\(\int \frac {(c+d x)^2 (a x+b x^2)^{3/2}}{x^3} \, dx\) [72]

Optimal result

Integrand size = 24, antiderivative size = 183 \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=-\frac {5 \left (a^2 d^2-12 b c (2 b c+a d)\right ) \sqrt {a x+b x^2}}{24 b}-\frac {\left (a^2 d^2-12 b c (2 b c+a d)\right ) x \sqrt {a x+b x^2}}{12 a}-\frac {2 c^2 \left (a x+b x^2\right )^{5/2}}{a x^3}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}-\frac {a \left (a^2 d^2-12 b c (2 b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{8 b^{3/2}} \] Output:

-5/24*(a^2*d^2-12*b*c*(a*d+2*b*c))*(b*x^2+a*x)^(1/2)/b-1/12*(a^2*d^2-12*b* 
c*(a*d+2*b*c))*x*(b*x^2+a*x)^(1/2)/a-2*c^2*(b*x^2+a*x)^(5/2)/a/x^3+1/3*d^2 
*(b*x^2+a*x)^(5/2)/b/x^2-1/8*a*(a^2*d^2-12*b*c*(a*d+2*b*c))*arctanh(b^(1/2 
)*x/(b*x^2+a*x)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\frac {(x (a+b x))^{3/2} \left (\frac {\sqrt {b} \left (3 a^2 d^2 x+8 b^2 x \left (3 c^2+3 c d x+d^2 x^2\right )+2 a b \left (-24 c^2+30 c d x+7 d^2 x^2\right )\right )}{a+b x}+\frac {6 a \left (-24 b^2 c^2-12 a b c d+a^2 d^2\right ) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{(a+b x)^{3/2}}\right )}{24 b^{3/2} x^2} \] Input:

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

Output:

((x*(a + b*x))^(3/2)*((Sqrt[b]*(3*a^2*d^2*x + 8*b^2*x*(3*c^2 + 3*c*d*x + d 
^2*x^2) + 2*a*b*(-24*c^2 + 30*c*d*x + 7*d^2*x^2)))/(a + b*x) + (6*a*(-24*b 
^2*c^2 - 12*a*b*c*d + a^2*d^2)*Sqrt[x]*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] 
- Sqrt[a + b*x])])/(a + b*x)^(3/2)))/(24*b^(3/2)*x^2)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.85, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1262, 27, 1220, 1131, 1131, 1091, 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.

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

\(\Big \downarrow \) 1262

\(\displaystyle \frac {\int \frac {\left (6 b c^2+d (12 b c-a d) x\right ) \left (b x^2+a x\right )^{3/2}}{2 x^3}dx}{3 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (6 b c^2+d (12 b c-a d) x\right ) \left (b x^2+a x\right )^{3/2}}{x^3}dx}{6 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {\frac {\left (-a^2 d^2+12 a b c d+24 b^2 c^2\right ) \int \frac {\left (b x^2+a x\right )^{3/2}}{x^2}dx}{a}-\frac {12 b c^2 \left (a x+b x^2\right )^{5/2}}{a x^3}}{6 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

\(\Big \downarrow \) 1131

\(\displaystyle \frac {\frac {\left (-a^2 d^2+12 a b c d+24 b^2 c^2\right ) \left (\frac {3}{4} a \int \frac {\sqrt {b x^2+a x}}{x}dx+\frac {\left (a x+b x^2\right )^{3/2}}{2 x}\right )}{a}-\frac {12 b c^2 \left (a x+b x^2\right )^{5/2}}{a x^3}}{6 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

\(\Big \downarrow \) 1131

\(\displaystyle \frac {\frac {\left (-a^2 d^2+12 a b c d+24 b^2 c^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a x}}dx+\sqrt {a x+b x^2}\right )+\frac {\left (a x+b x^2\right )^{3/2}}{2 x}\right )}{a}-\frac {12 b c^2 \left (a x+b x^2\right )^{5/2}}{a x^3}}{6 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {\frac {\left (-a^2 d^2+12 a b c d+24 b^2 c^2\right ) \left (\frac {3}{4} a \left (a \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}+\sqrt {a x+b x^2}\right )+\frac {\left (a x+b x^2\right )^{3/2}}{2 x}\right )}{a}-\frac {12 b c^2 \left (a x+b x^2\right )^{5/2}}{a x^3}}{6 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b}}+\sqrt {a x+b x^2}\right )+\frac {\left (a x+b x^2\right )^{3/2}}{2 x}\right ) \left (-a^2 d^2+12 a b c d+24 b^2 c^2\right )}{a}-\frac {12 b c^2 \left (a x+b x^2\right )^{5/2}}{a x^3}}{6 b}+\frac {d^2 \left (a x+b x^2\right )^{5/2}}{3 b x^2}\)

Input:

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

Output:

(d^2*(a*x + b*x^2)^(5/2))/(3*b*x^2) + ((-12*b*c^2*(a*x + b*x^2)^(5/2))/(a* 
x^3) + ((24*b^2*c^2 + 12*a*b*c*d - a^2*d^2)*((a*x + b*x^2)^(3/2)/(2*x) + ( 
3*a*(Sqrt[a*x + b*x^2] + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/Sqrt[b 
]))/4))/a)/(6*b)
 

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 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 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1131

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x 
] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1)))   Int[(d + e*x)^(m + 1)*(a + 
 b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b 
*d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne 
Q[m + 2*p + 1, 0] && IntegerQ[2*p]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

rule 1262

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b 
*x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m 
 + n + 2*p + 1))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m 
 + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n + e*g^n*( 
m + p + n)*(d + e*x)^(n - 2)*(b*d - 2*a*e + (2*c*d - b*e)*x), x], x], x] /; 
 FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && 
IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(-\frac {a x \left (a^{2} d^{2}-12 a b c d -24 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )-\left (-16 \left (-\frac {7}{24} d^{2} x^{2}-\frac {5}{4} c d x +c^{2}\right ) a \,b^{\frac {3}{2}}+x \left (\left (\frac {8}{3} d^{2} x^{2}+8 c d x +8 c^{2}\right ) b^{\frac {5}{2}}+a^{2} d^{2} \sqrt {b}\right )\right ) \sqrt {x \left (b x +a \right )}}{8 b^{\frac {3}{2}} x}\)	\(123\)
risch	\(\frac {\left (b x +a \right ) \left (8 x^{3} b^{2} d^{2}+14 a b \,d^{2} x^{2}+24 b^{2} c \,x^{2} d +3 a^{2} d^{2} x +60 a b c d x +24 x \,b^{2} c^{2}-48 a b \,c^{2}\right )}{24 b \sqrt {x \left (b x +a \right )}}-\frac {\left (a^{2} d^{2}-12 a b c d -24 b^{2} c^{2}\right ) a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{16 b^{\frac {3}{2}}}\)	\(137\)
default	\(c^{2} \left (-\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{a \,x^{3}}+\frac {4 b \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{a \,x^{2}}-\frac {6 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2}\right )}{a}\right )}{a}\right )+d^{2} \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2}\right )+2 c d \left (\frac {2 \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}{a \,x^{2}}-\frac {6 b \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2}\right )}{a}\right )\)	\(308\)

Input:

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

Output:

-1/8/b^(3/2)*(a*x*(a^2*d^2-12*a*b*c*d-24*b^2*c^2)*arctanh((x*(b*x+a))^(1/2 
)/x/b^(1/2))-(-16*(-7/24*d^2*x^2-5/4*c*d*x+c^2)*a*b^(3/2)+x*((8/3*d^2*x^2+ 
8*c*d*x+8*c^2)*b^(5/2)+a^2*d^2*b^(1/2)))*(x*(b*x+a))^(1/2))/x
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.64 \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\left [-\frac {3 \, {\left (24 \, a b^{2} c^{2} + 12 \, a^{2} b c d - a^{3} d^{2}\right )} \sqrt {b} x \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (8 \, b^{3} d^{2} x^{3} - 48 \, a b^{2} c^{2} + 2 \, {\left (12 \, b^{3} c d + 7 \, a b^{2} d^{2}\right )} x^{2} + 3 \, {\left (8 \, b^{3} c^{2} + 20 \, a b^{2} c d + a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a x}}{48 \, b^{2} x}, -\frac {3 \, {\left (24 \, a b^{2} c^{2} + 12 \, a^{2} b c d - a^{3} d^{2}\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (8 \, b^{3} d^{2} x^{3} - 48 \, a b^{2} c^{2} + 2 \, {\left (12 \, b^{3} c d + 7 \, a b^{2} d^{2}\right )} x^{2} + 3 \, {\left (8 \, b^{3} c^{2} + 20 \, a b^{2} c d + a^{2} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a x}}{24 \, b^{2} x}\right ] \] Input:

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

Output:

[-1/48*(3*(24*a*b^2*c^2 + 12*a^2*b*c*d - a^3*d^2)*sqrt(b)*x*log(2*b*x + a 
- 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(8*b^3*d^2*x^3 - 48*a*b^2*c^2 + 2*(12*b 
^3*c*d + 7*a*b^2*d^2)*x^2 + 3*(8*b^3*c^2 + 20*a*b^2*c*d + a^2*b*d^2)*x)*sq 
rt(b*x^2 + a*x))/(b^2*x), -1/24*(3*(24*a*b^2*c^2 + 12*a^2*b*c*d - a^3*d^2) 
*sqrt(-b)*x*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a)) - (8*b^3*d^2*x^3 
- 48*a*b^2*c^2 + 2*(12*b^3*c*d + 7*a*b^2*d^2)*x^2 + 3*(8*b^3*c^2 + 20*a*b^ 
2*c*d + a^2*b*d^2)*x)*sqrt(b*x^2 + a*x))/(b^2*x)]
 

Sympy [F]

\[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\int \frac {\left (x \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )^{2}}{x^{3}}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.24 \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\frac {1}{4} \, \sqrt {b x^{2} + a x} a d^{2} x + \frac {3}{2} \, a \sqrt {b} c^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \frac {3 \, a^{2} c d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{4 \, \sqrt {b}} - \frac {a^{3} d^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3}{2} \, \sqrt {b x^{2} + a x} a c d + \frac {1}{3} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} d^{2} + \frac {\sqrt {b x^{2} + a x} a^{2} d^{2}}{8 \, b} - \frac {3 \, \sqrt {b x^{2} + a x} a c^{2}}{x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} c d}{x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} c^{2}}{x^{2}} \] Input:

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

Output:

1/4*sqrt(b*x^2 + a*x)*a*d^2*x + 3/2*a*sqrt(b)*c^2*log(2*b*x + a + 2*sqrt(b 
*x^2 + a*x)*sqrt(b)) + 3/4*a^2*c*d*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqr 
t(b))/sqrt(b) - 1/16*a^3*d^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/ 
b^(3/2) + 3/2*sqrt(b*x^2 + a*x)*a*c*d + 1/3*(b*x^2 + a*x)^(3/2)*d^2 + 1/8* 
sqrt(b*x^2 + a*x)*a^2*d^2/b - 3*sqrt(b*x^2 + a*x)*a*c^2/x + (b*x^2 + a*x)^ 
(3/2)*c*d/x + (b*x^2 + a*x)^(3/2)*c^2/x^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\frac {2 \, a^{2} c^{2}}{\sqrt {b} x - \sqrt {b x^{2} + a x}} + \frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, b d^{2} x + \frac {12 \, b^{3} c d + 7 \, a b^{2} d^{2}}{b^{2}}\right )} x + \frac {3 \, {\left (8 \, b^{3} c^{2} + 20 \, a b^{2} c d + a^{2} b d^{2}\right )}}{b^{2}}\right )} - \frac {{\left (24 \, a b^{2} c^{2} + 12 \, a^{2} b c d - a^{3} d^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, b^{\frac {3}{2}}} \] Input:

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

Output:

2*a^2*c^2/(sqrt(b)*x - sqrt(b*x^2 + a*x)) + 1/24*sqrt(b*x^2 + a*x)*(2*(4*b 
*d^2*x + (12*b^3*c*d + 7*a*b^2*d^2)/b^2)*x + 3*(8*b^3*c^2 + 20*a*b^2*c*d + 
 a^2*b*d^2)/b^2) - 1/16*(24*a*b^2*c^2 + 12*a^2*b*c*d - a^3*d^2)*log(abs(2* 
(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\int \frac {{\left (b\,x^2+a\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^2}{x^3} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.40 \[ \int \frac {(c+d x)^2 \left (a x+b x^2\right )^{3/2}}{x^3} \, dx=\frac {24 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b \,d^{2} x -384 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} c^{2}+480 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} c d x +112 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{2} d^{2} x^{2}+192 \sqrt {x}\, \sqrt {b x +a}\, b^{3} c^{2} x +192 \sqrt {x}\, \sqrt {b x +a}\, b^{3} c d \,x^{2}+64 \sqrt {x}\, \sqrt {b x +a}\, b^{3} d^{2} x^{3}-24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} d^{2} x +288 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b c d x +576 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a \,b^{2} c^{2} x +3 \sqrt {b}\, a^{3} d^{2} x -96 \sqrt {b}\, a^{2} b c d x -432 \sqrt {b}\, a \,b^{2} c^{2} x}{192 b^{2} x} \] Input:

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

Output:

(24*sqrt(x)*sqrt(a + b*x)*a**2*b*d**2*x - 384*sqrt(x)*sqrt(a + b*x)*a*b**2 
*c**2 + 480*sqrt(x)*sqrt(a + b*x)*a*b**2*c*d*x + 112*sqrt(x)*sqrt(a + b*x) 
*a*b**2*d**2*x**2 + 192*sqrt(x)*sqrt(a + b*x)*b**3*c**2*x + 192*sqrt(x)*sq 
rt(a + b*x)*b**3*c*d*x**2 + 64*sqrt(x)*sqrt(a + b*x)*b**3*d**2*x**3 - 24*s 
qrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**3*d**2*x + 288*sq 
rt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**2*b*c*d*x + 576*sq 
rt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a*b**2*c**2*x + 3*sqr 
t(b)*a**3*d**2*x - 96*sqrt(b)*a**2*b*c*d*x - 432*sqrt(b)*a*b**2*c**2*x)/(1 
92*b**2*x)