\(\int (c+\frac {d}{x^2})^{3/2} x^7 (a+b x) \, dx\) [13]

Optimal result

Integrand size = 20, antiderivative size = 192 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=-\frac {3 a d^3 \sqrt {c+\frac {d}{x^2}} x^2}{128 c^2}+\frac {a d^2 \sqrt {c+\frac {d}{x^2}} x^4}{64 c}+\frac {8 b d^2 \left (c+\frac {d}{x^2}\right )^{5/2} x^5}{315 c^3}+\frac {3}{16} a d \sqrt {c+\frac {d}{x^2}} x^6-\frac {4 b d \left (c+\frac {d}{x^2}\right )^{5/2} x^7}{63 c^2}+\frac {1}{8} a c \sqrt {c+\frac {d}{x^2}} x^8+\frac {b \left (c+\frac {d}{x^2}\right )^{5/2} x^9}{9 c}+\frac {3 a d^4 \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{128 c^{5/2}} \] Output:

-3/128*a*d^3*(c+d/x^2)^(1/2)*x^2/c^2+1/64*a*d^2*(c+d/x^2)^(1/2)*x^4/c+8/31 
5*b*d^2*(c+d/x^2)^(5/2)*x^5/c^3+3/16*a*d*(c+d/x^2)^(1/2)*x^6-4/63*b*d*(c+d 
/x^2)^(5/2)*x^7/c^2+1/8*a*c*(c+d/x^2)^(1/2)*x^8+1/9*b*(c+d/x^2)^(5/2)*x^9/ 
c+3/128*a*d^4*arctanh((c+d/x^2)^(1/2)/c^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.76 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (\sqrt {d+c x^2} \left (128 b \left (d+c x^2\right )^2 \left (8 d^2-20 c d x^2+35 c^2 x^4\right )+315 a c x \left (-3 d^3+2 c d^2 x^2+24 c^2 d x^4+16 c^3 x^6\right )\right )-945 a \sqrt {c} d^4 \log \left (-\sqrt {c} x+\sqrt {d+c x^2}\right )\right )}{40320 c^3 \sqrt {d+c x^2}} \] Input:

Integrate[(c + d/x^2)^(3/2)*x^7*(a + b*x),x]
 

Output:

(Sqrt[c + d/x^2]*x*(Sqrt[d + c*x^2]*(128*b*(d + c*x^2)^2*(8*d^2 - 20*c*d*x 
^2 + 35*c^2*x^4) + 315*a*c*x*(-3*d^3 + 2*c*d^2*x^2 + 24*c^2*d*x^4 + 16*c^3 
*x^6)) - 945*a*Sqrt[c]*d^4*Log[-(Sqrt[c]*x) + Sqrt[d + c*x^2]]))/(40320*c^ 
3*Sqrt[d + c*x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {1892, 1803, 539, 25, 539, 27, 539, 25, 539, 27, 534, 243, 51, 51, 73, 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.

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

\(\Big \downarrow \) 1892

\(\displaystyle \int x^8 \left (\frac {a}{x}+b\right ) \left (c+\frac {d}{x^2}\right )^{3/2}dx\)

\(\Big \downarrow \) 1803

\(\displaystyle -\int \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {a}{x}+b\right ) x^{10}d\frac {1}{x}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\int -\left (c+\frac {d}{x^2}\right )^{3/2} \left (9 a c-\frac {4 b d}{x}\right ) x^9d\frac {1}{x}}{9 c}+\frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {\int \left (c+\frac {d}{x^2}\right )^{3/2} \left (9 a c-\frac {4 b d}{x}\right ) x^9d\frac {1}{x}}{9 c}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {\int c d \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {27 a}{x}+32 b\right ) x^8d\frac {1}{x}}{8 c}-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \int \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {27 a}{x}+32 b\right ) x^8d\frac {1}{x}-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (-\frac {\int -\left (c+\frac {d}{x^2}\right )^{3/2} \left (189 a c-\frac {64 b d}{x}\right ) x^7d\frac {1}{x}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {\int \left (c+\frac {d}{x^2}\right )^{3/2} \left (189 a c-\frac {64 b d}{x}\right ) x^7d\frac {1}{x}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {\int 3 c d \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {63 a}{x}+128 b\right ) x^6d\frac {1}{x}}{6 c}-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \int \left (c+\frac {d}{x^2}\right )^{3/2} \left (\frac {63 a}{x}+128 b\right ) x^6d\frac {1}{x}-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \left (63 a \int \left (c+\frac {d}{x^2}\right )^{3/2} x^5d\frac {1}{x}-\frac {128 b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\right )-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \left (\frac {63}{2} a \int \left (c+\frac {d}{x^2}\right )^{3/2} x^3d\frac {1}{x^2}-\frac {128 b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\right )-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \left (\frac {63}{2} a \left (\frac {3}{4} d \int \sqrt {c+\frac {d}{x^2}} x^2d\frac {1}{x^2}-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {128 b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\right )-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \left (\frac {63}{2} a \left (\frac {3}{4} d \left (\frac {1}{2} d \int \frac {x}{\sqrt {c+\frac {d}{x^2}}}d\frac {1}{x^2}-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {128 b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\right )-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \left (\frac {63}{2} a \left (\frac {3}{4} d \left (\int \frac {1}{\frac {\sqrt {c+\frac {d}{x^2}}}{d}-\frac {c}{d}}d\sqrt {c+\frac {d}{x^2}}-x \sqrt {c+\frac {d}{x^2}}\right )-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {128 b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\right )-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {b x^9 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}-\frac {-\frac {1}{8} d \left (\frac {-\frac {1}{2} d \left (\frac {63}{2} a \left (\frac {3}{4} d \left (x \left (-\sqrt {c+\frac {d}{x^2}}\right )-\frac {d \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{\sqrt {c}}\right )-\frac {1}{2} x^2 \left (c+\frac {d}{x^2}\right )^{3/2}\right )-\frac {128 b x^5 \left (c+\frac {d}{x^2}\right )^{5/2}}{5 c}\right )-\frac {63}{2} a x^6 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}-\frac {32 b x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\right )-\frac {9}{8} a x^8 \left (c+\frac {d}{x^2}\right )^{5/2}}{9 c}\)

Input:

Int[(c + d/x^2)^(3/2)*x^7*(a + b*x),x]
 

Output:

(b*(c + d/x^2)^(5/2)*x^9)/(9*c) - ((-9*a*(c + d/x^2)^(5/2)*x^8)/8 - (d*((- 
32*b*(c + d/x^2)^(5/2)*x^7)/(7*c) + ((-63*a*(c + d/x^2)^(5/2)*x^6)/2 - (d* 
((-128*b*(c + d/x^2)^(5/2)*x^5)/(5*c) + (63*a*(-1/2*((c + d/x^2)^(3/2)*x^2 
) + (3*d*(-(Sqrt[c + d/x^2]*x) - (d*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/Sqrt 
[c]))/4))/2))/2)/(7*c)))/8)/(9*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 51

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, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

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

rule 539

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

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x 
)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && 
 EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 1892

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^ 
(p_.), x_Symbol] :> Int[x^(m + mn*q)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; F 
reeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n 
2] ||  !IntegerQ[p])
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\left (4480 b \,c^{4} x^{8}+5040 a \,c^{4} x^{7}+6400 b \,c^{3} d \,x^{6}+7560 a \,c^{3} d \,x^{5}+384 b \,c^{2} d^{2} x^{4}+630 a \,c^{2} d^{2} x^{3}-512 b c \,d^{3} x^{2}-945 a c \,d^{3} x +1024 b \,d^{4}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{40320 c^{3}}+\frac {3 a \,d^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{128 c^{\frac {5}{2}} \sqrt {c \,x^{2}+d}}\)	\(156\)
default	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (4480 \left (c \,x^{2}+d \right )^{\frac {5}{2}} c^{\frac {5}{2}} b \,x^{4}+5040 \left (c \,x^{2}+d \right )^{\frac {5}{2}} c^{\frac {5}{2}} a \,x^{3}-2560 \left (c \,x^{2}+d \right )^{\frac {5}{2}} c^{\frac {3}{2}} b d \,x^{2}-2520 \left (c \,x^{2}+d \right )^{\frac {5}{2}} c^{\frac {3}{2}} a d x +1024 \left (c \,x^{2}+d \right )^{\frac {5}{2}} \sqrt {c}\, b \,d^{2}+630 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c^{\frac {3}{2}} a \,d^{2} x +945 \sqrt {c \,x^{2}+d}\, c^{\frac {3}{2}} a \,d^{3} x +945 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) a c \,d^{4}\right )}{40320 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c^{\frac {7}{2}}}\)	\(183\)

Input:

int((c+d/x^2)^(3/2)*x^7*(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

1/40320*(4480*b*c^4*x^8+5040*a*c^4*x^7+6400*b*c^3*d*x^6+7560*a*c^3*d*x^5+3 
84*b*c^2*d^2*x^4+630*a*c^2*d^2*x^3-512*b*c*d^3*x^2-945*a*c*d^3*x+1024*b*d^ 
4)/c^3*x*((c*x^2+d)/x^2)^(1/2)+3/128*a/c^(5/2)*d^4*ln(c^(1/2)*x+(c*x^2+d)^ 
(1/2))/(c*x^2+d)^(1/2)*x*((c*x^2+d)/x^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.60 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=\left [\frac {945 \, a \sqrt {c} d^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) + 2 \, {\left (4480 \, b c^{4} x^{9} + 5040 \, a c^{4} x^{8} + 6400 \, b c^{3} d x^{7} + 7560 \, a c^{3} d x^{6} + 384 \, b c^{2} d^{2} x^{5} + 630 \, a c^{2} d^{2} x^{4} - 512 \, b c d^{3} x^{3} - 945 \, a c d^{3} x^{2} + 1024 \, b d^{4} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{80640 \, c^{3}}, -\frac {945 \, a \sqrt {-c} d^{4} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (4480 \, b c^{4} x^{9} + 5040 \, a c^{4} x^{8} + 6400 \, b c^{3} d x^{7} + 7560 \, a c^{3} d x^{6} + 384 \, b c^{2} d^{2} x^{5} + 630 \, a c^{2} d^{2} x^{4} - 512 \, b c d^{3} x^{3} - 945 \, a c d^{3} x^{2} + 1024 \, b d^{4} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{40320 \, c^{3}}\right ] \] Input:

integrate((c+d/x^2)^(3/2)*x^7*(b*x+a),x, algorithm="fricas")
 

Output:

[1/80640*(945*a*sqrt(c)*d^4*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/ 
x^2) - d) + 2*(4480*b*c^4*x^9 + 5040*a*c^4*x^8 + 6400*b*c^3*d*x^7 + 7560*a 
*c^3*d*x^6 + 384*b*c^2*d^2*x^5 + 630*a*c^2*d^2*x^4 - 512*b*c*d^3*x^3 - 945 
*a*c*d^3*x^2 + 1024*b*d^4*x)*sqrt((c*x^2 + d)/x^2))/c^3, -1/40320*(945*a*s 
qrt(-c)*d^4*arctan(sqrt(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (4480 
*b*c^4*x^9 + 5040*a*c^4*x^8 + 6400*b*c^3*d*x^7 + 7560*a*c^3*d*x^6 + 384*b* 
c^2*d^2*x^5 + 630*a*c^2*d^2*x^4 - 512*b*c*d^3*x^3 - 945*a*c*d^3*x^2 + 1024 
*b*d^4*x)*sqrt((c*x^2 + d)/x^2))/c^3]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1073 vs. \(2 (184) = 368\).

Time = 25.41 (sec) , antiderivative size = 1073, normalized size of antiderivative = 5.59 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx =\text {Too large to display} \] Input:

integrate((c+d/x**2)**(3/2)*x**7*(b*x+a),x)
 

Output:

a*c**2*x**9/(8*sqrt(d)*sqrt(c*x**2/d + 1)) + 5*a*c*sqrt(d)*x**7/(16*sqrt(c 
*x**2/d + 1)) + 13*a*d**(3/2)*x**5/(64*sqrt(c*x**2/d + 1)) - a*d**(5/2)*x* 
*3/(128*c*sqrt(c*x**2/d + 1)) - 3*a*d**(7/2)*x/(128*c**2*sqrt(c*x**2/d + 1 
)) + 3*a*d**4*asinh(sqrt(c)*x/sqrt(d))/(128*c**(5/2)) + 35*b*c**8*d**(19/2 
)*x**14*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945 
*c**5*d**11*x**2 + 315*c**4*d**12) + 110*b*c**7*d**(21/2)*x**12*sqrt(c*x** 
2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 
 315*c**4*d**12) + 114*b*c**6*d**(23/2)*x**10*sqrt(c*x**2/d + 1)/(315*c**7 
*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 
 40*b*c**5*d**(25/2)*x**8*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c** 
6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 15*b*c**5*d**(11/2) 
*x**10*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c 
**3*d**6) - 5*b*c**4*d**(27/2)*x**6*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 
 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 33*b*c**4 
*d**(13/2)*x**8*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x** 
2 + 105*c**3*d**6) - 30*b*c**3*d**(29/2)*x**4*sqrt(c*x**2/d + 1)/(315*c**7 
*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 
 17*b*c**3*d**(15/2)*x**6*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c** 
4*d**5*x**2 + 105*c**3*d**6) - 40*b*c**2*d**(31/2)*x**2*sqrt(c*x**2/d + 1) 
/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*...
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.16 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=-\frac {1}{256} \, {\left (\frac {3 \, d^{4} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} d^{4} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c d^{4} - 11 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} d^{4} + 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{3} d^{4}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{4} c^{2} - 4 \, {\left (c + \frac {d}{x^{2}}\right )}^{3} c^{3} + 6 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} c^{4} - 4 \, {\left (c + \frac {d}{x^{2}}\right )} c^{5} + c^{6}}\right )} a + \frac {{\left (35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} x^{9} - 90 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} d x^{7} + 63 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{2} x^{5}\right )} b}{315 \, c^{3}} \] Input:

integrate((c+d/x^2)^(3/2)*x^7*(b*x+a),x, algorithm="maxima")
 

Output:

-1/256*(3*d^4*log((sqrt(c + d/x^2) - sqrt(c))/(sqrt(c + d/x^2) + sqrt(c))) 
/c^(5/2) + 2*(3*(c + d/x^2)^(7/2)*d^4 - 11*(c + d/x^2)^(5/2)*c*d^4 - 11*(c 
 + d/x^2)^(3/2)*c^2*d^4 + 3*sqrt(c + d/x^2)*c^3*d^4)/((c + d/x^2)^4*c^2 - 
4*(c + d/x^2)^3*c^3 + 6*(c + d/x^2)^2*c^4 - 4*(c + d/x^2)*c^5 + c^6))*a + 
1/315*(35*(c + d/x^2)^(9/2)*x^9 - 90*(c + d/x^2)^(7/2)*d*x^7 + 63*(c + d/x 
^2)^(5/2)*d^2*x^5)*b/c^3
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.92 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=-\frac {3 \, a d^{4} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + d} \right |}\right ) \mathrm {sgn}\left (x\right )}{128 \, c^{\frac {5}{2}}} + \frac {1}{40320} \, \sqrt {c x^{2} + d} {\left (\frac {1024 \, b d^{4} \mathrm {sgn}\left (x\right )}{c^{3}} - {\left (\frac {945 \, a d^{3} \mathrm {sgn}\left (x\right )}{c^{2}} + 2 \, {\left (\frac {256 \, b d^{3} \mathrm {sgn}\left (x\right )}{c^{2}} - {\left (\frac {315 \, a d^{2} \mathrm {sgn}\left (x\right )}{c} + 4 \, {\left (\frac {48 \, b d^{2} \mathrm {sgn}\left (x\right )}{c} + 5 \, {\left (189 \, a d \mathrm {sgn}\left (x\right ) + 2 \, {\left (80 \, b d \mathrm {sgn}\left (x\right ) + 7 \, {\left (8 \, b c x \mathrm {sgn}\left (x\right ) + 9 \, a c \mathrm {sgn}\left (x\right )\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} + \frac {{\left (945 \, a c d^{4} \log \left ({\left | d \right |}\right ) - 2048 \, b \sqrt {c} d^{\frac {9}{2}}\right )} \mathrm {sgn}\left (x\right )}{80640 \, c^{\frac {7}{2}}} \] Input:

integrate((c+d/x^2)^(3/2)*x^7*(b*x+a),x, algorithm="giac")
 

Output:

-3/128*a*d^4*log(abs(-sqrt(c)*x + sqrt(c*x^2 + d)))*sgn(x)/c^(5/2) + 1/403 
20*sqrt(c*x^2 + d)*(1024*b*d^4*sgn(x)/c^3 - (945*a*d^3*sgn(x)/c^2 + 2*(256 
*b*d^3*sgn(x)/c^2 - (315*a*d^2*sgn(x)/c + 4*(48*b*d^2*sgn(x)/c + 5*(189*a* 
d*sgn(x) + 2*(80*b*d*sgn(x) + 7*(8*b*c*x*sgn(x) + 9*a*c*sgn(x))*x)*x)*x)*x 
)*x)*x)*x) + 1/80640*(945*a*c*d^4*log(abs(d)) - 2048*b*sqrt(c)*d^(9/2))*sg 
n(x)/c^(7/2)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 7.39 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.80 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {b\,c\,x^9}{9}+\frac {10\,b\,d\,x^7}{63}+\frac {b\,d^2\,x^5}{105\,c}-\frac {4\,b\,d^3\,x^3}{315\,c^2}+\frac {8\,b\,d^4\,x}{315\,c^3}\right )+\frac {11\,a\,x^8\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{128}+\frac {11\,a\,x^8\,{\left (c+\frac {d}{x^2}\right )}^{5/2}}{128\,c}-\frac {3\,a\,x^8\,{\left (c+\frac {d}{x^2}\right )}^{7/2}}{128\,c^2}-\frac {3\,a\,c\,x^8\,\sqrt {c+\frac {d}{x^2}}}{128}-\frac {a\,d^4\,\mathrm {atan}\left (\frac {\sqrt {c+\frac {d}{x^2}}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,3{}\mathrm {i}}{128\,c^{5/2}} \] Input:

int(x^7*(c + d/x^2)^(3/2)*(a + b*x),x)
 

Output:

(c + d/x^2)^(1/2)*((b*c*x^9)/9 + (10*b*d*x^7)/63 + (b*d^2*x^5)/(105*c) - ( 
4*b*d^3*x^3)/(315*c^2) + (8*b*d^4*x)/(315*c^3)) + (11*a*x^8*(c + d/x^2)^(3 
/2))/128 + (11*a*x^8*(c + d/x^2)^(5/2))/(128*c) - (3*a*x^8*(c + d/x^2)^(7/ 
2))/(128*c^2) - (a*d^4*atan(((c + d/x^2)^(1/2)*1i)/c^(1/2))*3i)/(128*c^(5/ 
2)) - (3*a*c*x^8*(c + d/x^2)^(1/2))/128
 

Reduce [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.99 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^7 (a+b x) \, dx=\frac {5040 \sqrt {c \,x^{2}+d}\, a \,c^{4} x^{7}+7560 \sqrt {c \,x^{2}+d}\, a \,c^{3} d \,x^{5}+630 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{2} x^{3}-945 \sqrt {c \,x^{2}+d}\, a c \,d^{3} x +4480 \sqrt {c \,x^{2}+d}\, b \,c^{4} x^{8}+6400 \sqrt {c \,x^{2}+d}\, b \,c^{3} d \,x^{6}+384 \sqrt {c \,x^{2}+d}\, b \,c^{2} d^{2} x^{4}-512 \sqrt {c \,x^{2}+d}\, b c \,d^{3} x^{2}+1024 \sqrt {c \,x^{2}+d}\, b \,d^{4}+945 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) a \,d^{4}}{40320 c^{3}} \] Input:

int((c+d/x^2)^(3/2)*x^7*(b*x+a),x)
 

Output:

(5040*sqrt(c*x**2 + d)*a*c**4*x**7 + 7560*sqrt(c*x**2 + d)*a*c**3*d*x**5 + 
 630*sqrt(c*x**2 + d)*a*c**2*d**2*x**3 - 945*sqrt(c*x**2 + d)*a*c*d**3*x + 
 4480*sqrt(c*x**2 + d)*b*c**4*x**8 + 6400*sqrt(c*x**2 + d)*b*c**3*d*x**6 + 
 384*sqrt(c*x**2 + d)*b*c**2*d**2*x**4 - 512*sqrt(c*x**2 + d)*b*c*d**3*x** 
2 + 1024*sqrt(c*x**2 + d)*b*d**4 + 945*sqrt(c)*log((sqrt(c*x**2 + d) + sqr 
t(c)*x)/sqrt(d))*a*d**4)/(40320*c**3)