Integrand size = 24, antiderivative size = 115 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=-\frac {2 b c \sqrt {c+d x}}{e^3 \sqrt {e x}}+\frac {b d \sqrt {e x} \sqrt {c+d x}}{e^4}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}+\frac {3 b c \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{e^{7/2}} \] Output:
-2*b*c*(d*x+c)^(1/2)/e^3/(e*x)^(1/2)+b*d*(e*x)^(1/2)*(d*x+c)^(1/2)/e^4-2/5 *a*(d*x+c)^(5/2)/c/e/(e*x)^(5/2)+3*b*c*d^(1/2)*arctanh(d^(1/2)*(e*x)^(1/2) /e^(1/2)/(d*x+c)^(1/2))/e^(7/2)
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=\frac {x \left (-\sqrt {c+d x} \left (5 b c x^2 (2 c-d x)+2 a (c+d x)^2\right )-15 b c^2 \sqrt {d} x^{5/2} \log \left (-\sqrt {d} \sqrt {x}+\sqrt {c+d x}\right )\right )}{5 c (e x)^{7/2}} \] Input:
Integrate[((c + d*x)^(3/2)*(a + b*x^2))/(e*x)^(7/2),x]
Output:
(x*(-(Sqrt[c + d*x]*(5*b*c*x^2*(2*c - d*x) + 2*a*(c + d*x)^2)) - 15*b*c^2* Sqrt[d]*x^(5/2)*Log[-(Sqrt[d]*Sqrt[x]) + Sqrt[c + d*x]]))/(5*c*(e*x)^(7/2) )
Time = 0.38 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {520, 8, 27, 57, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^{3/2}}{(e x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 520 |
\(\displaystyle -\frac {2 \int -\frac {5 b c x (c+d x)^{3/2}}{2 (e x)^{5/2}}dx}{5 c e}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {2 \int -\frac {5 b c (c+d x)^{3/2}}{2 (e x)^{3/2}}dx}{5 c e^2}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {(c+d x)^{3/2}}{(e x)^{3/2}}dx}{e^2}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {b \left (\frac {3 d \int \frac {\sqrt {c+d x}}{\sqrt {e x}}dx}{e}-\frac {2 (c+d x)^{3/2}}{e \sqrt {e x}}\right )}{e^2}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {b \left (\frac {3 d \left (\frac {1}{2} c \int \frac {1}{\sqrt {e x} \sqrt {c+d x}}dx+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{e}-\frac {2 (c+d x)^{3/2}}{e \sqrt {e x}}\right )}{e^2}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {b \left (\frac {3 d \left (c \int \frac {1}{e-\frac {d e x}{c+d x}}d\frac {\sqrt {e x}}{\sqrt {c+d x}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{e}-\frac {2 (c+d x)^{3/2}}{e \sqrt {e x}}\right )}{e^2}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {b \left (\frac {3 d \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c+d x}}\right )}{\sqrt {d} \sqrt {e}}+\frac {\sqrt {e x} \sqrt {c+d x}}{e}\right )}{e}-\frac {2 (c+d x)^{3/2}}{e \sqrt {e x}}\right )}{e^2}-\frac {2 a (c+d x)^{5/2}}{5 c e (e x)^{5/2}}\) |
Input:
Int[((c + d*x)^(3/2)*(a + b*x^2))/(e*x)^(7/2),x]
Output:
(-2*a*(c + d*x)^(5/2))/(5*c*e*(e*x)^(5/2)) + (b*((-2*(c + d*x)^(3/2))/(e*S qrt[e*x]) + (3*d*((Sqrt[e*x]*Sqrt[c + d*x])/e + (c*ArcTanh[(Sqrt[d]*Sqrt[e *x])/(Sqrt[e]*Sqrt[c + d*x])])/(Sqrt[d]*Sqrt[e])))/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 + 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[((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 = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-\frac {\sqrt {d x +c}\, \left (-5 b c d \,x^{3}+2 a \,d^{2} x^{2}+10 b \,c^{2} x^{2}+4 a d x c +2 a \,c^{2}\right )}{5 x^{2} c \,e^{3} \sqrt {e x}}+\frac {3 d b c \ln \left (\frac {\frac {1}{2} c e +d e x}{\sqrt {d e}}+\sqrt {d e \,x^{2}+c e x}\right ) \sqrt {\left (d x +c \right ) e x}}{2 \sqrt {d e}\, e^{3} \sqrt {e x}\, \sqrt {d x +c}}\) | \(129\) |
default | \(\frac {\sqrt {d x +c}\, \left (15 \ln \left (\frac {2 d e x +2 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}+c e}{2 \sqrt {d e}}\right ) b \,c^{2} d e \,x^{3}+10 \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\, b c d \,x^{3}-4 a \,d^{2} x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-20 b \,c^{2} x^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-8 a c d x \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}-4 a \,c^{2} \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}\right )}{10 e^{3} x^{2} c \sqrt {e x}\, \sqrt {\left (d x +c \right ) e x}\, \sqrt {d e}}\) | \(198\) |
Input:
int((d*x+c)^(3/2)*(b*x^2+a)/(e*x)^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/5*(d*x+c)^(1/2)*(-5*b*c*d*x^3+2*a*d^2*x^2+10*b*c^2*x^2+4*a*c*d*x+2*a*c^ 2)/x^2/c/e^3/(e*x)^(1/2)+3/2*d*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^3*((d*x+c)*e*x)^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/ 2)
Time = 0.10 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.86 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=\left [\frac {15 \, b c^{2} e x^{3} \sqrt {\frac {d}{e}} \log \left (2 \, d x + 2 \, \sqrt {d x + c} \sqrt {e x} \sqrt {\frac {d}{e}} + c\right ) + 2 \, {\left (5 \, b c d x^{3} - 4 \, a c d x - 2 \, a c^{2} - 2 \, {\left (5 \, b c^{2} + a d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {e x}}{10 \, c e^{4} x^{3}}, -\frac {15 \, b c^{2} e x^{3} \sqrt {-\frac {d}{e}} \arctan \left (\frac {\sqrt {e x} \sqrt {-\frac {d}{e}}}{\sqrt {d x + c}}\right ) - {\left (5 \, b c d x^{3} - 4 \, a c d x - 2 \, a c^{2} - 2 \, {\left (5 \, b c^{2} + a d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {e x}}{5 \, c e^{4} x^{3}}\right ] \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)/(e*x)^(7/2),x, algorithm="fricas")
Output:
[1/10*(15*b*c^2*e*x^3*sqrt(d/e)*log(2*d*x + 2*sqrt(d*x + c)*sqrt(e*x)*sqrt (d/e) + c) + 2*(5*b*c*d*x^3 - 4*a*c*d*x - 2*a*c^2 - 2*(5*b*c^2 + a*d^2)*x^ 2)*sqrt(d*x + c)*sqrt(e*x))/(c*e^4*x^3), -1/5*(15*b*c^2*e*x^3*sqrt(-d/e)*a rctan(sqrt(e*x)*sqrt(-d/e)/sqrt(d*x + c)) - (5*b*c*d*x^3 - 4*a*c*d*x - 2*a *c^2 - 2*(5*b*c^2 + a*d^2)*x^2)*sqrt(d*x + c)*sqrt(e*x))/(c*e^4*x^3)]
Time = 21.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.77 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=- \frac {2 a c \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{5 e^{\frac {7}{2}} x^{2}} - \frac {4 a d^{\frac {3}{2}} \sqrt {\frac {c}{d x} + 1}}{5 e^{\frac {7}{2}} x} - \frac {2 a d^{\frac {5}{2}} \sqrt {\frac {c}{d x} + 1}}{5 c e^{\frac {7}{2}}} - \frac {2 b c^{\frac {3}{2}}}{e^{\frac {7}{2}} \sqrt {x} \sqrt {1 + \frac {d x}{c}}} + \frac {b \sqrt {c} d \sqrt {x} \sqrt {1 + \frac {d x}{c}}}{e^{\frac {7}{2}}} - \frac {2 b \sqrt {c} d \sqrt {x}}{e^{\frac {7}{2}} \sqrt {1 + \frac {d x}{c}}} + \frac {3 b c \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}} \right )}}{e^{\frac {7}{2}}} \] Input:
integrate((d*x+c)**(3/2)*(b*x**2+a)/(e*x)**(7/2),x)
Output:
-2*a*c*sqrt(d)*sqrt(c/(d*x) + 1)/(5*e**(7/2)*x**2) - 4*a*d**(3/2)*sqrt(c/( d*x) + 1)/(5*e**(7/2)*x) - 2*a*d**(5/2)*sqrt(c/(d*x) + 1)/(5*c*e**(7/2)) - 2*b*c**(3/2)/(e**(7/2)*sqrt(x)*sqrt(1 + d*x/c)) + b*sqrt(c)*d*sqrt(x)*sqr t(1 + d*x/c)/e**(7/2) - 2*b*sqrt(c)*d*sqrt(x)/(e**(7/2)*sqrt(1 + d*x/c)) + 3*b*c*sqrt(d)*asinh(sqrt(d)*sqrt(x)/sqrt(c))/e**(7/2)
Exception generated. \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)/(e*x)^(7/2),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
Time = 0.19 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.44 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=-\frac {{\left (\frac {15 \, b c d \log \left ({\left | -\sqrt {d e} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d e - c d e} \right |}\right )}{\sqrt {d e}} + \frac {{\left (15 \, b c^{3} d^{3} e^{2} - {\left (35 \, b c^{2} d^{3} e^{2} + {\left (5 \, {\left (d x + c\right )} b d^{3} e^{2} - \frac {25 \, b c^{3} d^{5} e^{4} + 2 \, a c d^{7} e^{4}}{c^{2} d^{2} e^{2}}\right )} {\left (d x + c\right )}\right )} {\left (d x + c\right )}\right )} \sqrt {d x + c}}{{\left ({\left (d x + c\right )} d e - c d e\right )}^{\frac {5}{2}}}\right )} d}{5 \, e^{3} {\left | d \right |}} \] Input:
integrate((d*x+c)^(3/2)*(b*x^2+a)/(e*x)^(7/2),x, algorithm="giac")
Output:
-1/5*(15*b*c*d*log(abs(-sqrt(d*e)*sqrt(d*x + c) + sqrt((d*x + c)*d*e - c*d *e)))/sqrt(d*e) + (15*b*c^3*d^3*e^2 - (35*b*c^2*d^3*e^2 + (5*(d*x + c)*b*d ^3*e^2 - (25*b*c^3*d^5*e^4 + 2*a*c*d^7*e^4)/(c^2*d^2*e^2))*(d*x + c))*(d*x + c))*sqrt(d*x + c)/((d*x + c)*d*e - c*d*e)^(5/2))*d/(e^3*abs(d))
Timed out. \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=\int \frac {\left (b\,x^2+a\right )\,{\left (c+d\,x\right )}^{3/2}}{{\left (e\,x\right )}^{7/2}} \,d x \] Input:
int(((a + b*x^2)*(c + d*x)^(3/2))/(e*x)^(7/2),x)
Output:
int(((a + b*x^2)*(c + d*x)^(3/2))/(e*x)^(7/2), x)
Time = 0.21 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.24 \[ \int \frac {(c+d x)^{3/2} \left (a+b x^2\right )}{(e x)^{7/2}} \, dx=\frac {\sqrt {e}\, \left (-8 \sqrt {x}\, \sqrt {d x +c}\, a \,c^{2}-16 \sqrt {x}\, \sqrt {d x +c}\, a c d x -8 \sqrt {x}\, \sqrt {d x +c}\, a \,d^{2} x^{2}-40 \sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} x^{2}+20 \sqrt {x}\, \sqrt {d x +c}\, b c d \,x^{3}+60 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d x +c}+\sqrt {x}\, \sqrt {d}}{\sqrt {c}}\right ) b \,c^{2} x^{3}-8 \sqrt {d}\, a \,d^{2} x^{3}+33 \sqrt {d}\, b \,c^{2} x^{3}\right )}{20 c \,e^{4} x^{3}} \] Input:
int((d*x+c)^(3/2)*(b*x^2+a)/(e*x)^(7/2),x)
Output:
(sqrt(e)*( - 8*sqrt(x)*sqrt(c + d*x)*a*c**2 - 16*sqrt(x)*sqrt(c + d*x)*a*c *d*x - 8*sqrt(x)*sqrt(c + d*x)*a*d**2*x**2 - 40*sqrt(x)*sqrt(c + d*x)*b*c* *2*x**2 + 20*sqrt(x)*sqrt(c + d*x)*b*c*d*x**3 + 60*sqrt(d)*log((sqrt(c + d *x) + sqrt(x)*sqrt(d))/sqrt(c))*b*c**2*x**3 - 8*sqrt(d)*a*d**2*x**3 + 33*s qrt(d)*b*c**2*x**3))/(20*c*e**4*x**3)