Integrand size = 24, antiderivative size = 111 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 (B d-A e) (a+b x)^{3/2}}{3 e (b d-a e) (d+e x)^{3/2}}-\frac {2 B \sqrt {a+b x}}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}} \]
-2/3*(-A*e+B*d)*(b*x+a)^(3/2)/e/(-a*e+b*d)/(e*x+d)^(3/2)+2*B*arctanh(e^(1/ 2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2))*b^(1/2)/e^(5/2)-2*B*(b*x+a)^(1/2)/ e^2/(e*x+d)^(1/2)
Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {a+b x} \left (A b e^2 x-b B d (3 d+4 e x)+a e (2 B d+A e+3 B e x)\right )}{3 e^2 (-b d+a e) (d+e x)^{3/2}}+\frac {2 \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{5/2}} \]
(-2*Sqrt[a + b*x]*(A*b*e^2*x - b*B*d*(3*d + 4*e*x) + a*e*(2*B*d + A*e + 3* B*e*x)))/(3*e^2*(-(b*d) + a*e)*(d + e*x)^(3/2)) + (2*Sqrt[b]*B*ArcTanh[(Sq rt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(5/2)
Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 57, 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.
\(\displaystyle \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {B \int \frac {\sqrt {a+b x}}{(d+e x)^{3/2}}dx}{e}-\frac {2 (a+b x)^{3/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {B \left (\frac {b \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{e}-\frac {2 \sqrt {a+b x}}{e \sqrt {d+e x}}\right )}{e}-\frac {2 (a+b x)^{3/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {B \left (\frac {2 b \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{e}-\frac {2 \sqrt {a+b x}}{e \sqrt {d+e x}}\right )}{e}-\frac {2 (a+b x)^{3/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {B \left (\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{3/2}}-\frac {2 \sqrt {a+b x}}{e \sqrt {d+e x}}\right )}{e}-\frac {2 (a+b x)^{3/2} (B d-A e)}{3 e (d+e x)^{3/2} (b d-a e)}\) |
(-2*(B*d - A*e)*(a + b*x)^(3/2))/(3*e*(b*d - a*e)*(d + e*x)^(3/2)) + (B*(( -2*Sqrt[a + b*x])/(e*Sqrt[d + e*x]) + (2*Sqrt[b]*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/e^(3/2)))/e
3.23.5.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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. \(502\) vs. \(2(89)=178\).
Time = 3.53 (sec) , antiderivative size = 503, normalized size of antiderivative = 4.53
method | result | size |
default | \(-\frac {\left (-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b \,e^{3} x^{2}+3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d \,e^{2} x^{2}-6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b d \,e^{2} x +6 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d^{2} e x +2 A b \,e^{2} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a b \,d^{2} e +3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{2} d^{3}+6 B a \,e^{2} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-8 B b d e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+2 A a \,e^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+4 B a d e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-6 B b \,d^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\right ) \sqrt {b x +a}}{3 \sqrt {b e}\, \left (a e -b d \right ) \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, e^{2} \left (e x +d \right )^{\frac {3}{2}}}\) | \(503\) |
-1/3*(-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^3*x^2+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^2*d*e^2*x^2-6*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*e^2*x+6*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 ^2*e*x+2*A*b*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-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*d^2*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^2*d^3+6*B*a*e^2*x*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-8*B*b*d*e*x*((b *x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+2*A*a*e^2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^( 1/2)+4*B*a*d*e*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)-6*B*b*d^2*((b*x+a)*(e*x +d))^(1/2)*(b*e)^(1/2))*(b*x+a)^(1/2)/(b*e)^(1/2)/(a*e-b*d)/((b*x+a)*(e*x+ d))^(1/2)/e^2/(e*x+d)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (89) = 178\).
Time = 0.75 (sec) , antiderivative size = 515, normalized size of antiderivative = 4.64 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=\left [\frac {3 \, {\left (B b d^{3} - B a d^{2} e + {\left (B b d e^{2} - B a e^{3}\right )} x^{2} + 2 \, {\left (B b d^{2} e - B a d e^{2}\right )} x\right )} \sqrt {\frac {b}{e}} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e^{2} x + b d e + a e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {\frac {b}{e}} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (3 \, B b d^{2} - 2 \, B a d e - A a e^{2} + {\left (4 \, B b d e - {\left (3 \, B a + A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{6 \, {\left (b d^{3} e^{2} - a d^{2} e^{3} + {\left (b d e^{4} - a e^{5}\right )} x^{2} + 2 \, {\left (b d^{2} e^{3} - a d e^{4}\right )} x\right )}}, -\frac {3 \, {\left (B b d^{3} - B a d^{2} e + {\left (B b d e^{2} - B a e^{3}\right )} x^{2} + 2 \, {\left (B b d^{2} e - B a d e^{2}\right )} x\right )} \sqrt {-\frac {b}{e}} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {b x + a} \sqrt {e x + d} \sqrt {-\frac {b}{e}}}{2 \, {\left (b^{2} e x^{2} + a b d + {\left (b^{2} d + a b e\right )} x\right )}}\right ) + 2 \, {\left (3 \, B b d^{2} - 2 \, B a d e - A a e^{2} + {\left (4 \, B b d e - {\left (3 \, B a + A b\right )} e^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {e x + d}}{3 \, {\left (b d^{3} e^{2} - a d^{2} e^{3} + {\left (b d e^{4} - a e^{5}\right )} x^{2} + 2 \, {\left (b d^{2} e^{3} - a d e^{4}\right )} x\right )}}\right ] \]
[1/6*(3*(B*b*d^3 - B*a*d^2*e + (B*b*d*e^2 - B*a*e^3)*x^2 + 2*(B*b*d^2*e - B*a*d*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^2*x + b*d*e + a*e^2)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(b/e) + 8* (b^2*d*e + a*b*e^2)*x) - 4*(3*B*b*d^2 - 2*B*a*d*e - A*a*e^2 + (4*B*b*d*e - (3*B*a + A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b*d^3*e^2 - a*d^2*e^3 + (b*d*e^4 - a*e^5)*x^2 + 2*(b*d^2*e^3 - a*d*e^4)*x), -1/3*(3*(B*b*d^3 - B*a*d^2*e + (B*b*d*e^2 - B*a*e^3)*x^2 + 2*(B*b*d^2*e - B*a*d*e^2)*x)*sqrt( -b/e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sqrt(b*x + a)*sqrt(e*x + d)*sqrt(-b /e)/(b^2*e*x^2 + a*b*d + (b^2*d + a*b*e)*x)) + 2*(3*B*b*d^2 - 2*B*a*d*e - A*a*e^2 + (4*B*b*d*e - (3*B*a + A*b)*e^2)*x)*sqrt(b*x + a)*sqrt(e*x + d))/ (b*d^3*e^2 - a*d^2*e^3 + (b*d*e^4 - a*e^5)*x^2 + 2*(b*d^2*e^3 - a*d*e^4)*x )]
\[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \sqrt {a + b x}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (89) = 178\).
Time = 0.36 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=-\frac {2 \, B {\left | b \right |} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} e^{2}} - \frac {2 \, \sqrt {b x + a} {\left (\frac {{\left (4 \, B b^{4} d e^{2} {\left | b \right |} - 3 \, B a b^{3} e^{3} {\left | b \right |} - A b^{4} e^{3} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{3} d e^{3} - a b^{2} e^{4}} + \frac {3 \, {\left (B b^{5} d^{2} e {\left | b \right |} - 2 \, B a b^{4} d e^{2} {\left | b \right |} + B a^{2} b^{3} e^{3} {\left | b \right |}\right )}}{b^{3} d e^{3} - a b^{2} e^{4}}\right )}}{3 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {3}{2}}} \]
-2*B*abs(b)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*e^2) - 2/3*sqrt(b*x + a)*((4*B*b^4*d*e^2*abs(b) - 3* B*a*b^3*e^3*abs(b) - A*b^4*e^3*abs(b))*(b*x + a)/(b^3*d*e^3 - a*b^2*e^4) + 3*(B*b^5*d^2*e*abs(b) - 2*B*a*b^4*d*e^2*abs(b) + B*a^2*b^3*e^3*abs(b))/(b ^3*d*e^3 - a*b^2*e^4))/(b^2*d + (b*x + a)*b*e - a*b*e)^(3/2)
Timed out. \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,\sqrt {a+b\,x}}{{\left (d+e\,x\right )}^{5/2}} \,d x \]