\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{x^4} \, dx\) [252]

Optimal result

Integrand size = 24, antiderivative size = 105 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 a c \sqrt {a x^2+b x^3}}{x}+\frac {2 c \left (a x^2+b x^3\right )^{3/2}}{3 x^3}+\frac {2 d \left (a x^2+b x^3\right )^{5/2}}{5 b x^5}-2 a^{3/2} c \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 x \left ((a+b x) \left (3 a^2 d+b^2 x (5 c+3 d x)+a b (20 c+6 d x)\right )-15 a^{3/2} b c \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{15 b \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(2*x*((a + b*x)*(3*a^2*d + b^2*x*(5*c + 3*d*x) + a*b*(20*c + 6*d*x)) - 15* 
a^(3/2)*b*c*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(15*b*Sqrt[x^2* 
(a + b*x)])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1945, 1927, 1927, 1914, 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.

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

\(\Big \downarrow \) 1945

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1927

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

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 1914

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) 
Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, 
n}, x] && NeQ[n, 2]
 

rule 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1945

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* 
p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1))   Int[(e* 
x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, 
x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p 
*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {2 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 d \left (b x +a \right )^{\frac {5}{2}}-15 c \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+5 \left (b x +a \right )^{\frac {3}{2}} b c +15 \sqrt {b x +a}\, a b c \right )}{15 x^{3} \left (b x +a \right )^{\frac {3}{2}} b}\)	\(82\)
pseudoelliptic	\(-\frac {3 \left (b^{2} x^{3} \left (a d -\frac {b c}{6}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {4 \left (\frac {7 x \left (\frac {15 d x}{7}+c \right ) b \,a^{\frac {3}{2}}}{4}+\left (\frac {3 d x}{2}+c \right ) a^{\frac {5}{2}}+\frac {3 \sqrt {a}\, b^{2} c \,x^{2}}{8}\right ) \sqrt {b x +a}}{9}\right )}{4 a^{\frac {3}{2}} x^{3}}\)	\(82\)

Input:

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

Output:

2/15*(b*x^3+a*x^2)^(3/2)*(3*d*(b*x+a)^(5/2)-15*c*a^(3/2)*b*arctanh((b*x+a) 
^(1/2)/a^(1/2))+5*(b*x+a)^(3/2)*b*c+15*(b*x+a)^(1/2)*a*b*c)/x^3/(b*x+a)^(3 
/2)/b
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.91 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\left [\frac {15 \, a^{\frac {3}{2}} b c x \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (3 \, b^{2} d x^{2} + 20 \, a b c + 3 \, a^{2} d + {\left (5 \, b^{2} c + 6 \, a b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{15 \, b x}, \frac {2 \, {\left (15 \, \sqrt {-a} a b c x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (3 \, b^{2} d x^{2} + 20 \, a b c + 3 \, a^{2} d + {\left (5 \, b^{2} c + 6 \, a b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}\right )}}{15 \, b x}\right ] \] Input:

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

Output:

[1/15*(15*a^(3/2)*b*c*x*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a) 
)/x^2) + 2*(3*b^2*d*x^2 + 20*a*b*c + 3*a^2*d + (5*b^2*c + 6*a*b*d)*x)*sqrt 
(b*x^3 + a*x^2))/(b*x), 2/15*(15*sqrt(-a)*a*b*c*x*arctan(sqrt(b*x^3 + a*x^ 
2)*sqrt(-a)/(b*x^2 + a*x)) + (3*b^2*d*x^2 + 20*a*b*c + 3*a^2*d + (5*b^2*c 
+ 6*a*b*d)*x)*sqrt(b*x^3 + a*x^2))/(b*x)]
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{4}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{x^{4}} \,d x } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.27 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {2 \, a^{2} c \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} - \frac {2 \, {\left (15 \, a^{2} b c \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + 20 \, \sqrt {-a} a^{\frac {3}{2}} b c + 3 \, \sqrt {-a} a^{\frac {5}{2}} d\right )} \mathrm {sgn}\left (x\right )}{15 \, \sqrt {-a} b} + \frac {2 \, {\left (5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{5} c \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {b x + a} a b^{5} c \mathrm {sgn}\left (x\right ) + 3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} d \mathrm {sgn}\left (x\right )\right )}}{15 \, b^{5}} \] Input:

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

Output:

2*a^2*c*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/sqrt(-a) - 2/15*(15*a^2*b*c* 
arctan(sqrt(a)/sqrt(-a)) + 20*sqrt(-a)*a^(3/2)*b*c + 3*sqrt(-a)*a^(5/2)*d) 
*sgn(x)/(sqrt(-a)*b) + 2/15*(5*(b*x + a)^(3/2)*b^5*c*sgn(x) + 15*sqrt(b*x 
+ a)*a*b^5*c*sgn(x) + 3*(b*x + a)^(5/2)*b^4*d*sgn(x))/b^5
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}\,\left (c+d\,x\right )}{x^4} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^4} \, dx=\frac {6 \sqrt {b x +a}\, a^{2} d +40 \sqrt {b x +a}\, a b c +12 \sqrt {b x +a}\, a b d x +10 \sqrt {b x +a}\, b^{2} c x +6 \sqrt {b x +a}\, b^{2} d \,x^{2}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b c -15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b c}{15 b} \] Input:

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

Output:

(6*sqrt(a + b*x)*a**2*d + 40*sqrt(a + b*x)*a*b*c + 12*sqrt(a + b*x)*a*b*d* 
x + 10*sqrt(a + b*x)*b**2*c*x + 6*sqrt(a + b*x)*b**2*d*x**2 + 15*sqrt(a)*l 
og(sqrt(a + b*x) - sqrt(a))*a*b*c - 15*sqrt(a)*log(sqrt(a + b*x) + sqrt(a) 
)*a*b*c)/(15*b)