Integrand size = 24, antiderivative size = 90 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {b \sqrt {e x} \sqrt {c+d x}}{e^3}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}+\frac {b c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} e^{5/2}} \] Output:
b*(e*x)^(1/2)*(d*x+c)^(1/2)/e^3-2/3*a*(d*x+c)^(3/2)/c/e/(e*x)^(3/2)+b*c*ar ctanh(d^(1/2)*(e*x)^(1/2)/e^(1/2)/(d*x+c)^(1/2))/d^(1/2)/e^(5/2)
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {x \left (\sqrt {c+d x} \left (3 b c x^2-2 a (c+d x)\right )-\frac {3 b c^2 x^{3/2} \log \left (-\sqrt {d} \sqrt {x}+\sqrt {c+d x}\right )}{\sqrt {d}}\right )}{3 c (e x)^{5/2}} \] Input:
Integrate[(Sqrt[c + d*x]*(a + b*x^2))/(e*x)^(5/2),x]
Output:
(x*(Sqrt[c + d*x]*(3*b*c*x^2 - 2*a*(c + d*x)) - (3*b*c^2*x^(3/2)*Log[-(Sqr t[d]*Sqrt[x]) + Sqrt[c + d*x]])/Sqrt[d]))/(3*c*(e*x)^(5/2))
Time = 0.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {520, 8, 27, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \sqrt {c+d x}}{(e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 520 |
\(\displaystyle -\frac {2 \int -\frac {3 b c x \sqrt {c+d x}}{2 (e x)^{3/2}}dx}{3 c e}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {2 \int -\frac {3 b c \sqrt {c+d x}}{2 \sqrt {e x}}dx}{3 c e^2}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {\sqrt {c+d x}}{\sqrt {e x}}dx}{e^2}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {b \left (\frac {1}{2} c \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{e^2}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {b \left (c \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{e^2}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {b \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{e^2}-\frac {2 a (c+d x)^{3/2}}{3 c e (e x)^{3/2}}\) |
Input:
Int[(Sqrt[c + d*x]*(a + b*x^2))/(e*x)^(5/2),x]
Output:
(-2*a*(c + d*x)^(3/2))/(3*c*e*(e*x)^(3/2)) + (b*((Sqrt[e*x]*Sqrt[c + d*x]) /e + (c*ArcTanh[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[c + d*x])])/(Sqrt[d]*Sqr t[e])))/e^2
Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m Int[u*(a* x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
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_)^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, e*x, x], R = Pol ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c)) Int[(e*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] && !IntegerQ[n]
Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.18
method | result | size |
risch | \(-\frac {\sqrt {d x +c}\, \left (-3 b c \,x^{2}+2 a d x +2 a c \right )}{3 x c \,e^{2} \sqrt {e x}}+\frac {b c \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{2 \sqrt {d e}\, e^{2} \sqrt {e x}\, \sqrt {d x +c}}\) | \(106\) |
default | \(\frac {\sqrt {d x +c}\, \left (3 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{2} e \,x^{2}+6 \sqrt {d e}\, \sqrt {\left (d x +c \right ) e x}\, b c \,x^{2}-4 a d x \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-4 a c \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\right )}{6 e^{2} x \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, c \sqrt {d e}}\) | \(145\) |
Input:
int((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/3*(d*x+c)^(1/2)*(-3*b*c*x^2+2*a*d*x+2*a*c)/x/c/e^2/(e*x)^(1/2)+1/2*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)/e^2*((d* x+c)*e*x)^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.10 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\left [\frac {3 \, \sqrt {d e} b c^{2} x^{2} \log \left (2 \, d e x + c e + 2 \, \sqrt {d e} \sqrt {d x + c} \sqrt {e x}\right ) + 2 \, {\left (3 \, b c d x^{2} - 2 \, a d^{2} x - 2 \, a c d\right )} \sqrt {d x + c} \sqrt {e x}}{6 \, c d e^{3} x^{2}}, -\frac {3 \, \sqrt {-d e} b c^{2} x^{2} \arctan \left (\frac {\sqrt {-d e} \sqrt {d x + c} \sqrt {e x}}{d e x + c e}\right ) - {\left (3 \, b c d x^{2} - 2 \, a d^{2} x - 2 \, a c d\right )} \sqrt {d x + c} \sqrt {e x}}{3 \, c d e^{3} x^{2}}\right ] \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(5/2),x, algorithm="fricas")
Output:
[1/6*(3*sqrt(d*e)*b*c^2*x^2*log(2*d*e*x + c*e + 2*sqrt(d*e)*sqrt(d*x + c)* sqrt(e*x)) + 2*(3*b*c*d*x^2 - 2*a*d^2*x - 2*a*c*d)*sqrt(d*x + c)*sqrt(e*x) )/(c*d*e^3*x^2), -1/3*(3*sqrt(-d*e)*b*c^2*x^2*arctan(sqrt(-d*e)*sqrt(d*x + c)*sqrt(e*x)/(d*e*x + c*e)) - (3*b*c*d*x^2 - 2*a*d^2*x - 2*a*c*d)*sqrt(d* x + c)*sqrt(e*x))/(c*d*e^3*x^2)]
Time = 4.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=- \frac {2 a \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{3 e^{\frac {5}{2}} x} - \frac {2 a d^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}}{3 c e^{\frac {5}{2}}} + \frac {b \sqrt {c} \sqrt {x} \sqrt {1 + \frac {d x}{c}}}{e^{\frac {5}{2}}} + \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{\sqrt {d} e^{\frac {5}{2}}} \] Input:
integrate((d*x+c)**(1/2)*(b*x**2+a)/(e*x)**(5/2),x)
Output:
-2*a*sqrt(d)*sqrt(c/(d*x) + 1)/(3*e**(5/2)*x) - 2*a*d**(3/2)*sqrt(c/(d*x) + 1)/(3*c*e**(5/2)) + b*sqrt(c)*sqrt(x)*sqrt(1 + d*x/c)/e**(5/2) + b*c*asi nh(sqrt(d)*sqrt(x)/sqrt(c))/(sqrt(d)*e**(5/2))
Exception generated. \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(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. 145 vs. \(2 (68) = 136\).
Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.61 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=-\frac {d^{3} {\left (\frac {3 \, b c \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e} d^{2}} - \frac {{\left (\frac {3 \, b c^{2} e}{d} + {\left (\frac {3 \, {\left (d x + c\right )} b e}{d} - \frac {2 \, {\left (3 \, b c^{2} d^{3} e^{2} + a d^{5} e^{2}\right )}}{c d^{4} e}\right )} {\left (d x + c\right )}\right )} \sqrt {d x + c}}{{\left ({\left (d x + c\right )} d e - c d e\right )}^{\frac {3}{2}}}\right )}}{3 \, e^{2} {\left | d \right |}} \] Input:
integrate((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(5/2),x, algorithm="giac")
Output:
-1/3*d^3*(3*b*c*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c)*d*e - c* d*e)))/(sqrt(d*e)*d^2) - (3*b*c^2*e/d + (3*(d*x + c)*b*e/d - 2*(3*b*c^2*d^ 3*e^2 + a*d^5*e^2)/(c*d^4*e))*(d*x + c))*sqrt(d*x + c)/((d*x + c)*d*e - c* d*e)^(3/2))/(e^2*abs(d))
Timed out. \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\int \frac {\left (b\,x^2+a\right )\,\sqrt {c+d\,x}}{{\left (e\,x\right )}^{5/2}} \,d x \] Input:
int(((a + b*x^2)*(c + d*x)^(1/2))/(e*x)^(5/2),x)
Output:
int(((a + b*x^2)*(c + d*x)^(1/2))/(e*x)^(5/2), x)
Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {c+d x} \left (a+b x^2\right )}{(e x)^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-4 \sqrt {x}\, \sqrt {d x +c}\, a c d -4 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{2} x +6 \sqrt {x}\, \sqrt {d x +c}\, b c d \,x^{2}+6 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{2} x^{2}-4 \sqrt {d}\, a \,d^{2} x^{2}+\sqrt {d}\, b \,c^{2} x^{2}\right )}{6 c d \,e^{3} x^{2}} \] Input:
int((d*x+c)^(1/2)*(b*x^2+a)/(e*x)^(5/2),x)
Output:
(sqrt(e)*( - 4*sqrt(x)*sqrt(c + d*x)*a*c*d - 4*sqrt(x)*sqrt(c + d*x)*a*d** 2*x + 6*sqrt(x)*sqrt(c + d*x)*b*c*d*x**2 + 6*sqrt(d)*log((sqrt(c + d*x) + sqrt(x)*sqrt(d))/sqrt(c))*b*c**2*x**2 - 4*sqrt(d)*a*d**2*x**2 + sqrt(d)*b* c**2*x**2))/(6*c*d*e**3*x**2)