\(\int (c+\frac {d}{x^2})^{3/2} x^4 (a+b x) \, dx\) [16]

Optimal result

Integrand size = 20, antiderivative size = 119 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\frac {b d^2 \sqrt {c+\frac {d}{x^2}} x^2}{16 c}+\frac {7}{24} b d \sqrt {c+\frac {d}{x^2}} x^4+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^5}{5 c}+\frac {1}{6} b c \sqrt {c+\frac {d}{x^2}} x^6-\frac {b d^3 \text {arctanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{16 c^{3/2}} \] Output:

1/16*b*d^2*(c+d/x^2)^(1/2)*x^2/c+7/24*b*d*(c+d/x^2)^(1/2)*x^4+1/5*a*(c+d/x 
^2)^(5/2)*x^5/c+1/6*b*c*(c+d/x^2)^(1/2)*x^6-1/16*b*d^3*arctanh((c+d/x^2)^( 
1/2)/c^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (\sqrt {c} \sqrt {d+c x^2} \left (48 a \left (d+c x^2\right )^2+5 b x \left (3 d^2+14 c d x^2+8 c^2 x^4\right )\right )+15 b d^3 \log \left (-\sqrt {c} x+\sqrt {d+c x^2}\right )\right )}{240 c^{3/2} \sqrt {d+c x^2}} \] Input:

Integrate[(c + d/x^2)^(3/2)*x^4*(a + b*x),x]
 

Output:

(Sqrt[c + d/x^2]*x*(Sqrt[c]*Sqrt[d + c*x^2]*(48*a*(d + c*x^2)^2 + 5*b*x*(3 
*d^2 + 14*c*d*x^2 + 8*c^2*x^4)) + 15*b*d^3*Log[-(Sqrt[c]*x) + Sqrt[d + c*x 
^2]]))/(240*c^(3/2)*Sqrt[d + c*x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.50 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1892, 1803, 539, 25, 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.

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

\(\Big \downarrow \) 1892

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(c + d/x^2)^(3/2)*x^4*(a + b*x),x]
 

Output:

(b*(c + d/x^2)^(5/2)*x^6)/(6*c) - ((-6*a*(c + d/x^2)^(5/2)*x^5)/5 - (b*d*( 
-1/2*((c + d/x^2)^(3/2)*x^2) + (3*d*(-(Sqrt[c + d/x^2]*x) - (d*ArcTanh[Sqr 
t[c + d/x^2]/Sqrt[c]])/Sqrt[c]))/4))/2)/(6*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {\left (40 b \,c^{2} x^{5}+48 a \,c^{2} x^{4}+70 b c d \,x^{3}+96 a d \,x^{2} c +15 b x \,d^{2}+48 a \,d^{2}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{240 c}-\frac {b \,d^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) x \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}{16 c^{\frac {3}{2}} \sqrt {c \,x^{2}+d}}\)	\(117\)
default	\(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (40 \sqrt {c}\, \left (c \,x^{2}+d \right )^{\frac {5}{2}} b x +48 a \left (c \,x^{2}+d \right )^{\frac {5}{2}} \sqrt {c}-10 \sqrt {c}\, \left (c \,x^{2}+d \right )^{\frac {3}{2}} b d x -15 \sqrt {c}\, \sqrt {c \,x^{2}+d}\, b \,d^{2} x -15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right ) b \,d^{3}\right )}{240 \left (c \,x^{2}+d \right )^{\frac {3}{2}} c^{\frac {3}{2}}}\)	\(121\)

Input:

int((c+d/x^2)^(3/2)*x^4*(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

1/240*(40*b*c^2*x^5+48*a*c^2*x^4+70*b*c*d*x^3+96*a*c*d*x^2+15*b*d^2*x+48*a 
*d^2)/c*x*((c*x^2+d)/x^2)^(1/2)-1/16*b*d^3/c^(3/2)*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 = 0.09 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.02 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\left [\frac {15 \, b \sqrt {c} d^{3} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) + 2 \, {\left (40 \, b c^{3} x^{6} + 48 \, a c^{3} x^{5} + 70 \, b c^{2} d x^{4} + 96 \, a c^{2} d x^{3} + 15 \, b c d^{2} x^{2} + 48 \, a c d^{2} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{480 \, c^{2}}, \frac {15 \, b \sqrt {-c} d^{3} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (40 \, b c^{3} x^{6} + 48 \, a c^{3} x^{5} + 70 \, b c^{2} d x^{4} + 96 \, a c^{2} d x^{3} + 15 \, b c d^{2} x^{2} + 48 \, a c d^{2} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{240 \, c^{2}}\right ] \] Input:

integrate((c+d/x^2)^(3/2)*x^4*(b*x+a),x, algorithm="fricas")
 

Output:

[1/480*(15*b*sqrt(c)*d^3*log(-2*c*x^2 + 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2 
) - d) + 2*(40*b*c^3*x^6 + 48*a*c^3*x^5 + 70*b*c^2*d*x^4 + 96*a*c^2*d*x^3 
+ 15*b*c*d^2*x^2 + 48*a*c*d^2*x)*sqrt((c*x^2 + d)/x^2))/c^2, 1/240*(15*b*s 
qrt(-c)*d^3*arctan(sqrt(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (40*b 
*c^3*x^6 + 48*a*c^3*x^5 + 70*b*c^2*d*x^4 + 96*a*c^2*d*x^3 + 15*b*c*d^2*x^2 
 + 48*a*c*d^2*x)*sqrt((c*x^2 + d)/x^2))/c^2]
 

Sympy [A] (verification not implemented)

Time = 7.87 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.69 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\frac {a c \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {2 a d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {a d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{5 c} + \frac {b c^{2} x^{7}}{6 \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {11 b c \sqrt {d} x^{5}}{24 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {17 b d^{\frac {3}{2}} x^{3}}{48 \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {b d^{\frac {5}{2}} x}{16 c \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {b d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{16 c^{\frac {3}{2}}} \] Input:

integrate((c+d/x**2)**(3/2)*x**4*(b*x+a),x)
 

Output:

a*c*sqrt(d)*x**4*sqrt(c*x**2/d + 1)/5 + 2*a*d**(3/2)*x**2*sqrt(c*x**2/d + 
1)/5 + a*d**(5/2)*sqrt(c*x**2/d + 1)/(5*c) + b*c**2*x**7/(6*sqrt(d)*sqrt(c 
*x**2/d + 1)) + 11*b*c*sqrt(d)*x**5/(24*sqrt(c*x**2/d + 1)) + 17*b*d**(3/2 
)*x**3/(48*sqrt(c*x**2/d + 1)) + b*d**(5/2)*x/(16*c*sqrt(c*x**2/d + 1)) - 
b*d**3*asinh(sqrt(c)*x/sqrt(d))/(16*c**(3/2))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.32 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5}}{5 \, c} + \frac {1}{96} \, {\left (\frac {3 \, d^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{3} + 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c d^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{2} d^{3}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} c - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} c^{2} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} c^{3} - c^{4}}\right )} b \] Input:

integrate((c+d/x^2)^(3/2)*x^4*(b*x+a),x, algorithm="maxima")
 

Output:

1/5*a*(c + d/x^2)^(5/2)*x^5/c + 1/96*(3*d^3*log((sqrt(c + d/x^2) - sqrt(c) 
)/(sqrt(c + d/x^2) + sqrt(c)))/c^(3/2) + 2*(3*(c + d/x^2)^(5/2)*d^3 + 8*(c 
 + d/x^2)^(3/2)*c*d^3 - 3*sqrt(c + d/x^2)*c^2*d^3)/((c + d/x^2)^3*c - 3*(c 
 + d/x^2)^2*c^2 + 3*(c + d/x^2)*c^3 - c^4))*b
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\frac {b d^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + d} \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {3}{2}}} + \frac {1}{240} \, \sqrt {c x^{2} + d} {\left (\frac {48 \, a d^{2} \mathrm {sgn}\left (x\right )}{c} + {\left (\frac {15 \, b d^{2} \mathrm {sgn}\left (x\right )}{c} + 2 \, {\left (48 \, a d \mathrm {sgn}\left (x\right ) + {\left (35 \, b d \mathrm {sgn}\left (x\right ) + 4 \, {\left (5 \, b c x \mathrm {sgn}\left (x\right ) + 6 \, a c \mathrm {sgn}\left (x\right )\right )} x\right )} x\right )} x\right )} x\right )} - \frac {{\left (5 \, b d^{3} \log \left ({\left | d \right |}\right ) + 32 \, a \sqrt {c} d^{\frac {5}{2}}\right )} \mathrm {sgn}\left (x\right )}{160 \, c^{\frac {3}{2}}} \] Input:

integrate((c+d/x^2)^(3/2)*x^4*(b*x+a),x, algorithm="giac")
 

Output:

1/16*b*d^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + d)))*sgn(x)/c^(3/2) + 1/240*s 
qrt(c*x^2 + d)*(48*a*d^2*sgn(x)/c + (15*b*d^2*sgn(x)/c + 2*(48*a*d*sgn(x) 
+ (35*b*d*sgn(x) + 4*(5*b*c*x*sgn(x) + 6*a*c*sgn(x))*x)*x)*x)*x) - 1/160*( 
5*b*d^3*log(abs(d)) + 32*a*sqrt(c)*d^(5/2))*sgn(x)/c^(3/2)
 

Mupad [B] (verification not implemented)

Time = 7.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.93 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {a\,c\,x^5}{5}+\frac {2\,a\,d\,x^3}{5}+\frac {a\,d^2\,x}{5\,c}\right )+\frac {b\,x^6\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{6}+\frac {b\,x^6\,{\left (c+\frac {d}{x^2}\right )}^{5/2}}{16\,c}-\frac {b\,c\,x^6\,\sqrt {c+\frac {d}{x^2}}}{16}+\frac {b\,d^3\,\mathrm {atan}\left (\frac {\sqrt {c+\frac {d}{x^2}}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{16\,c^{3/2}} \] Input:

int(x^4*(c + d/x^2)^(3/2)*(a + b*x),x)
 

Output:

(c + d/x^2)^(1/2)*((a*c*x^5)/5 + (2*a*d*x^3)/5 + (a*d^2*x)/(5*c)) + (b*x^6 
*(c + d/x^2)^(3/2))/6 + (b*x^6*(c + d/x^2)^(5/2))/(16*c) + (b*d^3*atan(((c 
 + d/x^2)^(1/2)*1i)/c^(1/2))*1i)/(16*c^(3/2)) - (b*c*x^6*(c + d/x^2)^(1/2) 
)/16
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13 \[ \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4 (a+b x) \, dx=\frac {48 \sqrt {c \,x^{2}+d}\, a \,c^{3} x^{4}+96 \sqrt {c \,x^{2}+d}\, a \,c^{2} d \,x^{2}+48 \sqrt {c \,x^{2}+d}\, a c \,d^{2}+40 \sqrt {c \,x^{2}+d}\, b \,c^{3} x^{5}+70 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{3}+15 \sqrt {c \,x^{2}+d}\, b c \,d^{2} x -15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+d}+\sqrt {c}\, x}{\sqrt {d}}\right ) b \,d^{3}}{240 c^{2}} \] Input:

int((c+d/x^2)^(3/2)*x^4*(b*x+a),x)
 

Output:

(48*sqrt(c*x**2 + d)*a*c**3*x**4 + 96*sqrt(c*x**2 + d)*a*c**2*d*x**2 + 48* 
sqrt(c*x**2 + d)*a*c*d**2 + 40*sqrt(c*x**2 + d)*b*c**3*x**5 + 70*sqrt(c*x* 
*2 + d)*b*c**2*d*x**3 + 15*sqrt(c*x**2 + d)*b*c*d**2*x - 15*sqrt(c)*log((s 
qrt(c*x**2 + d) + sqrt(c)*x)/sqrt(d))*b*d**3)/(240*c**2)