Integrand size = 20, antiderivative size = 104 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=2 a c^2 \sqrt {c+d x}+\frac {2}{3} a c (c+d x)^{3/2}+\frac {2}{5} a (c+d x)^{5/2}-\frac {2 b c (c+d x)^{7/2}}{7 d^2}+\frac {2 b (c+d x)^{9/2}}{9 d^2}-2 a c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Output:
2*a*c^2*(d*x+c)^(1/2)+2/3*a*c*(d*x+c)^(3/2)+2/5*a*(d*x+c)^(5/2)-2/7*b*c*(d *x+c)^(7/2)/d^2+2/9*b*(d*x+c)^(9/2)/d^2-2*a*c^(5/2)*arctanh((d*x+c)^(1/2)/ c^(1/2))
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=\frac {2 b (c+d x)^{7/2} (-2 c+7 d x)}{63 d^2}+\frac {2}{15} a \sqrt {c+d x} \left (23 c^2+11 c d x+3 d^2 x^2\right )-2 a c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right ) \] Input:
Integrate[((c + d*x)^(5/2)*(a + b*x^2))/x,x]
Output:
(2*b*(c + d*x)^(7/2)*(-2*c + 7*d*x))/(63*d^2) + (2*a*Sqrt[c + d*x]*(23*c^2 + 11*c*d*x + 3*d^2*x^2))/15 - 2*a*c^(5/2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {517, 25, 1584, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) (c+d x)^{5/2}}{x} \, dx\) |
\(\Big \downarrow \) 517 |
\(\displaystyle \frac {2 \int \frac {(c+d x)^3 \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )}{d x}d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 \int -\frac {(c+d x)^3 \left (b c^2-2 b (c+d x) c+a d^2+b (c+d x)^2\right )}{d x}d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle -\frac {2 \int \left (-b (c+d x)^4+b c (c+d x)^3-a d^2 (c+d x)^2-a c d^2 (c+d x)-a c^2 d^2-\frac {a c^3 d}{x}\right )d\sqrt {c+d x}}{d^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (-a c^{5/2} d^2 \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+a c^2 d^2 \sqrt {c+d x}+\frac {1}{5} a d^2 (c+d x)^{5/2}+\frac {1}{3} a c d^2 (c+d x)^{3/2}+\frac {1}{9} b (c+d x)^{9/2}-\frac {1}{7} b c (c+d x)^{7/2}\right )}{d^2}\) |
Input:
Int[((c + d*x)^(5/2)*(a + b*x^2))/x,x]
Output:
(2*(a*c^2*d^2*Sqrt[c + d*x] + (a*c*d^2*(c + d*x)^(3/2))/3 + (a*d^2*(c + d* x)^(5/2))/5 - (b*c*(c + d*x)^(7/2))/7 + (b*(c + d*x)^(9/2))/9 - a*c^(5/2)* d^2*ArcTanh[Sqrt[c + d*x]/Sqrt[c]]))/d^2
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {9}{2}}}{9}-\frac {2 b c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a c \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a \,c^{2} d^{2} \sqrt {d x +c}-2 a \,c^{\frac {5}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{2}}\) | \(91\) |
default | \(\frac {\frac {2 b \left (d x +c \right )^{\frac {9}{2}}}{9}-\frac {2 b c \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 a \,d^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a c \,d^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a \,c^{2} d^{2} \sqrt {d x +c}-2 a \,c^{\frac {5}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{d^{2}}\) | \(91\) |
pseudoelliptic | \(\frac {-2 a \,c^{\frac {5}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )+\frac {46 \sqrt {d x +c}\, \left (\frac {3 x^{2} \left (\frac {5 b \,x^{2}}{9}+a \right ) d^{4}}{23}+\frac {11 \left (\frac {95 b \,x^{2}}{231}+a \right ) x c \,d^{3}}{23}+c^{2} \left (\frac {25 b \,x^{2}}{161}+a \right ) d^{2}+\frac {5 b \,c^{3} d x}{483}-\frac {10 b \,c^{4}}{483}\right )}{15}}{d^{2}}\) | \(98\) |
Input:
int((d*x+c)^(5/2)*(b*x^2+a)/x,x,method=_RETURNVERBOSE)
Output:
2/d^2*(1/9*b*(d*x+c)^(9/2)-1/7*b*c*(d*x+c)^(7/2)+1/5*a*d^2*(d*x+c)^(5/2)+1 /3*a*c*d^2*(d*x+c)^(3/2)+a*c^2*d^2*(d*x+c)^(1/2)-a*c^(5/2)*d^2*arctanh((d* x+c)^(1/2)/c^(1/2)))
Time = 0.10 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.28 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=\left [\frac {315 \, a c^{\frac {5}{2}} d^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (35 \, b d^{4} x^{4} + 95 \, b c d^{3} x^{3} - 10 \, b c^{4} + 483 \, a c^{2} d^{2} + 3 \, {\left (25 \, b c^{2} d^{2} + 21 \, a d^{4}\right )} x^{2} + {\left (5 \, b c^{3} d + 231 \, a c d^{3}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{2}}, \frac {2 \, {\left (315 \, a \sqrt {-c} c^{2} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (35 \, b d^{4} x^{4} + 95 \, b c d^{3} x^{3} - 10 \, b c^{4} + 483 \, a c^{2} d^{2} + 3 \, {\left (25 \, b c^{2} d^{2} + 21 \, a d^{4}\right )} x^{2} + {\left (5 \, b c^{3} d + 231 \, a c d^{3}\right )} x\right )} \sqrt {d x + c}\right )}}{315 \, d^{2}}\right ] \] Input:
integrate((d*x+c)^(5/2)*(b*x^2+a)/x,x, algorithm="fricas")
Output:
[1/315*(315*a*c^(5/2)*d^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2 *(35*b*d^4*x^4 + 95*b*c*d^3*x^3 - 10*b*c^4 + 483*a*c^2*d^2 + 3*(25*b*c^2*d ^2 + 21*a*d^4)*x^2 + (5*b*c^3*d + 231*a*c*d^3)*x)*sqrt(d*x + c))/d^2, 2/31 5*(315*a*sqrt(-c)*c^2*d^2*arctan(sqrt(-c)/sqrt(d*x + c)) + (35*b*d^4*x^4 + 95*b*c*d^3*x^3 - 10*b*c^4 + 483*a*c^2*d^2 + 3*(25*b*c^2*d^2 + 21*a*d^4)*x ^2 + (5*b*c^3*d + 231*a*c*d^3)*x)*sqrt(d*x + c))/d^2]
Time = 2.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.26 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=\begin {cases} \frac {2 a c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a c^{2} \sqrt {c + d x} + \frac {2 a c \left (c + d x\right )^{\frac {3}{2}}}{3} + \frac {2 a \left (c + d x\right )^{\frac {5}{2}}}{5} - \frac {2 b c \left (c + d x\right )^{\frac {7}{2}}}{7 d^{2}} + \frac {2 b \left (c + d x\right )^{\frac {9}{2}}}{9 d^{2}} & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (\frac {a \log {\left (b x^{2} \right )}}{2} + \frac {b x^{2}}{2}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((d*x+c)**(5/2)*(b*x**2+a)/x,x)
Output:
Piecewise((2*a*c**3*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + 2*a*c**2*sqrt( c + d*x) + 2*a*c*(c + d*x)**(3/2)/3 + 2*a*(c + d*x)**(5/2)/5 - 2*b*c*(c + d*x)**(7/2)/(7*d**2) + 2*b*(c + d*x)**(9/2)/(9*d**2), Ne(d, 0)), (c**(5/2) *(a*log(b*x**2)/2 + b*x**2/2), True))
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=a c^{\frac {5}{2}} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right ) + \frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b - 45 \, {\left (d x + c\right )}^{\frac {7}{2}} b c + 63 \, {\left (d x + c\right )}^{\frac {5}{2}} a d^{2} + 105 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 315 \, \sqrt {d x + c} a c^{2} d^{2}\right )}}{315 \, d^{2}} \] Input:
integrate((d*x+c)^(5/2)*(b*x^2+a)/x,x, algorithm="maxima")
Output:
a*c^(5/2)*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c))) + 2/315 *(35*(d*x + c)^(9/2)*b - 45*(d*x + c)^(7/2)*b*c + 63*(d*x + c)^(5/2)*a*d^2 + 105*(d*x + c)^(3/2)*a*c*d^2 + 315*sqrt(d*x + c)*a*c^2*d^2)/d^2
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=\frac {2 \, a c^{3} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b d^{16} - 45 \, {\left (d x + c\right )}^{\frac {7}{2}} b c d^{16} + 63 \, {\left (d x + c\right )}^{\frac {5}{2}} a d^{18} + 105 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{18} + 315 \, \sqrt {d x + c} a c^{2} d^{18}\right )}}{315 \, d^{18}} \] Input:
integrate((d*x+c)^(5/2)*(b*x^2+a)/x,x, algorithm="giac")
Output:
2*a*c^3*arctan(sqrt(d*x + c)/sqrt(-c))/sqrt(-c) + 2/315*(35*(d*x + c)^(9/2 )*b*d^16 - 45*(d*x + c)^(7/2)*b*c*d^16 + 63*(d*x + c)^(5/2)*a*d^18 + 105*( d*x + c)^(3/2)*a*c*d^18 + 315*sqrt(d*x + c)*a*c^2*d^18)/d^18
Time = 0.06 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.54 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=\left (\frac {2\,b\,c^2+2\,a\,d^2}{5\,d^2}-\frac {2\,b\,c^2}{5\,d^2}\right )\,{\left (c+d\,x\right )}^{5/2}+\frac {c\,\left (\frac {2\,b\,c^2+2\,a\,d^2}{d^2}-\frac {2\,b\,c^2}{d^2}\right )\,{\left (c+d\,x\right )}^{3/2}}{3}+c^2\,\left (\frac {2\,b\,c^2+2\,a\,d^2}{d^2}-\frac {2\,b\,c^2}{d^2}\right )\,\sqrt {c+d\,x}+\frac {2\,b\,{\left (c+d\,x\right )}^{9/2}}{9\,d^2}-\frac {2\,b\,c\,{\left (c+d\,x\right )}^{7/2}}{7\,d^2}+a\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c+d\,x}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,2{}\mathrm {i} \] Input:
int(((a + b*x^2)*(c + d*x)^(5/2))/x,x)
Output:
((2*a*d^2 + 2*b*c^2)/(5*d^2) - (2*b*c^2)/(5*d^2))*(c + d*x)^(5/2) + (c*((2 *a*d^2 + 2*b*c^2)/d^2 - (2*b*c^2)/d^2)*(c + d*x)^(3/2))/3 + a*c^(5/2)*atan (((c + d*x)^(1/2)*1i)/c^(1/2))*2i + c^2*((2*a*d^2 + 2*b*c^2)/d^2 - (2*b*c^ 2)/d^2)*(c + d*x)^(1/2) + (2*b*(c + d*x)^(9/2))/(9*d^2) - (2*b*c*(c + d*x) ^(7/2))/(7*d^2)
Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.62 \[ \int \frac {(c+d x)^{5/2} \left (a+b x^2\right )}{x} \, dx=\frac {966 \sqrt {d x +c}\, a \,c^{2} d^{2}+462 \sqrt {d x +c}\, a c \,d^{3} x +126 \sqrt {d x +c}\, a \,d^{4} x^{2}-20 \sqrt {d x +c}\, b \,c^{4}+10 \sqrt {d x +c}\, b \,c^{3} d x +150 \sqrt {d x +c}\, b \,c^{2} d^{2} x^{2}+190 \sqrt {d x +c}\, b c \,d^{3} x^{3}+70 \sqrt {d x +c}\, b \,d^{4} x^{4}+315 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a \,c^{2} d^{2}-315 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a \,c^{2} d^{2}}{315 d^{2}} \] Input:
int((d*x+c)^(5/2)*(b*x^2+a)/x,x)
Output:
(966*sqrt(c + d*x)*a*c**2*d**2 + 462*sqrt(c + d*x)*a*c*d**3*x + 126*sqrt(c + d*x)*a*d**4*x**2 - 20*sqrt(c + d*x)*b*c**4 + 10*sqrt(c + d*x)*b*c**3*d* x + 150*sqrt(c + d*x)*b*c**2*d**2*x**2 + 190*sqrt(c + d*x)*b*c*d**3*x**3 + 70*sqrt(c + d*x)*b*d**4*x**4 + 315*sqrt(c)*log(sqrt(c + d*x) - sqrt(c))*a *c**2*d**2 - 315*sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a*c**2*d**2)/(315*d* *2)