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )}{2 \, \sqrt {b e} b^{3}} + \frac {4 \, {\left (B a b d e {\left | b \right |} - A b^{2} d e {\left | b \right |} - B a^{2} e^{2} {\left | b \right |} + A a b e^{2} {\left | b \right |}\right )}}{{\left (b^{2} d - a b e - {\left (\sqrt {b e} \sqrt {b x + a} - \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}\right )}^{2}\right )} \sqrt {b e} b^{2}} \]
sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*B*abs(b)/b^4 - 1/2*(B*b* d*abs(b) - 3*B*a*e*abs(b) + 2*A*b*e*abs(b))*log((sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)/(sqrt(b*e)*b^3) + 4*(B*a*b*d*e*ab s(b) - A*b^2*d*e*abs(b) - B*a^2*e^2*abs(b) + A*a*b*e^2*abs(b))/((b^2*d - a *b*e - (sqrt(b*e)*sqrt(b*x + a) - sqrt(b^2*d + (b*x + a)*b*e - a*b*e))^2)* sqrt(b*e)*b^2)
Timed out. \[ \int \frac {(A+B x) \sqrt {d+e x}}{(a+b x)^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,\sqrt {d+e\,x}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]