\(\int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx\) [22]

Optimal result

Integrand size = 24, antiderivative size = 224 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=\frac {1}{64} \left (64 c^3-\frac {a d \left (48 b^2 c^2-24 a b c d+5 a^2 d^2\right )}{b^3}\right ) \sqrt {a x+b x^2}+\frac {d^2 (24 b c-5 a d) \left (a x+b x^2\right )^{3/2}}{24 b^2}+\frac {d \left (48 b^2 c^2-24 a b c d+5 a^2 d^2\right ) \left (a x+b x^2\right )^{3/2}}{32 b^3 x}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}+\frac {a \left (64 b^3 c^3-a d \left (48 b^2 c^2-24 a b c d+5 a^2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{7/2}} \] Output:

1/64*(64*c^3-a*d*(5*a^2*d^2-24*a*b*c*d+48*b^2*c^2)/b^3)*(b*x^2+a*x)^(1/2)+ 
1/24*d^2*(-5*a*d+24*b*c)*(b*x^2+a*x)^(3/2)/b^2+1/32*d*(5*a^2*d^2-24*a*b*c* 
d+48*b^2*c^2)*(b*x^2+a*x)^(3/2)/b^3/x+1/4*d^3*x*(b*x^2+a*x)^(3/2)/b+1/64*a 
*(64*b^3*c^3-a*d*(5*a^2*d^2-24*a*b*c*d+48*b^2*c^2))*arctanh(b^(1/2)*x/(b*x 
^2+a*x)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=\frac {\sqrt {x (a+b x)} \left (\sqrt {b} \left (15 a^3 d^3-2 a^2 b d^2 (36 c+5 d x)+8 a b^2 d \left (18 c^2+6 c d x+d^2 x^2\right )+48 b^3 \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )\right )+\frac {3 a \left (-64 b^3 c^3+48 a b^2 c^2 d-24 a^2 b c d^2+5 a^3 d^3\right ) \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {x} \sqrt {a+b x}}\right )}{192 b^{7/2}} \] Input:

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

Output:

(Sqrt[x*(a + b*x)]*(Sqrt[b]*(15*a^3*d^3 - 2*a^2*b*d^2*(36*c + 5*d*x) + 8*a 
*b^2*d*(18*c^2 + 6*c*d*x + d^2*x^2) + 48*b^3*(4*c^3 + 6*c^2*d*x + 4*c*d^2* 
x^2 + d^3*x^3)) + (3*a*(-64*b^3*c^3 + 48*a*b^2*c^2*d - 24*a^2*b*c*d^2 + 5* 
a^3*d^3)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(Sqrt[x]*Sqrt[a + b*x])) 
)/(192*b^(7/2))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1262, 27, 2169, 27, 1221, 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.

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

\(\Big \downarrow \) 1262

\(\displaystyle \frac {\int \frac {\sqrt {b x^2+a x} \left (8 b c^3+24 b d x c^2+d^2 (24 b c-5 a d) x^2\right )}{2 x}dx}{4 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {b x^2+a x} \left (8 b c^3+24 b d x c^2+d^2 (24 b c-5 a d) x^2\right )}{x}dx}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 2169

\(\displaystyle \frac {\frac {\int \frac {3 \left (16 b^2 c^3+d \left (48 b^2 c^2-24 a b d c+5 a^2 d^2\right ) x\right ) \sqrt {b x^2+a x}}{2 x}dx}{3 b}+\frac {d^2 \left (a x+b x^2\right )^{3/2} (24 b c-5 a d)}{3 b}}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (16 b^2 c^3+d \left (48 b^2 c^2-24 a b d c+5 a^2 d^2\right ) x\right ) \sqrt {b x^2+a x}}{x}dx}{2 b}+\frac {d^2 \left (a x+b x^2\right )^{3/2} (24 b c-5 a d)}{3 b}}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 1221

\(\displaystyle \frac {\frac {\frac {\left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right ) \int \frac {\sqrt {b x^2+a x}}{x}dx}{4 b}+\frac {d \left (a x+b x^2\right )^{3/2} \left (5 a^2 d^2-24 a b c d+48 b^2 c^2\right )}{2 b x}}{2 b}+\frac {d^2 \left (a x+b x^2\right )^{3/2} (24 b c-5 a d)}{3 b}}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 1131

\(\displaystyle \frac {\frac {\frac {\left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a x}}dx+\sqrt {a x+b x^2}\right )}{4 b}+\frac {d \left (a x+b x^2\right )^{3/2} \left (5 a^2 d^2-24 a b c d+48 b^2 c^2\right )}{2 b x}}{2 b}+\frac {d^2 \left (a x+b x^2\right )^{3/2} (24 b c-5 a d)}{3 b}}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {\frac {\frac {\left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right ) \left (a \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}+\sqrt {a x+b x^2}\right )}{4 b}+\frac {d \left (a x+b x^2\right )^{3/2} \left (5 a^2 d^2-24 a b c d+48 b^2 c^2\right )}{2 b x}}{2 b}+\frac {d^2 \left (a x+b x^2\right )^{3/2} (24 b c-5 a d)}{3 b}}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {d \left (a x+b x^2\right )^{3/2} \left (5 a^2 d^2-24 a b c d+48 b^2 c^2\right )}{2 b x}+\frac {\left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b}}+\sqrt {a x+b x^2}\right ) \left (-5 a^3 d^3+24 a^2 b c d^2-48 a b^2 c^2 d+64 b^3 c^3\right )}{4 b}}{2 b}+\frac {d^2 \left (a x+b x^2\right )^{3/2} (24 b c-5 a d)}{3 b}}{8 b}+\frac {d^3 x \left (a x+b x^2\right )^{3/2}}{4 b}\)

Input:

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

Output:

(d^3*x*(a*x + b*x^2)^(3/2))/(4*b) + ((d^2*(24*b*c - 5*a*d)*(a*x + b*x^2)^( 
3/2))/(3*b) + ((d*(48*b^2*c^2 - 24*a*b*c*d + 5*a^2*d^2)*(a*x + b*x^2)^(3/2 
))/(2*b*x) + ((64*b^3*c^3 - 48*a*b^2*c^2*d + 24*a^2*b*c*d^2 - 5*a^3*d^3)*( 
Sqrt[a*x + b*x^2] + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/Sqrt[b]))/( 
4*b))/(2*b))/(8*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 1221

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

rule 2169

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S 
imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q 
+ 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + 
b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 
1)*(d + e*x)^q + e*f*(m + p + q)*(d + e*x)^(q - 2)*(b*d - 2*a*e + (2*c*d - 
b*e)*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, 
 p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2 
, 0]
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(-\frac {5 \left (a \left (a^{3} d^{3}-\frac {24}{5} a^{2} b c \,d^{2}+\frac {48}{5} a \,b^{2} c^{2} d -\frac {64}{5} b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )-\sqrt {x \left (b x +a \right )}\, \left (\frac {64 \left (\frac {d x}{2}+c \right ) \left (\frac {1}{2} d^{2} x^{2}+c d x +c^{2}\right ) b^{\frac {7}{2}}}{5}+d a \left (\left (\frac {8}{15} d^{2} x^{2}+\frac {16}{5} c d x +\frac {48}{5} c^{2}\right ) b^{\frac {5}{2}}+d \left (\left (-\frac {2 d x}{3}-\frac {24 c}{5}\right ) b^{\frac {3}{2}}+\sqrt {b}\, a d \right ) a \right )\right )\right )}{64 b^{\frac {7}{2}}}\)	\(150\)
risch	\(\frac {\left (48 b^{3} d^{3} x^{3}+8 a \,b^{2} d^{3} x^{2}+192 b^{3} c \,d^{2} x^{2}-10 a^{2} b \,d^{3} x +48 a \,b^{2} c \,d^{2} x +288 b^{3} c^{2} d x +15 a^{3} d^{3}-72 a^{2} b c \,d^{2}+144 a \,b^{2} c^{2} d +192 b^{3} c^{3}\right ) x \left (b x +a \right )}{192 b^{3} \sqrt {x \left (b x +a \right )}}-\frac {a \left (5 a^{3} d^{3}-24 a^{2} b c \,d^{2}+48 a \,b^{2} c^{2} d -64 b^{3} c^{3}\right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 b^{\frac {7}{2}}}\)	\(192\)
default	\(c^{3} \left (\sqrt {b \,x^{2}+a x}+\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b}}\right )+d^{3} \left (\frac {x \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{4 b}-\frac {5 a \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2 b}\right )}{8 b}\right )+3 c^{2} d \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )+3 c \,d^{2} \left (\frac {\left (b \,x^{2}+a x \right )^{\frac {3}{2}}}{3 b}-\frac {a \left (\frac {\left (2 b x +a \right ) \sqrt {b \,x^{2}+a x}}{4 b}-\frac {a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {3}{2}}}\right )}{2 b}\right )\)	\(299\)

Input:

int((d*x+c)^3*(b*x^2+a*x)^(1/2)/x,x,method=_RETURNVERBOSE)
 

Output:

-5/64*(a*(a^3*d^3-24/5*a^2*b*c*d^2+48/5*a*b^2*c^2*d-64/5*b^3*c^3)*arctanh( 
(x*(b*x+a))^(1/2)/x/b^(1/2))-(x*(b*x+a))^(1/2)*(64/5*(1/2*d*x+c)*(1/2*d^2* 
x^2+c*d*x+c^2)*b^(7/2)+d*a*((8/15*d^2*x^2+16/5*c*d*x+48/5*c^2)*b^(5/2)+d*( 
(-2/3*d*x-24/5*c)*b^(3/2)+b^(1/2)*a*d)*a)))/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.76 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=\left [-\frac {3 \, {\left (64 \, a b^{3} c^{3} - 48 \, a^{2} b^{2} c^{2} d + 24 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (48 \, b^{4} d^{3} x^{3} + 192 \, b^{4} c^{3} + 144 \, a b^{3} c^{2} d - 72 \, a^{2} b^{2} c d^{2} + 15 \, a^{3} b d^{3} + 8 \, {\left (24 \, b^{4} c d^{2} + a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (144 \, b^{4} c^{2} d + 24 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x^{2} + a x}}{384 \, b^{4}}, -\frac {3 \, {\left (64 \, a b^{3} c^{3} - 48 \, a^{2} b^{2} c^{2} d + 24 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (48 \, b^{4} d^{3} x^{3} + 192 \, b^{4} c^{3} + 144 \, a b^{3} c^{2} d - 72 \, a^{2} b^{2} c d^{2} + 15 \, a^{3} b d^{3} + 8 \, {\left (24 \, b^{4} c d^{2} + a b^{3} d^{3}\right )} x^{2} + 2 \, {\left (144 \, b^{4} c^{2} d + 24 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x\right )} \sqrt {b x^{2} + a x}}{192 \, b^{4}}\right ] \] Input:

integrate((d*x+c)^3*(b*x^2+a*x)^(1/2)/x,x, algorithm="fricas")
 

Output:

[-1/384*(3*(64*a*b^3*c^3 - 48*a^2*b^2*c^2*d + 24*a^3*b*c*d^2 - 5*a^4*d^3)* 
sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(48*b^4*d^3*x^3 + 
 192*b^4*c^3 + 144*a*b^3*c^2*d - 72*a^2*b^2*c*d^2 + 15*a^3*b*d^3 + 8*(24*b 
^4*c*d^2 + a*b^3*d^3)*x^2 + 2*(144*b^4*c^2*d + 24*a*b^3*c*d^2 - 5*a^2*b^2* 
d^3)*x)*sqrt(b*x^2 + a*x))/b^4, -1/192*(3*(64*a*b^3*c^3 - 48*a^2*b^2*c^2*d 
 + 24*a^3*b*c*d^2 - 5*a^4*d^3)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/ 
(b*x + a)) - (48*b^4*d^3*x^3 + 192*b^4*c^3 + 144*a*b^3*c^2*d - 72*a^2*b^2* 
c*d^2 + 15*a^3*b*d^3 + 8*(24*b^4*c*d^2 + a*b^3*d^3)*x^2 + 2*(144*b^4*c^2*d 
 + 24*a*b^3*c*d^2 - 5*a^2*b^2*d^3)*x)*sqrt(b*x^2 + a*x))/b^4]
 

Sympy [F]

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

integrate((d*x+c)**3*(b*x**2+a*x)**(1/2)/x,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.52 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=\frac {3}{2} \, \sqrt {b x^{2} + a x} c^{2} d x - \frac {3 \, \sqrt {b x^{2} + a x} a c d^{2} x}{4 \, b} + \frac {5 \, \sqrt {b x^{2} + a x} a^{2} d^{3} x}{32 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} d^{3} x}{4 \, b} + \frac {a c^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, \sqrt {b}} - \frac {3 \, a^{2} c^{2} d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} + \frac {3 \, a^{3} c d^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} - \frac {5 \, a^{4} d^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {7}{2}}} + \sqrt {b x^{2} + a x} c^{3} + \frac {3 \, \sqrt {b x^{2} + a x} a c^{2} d}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a x} a^{2} c d^{2}}{8 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} c d^{2}}{b} + \frac {5 \, \sqrt {b x^{2} + a x} a^{3} d^{3}}{64 \, b^{3}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a d^{3}}{24 \, b^{2}} \] Input:

integrate((d*x+c)^3*(b*x^2+a*x)^(1/2)/x,x, algorithm="maxima")
 

Output:

3/2*sqrt(b*x^2 + a*x)*c^2*d*x - 3/4*sqrt(b*x^2 + a*x)*a*c*d^2*x/b + 5/32*s 
qrt(b*x^2 + a*x)*a^2*d^3*x/b^2 + 1/4*(b*x^2 + a*x)^(3/2)*d^3*x/b + 1/2*a*c 
^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) - 3/8*a^2*c^2*d*lo 
g(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) + 3/16*a^3*c*d^2*log(2* 
b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 5/128*a^4*d^3*log(2*b*x + 
 a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + sqrt(b*x^2 + a*x)*c^3 + 3/4*sq 
rt(b*x^2 + a*x)*a*c^2*d/b - 3/8*sqrt(b*x^2 + a*x)*a^2*c*d^2/b^2 + (b*x^2 + 
 a*x)^(3/2)*c*d^2/b + 5/64*sqrt(b*x^2 + a*x)*a^3*d^3/b^3 - 5/24*(b*x^2 + a 
*x)^(3/2)*a*d^3/b^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=\frac {1}{192} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, {\left (6 \, d^{3} x + \frac {24 \, b^{3} c d^{2} + a b^{2} d^{3}}{b^{3}}\right )} x + \frac {144 \, b^{3} c^{2} d + 24 \, a b^{2} c d^{2} - 5 \, a^{2} b d^{3}}{b^{3}}\right )} x + \frac {3 \, {\left (64 \, b^{3} c^{3} + 48 \, a b^{2} c^{2} d - 24 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )}}{b^{3}}\right )} - \frac {{\left (64 \, a b^{3} c^{3} - 48 \, a^{2} b^{2} c^{2} d + 24 \, a^{3} b c d^{2} - 5 \, a^{4} d^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{128 \, b^{\frac {7}{2}}} \] Input:

integrate((d*x+c)^3*(b*x^2+a*x)^(1/2)/x,x, algorithm="giac")
 

Output:

1/192*sqrt(b*x^2 + a*x)*(2*(4*(6*d^3*x + (24*b^3*c*d^2 + a*b^2*d^3)/b^3)*x 
 + (144*b^3*c^2*d + 24*a*b^2*c*d^2 - 5*a^2*b*d^3)/b^3)*x + 3*(64*b^3*c^3 + 
 48*a*b^2*c^2*d - 24*a^2*b*c*d^2 + 5*a^3*d^3)/b^3) - 1/128*(64*a*b^3*c^3 - 
 48*a^2*b^2*c^2*d + 24*a^3*b*c*d^2 - 5*a^4*d^3)*log(abs(2*(sqrt(b)*x - sqr 
t(b*x^2 + a*x))*sqrt(b) + a))/b^(7/2)
 

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=c^3\,\sqrt {b\,x^2+a\,x}+\frac {a\,c^3\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{2\,\sqrt {b}}-\frac {5\,a\,d^3\,\left (\frac {a^3\,\ln \left (\frac {a+2\,b\,x}{\sqrt {b}}+2\,\sqrt {b\,x^2+a\,x}\right )}{16\,b^{5/2}}+\frac {\sqrt {b\,x^2+a\,x}\,\left (-3\,a^2+2\,a\,b\,x+8\,b^2\,x^2\right )}{24\,b^2}\right )}{8\,b}+3\,c^2\,d\,\sqrt {b\,x^2+a\,x}\,\left (\frac {x}{2}+\frac {a}{4\,b}\right )+\frac {d^3\,x\,{\left (b\,x^2+a\,x\right )}^{3/2}}{4\,b}-\frac {3\,a^2\,c^2\,d\,\ln \left (\frac {\frac {a}{2}+b\,x}{\sqrt {b}}+\sqrt {b\,x^2+a\,x}\right )}{8\,b^{3/2}}+\frac {3\,a^3\,c\,d^2\,\ln \left (\frac {a+2\,b\,x}{\sqrt {b}}+2\,\sqrt {b\,x^2+a\,x}\right )}{16\,b^{5/2}}+\frac {c\,d^2\,\sqrt {b\,x^2+a\,x}\,\left (-3\,a^2+2\,a\,b\,x+8\,b^2\,x^2\right )}{8\,b^2} \] Input:

int(((a*x + b*x^2)^(1/2)*(c + d*x)^3)/x,x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.39 \[ \int \frac {(c+d x)^3 \sqrt {a x+b x^2}}{x} \, dx=\frac {15 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b \,d^{3}-72 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} c \,d^{2}-10 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} d^{3} x +144 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c^{2} d +48 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c \,d^{2} x +8 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} d^{3} x^{2}+192 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c^{3}+288 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c^{2} d x +192 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c \,d^{2} x^{2}+48 \sqrt {x}\, \sqrt {b x +a}\, b^{4} d^{3} x^{3}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{4} d^{3}+72 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} b c \,d^{2}-144 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b^{2} c^{2} d +192 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a \,b^{3} c^{3}}{192 b^{4}} \] Input:

int((d*x+c)^3*(b*x^2+a*x)^(1/2)/x,x)
 

Output:

(15*sqrt(x)*sqrt(a + b*x)*a**3*b*d**3 - 72*sqrt(x)*sqrt(a + b*x)*a**2*b**2 
*c*d**2 - 10*sqrt(x)*sqrt(a + b*x)*a**2*b**2*d**3*x + 144*sqrt(x)*sqrt(a + 
 b*x)*a*b**3*c**2*d + 48*sqrt(x)*sqrt(a + b*x)*a*b**3*c*d**2*x + 8*sqrt(x) 
*sqrt(a + b*x)*a*b**3*d**3*x**2 + 192*sqrt(x)*sqrt(a + b*x)*b**4*c**3 + 28 
8*sqrt(x)*sqrt(a + b*x)*b**4*c**2*d*x + 192*sqrt(x)*sqrt(a + b*x)*b**4*c*d 
**2*x**2 + 48*sqrt(x)*sqrt(a + b*x)*b**4*d**3*x**3 - 15*sqrt(b)*log((sqrt( 
a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**4*d**3 + 72*sqrt(b)*log((sqrt(a + 
b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**3*b*c*d**2 - 144*sqrt(b)*log((sqrt(a + 
 b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**2*b**2*c**2*d + 192*sqrt(b)*log((sqrt 
(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a*b**3*c**3)/(192*b**4)