\(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x} \, dx\) [865]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=a^2 c^2 \sqrt {c+d x^2}+\frac {1}{3} a^2 c \left (c+d x^2\right )^{3/2}+\frac {1}{5} a^2 \left (c+d x^2\right )^{5/2}-\frac {b (b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^2}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}-a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \] Output:

a^2*c^2*(d*x^2+c)^(1/2)+1/3*a^2*c*(d*x^2+c)^(3/2)+1/5*a^2*(d*x^2+c)^(5/2)- 
1/7*b*(-2*a*d+b*c)*(d*x^2+c)^(7/2)/d^2+1/9*b^2*(d*x^2+c)^(9/2)/d^2-a^2*c^( 
5/2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {\sqrt {c+d x^2} \left (90 a b d \left (c+d x^2\right )^3-5 b^2 \left (2 c-7 d x^2\right ) \left (c+d x^2\right )^3+21 a^2 d^2 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{315 d^2}-a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \] Input:

Integrate[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x,x]
 

Output:

(Sqrt[c + d*x^2]*(90*a*b*d*(c + d*x^2)^3 - 5*b^2*(2*c - 7*d*x^2)*(c + d*x^ 
2)^3 + 21*a^2*d^2*(23*c^2 + 11*c*d*x^2 + 3*d^2*x^4)))/(315*d^2) - a^2*c^(5 
/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {354, 99, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^2 \left (d x^2+c\right )^{5/2}}{x^2}dx^2\)

\(\Big \downarrow \) 99

\(\displaystyle \frac {1}{2} \int \left (\frac {b^2 \left (d x^2+c\right )^{7/2}}{d}-\frac {b (b c-2 a d) \left (d x^2+c\right )^{5/2}}{d}+\frac {a^2 \left (d x^2+c\right )^{5/2}}{x^2}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-2 a^2 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )+2 a^2 c^2 \sqrt {c+d x^2}+\frac {2}{5} a^2 \left (c+d x^2\right )^{5/2}+\frac {2}{3} a^2 c \left (c+d x^2\right )^{3/2}-\frac {2 b \left (c+d x^2\right )^{7/2} (b c-2 a d)}{7 d^2}+\frac {2 b^2 \left (c+d x^2\right )^{9/2}}{9 d^2}\right )\)

Input:

Int[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x,x]
 

Output:

(2*a^2*c^2*Sqrt[c + d*x^2] + (2*a^2*c*(c + d*x^2)^(3/2))/3 + (2*a^2*(c + d 
*x^2)^(5/2))/5 - (2*b*(b*c - 2*a*d)*(c + d*x^2)^(7/2))/(7*d^2) + (2*b^2*(c 
 + d*x^2)^(9/2))/(9*d^2) - 2*a^2*c^(5/2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]) 
/2
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95

method	result	size
default	\(a^{2} \left (\frac {\left (x^{2} d +c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (x^{2} d +c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {x^{2} d +c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )\right )\right )\right )+b^{2} \left (\frac {x^{2} \left (x^{2} d +c \right )^{\frac {7}{2}}}{9 d}-\frac {2 c \left (x^{2} d +c \right )^{\frac {7}{2}}}{63 d^{2}}\right )+\frac {2 a b \left (x^{2} d +c \right )^{\frac {7}{2}}}{7 d}\)	\(125\)
pseudoelliptic	\(\frac {-15 a^{2} c^{\frac {5}{2}} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )+23 \left (\frac {3 \left (\frac {5}{9} b^{2} x^{4}+\frac {10}{7} a b \,x^{2}+a^{2}\right ) x^{4} d^{4}}{23}+\frac {11 c \,x^{2} \left (\frac {95}{231} b^{2} x^{4}+\frac {90}{77} a b \,x^{2}+a^{2}\right ) d^{3}}{23}+c^{2} \left (\frac {25}{161} b^{2} x^{4}+\frac {90}{161} a b \,x^{2}+a^{2}\right ) d^{2}+\frac {30 \left (\frac {b \,x^{2}}{18}+a \right ) c^{3} b d}{161}-\frac {10 b^{2} c^{4}}{483}\right ) \sqrt {x^{2} d +c}}{15 d^{2}}\)	\(148\)

Input:

int((b*x^2+a)^2*(d*x^2+c)^(5/2)/x,x,method=_RETURNVERBOSE)
 

Output:

a^2*(1/5*(d*x^2+c)^(5/2)+c*(1/3*(d*x^2+c)^(3/2)+c*((d*x^2+c)^(1/2)-c^(1/2) 
*ln((2*c+2*c^(1/2)*(d*x^2+c)^(1/2))/x))))+b^2*(1/9*x^2*(d*x^2+c)^(7/2)/d-2 
/63*c/d^2*(d*x^2+c)^(7/2))+2/7*a*b/d*(d*x^2+c)^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.75 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\left [\frac {315 \, a^{2} c^{\frac {5}{2}} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (35 \, b^{2} d^{4} x^{8} + 5 \, {\left (19 \, b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 10 \, b^{2} c^{4} + 90 \, a b c^{3} d + 483 \, a^{2} c^{2} d^{2} + 3 \, {\left (25 \, b^{2} c^{2} d^{2} + 90 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} + {\left (5 \, b^{2} c^{3} d + 270 \, a b c^{2} d^{2} + 231 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{630 \, d^{2}}, \frac {315 \, a^{2} \sqrt {-c} c^{2} d^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) + {\left (35 \, b^{2} d^{4} x^{8} + 5 \, {\left (19 \, b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 10 \, b^{2} c^{4} + 90 \, a b c^{3} d + 483 \, a^{2} c^{2} d^{2} + 3 \, {\left (25 \, b^{2} c^{2} d^{2} + 90 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{4} + {\left (5 \, b^{2} c^{3} d + 270 \, a b c^{2} d^{2} + 231 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{315 \, d^{2}}\right ] \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x,x, algorithm="fricas")
 

Output:

[1/630*(315*a^2*c^(5/2)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c) 
/x^2) + 2*(35*b^2*d^4*x^8 + 5*(19*b^2*c*d^3 + 18*a*b*d^4)*x^6 - 10*b^2*c^4 
 + 90*a*b*c^3*d + 483*a^2*c^2*d^2 + 3*(25*b^2*c^2*d^2 + 90*a*b*c*d^3 + 21* 
a^2*d^4)*x^4 + (5*b^2*c^3*d + 270*a*b*c^2*d^2 + 231*a^2*c*d^3)*x^2)*sqrt(d 
*x^2 + c))/d^2, 1/315*(315*a^2*sqrt(-c)*c^2*d^2*arctan(sqrt(d*x^2 + c)*sqr 
t(-c)/c) + (35*b^2*d^4*x^8 + 5*(19*b^2*c*d^3 + 18*a*b*d^4)*x^6 - 10*b^2*c^ 
4 + 90*a*b*c^3*d + 483*a^2*c^2*d^2 + 3*(25*b^2*c^2*d^2 + 90*a*b*c*d^3 + 21 
*a^2*d^4)*x^4 + (5*b^2*c^3*d + 270*a*b*c^2*d^2 + 231*a^2*c*d^3)*x^2)*sqrt( 
d*x^2 + c))/d^2]
 

Sympy [A] (verification not implemented)

Time = 21.77 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {\begin {cases} \frac {2 a^{2} c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + 2 a^{2} c^{2} \sqrt {c + d x^{2}} + \frac {2 a^{2} c \left (c + d x^{2}\right )^{\frac {3}{2}}}{3} + \frac {2 a^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}{5} + \frac {2 b^{2} \left (c + d x^{2}\right )^{\frac {9}{2}}}{9 d^{2}} + \frac {2 \left (c + d x^{2}\right )^{\frac {7}{2}} \cdot \left (2 a b d - b^{2} c\right )}{7 d^{2}} & \text {for}\: d \neq 0 \\a^{2} c^{\frac {5}{2}} \log {\left (x^{2} \right )} + 2 a b c^{\frac {5}{2}} x^{2} + \frac {b^{2} c^{\frac {5}{2}} x^{4}}{2} & \text {otherwise} \end {cases}}{2} \] Input:

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x,x)
 

Output:

Piecewise((2*a**2*c**3*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + 2*a**2*c 
**2*sqrt(c + d*x**2) + 2*a**2*c*(c + d*x**2)**(3/2)/3 + 2*a**2*(c + d*x**2 
)**(5/2)/5 + 2*b**2*(c + d*x**2)**(9/2)/(9*d**2) + 2*(c + d*x**2)**(7/2)*( 
2*a*b*d - b**2*c)/(7*d**2), Ne(d, 0)), (a**2*c**(5/2)*log(x**2) + 2*a*b*c* 
*(5/2)*x**2 + b**2*c**(5/2)*x**4/2, True))/2
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x^{2}}{9 \, d} - a^{2} c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c + \sqrt {d x^{2} + c} a^{2} c^{2} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c}{63 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{7 \, d} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x,x, algorithm="maxima")
 

Output:

1/9*(d*x^2 + c)^(7/2)*b^2*x^2/d - a^2*c^(5/2)*arcsinh(c/(sqrt(c*d)*abs(x)) 
) + 1/5*(d*x^2 + c)^(5/2)*a^2 + 1/3*(d*x^2 + c)^(3/2)*a^2*c + sqrt(d*x^2 + 
 c)*a^2*c^2 - 2/63*(d*x^2 + c)^(7/2)*b^2*c/d^2 + 2/7*(d*x^2 + c)^(7/2)*a*b 
/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {a^{2} c^{3} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {35 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} b^{2} d^{16} - 45 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c d^{16} + 90 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b d^{17} + 63 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{18} + 105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{18} + 315 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{18}}{315 \, d^{18}} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x,x, algorithm="giac")
 

Output:

a^2*c^3*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/315*(35*(d*x^2 + c)^ 
(9/2)*b^2*d^16 - 45*(d*x^2 + c)^(7/2)*b^2*c*d^16 + 90*(d*x^2 + c)^(7/2)*a* 
b*d^17 + 63*(d*x^2 + c)^(5/2)*a^2*d^18 + 105*(d*x^2 + c)^(3/2)*a^2*c*d^18 
+ 315*sqrt(d*x^2 + c)*a^2*c^2*d^18)/d^18
 

Mupad [B] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.89 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx={\left (d\,x^2+c\right )}^{5/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{5\,d^2}-\frac {c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )}{5}\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{7\,d^2}-\frac {b^2\,c}{7\,d^2}\right )\,{\left (d\,x^2+c\right )}^{7/2}+c^2\,\sqrt {d\,x^2+c}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{9/2}}{9\,d^2}+\frac {c\,{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {{\left (a\,d-b\,c\right )}^2}{d^2}-c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\right )}{3}+a^2\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i} \] Input:

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x,x)
 

Output:

(c + d*x^2)^(5/2)*((a*d - b*c)^2/(5*d^2) - (c*((2*b^2*c - 2*a*b*d)/d^2 - ( 
b^2*c)/d^2))/5) - ((2*b^2*c - 2*a*b*d)/(7*d^2) - (b^2*c)/(7*d^2))*(c + d*x 
^2)^(7/2) + a^2*c^(5/2)*atan(((c + d*x^2)^(1/2)*1i)/c^(1/2))*1i + c^2*(c + 
 d*x^2)^(1/2)*((a*d - b*c)^2/d^2 - c*((2*b^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^ 
2)) + (b^2*(c + d*x^2)^(9/2))/(9*d^2) + (c*(c + d*x^2)^(3/2)*((a*d - b*c)^ 
2/d^2 - c*((2*b^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^2)))/3
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.31 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x} \, dx=\frac {483 \sqrt {d \,x^{2}+c}\, a^{2} c^{2} d^{2}+231 \sqrt {d \,x^{2}+c}\, a^{2} c \,d^{3} x^{2}+63 \sqrt {d \,x^{2}+c}\, a^{2} d^{4} x^{4}+90 \sqrt {d \,x^{2}+c}\, a b \,c^{3} d +270 \sqrt {d \,x^{2}+c}\, a b \,c^{2} d^{2} x^{2}+270 \sqrt {d \,x^{2}+c}\, a b c \,d^{3} x^{4}+90 \sqrt {d \,x^{2}+c}\, a b \,d^{4} x^{6}-10 \sqrt {d \,x^{2}+c}\, b^{2} c^{4}+5 \sqrt {d \,x^{2}+c}\, b^{2} c^{3} d \,x^{2}+75 \sqrt {d \,x^{2}+c}\, b^{2} c^{2} d^{2} x^{4}+95 \sqrt {d \,x^{2}+c}\, b^{2} c \,d^{3} x^{6}+35 \sqrt {d \,x^{2}+c}\, b^{2} d^{4} x^{8}+315 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} c^{2} d^{2}-315 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} c^{2} d^{2}}{315 d^{2}} \] Input:

int((b*x^2+a)^2*(d*x^2+c)^(5/2)/x,x)
 

Output:

(483*sqrt(c + d*x**2)*a**2*c**2*d**2 + 231*sqrt(c + d*x**2)*a**2*c*d**3*x* 
*2 + 63*sqrt(c + d*x**2)*a**2*d**4*x**4 + 90*sqrt(c + d*x**2)*a*b*c**3*d + 
 270*sqrt(c + d*x**2)*a*b*c**2*d**2*x**2 + 270*sqrt(c + d*x**2)*a*b*c*d**3 
*x**4 + 90*sqrt(c + d*x**2)*a*b*d**4*x**6 - 10*sqrt(c + d*x**2)*b**2*c**4 
+ 5*sqrt(c + d*x**2)*b**2*c**3*d*x**2 + 75*sqrt(c + d*x**2)*b**2*c**2*d**2 
*x**4 + 95*sqrt(c + d*x**2)*b**2*c*d**3*x**6 + 35*sqrt(c + d*x**2)*b**2*d* 
*4*x**8 + 315*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x)/sqrt(c) 
)*a**2*c**2*d**2 - 315*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x 
)/sqrt(c))*a**2*c**2*d**2)/(315*d**2)