Integrand size = 24, antiderivative size = 132 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {\left (5 b c^2+2 a d^2\right ) (e x)^{3/2}}{3 c d^2 e (c+d x)^{3/2}}+\frac {b (e x)^{5/2}}{d e^2 (c+d x)^{3/2}}+\frac {5 b c \sqrt {e x}}{d^3 \sqrt {c+d x}}-\frac {5 b c \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{7/2}} \] Output:
1/3*(2*a*d^2+5*b*c^2)*(e*x)^(3/2)/c/d^2/e/(d*x+c)^(3/2)+b*(e*x)^(5/2)/d/e^ 2/(d*x+c)^(3/2)+5*b*c*(e*x)^(1/2)/d^3/(d*x+c)^(1/2)-5*b*c*e^(1/2)*arctanh( d^(1/2)*(e*x)^(1/2)/e^(1/2)/(d*x+c)^(1/2))/d^(7/2)
Time = 0.40 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {e x} \left (\frac {\sqrt {d} \left (2 a d^3 x+b c \left (15 c^2+20 c d x+3 d^2 x^2\right )\right )}{(c+d x)^{3/2}}+\frac {30 b c^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}-\sqrt {c+d x}}\right )}{\sqrt {x}}\right )}{3 c d^{7/2}} \] Input:
Integrate[(Sqrt[e*x]*(a + b*x^2))/(c + d*x)^(5/2),x]
Output:
(Sqrt[e*x]*((Sqrt[d]*(2*a*d^3*x + b*c*(15*c^2 + 20*c*d*x + 3*d^2*x^2)))/(c + d*x)^(3/2) + (30*b*c^2*ArcTanh[(Sqrt[d]*Sqrt[x])/(Sqrt[c] - Sqrt[c + d* x])])/Sqrt[x]))/(3*c*d^(7/2))
Time = 0.38 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {519, 27, 87, 60, 65, 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.
\(\displaystyle \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 519 |
\(\displaystyle \frac {2 (e x)^{3/2} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}-\frac {2 \int \frac {3 b c \sqrt {e x} (c-d x)}{2 d^2 (c+d x)^{3/2}}dx}{3 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 (e x)^{3/2} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}-\frac {b \int \frac {\sqrt {e x} (c-d x)}{(c+d x)^{3/2}}dx}{d^2}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {2 (e x)^{3/2} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}-\frac {b \left (\frac {4 (e x)^{3/2}}{e \sqrt {c+d x}}-5 \int \frac {\sqrt {e x}}{\sqrt {c+d x}}dx\right )}{d^2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 (e x)^{3/2} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}-\frac {b \left (\frac {4 (e x)^{3/2}}{e \sqrt {c+d x}}-5 \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx}{2 d}\right )\right )}{d^2}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {2 (e x)^{3/2} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}-\frac {b \left (\frac {4 (e x)^{3/2}}{e \sqrt {c+d x}}-5 \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c e \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}}{d}\right )\right )}{d^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 (e x)^{3/2} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}-\frac {b \left (\frac {4 (e x)^{3/2}}{e \sqrt {c+d x}}-5 \left (\frac {\sqrt {e x} \sqrt {c+d x}}{d}-\frac {c \sqrt {e} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{d^{3/2}}\right )\right )}{d^2}\) |
Input:
Int[(Sqrt[e*x]*(a + b*x^2))/(c + d*x)^(5/2),x]
Output:
(2*(a + (b*c^2)/d^2)*(e*x)^(3/2))/(3*c*e*(c + d*x)^(3/2)) - (b*((4*(e*x)^( 3/2))/(e*Sqrt[c + d*x]) - 5*((Sqrt[e*x]*Sqrt[c + d*x])/d - (c*Sqrt[e]*ArcT anh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[c + d*x])])/d^(3/2))))/d^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 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]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, c + d*x, x], R = PolynomialRemainder[(a + b*x^2)^p, c + d*x, x]}, Simp[(-R)*(e*x)^(m + 1)*( (c + d*x)^(n + 1)/(c*e*(n + 1))), x] + Simp[1/(c*(n + 1)) Int[(e*x)^m*(c + d*x)^(n + 1)*ExpandToSum[c*(n + 1)*Qx + R*(m + n + 2), x], x], x]] /; Fre eQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0] && LtQ[n, -1] && !IntegerQ[m]
Leaf count of result is larger than twice the leaf count of optimal. \(252\) vs. \(2(106)=212\).
Time = 0.26 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.92
method | result | size |
risch | \(\frac {b x \sqrt {d x +c}\, e}{d^{3} \sqrt {e x}}-\frac {\left (\frac {2 \left (-2 a \,d^{2}-6 b \,c^{2}\right ) \sqrt {d e \left (x +\frac {c}{d}\right )^{2}-c e \left (x +\frac {c}{d}\right )}}{d c e \left (x +\frac {c}{d}\right )}+\frac {5 b c \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right )}{\sqrt {d e}}+\frac {2 c \left (a \,d^{2}+b \,c^{2}\right ) \left (\frac {2 \sqrt {d e \left (x +\frac {c}{d}\right )^{2}-c e \left (x +\frac {c}{d}\right )}}{3 c e \left (x +\frac {c}{d}\right )^{2}}+\frac {4 d \sqrt {d e \left (x +\frac {c}{d}\right )^{2}-c e \left (x +\frac {c}{d}\right )}}{3 e \,c^{2} \left (x +\frac {c}{d}\right )}\right )}{d^{2}}\right ) e \sqrt {\left (d x +c \right ) e x}}{2 d^{3} \sqrt {e x}\, \sqrt {d x +c}}\) | \(253\) |
default | \(-\frac {\left (15 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{2} d^{2} e \,x^{2}+30 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{3} d e x -6 b c \,d^{2} x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+15 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{4} e -4 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, a \,d^{3} x -40 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b \,c^{2} d x -30 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b \,c^{3}\right ) \sqrt {e x}}{6 c \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, d^{3} \left (d x +c \right )^{\frac {3}{2}}}\) | \(259\) |
Input:
int((e*x)^(1/2)*(b*x^2+a)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
b/d^3*x*(d*x+c)^(1/2)*e/(e*x)^(1/2)-1/2/d^3*(2*(-2*a*d^2-6*b*c^2)/d/c/e/(x +c/d)*(d*e*(x+c/d)^2-c*e*(x+c/d))^(1/2)+5*b*c*ln((1/2*c*e+d*e*x)/(d*e)^(1/ 2)+(d*e*x^2+c*e*x)^(1/2))/(d*e)^(1/2)+2*c*(a*d^2+b*c^2)/d^2*(2/3/c/e/(x+c/ d)^2*(d*e*(x+c/d)^2-c*e*(x+c/d))^(1/2)+4/3*d/e/c^2/(x+c/d)*(d*e*(x+c/d)^2- c*e*(x+c/d))^(1/2)))*e*((d*x+c)*e*x)^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.18 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\left [\frac {15 \, {\left (b c^{2} d^{2} x^{2} + 2 \, b c^{3} d x + b c^{4}\right )} \sqrt {\frac {e}{d}} \log \left (2 \, d e x - 2 \, \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}} + c e\right ) + 2 \, {\left (3 \, b c d^{2} x^{2} + 15 \, b c^{3} + 2 \, {\left (10 \, b c^{2} d + a d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{6 \, {\left (c d^{5} x^{2} + 2 \, c^{2} d^{4} x + c^{3} d^{3}\right )}}, \frac {15 \, {\left (b c^{2} d^{2} x^{2} + 2 \, b c^{3} d x + b c^{4}\right )} \sqrt {-\frac {e}{d}} \arctan \left (\frac {\sqrt {d x + c} \sqrt {e x} d \sqrt {-\frac {e}{d}}}{d e x + c e}\right ) + {\left (3 \, b c d^{2} x^{2} + 15 \, b c^{3} + 2 \, {\left (10 \, b c^{2} d + a d^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {e x}}{3 \, {\left (c d^{5} x^{2} + 2 \, c^{2} d^{4} x + c^{3} d^{3}\right )}}\right ] \] Input:
integrate((e*x)^(1/2)*(b*x^2+a)/(d*x+c)^(5/2),x, algorithm="fricas")
Output:
[1/6*(15*(b*c^2*d^2*x^2 + 2*b*c^3*d*x + b*c^4)*sqrt(e/d)*log(2*d*e*x - 2*s qrt(d*x + c)*sqrt(e*x)*d*sqrt(e/d) + c*e) + 2*(3*b*c*d^2*x^2 + 15*b*c^3 + 2*(10*b*c^2*d + a*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/(c*d^5*x^2 + 2*c^2*d^4* x + c^3*d^3), 1/3*(15*(b*c^2*d^2*x^2 + 2*b*c^3*d*x + b*c^4)*sqrt(-e/d)*arc tan(sqrt(d*x + c)*sqrt(e*x)*d*sqrt(-e/d)/(d*e*x + c*e)) + (3*b*c*d^2*x^2 + 15*b*c^3 + 2*(10*b*c^2*d + a*d^3)*x)*sqrt(d*x + c)*sqrt(e*x))/(c*d^5*x^2 + 2*c^2*d^4*x + c^3*d^3)]
Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (126) = 252\).
Time = 7.09 (sec) , antiderivative size = 474, normalized size of antiderivative = 3.59 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {2 a \sqrt {e} x^{\frac {3}{2}}}{3 c^{\frac {5}{2}} \sqrt {1 + \frac {d x}{c}} + 3 c^{\frac {3}{2}} d x \sqrt {1 + \frac {d x}{c}}} + b \left (- \frac {15 c^{\frac {81}{2}} d^{22} \sqrt {e} x^{\frac {51}{2}} \sqrt {1 + \frac {d x}{c}} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{3 c^{\frac {79}{2}} d^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {d x}{c}} + 3 c^{\frac {77}{2}} d^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {d x}{c}}} - \frac {15 c^{\frac {79}{2}} d^{23} \sqrt {e} x^{\frac {53}{2}} \sqrt {1 + \frac {d x}{c}} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{3 c^{\frac {79}{2}} d^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {d x}{c}} + 3 c^{\frac {77}{2}} d^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {15 c^{40} d^{\frac {45}{2}} \sqrt {e} x^{26}}{3 c^{\frac {79}{2}} d^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {d x}{c}} + 3 c^{\frac {77}{2}} d^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {20 c^{39} d^{\frac {47}{2}} \sqrt {e} x^{27}}{3 c^{\frac {79}{2}} d^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {d x}{c}} + 3 c^{\frac {77}{2}} d^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {3 c^{38} d^{\frac {49}{2}} \sqrt {e} x^{28}}{3 c^{\frac {79}{2}} d^{\frac {51}{2}} x^{\frac {51}{2}} \sqrt {1 + \frac {d x}{c}} + 3 c^{\frac {77}{2}} d^{\frac {53}{2}} x^{\frac {53}{2}} \sqrt {1 + \frac {d x}{c}}}\right ) \] Input:
integrate((e*x)**(1/2)*(b*x**2+a)/(d*x+c)**(5/2),x)
Output:
2*a*sqrt(e)*x**(3/2)/(3*c**(5/2)*sqrt(1 + d*x/c) + 3*c**(3/2)*d*x*sqrt(1 + d*x/c)) + b*(-15*c**(81/2)*d**22*sqrt(e)*x**(51/2)*sqrt(1 + d*x/c)*asinh( sqrt(d)*sqrt(x)/sqrt(c))/(3*c**(79/2)*d**(51/2)*x**(51/2)*sqrt(1 + d*x/c) + 3*c**(77/2)*d**(53/2)*x**(53/2)*sqrt(1 + d*x/c)) - 15*c**(79/2)*d**23*sq rt(e)*x**(53/2)*sqrt(1 + d*x/c)*asinh(sqrt(d)*sqrt(x)/sqrt(c))/(3*c**(79/2 )*d**(51/2)*x**(51/2)*sqrt(1 + d*x/c) + 3*c**(77/2)*d**(53/2)*x**(53/2)*sq rt(1 + d*x/c)) + 15*c**40*d**(45/2)*sqrt(e)*x**26/(3*c**(79/2)*d**(51/2)*x **(51/2)*sqrt(1 + d*x/c) + 3*c**(77/2)*d**(53/2)*x**(53/2)*sqrt(1 + d*x/c) ) + 20*c**39*d**(47/2)*sqrt(e)*x**27/(3*c**(79/2)*d**(51/2)*x**(51/2)*sqrt (1 + d*x/c) + 3*c**(77/2)*d**(53/2)*x**(53/2)*sqrt(1 + d*x/c)) + 3*c**38*d **(49/2)*sqrt(e)*x**28/(3*c**(79/2)*d**(51/2)*x**(51/2)*sqrt(1 + d*x/c) + 3*c**(77/2)*d**(53/2)*x**(53/2)*sqrt(1 + d*x/c)))
Exception generated. \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x)^(1/2)*(b*x^2+a)/(d*x+c)^(5/2),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(e>0)', see `assume?` for more de tails)Is e
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (106) = 212\).
Time = 0.21 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.36 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {5 \, \sqrt {d e} b c {\left | d \right |} \log \left ({\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )}{2 \, d^{5}} + \frac {\sqrt {{\left (d x + c\right )} d e - c d e} \sqrt {d x + c} b {\left | d \right |}}{d^{5}} + \frac {4 \, {\left (7 \, \sqrt {d e} b c^{4} d^{2} e^{3} {\left | d \right |} + \sqrt {d e} a c^{2} d^{4} e^{3} {\left | d \right |} + 12 \, \sqrt {d e} {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2} b c^{3} d e^{2} {\left | d \right |} + 9 \, \sqrt {d e} {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{4} b c^{2} e {\left | d \right |} + 3 \, \sqrt {d e} {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{4} a d^{2} e {\left | d \right |}\right )}}{3 \, {\left (c d e + {\left (\sqrt {d e} \sqrt {d x + c} - \sqrt {{\left (d x + c\right )} d e - c d e}\right )}^{2}\right )}^{3} d^{4}} \] Input:
integrate((e*x)^(1/2)*(b*x^2+a)/(d*x+c)^(5/2),x, algorithm="giac")
Output:
5/2*sqrt(d*e)*b*c*abs(d)*log((sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)/d^5 + sqrt((d*x + c)*d*e - c*d*e)*sqrt(d*x + c)*b*abs(d)/d^5 + 4/3*(7*sqrt(d*e)*b*c^4*d^2*e^3*abs(d) + sqrt(d*e)*a*c^2*d^4*e^3*abs(d) + 12*sqrt(d*e)*(sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2*b *c^3*d*e^2*abs(d) + 9*sqrt(d*e)*(sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)* d*e - c*d*e))^4*b*c^2*e*abs(d) + 3*sqrt(d*e)*(sqrt(d*e)*sqrt(d*x + c) - sq rt((d*x + c)*d*e - c*d*e))^4*a*d^2*e*abs(d))/((c*d*e + (sqrt(d*e)*sqrt(d*x + c) - sqrt((d*x + c)*d*e - c*d*e))^2)^3*d^4)
Timed out. \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\int \frac {\sqrt {e\,x}\,\left (b\,x^2+a\right )}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int(((e*x)^(1/2)*(a + b*x^2))/(c + d*x)^(5/2),x)
Output:
int(((e*x)^(1/2)*(a + b*x^2))/(c + d*x)^(5/2), x)
Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt {e x} \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-30 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{3}-30 \sqrt {d}\, \sqrt {d x +c}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{2} d x +4 \sqrt {d}\, \sqrt {d x +c}\, a c \,d^{2}+4 \sqrt {d}\, \sqrt {d x +c}\, a \,d^{3} x -5 \sqrt {d}\, \sqrt {d x +c}\, b \,c^{3}-5 \sqrt {d}\, \sqrt {d x +c}\, b \,c^{2} d x +4 \sqrt {x}\, a \,d^{4} x +30 \sqrt {x}\, b \,c^{3} d +40 \sqrt {x}\, b \,c^{2} d^{2} x +6 \sqrt {x}\, b c \,d^{3} x^{2}\right )}{6 \sqrt {d x +c}\, c \,d^{4} \left (d x +c \right )} \] Input:
int((e*x)^(1/2)*(b*x^2+a)/(d*x+c)^(5/2),x)
Output:
(sqrt(e)*( - 30*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d) )/sqrt(c))*b*c**3 - 30*sqrt(d)*sqrt(c + d*x)*log((sqrt(c + d*x) + sqrt(x)* sqrt(d))/sqrt(c))*b*c**2*d*x + 4*sqrt(d)*sqrt(c + d*x)*a*c*d**2 + 4*sqrt(d )*sqrt(c + d*x)*a*d**3*x - 5*sqrt(d)*sqrt(c + d*x)*b*c**3 - 5*sqrt(d)*sqrt (c + d*x)*b*c**2*d*x + 4*sqrt(x)*a*d**4*x + 30*sqrt(x)*b*c**3*d + 40*sqrt( x)*b*c**2*d**2*x + 6*sqrt(x)*b*c*d**3*x**2))/(6*sqrt(c + d*x)*c*d**4*(c + d*x))