\(\int \frac {(a x+b x^2+c x^3)^{3/2}}{(d x)^{3/2}} \, dx\) [6]

Optimal result

Integrand size = 26, antiderivative size = 158 \[ \int \frac {\left (a x+b x^2+c x^3\right )^{3/2}}{(d x)^{3/2}} \, dx=-\frac {3 \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x+b x^2+c x^3}}{64 c^2 d \sqrt {d x}}+\frac {(b+2 c x) \left (a x+b x^2+c x^3\right )^{3/2}}{8 c (d x)^{3/2}}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {d x} (b+2 c x)}{2 \sqrt {c} \sqrt {d} \sqrt {a x+b x^2+c x^3}}\right )}{128 c^{5/2} d^{3/2}} \] Output:

-3/64*(-4*a*c+b^2)*(2*c*x+b)*(c*x^3+b*x^2+a*x)^(1/2)/c^2/d/(d*x)^(1/2)+1/8 
*(2*c*x+b)*(c*x^3+b*x^2+a*x)^(3/2)/c/(d*x)^(3/2)+3/128*(-4*a*c+b^2)^2*arct 
anh(1/2*(d*x)^(1/2)*(2*c*x+b)/c^(1/2)/d^(1/2)/(c*x^3+b*x^2+a*x)^(1/2))/c^( 
5/2)/d^(3/2)
 

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a x+b x^2+c x^3\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {(x (a+x (b+c x)))^{3/2} \left (\frac {\sqrt {c} (b+2 c x) \left (-3 b^2+8 b c x+4 c \left (5 a+2 c x^2\right )\right )}{a+x (b+c x)}+\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{(a+x (b+c x))^{3/2}}\right )}{64 c^{5/2} (d x)^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2467, 30, 1087, 1087, 1092, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 1087

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {\sqrt {a x+b x^2+c x^3} \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{16 c}\right )}{d \sqrt {d x} \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\sqrt {a x+b x^2+c x^3} \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{16 c}\right )}{d \sqrt {d x} \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a x+b x^2+c x^3} \left (\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{d \sqrt {d x} \sqrt {a+b x+c x^2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

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 1087

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* 
p + 1)))   Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {\left (16 c^{3} x^{3}+24 b \,c^{2} x^{2}+40 a \,c^{2} x +2 b^{2} c x +20 a b c -3 b^{3}\right ) \left (c \,x^{2}+b x +a \right ) x}{64 c^{2} d \sqrt {d x}\, \sqrt {x \left (c \,x^{2}+b x +a \right )}}+\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {1}{2} b d +x c d}{\sqrt {c d}}+\sqrt {c d \,x^{2}+b d x +a d}\right ) \sqrt {d \left (c \,x^{2}+b x +a \right )}\, x}{128 c^{2} \sqrt {c d}\, d \sqrt {d x}\, \sqrt {x \left (c \,x^{2}+b x +a \right )}}\)	\(181\)
default	\(\frac {\sqrt {x \left (c \,x^{2}+b x +a \right )}\, \left (32 c^{3} x^{3} \sqrt {c d}\, \sqrt {d \left (c \,x^{2}+b x +a \right )}+48 b \,c^{2} x^{2} \sqrt {c d}\, \sqrt {d \left (c \,x^{2}+b x +a \right )}+48 \ln \left (\frac {2 x c d +2 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}+b d}{2 \sqrt {c d}}\right ) a^{2} c^{2} d -24 \ln \left (\frac {2 x c d +2 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}+b d}{2 \sqrt {c d}}\right ) a \,b^{2} c d +3 \ln \left (\frac {2 x c d +2 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}+b d}{2 \sqrt {c d}}\right ) b^{4} d +80 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}\, a \,c^{2} x +4 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}\, b^{2} c x +40 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}\, a b c -6 \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}\, b^{3}\right )}{128 d \sqrt {d x}\, c^{2} \sqrt {d \left (c \,x^{2}+b x +a \right )}\, \sqrt {c d}}\)	\(340\)

Input:

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

Output:

1/64*(16*c^3*x^3+24*b*c^2*x^2+40*a*c^2*x+2*b^2*c*x+20*a*b*c-3*b^3)*(c*x^2+ 
b*x+a)/c^2/d/(d*x)^(1/2)/(x*(c*x^2+b*x+a))^(1/2)*x+3/128*(16*a^2*c^2-8*a*b 
^2*c+b^4)/c^2*ln((1/2*b*d+x*c*d)/(c*d)^(1/2)+(c*d*x^2+b*d*x+a*d)^(1/2))/(c 
*d)^(1/2)/d*(d*(c*x^2+b*x+a))^(1/2)/(d*x)^(1/2)/(x*(c*x^2+b*x+a))^(1/2)*x
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.21 \[ \int \frac {\left (a x+b x^2+c x^3\right )^{3/2}}{(d x)^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c d} x \log \left (\frac {8 \, c^{2} d x^{3} + 8 \, b c d x^{2} + {\left (b^{2} + 4 \, a c\right )} d x + 4 \, \sqrt {c x^{3} + b x^{2} + a x} \sqrt {c d} {\left (2 \, c x + b\right )} \sqrt {d x}}{x}\right ) + 4 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{3} + b x^{2} + a x} \sqrt {d x}}{256 \, c^{3} d^{2} x}, -\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c d} x \arctan \left (\frac {\sqrt {c x^{3} + b x^{2} + a x} \sqrt {-c d} {\left (2 \, c x + b\right )} \sqrt {d x}}{2 \, {\left (c^{2} d x^{3} + b c d x^{2} + a c d x\right )}}\right ) - 2 \, {\left (16 \, c^{4} x^{3} + 24 \, b c^{3} x^{2} - 3 \, b^{3} c + 20 \, a b c^{2} + 2 \, {\left (b^{2} c^{2} + 20 \, a c^{3}\right )} x\right )} \sqrt {c x^{3} + b x^{2} + a x} \sqrt {d x}}{128 \, c^{3} d^{2} x}\right ] \] Input:

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

Output:

[1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(c*d)*x*log((8*c^2*d*x^3 + 8* 
b*c*d*x^2 + (b^2 + 4*a*c)*d*x + 4*sqrt(c*x^3 + b*x^2 + a*x)*sqrt(c*d)*(2*c 
*x + b)*sqrt(d*x))/x) + 4*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^ 
2 + 2*(b^2*c^2 + 20*a*c^3)*x)*sqrt(c*x^3 + b*x^2 + a*x)*sqrt(d*x))/(c^3*d^ 
2*x), -1/128*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-c*d)*x*arctan(1/2*sqr 
t(c*x^3 + b*x^2 + a*x)*sqrt(-c*d)*(2*c*x + b)*sqrt(d*x)/(c^2*d*x^3 + b*c*d 
*x^2 + a*c*d*x)) - 2*(16*c^4*x^3 + 24*b*c^3*x^2 - 3*b^3*c + 20*a*b*c^2 + 2 
*(b^2*c^2 + 20*a*c^3)*x)*sqrt(c*x^3 + b*x^2 + a*x)*sqrt(d*x))/(c^3*d^2*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 808 vs. \(2 (130) = 260\).

Time = 0.32 (sec) , antiderivative size = 808, normalized size of antiderivative = 5.11 \[ \int \frac {\left (a x+b x^2+c x^3\right )^{3/2}}{(d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/384*(48*(2*sqrt(c*d^3*x^2 + b*d^3*x + a*d^3)*(2*d*x + b*d/c) + (b^2*d^3 
- 4*a*c*d^3)*log(abs(-b*d^2 - 2*(sqrt(c*d)*d*x - sqrt(c*d^3*x^2 + b*d^3*x 
+ a*d^3))*sqrt(c*d)))/(sqrt(c*d)*c) - (b^2*d^3*log(abs(-b*d^2 + 2*sqrt(a*d 
^3)*sqrt(c*d))) - 4*a*c*d^3*log(abs(-b*d^2 + 2*sqrt(a*d^3)*sqrt(c*d))) + 2 
*sqrt(a*d^3)*sqrt(c*d)*b*d)/(sqrt(c*d)*c))*a*abs(d)^2/d^4 + 8*(2*sqrt(c*d^ 
3*x^2 + b*d^3*x + a*d^3)*(2*(4*d*x + b*d/c)*d*x - (3*b^2*d^2 - 8*a*c*d^2)/ 
c^2) - 3*(b^3*d^4 - 4*a*b*c*d^4)*log(abs(-b*d^2 - 2*(sqrt(c*d)*d*x - sqrt( 
c*d^3*x^2 + b*d^3*x + a*d^3))*sqrt(c*d)))/(sqrt(c*d)*c^2) + (3*b^3*d^4*log 
(abs(-b*d^2 + 2*sqrt(a*d^3)*sqrt(c*d))) - 12*a*b*c*d^4*log(abs(-b*d^2 + 2* 
sqrt(a*d^3)*sqrt(c*d))) + 6*sqrt(a*d^3)*sqrt(c*d)*b^2*d^2 - 16*sqrt(a*d^3) 
*sqrt(c*d)*a*c*d^2)/(sqrt(c*d)*c^2))*b*abs(d)^2/d^5 + (2*sqrt(c*d^3*x^2 + 
b*d^3*x + a*d^3)*(2*(4*(6*d*x + b*d/c)*d*x - (5*b^2*c*d^2 - 12*a*c^2*d^2)/ 
c^3)*d*x + (15*b^3*d^3 - 52*a*b*c*d^3)/c^3) + 3*(5*b^4*d^5 - 24*a*b^2*c*d^ 
5 + 16*a^2*c^2*d^5)*log(abs(-b*d^2 - 2*(sqrt(c*d)*d*x - sqrt(c*d^3*x^2 + b 
*d^3*x + a*d^3))*sqrt(c*d)))/(sqrt(c*d)*c^3) - (15*b^4*d^5*log(abs(-b*d^2 
+ 2*sqrt(a*d^3)*sqrt(c*d))) - 72*a*b^2*c*d^5*log(abs(-b*d^2 + 2*sqrt(a*d^3 
)*sqrt(c*d))) + 48*a^2*c^2*d^5*log(abs(-b*d^2 + 2*sqrt(a*d^3)*sqrt(c*d))) 
+ 30*sqrt(a*d^3)*sqrt(c*d)*b^3*d^3 - 104*sqrt(a*d^3)*sqrt(c*d)*a*b*c*d^3)/ 
(sqrt(c*d)*c^3))*c*abs(d)^2/d^6)/d^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.63 \[ \int \frac {\left (a x+b x^2+c x^3\right )^{3/2}}{(d x)^{3/2}} \, dx=\frac {\sqrt {d}\, \left (40 \sqrt {c \,x^{2}+b x +a}\, a b \,c^{2}+80 \sqrt {c \,x^{2}+b x +a}\, a \,c^{3} x -6 \sqrt {c \,x^{2}+b x +a}\, b^{3} c +4 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{2} x +48 \sqrt {c \,x^{2}+b x +a}\, b \,c^{3} x^{2}+32 \sqrt {c \,x^{2}+b x +a}\, c^{4} x^{3}+48 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a^{2} c^{2}-24 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a \,b^{2} c +3 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{4}\right )}{128 c^{3} d^{2}} \] Input:

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

Output:

(sqrt(d)*(40*sqrt(a + b*x + c*x**2)*a*b*c**2 + 80*sqrt(a + b*x + c*x**2)*a 
*c**3*x - 6*sqrt(a + b*x + c*x**2)*b**3*c + 4*sqrt(a + b*x + c*x**2)*b**2* 
c**2*x + 48*sqrt(a + b*x + c*x**2)*b*c**3*x**2 + 32*sqrt(a + b*x + c*x**2) 
*c**4*x**3 + 48*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x) 
/sqrt(4*a*c - b**2))*a**2*c**2 - 24*sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + 
c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2*c + 3*sqrt(c)*log((2*sqrt( 
c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b**4))/(128*c** 
3*d**2)