\(\int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx\) [992]

Optimal result

Integrand size = 22, antiderivative size = 121 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=\frac {\left (b c^2-4 a d^2\right ) \sqrt {a+b x^2}}{8 a x^2}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}-\frac {2 c d \left (a+b x^2\right )^{3/2}}{3 a x^3}+\frac {b \left (b c^2-4 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}} \] Output:

1/8*(-4*a*d^2+b*c^2)*(b*x^2+a)^(1/2)/a/x^2-1/4*c^2*(b*x^2+a)^(3/2)/a/x^4-2 
/3*c*d*(b*x^2+a)^(3/2)/a/x^3+1/8*b*(-4*a*d^2+b*c^2)*arctanh((b*x^2+a)^(1/2 
)/a^(1/2))/a^(3/2)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=\frac {\sqrt {a+b x^2} \left (-6 a c^2-16 a c d x-3 b c^2 x^2-12 a d^2 x^2-16 b c d x^3\right )}{24 a x^4}+\frac {b \left (-b c^2+4 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{4 a^{3/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-6*a*c^2 - 16*a*c*d*x - 3*b*c^2*x^2 - 12*a*d^2*x^2 - 16* 
b*c*d*x^3))/(24*a*x^4) + (b*(-(b*c^2) + 4*a*d^2)*ArcTanh[(Sqrt[b]*x - Sqrt 
[a + b*x^2])/Sqrt[a]])/(4*a^(3/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {540, 25, 534, 243, 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.

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

\(\Big \downarrow \) 540

\(\displaystyle -\frac {\int -\frac {\left (8 a c d-\left (b c^2-4 a d^2\right ) x\right ) \sqrt {b x^2+a}}{x^4}dx}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (8 a c d-\left (b c^2-4 a d^2\right ) x\right ) \sqrt {b x^2+a}}{x^4}dx}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {-\left (b c^2-4 a d^2\right ) \int \frac {\sqrt {b x^2+a}}{x^3}dx-\frac {8 c d \left (a+b x^2\right )^{3/2}}{3 x^3}}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {-\frac {1}{2} \left (b c^2-4 a d^2\right ) \int \frac {\sqrt {b x^2+a}}{x^4}dx^2-\frac {8 c d \left (a+b x^2\right )^{3/2}}{3 x^3}}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {-\frac {1}{2} \left (b c^2-4 a d^2\right ) \left (\frac {1}{2} b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {8 c d \left (a+b x^2\right )^{3/2}}{3 x^3}}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {1}{2} \left (b c^2-4 a d^2\right ) \left (\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {8 c d \left (a+b x^2\right )^{3/2}}{3 x^3}}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {1}{2} \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{x^2}\right ) \left (b c^2-4 a d^2\right )-\frac {8 c d \left (a+b x^2\right )^{3/2}}{3 x^3}}{4 a}-\frac {c^2 \left (a+b x^2\right )^{3/2}}{4 a x^4}\)

Input:

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

Output:

-1/4*(c^2*(a + b*x^2)^(3/2))/(a*x^4) + ((-8*c*d*(a + b*x^2)^(3/2))/(3*x^3) 
 - ((b*c^2 - 4*a*d^2)*(-(Sqrt[a + b*x^2]/x^2) - (b*ArcTanh[Sqrt[a + b*x^2] 
/Sqrt[a]])/Sqrt[a]))/2)/(4*a)
 

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 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*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*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (16 b c d \,x^{3}+12 a \,d^{2} x^{2}+3 b \,c^{2} x^{2}+16 a d x c +6 a \,c^{2}\right )}{24 x^{4} a}-\frac {\left (4 a \,d^{2}-b \,c^{2}\right ) b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{8 a^{\frac {3}{2}}}\)	\(100\)
default	\(c^{2} \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )+d^{2} \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )-\frac {2 c d \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 a \,x^{3}}\)	\(177\)

Input:

int((d*x+c)^2*(b*x^2+a)^(1/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(b*x^2+a)^(1/2)*(16*b*c*d*x^3+12*a*d^2*x^2+3*b*c^2*x^2+16*a*c*d*x+6* 
a*c^2)/x^4/a-1/8*(4*a*d^2-b*c^2)*b/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/ 
2))/x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.88 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=\left [-\frac {3 \, {\left (b^{2} c^{2} - 4 \, a b d^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (16 \, a b c d x^{3} + 16 \, a^{2} c d x + 6 \, a^{2} c^{2} + 3 \, {\left (a b c^{2} + 4 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a^{2} x^{4}}, -\frac {3 \, {\left (b^{2} c^{2} - 4 \, a b d^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (16 \, a b c d x^{3} + 16 \, a^{2} c d x + 6 \, a^{2} c^{2} + 3 \, {\left (a b c^{2} + 4 \, a^{2} d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{24 \, a^{2} x^{4}}\right ] \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(1/2)/x^5,x, algorithm="fricas")
 

Output:

[-1/48*(3*(b^2*c^2 - 4*a*b*d^2)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a 
)*sqrt(a) + 2*a)/x^2) + 2*(16*a*b*c*d*x^3 + 16*a^2*c*d*x + 6*a^2*c^2 + 3*( 
a*b*c^2 + 4*a^2*d^2)*x^2)*sqrt(b*x^2 + a))/(a^2*x^4), -1/24*(3*(b^2*c^2 - 
4*a*b*d^2)*sqrt(-a)*x^4*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (16*a*b*c*d*x 
^3 + 16*a^2*c*d*x + 6*a^2*c^2 + 3*(a*b*c^2 + 4*a^2*d^2)*x^2)*sqrt(b*x^2 + 
a))/(a^2*x^4)]
 

Sympy [A] (verification not implemented)

Time = 3.59 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.71 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=- \frac {a c^{2}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b} c^{2}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b} c d \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {\sqrt {b} d^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {b^{\frac {3}{2}} c^{2}}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 b^{\frac {3}{2}} c d \sqrt {\frac {a}{b x^{2}} + 1}}{3 a} - \frac {b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {a}} + \frac {b^{2} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} \] Input:

integrate((d*x+c)**2*(b*x**2+a)**(1/2)/x**5,x)
 

Output:

-a*c**2/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 3*sqrt(b)*c**2/(8*x**3*sqr 
t(a/(b*x**2) + 1)) - 2*sqrt(b)*c*d*sqrt(a/(b*x**2) + 1)/(3*x**2) - sqrt(b) 
*d**2*sqrt(a/(b*x**2) + 1)/(2*x) - b**(3/2)*c**2/(8*a*x*sqrt(a/(b*x**2) + 
1)) - 2*b**(3/2)*c*d*sqrt(a/(b*x**2) + 1)/(3*a) - b*d**2*asinh(sqrt(a)/(sq 
rt(b)*x))/(2*sqrt(a)) + b**2*c**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(3/2))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.35 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=\frac {b^{2} c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} - \frac {b d^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} - \frac {\sqrt {b x^{2} + a} b^{2} c^{2}}{8 \, a^{2}} + \frac {\sqrt {b x^{2} + a} b d^{2}}{2 \, a} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b c^{2}}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} d^{2}}{2 \, a x^{2}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c d}{3 \, a x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{2}}{4 \, a x^{4}} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(1/2)/x^5,x, algorithm="maxima")
 

Output:

1/8*b^2*c^2*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 1/2*b*d^2*arcsinh(a/(s 
qrt(a*b)*abs(x)))/sqrt(a) - 1/8*sqrt(b*x^2 + a)*b^2*c^2/a^2 + 1/2*sqrt(b*x 
^2 + a)*b*d^2/a + 1/8*(b*x^2 + a)^(3/2)*b*c^2/(a^2*x^2) - 1/2*(b*x^2 + a)^ 
(3/2)*d^2/(a*x^2) - 2/3*(b*x^2 + a)^(3/2)*c*d/(a*x^3) - 1/4*(b*x^2 + a)^(3 
/2)*c^2/(a*x^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 398 vs. \(2 (101) = 202\).

Time = 0.13 (sec) , antiderivative size = 398, normalized size of antiderivative = 3.29 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=-\frac {{\left (b^{2} c^{2} - 4 \, a b d^{2}\right )} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a} + \frac {3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{7} b^{2} c^{2} + 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{7} a b d^{2} + 48 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a b^{\frac {3}{2}} c d + 21 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{5} a b^{2} c^{2} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{5} a^{2} b d^{2} - 48 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} b^{\frac {3}{2}} c d + 21 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} a^{2} b^{2} c^{2} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} a^{3} b d^{2} + 16 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} b^{\frac {3}{2}} c d + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} a^{3} b^{2} c^{2} + 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} a^{4} b d^{2} - 16 \, a^{4} b^{\frac {3}{2}} c d}{12 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{4} a} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(1/2)/x^5,x, algorithm="giac")
 

Output:

-1/4*(b^2*c^2 - 4*a*b*d^2)*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a)) 
/(sqrt(-a)*a) + 1/12*(3*(sqrt(b)*x - sqrt(b*x^2 + a))^7*b^2*c^2 + 12*(sqrt 
(b)*x - sqrt(b*x^2 + a))^7*a*b*d^2 + 48*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a* 
b^(3/2)*c*d + 21*(sqrt(b)*x - sqrt(b*x^2 + a))^5*a*b^2*c^2 - 12*(sqrt(b)*x 
 - sqrt(b*x^2 + a))^5*a^2*b*d^2 - 48*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^2*b 
^(3/2)*c*d + 21*(sqrt(b)*x - sqrt(b*x^2 + a))^3*a^2*b^2*c^2 - 12*(sqrt(b)* 
x - sqrt(b*x^2 + a))^3*a^3*b*d^2 + 16*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3* 
b^(3/2)*c*d + 3*(sqrt(b)*x - sqrt(b*x^2 + a))*a^3*b^2*c^2 + 12*(sqrt(b)*x 
- sqrt(b*x^2 + a))*a^4*b*d^2 - 16*a^4*b^(3/2)*c*d)/(((sqrt(b)*x - sqrt(b*x 
^2 + a))^2 - a)^4*a)
 

Mupad [F(-1)]

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

int(((a + b*x^2)^(1/2)*(c + d*x)^2)/x^5,x)
                                                                                    
                                                                                    
 

Output:

int(((a + b*x^2)^(1/2)*(c + d*x)^2)/x^5, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.02 \[ \int \frac {(c+d x)^2 \sqrt {a+b x^2}}{x^5} \, dx=\frac {-6 \sqrt {b \,x^{2}+a}\, a^{2} c^{2}-16 \sqrt {b \,x^{2}+a}\, a^{2} c d x -12 \sqrt {b \,x^{2}+a}\, a^{2} d^{2} x^{2}-3 \sqrt {b \,x^{2}+a}\, a b \,c^{2} x^{2}-16 \sqrt {b \,x^{2}+a}\, a b c d \,x^{3}+12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,d^{2} x^{4}-3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} c^{2} x^{4}-12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,d^{2} x^{4}+3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} c^{2} x^{4}-8 \sqrt {b}\, a b c d \,x^{4}}{24 a^{2} x^{4}} \] Input:

int((d*x+c)^2*(b*x^2+a)^(1/2)/x^5,x)
 

Output:

( - 6*sqrt(a + b*x**2)*a**2*c**2 - 16*sqrt(a + b*x**2)*a**2*c*d*x - 12*sqr 
t(a + b*x**2)*a**2*d**2*x**2 - 3*sqrt(a + b*x**2)*a*b*c**2*x**2 - 16*sqrt( 
a + b*x**2)*a*b*c*d*x**3 + 12*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sq 
rt(b)*x)/sqrt(a))*a*b*d**2*x**4 - 3*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a 
) + sqrt(b)*x)/sqrt(a))*b**2*c**2*x**4 - 12*sqrt(a)*log((sqrt(a + b*x**2) 
+ sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*d**2*x**4 + 3*sqrt(a)*log((sqrt(a + b* 
x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**2*c**2*x**4 - 8*sqrt(b)*a*b*c*d*x 
**4)/(24*a**2*x**4)