\(\int \frac {(a+b x^2)^2 (c+d x^2)^{5/2}}{x^8} \, dx\) [874]

Optimal result

Integrand size = 24, antiderivative size = 176 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=-\frac {b c (5 b c+4 a d) \sqrt {c+d x^2}}{15 x^3}-\frac {7 b d (5 b c+4 a d) \sqrt {c+d x^2}}{15 x}+\frac {b d^2 (5 b c+4 a d) x \sqrt {c+d x^2}}{10 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}-\frac {2 a b \left (c+d x^2\right )^{7/2}}{5 c x^5}+\frac {1}{2} b d^{3/2} (5 b c+4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right ) \] Output:

-1/15*b*c*(4*a*d+5*b*c)*(d*x^2+c)^(1/2)/x^3-7/15*b*d*(4*a*d+5*b*c)*(d*x^2+ 
c)^(1/2)/x+1/10*b*d^2*(4*a*d+5*b*c)*x*(d*x^2+c)^(1/2)/c-1/7*a^2*(d*x^2+c)^ 
(7/2)/c/x^7-2/5*a*b*(d*x^2+c)^(7/2)/c/x^5+1/2*b*d^(3/2)*(4*a*d+5*b*c)*arct 
anh(d^(1/2)*x/(d*x^2+c)^(1/2))
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=-\frac {\sqrt {c+d x^2} \left (30 a^2 \left (c+d x^2\right )^3+35 b^2 c x^4 \left (2 c^2+14 c d x^2-3 d^2 x^4\right )+28 a b c x^2 \left (3 c^2+11 c d x^2+23 d^2 x^4\right )\right )}{210 c x^7}-\frac {1}{2} b d^{3/2} (5 b c+4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right ) \] Input:

Integrate[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^8,x]
 

Output:

-1/210*(Sqrt[c + d*x^2]*(30*a^2*(c + d*x^2)^3 + 35*b^2*c*x^4*(2*c^2 + 14*c 
*d*x^2 - 3*d^2*x^4) + 28*a*b*c*x^2*(3*c^2 + 11*c*d*x^2 + 23*d^2*x^4)))/(c* 
x^7) - (b*d^(3/2)*(5*b*c + 4*a*d)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/2
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {365, 27, 359, 247, 247, 211, 224, 219}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx\)

\(\Big \downarrow \) 365

\(\displaystyle \frac {\int \frac {7 b c \left (b x^2+2 a\right ) \left (d x^2+c\right )^{5/2}}{x^6}dx}{7 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle b \int \frac {\left (b x^2+2 a\right ) \left (d x^2+c\right )^{5/2}}{x^6}dx-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 359

\(\displaystyle b \left (\frac {(4 a d+5 b c) \int \frac {\left (d x^2+c\right )^{5/2}}{x^4}dx}{5 c}-\frac {2 a \left (c+d x^2\right )^{7/2}}{5 c x^5}\right )-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 247

\(\displaystyle b \left (\frac {(4 a d+5 b c) \left (\frac {5}{3} d \int \frac {\left (d x^2+c\right )^{3/2}}{x^2}dx-\frac {\left (c+d x^2\right )^{5/2}}{3 x^3}\right )}{5 c}-\frac {2 a \left (c+d x^2\right )^{7/2}}{5 c x^5}\right )-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 247

\(\displaystyle b \left (\frac {(4 a d+5 b c) \left (\frac {5}{3} d \left (3 d \int \sqrt {d x^2+c}dx-\frac {\left (c+d x^2\right )^{3/2}}{x}\right )-\frac {\left (c+d x^2\right )^{5/2}}{3 x^3}\right )}{5 c}-\frac {2 a \left (c+d x^2\right )^{7/2}}{5 c x^5}\right )-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 211

\(\displaystyle b \left (\frac {(4 a d+5 b c) \left (\frac {5}{3} d \left (3 d \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )-\frac {\left (c+d x^2\right )^{3/2}}{x}\right )-\frac {\left (c+d x^2\right )^{5/2}}{3 x^3}\right )}{5 c}-\frac {2 a \left (c+d x^2\right )^{7/2}}{5 c x^5}\right )-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 224

\(\displaystyle b \left (\frac {(4 a d+5 b c) \left (\frac {5}{3} d \left (3 d \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )-\frac {\left (c+d x^2\right )^{3/2}}{x}\right )-\frac {\left (c+d x^2\right )^{5/2}}{3 x^3}\right )}{5 c}-\frac {2 a \left (c+d x^2\right )^{7/2}}{5 c x^5}\right )-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

\(\Big \downarrow \) 219

\(\displaystyle b \left (\frac {(4 a d+5 b c) \left (\frac {5}{3} d \left (3 d \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )-\frac {\left (c+d x^2\right )^{3/2}}{x}\right )-\frac {\left (c+d x^2\right )^{5/2}}{3 x^3}\right )}{5 c}-\frac {2 a \left (c+d x^2\right )^{7/2}}{5 c x^5}\right )-\frac {a^2 \left (c+d x^2\right )^{7/2}}{7 c x^7}\)

Input:

Int[((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^8,x]
 

Output:

-1/7*(a^2*(c + d*x^2)^(7/2))/(c*x^7) + b*((-2*a*(c + d*x^2)^(7/2))/(5*c*x^ 
5) + ((5*b*c + 4*a*d)*(-1/3*(c + d*x^2)^(5/2)/x^3 + (5*d*(-((c + d*x^2)^(3 
/2)/x) + 3*d*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^ 
2]])/(2*Sqrt[d]))))/3))/(5*c))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

rule 365

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x 
_Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] 
- Simp[1/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p 
+ 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(-\frac {-14 c \,d^{2} \left (a d +\frac {5 b c}{4}\right ) b \,x^{7} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )+\left (3 c^{2} \left (\frac {49}{9} b^{2} x^{4}+\frac {154}{45} a b \,x^{2}+a^{2}\right ) x^{2} d^{\frac {3}{2}}+3 c \left (-\frac {7}{6} b^{2} x^{4}+\frac {322}{45} a b \,x^{2}+a^{2}\right ) x^{4} d^{\frac {5}{2}}+d^{\frac {7}{2}} a^{2} x^{6}+c^{3} \sqrt {d}\, \left (\frac {7}{3} b^{2} x^{4}+\frac {14}{5} a b \,x^{2}+a^{2}\right )\right ) \sqrt {x^{2} d +c}}{7 \sqrt {d}\, x^{7} c}\)	\(153\)
risch	\(-\frac {\sqrt {x^{2} d +c}\, \left (-105 x^{8} b^{2} c \,d^{2}+30 a^{2} d^{3} x^{6}+644 a b c \,d^{2} x^{6}+490 b^{2} c^{2} d \,x^{6}+90 a^{2} c \,d^{2} x^{4}+308 a b \,c^{2} d \,x^{4}+70 b^{2} c^{3} x^{4}+90 a^{2} c^{2} d \,x^{2}+84 a b \,c^{3} x^{2}+30 a^{2} c^{3}\right )}{210 x^{7} c}+\frac {\left (4 a d +5 b c \right ) b \,d^{\frac {3}{2}} \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2}\)	\(161\)
default	\(-\frac {a^{2} \left (x^{2} d +c \right )^{\frac {7}{2}}}{7 c \,x^{7}}+b^{2} \left (-\frac {\left (x^{2} d +c \right )^{\frac {7}{2}}}{3 c \,x^{3}}+\frac {4 d \left (-\frac {\left (x^{2} d +c \right )^{\frac {7}{2}}}{c x}+\frac {6 d \left (\frac {x \left (x^{2} d +c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (x^{2} d +c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{c}\right )}{3 c}\right )+2 a b \left (-\frac {\left (x^{2} d +c \right )^{\frac {7}{2}}}{5 c \,x^{5}}+\frac {2 d \left (-\frac {\left (x^{2} d +c \right )^{\frac {7}{2}}}{3 c \,x^{3}}+\frac {4 d \left (-\frac {\left (x^{2} d +c \right )^{\frac {7}{2}}}{c x}+\frac {6 d \left (\frac {x \left (x^{2} d +c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (x^{2} d +c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {x^{2} d +c}}{2}+\frac {c \ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{c}\right )}{3 c}\right )}{5 c}\right )\)	\(284\)

Input:

int((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^8,x,method=_RETURNVERBOSE)
 

Output:

-1/7*(-14*c*d^2*(a*d+5/4*b*c)*b*x^7*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+(3* 
c^2*(49/9*b^2*x^4+154/45*a*b*x^2+a^2)*x^2*d^(3/2)+3*c*(-7/6*b^2*x^4+322/45 
*a*b*x^2+a^2)*x^4*d^(5/2)+d^(7/2)*a^2*x^6+c^3*d^(1/2)*(7/3*b^2*x^4+14/5*a* 
b*x^2+a^2))*(d*x^2+c)^(1/2))/d^(1/2)/x^7/c
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=\left [\frac {105 \, {\left (5 \, b^{2} c^{2} d + 4 \, a b c d^{2}\right )} \sqrt {d} x^{7} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (105 \, b^{2} c d^{2} x^{8} - 2 \, {\left (245 \, b^{2} c^{2} d + 322 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} - 30 \, a^{2} c^{3} - 2 \, {\left (35 \, b^{2} c^{3} + 154 \, a b c^{2} d + 45 \, a^{2} c d^{2}\right )} x^{4} - 6 \, {\left (14 \, a b c^{3} + 15 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{420 \, c x^{7}}, -\frac {105 \, {\left (5 \, b^{2} c^{2} d + 4 \, a b c d^{2}\right )} \sqrt {-d} x^{7} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (105 \, b^{2} c d^{2} x^{8} - 2 \, {\left (245 \, b^{2} c^{2} d + 322 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} - 30 \, a^{2} c^{3} - 2 \, {\left (35 \, b^{2} c^{3} + 154 \, a b c^{2} d + 45 \, a^{2} c d^{2}\right )} x^{4} - 6 \, {\left (14 \, a b c^{3} + 15 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{210 \, c x^{7}}\right ] \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^8,x, algorithm="fricas")
 

Output:

[1/420*(105*(5*b^2*c^2*d + 4*a*b*c*d^2)*sqrt(d)*x^7*log(-2*d*x^2 - 2*sqrt( 
d*x^2 + c)*sqrt(d)*x - c) + 2*(105*b^2*c*d^2*x^8 - 2*(245*b^2*c^2*d + 322* 
a*b*c*d^2 + 15*a^2*d^3)*x^6 - 30*a^2*c^3 - 2*(35*b^2*c^3 + 154*a*b*c^2*d + 
 45*a^2*c*d^2)*x^4 - 6*(14*a*b*c^3 + 15*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/( 
c*x^7), -1/210*(105*(5*b^2*c^2*d + 4*a*b*c*d^2)*sqrt(-d)*x^7*arctan(sqrt(- 
d)*x/sqrt(d*x^2 + c)) - (105*b^2*c*d^2*x^8 - 2*(245*b^2*c^2*d + 322*a*b*c* 
d^2 + 15*a^2*d^3)*x^6 - 30*a^2*c^3 - 2*(35*b^2*c^3 + 154*a*b*c^2*d + 45*a^ 
2*c*d^2)*x^4 - 6*(14*a*b*c^3 + 15*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c*x^7) 
]
 

Sympy [A] (verification not implemented)

Time = 4.00 (sec) , antiderivative size = 828, normalized size of antiderivative = 4.70 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx =\text {Too large to display} \] Input:

integrate((b*x**2+a)**2*(d*x**2+c)**(5/2)/x**8,x)
 

Output:

-15*a**2*c**7*d**(9/2)*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4 
*d**5*x**8 + 105*c**3*d**6*x**10) - 33*a**2*c**6*d**(11/2)*x**2*sqrt(c/(d* 
x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6*x**10) 
 - 17*a**2*c**5*d**(13/2)*x**4*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 
210*c**4*d**5*x**8 + 105*c**3*d**6*x**10) - 3*a**2*c**4*d**(15/2)*x**6*sqr 
t(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6 
*x**10) - 12*a**2*c**3*d**(17/2)*x**8*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4* 
x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6*x**10) - 8*a**2*c**2*d**(19/2)*x 
**10*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c 
**3*d**6*x**10) - 2*a**2*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/(5*x**4) - 7*a**2 
*d**(5/2)*sqrt(c/(d*x**2) + 1)/(15*x**2) - a**2*d**(7/2)*sqrt(c/(d*x**2) + 
 1)/(15*c) - 2*a*b*sqrt(c)*d**2/(x*sqrt(1 + d*x**2/c)) - 2*a*b*c**2*sqrt(d 
)*sqrt(c/(d*x**2) + 1)/(5*x**4) - 22*a*b*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/( 
15*x**2) - 16*a*b*d**(5/2)*sqrt(c/(d*x**2) + 1)/15 + 2*a*b*d**(5/2)*asinh( 
sqrt(d)*x/sqrt(c)) - 2*a*b*d**3*x/(sqrt(c)*sqrt(1 + d*x**2/c)) - 2*b**2*c* 
*(3/2)*d/(x*sqrt(1 + d*x**2/c)) - 2*b**2*sqrt(c)*d**2*x/sqrt(1 + d*x**2/c) 
 - b**2*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - b**2*c*d**(3/2)*sqrt( 
c/(d*x**2) + 1)/3 + 2*b**2*c*d**(3/2)*asinh(sqrt(d)*x/sqrt(c)) + b**2*d**2 
*Piecewise((c*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), 
Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/2 + x*sqrt(c + d*x**2)/2, Ne(...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=\frac {5}{2} \, \sqrt {d x^{2} + c} b^{2} d^{2} x + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{2} x}{3 \, c} + \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{3} x}{3 \, c^{2}} + \frac {2 \, \sqrt {d x^{2} + c} a b d^{3} x}{c} + \frac {5}{2} \, b^{2} c d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) + 2 \, a b d^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d}{3 \, c x} - \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d^{2}}{15 \, c^{2} x} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2}}{3 \, c x^{3}} - \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b d}{15 \, c^{2} x^{3}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{5 \, c x^{5}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{7 \, c x^{7}} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^8,x, algorithm="maxima")
 

Output:

5/2*sqrt(d*x^2 + c)*b^2*d^2*x + 5/3*(d*x^2 + c)^(3/2)*b^2*d^2*x/c + 4/3*(d 
*x^2 + c)^(3/2)*a*b*d^3*x/c^2 + 2*sqrt(d*x^2 + c)*a*b*d^3*x/c + 5/2*b^2*c* 
d^(3/2)*arcsinh(d*x/sqrt(c*d)) + 2*a*b*d^(5/2)*arcsinh(d*x/sqrt(c*d)) - 4/ 
3*(d*x^2 + c)^(5/2)*b^2*d/(c*x) - 16/15*(d*x^2 + c)^(5/2)*a*b*d^2/(c^2*x) 
- 1/3*(d*x^2 + c)^(7/2)*b^2/(c*x^3) - 4/15*(d*x^2 + c)^(7/2)*a*b*d/(c^2*x^ 
3) - 2/5*(d*x^2 + c)^(7/2)*a*b/(c*x^5) - 1/7*(d*x^2 + c)^(7/2)*a^2/(c*x^7)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (148) = 296\).

Time = 0.16 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.16 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b^{2} d^{2} x - \frac {1}{4} \, {\left (5 \, b^{2} c d^{\frac {3}{2}} + 4 \, a b d^{\frac {5}{2}}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right ) + \frac {2 \, {\left (315 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{12} b^{2} c^{2} d^{\frac {3}{2}} + 630 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{12} a b c d^{\frac {5}{2}} + 105 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{12} a^{2} d^{\frac {7}{2}} - 1680 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{10} b^{2} c^{3} d^{\frac {3}{2}} - 2520 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{10} a b c^{2} d^{\frac {5}{2}} + 3815 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{4} d^{\frac {3}{2}} + 5110 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c^{3} d^{\frac {5}{2}} + 525 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} c^{2} d^{\frac {7}{2}} - 4760 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{5} d^{\frac {3}{2}} - 6160 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{4} d^{\frac {5}{2}} + 3465 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{6} d^{\frac {3}{2}} + 4242 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{5} d^{\frac {5}{2}} + 315 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{4} d^{\frac {7}{2}} - 1400 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{7} d^{\frac {3}{2}} - 1624 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{6} d^{\frac {5}{2}} + 245 \, b^{2} c^{8} d^{\frac {3}{2}} + 322 \, a b c^{7} d^{\frac {5}{2}} + 15 \, a^{2} c^{6} d^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{7}} \] Input:

integrate((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^8,x, algorithm="giac")
 

Output:

1/2*sqrt(d*x^2 + c)*b^2*d^2*x - 1/4*(5*b^2*c*d^(3/2) + 4*a*b*d^(5/2))*log( 
(sqrt(d)*x - sqrt(d*x^2 + c))^2) + 2/105*(315*(sqrt(d)*x - sqrt(d*x^2 + c) 
)^12*b^2*c^2*d^(3/2) + 630*(sqrt(d)*x - sqrt(d*x^2 + c))^12*a*b*c*d^(5/2) 
+ 105*(sqrt(d)*x - sqrt(d*x^2 + c))^12*a^2*d^(7/2) - 1680*(sqrt(d)*x - sqr 
t(d*x^2 + c))^10*b^2*c^3*d^(3/2) - 2520*(sqrt(d)*x - sqrt(d*x^2 + c))^10*a 
*b*c^2*d^(5/2) + 3815*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^4*d^(3/2) + 51 
10*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c^3*d^(5/2) + 525*(sqrt(d)*x - sqrt 
(d*x^2 + c))^8*a^2*c^2*d^(7/2) - 4760*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2* 
c^5*d^(3/2) - 6160*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^4*d^(5/2) + 3465* 
(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^6*d^(3/2) + 4242*(sqrt(d)*x - sqrt(d 
*x^2 + c))^4*a*b*c^5*d^(5/2) + 315*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^4 
*d^(7/2) - 1400*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^7*d^(3/2) - 1624*(sq 
rt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^6*d^(5/2) + 245*b^2*c^8*d^(3/2) + 322*a 
*b*c^7*d^(5/2) + 15*a^2*c^6*d^(7/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) 
^7
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}}{x^8} \,d x \] Input:

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^8,x)
                                                                                    
                                                                                    
 

Output:

int(((a + b*x^2)^2*(c + d*x^2)^(5/2))/x^8, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.73 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^8} \, dx=\frac {-30 \sqrt {d \,x^{2}+c}\, a^{2} c^{3}-90 \sqrt {d \,x^{2}+c}\, a^{2} c^{2} d \,x^{2}-90 \sqrt {d \,x^{2}+c}\, a^{2} c \,d^{2} x^{4}-30 \sqrt {d \,x^{2}+c}\, a^{2} d^{3} x^{6}-84 \sqrt {d \,x^{2}+c}\, a b \,c^{3} x^{2}-308 \sqrt {d \,x^{2}+c}\, a b \,c^{2} d \,x^{4}-644 \sqrt {d \,x^{2}+c}\, a b c \,d^{2} x^{6}-70 \sqrt {d \,x^{2}+c}\, b^{2} c^{3} x^{4}-490 \sqrt {d \,x^{2}+c}\, b^{2} c^{2} d \,x^{6}+105 \sqrt {d \,x^{2}+c}\, b^{2} c \,d^{2} x^{8}+420 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c \,d^{2} x^{7}+525 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{2} d \,x^{7}-30 \sqrt {d}\, a^{2} d^{3} x^{7}+284 \sqrt {d}\, a b c \,d^{2} x^{7}+385 \sqrt {d}\, b^{2} c^{2} d \,x^{7}}{210 c \,x^{7}} \] Input:

int((b*x^2+a)^2*(d*x^2+c)^(5/2)/x^8,x)
 

Output:

( - 30*sqrt(c + d*x**2)*a**2*c**3 - 90*sqrt(c + d*x**2)*a**2*c**2*d*x**2 - 
 90*sqrt(c + d*x**2)*a**2*c*d**2*x**4 - 30*sqrt(c + d*x**2)*a**2*d**3*x**6 
 - 84*sqrt(c + d*x**2)*a*b*c**3*x**2 - 308*sqrt(c + d*x**2)*a*b*c**2*d*x** 
4 - 644*sqrt(c + d*x**2)*a*b*c*d**2*x**6 - 70*sqrt(c + d*x**2)*b**2*c**3*x 
**4 - 490*sqrt(c + d*x**2)*b**2*c**2*d*x**6 + 105*sqrt(c + d*x**2)*b**2*c* 
d**2*x**8 + 420*sqrt(d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a*b*c* 
d**2*x**7 + 525*sqrt(d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*b**2*c 
**2*d*x**7 - 30*sqrt(d)*a**2*d**3*x**7 + 284*sqrt(d)*a*b*c*d**2*x**7 + 385 
*sqrt(d)*b**2*c**2*d*x**7)/(210*c*x**7)