\(\int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^4} \, dx\) [243]

Optimal result

Integrand size = 24, antiderivative size = 100 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^4} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{2 x^3}-\frac {(b c+4 a d) \sqrt {a x^2+b x^3}}{4 a x^2}+\frac {b (b c-4 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{4 a^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^4} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} (b c x+2 a (c+2 d x))+b (b c-4 a d) x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 a^{3/2} x^3 \sqrt {a+b x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1944, 1926, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/2*(c*(a*x^2 + b*x^3)^(3/2))/(a*x^5) - ((b*c - 4*a*d)*(-(Sqrt[a*x^2 + b* 
x^3]/x^2) - (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/Sqrt[a]))/(4*a)
 

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 1926

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 + j*p + 1))), x] - Simp[b*p 
*((n - j)/(c^n*(m + j*p + 1)))   Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), 
 x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Integer 
sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {b^{2} x^{3} \left (a d -\frac {b c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {4 \left (\frac {b x \left (3 d x +c \right ) a^{\frac {3}{2}}}{4}+\left (\frac {3 d x}{2}+c \right ) a^{\frac {5}{2}}-\frac {3 \sqrt {a}\, b^{2} c \,x^{2}}{8}\right ) \sqrt {b x +a}}{3}}{4 a^{\frac {5}{2}} x^{3}}\)	\(82\)
risch	\(-\frac {\left (4 a d x +c b x +2 a c \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3} a}-\frac {\left (4 a d -b c \right ) b \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{4 a^{\frac {3}{2}} x \sqrt {b x +a}}\)	\(83\)
default	\(-\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (4 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {5}{2}} d +\left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}} b c +4 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{2} d \,x^{2}-\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} c \,x^{2}-4 \sqrt {b x +a}\, a^{\frac {7}{2}} d +\sqrt {b x +a}\, a^{\frac {5}{2}} b c \right )}{4 b \,x^{3} \sqrt {b x +a}\, a^{\frac {5}{2}}}\)	\(131\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.90 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^4} \, dx=\left [-\frac {{\left (b^{2} c - 4 \, a b d\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} {\left (2 \, a^{2} c + {\left (a b c + 4 \, a^{2} d\right )} x\right )}}{8 \, a^{2} x^{3}}, -\frac {{\left (b^{2} c - 4 \, a b d\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (2 \, a^{2} c + {\left (a b c + 4 \, a^{2} d\right )} x\right )}}{4 \, a^{2} x^{3}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="fricas")
 

Output:

[-1/8*((b^2*c - 4*a*b*d)*sqrt(a)*x^3*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a 
*x^2)*sqrt(a))/x^2) + 2*sqrt(b*x^3 + a*x^2)*(2*a^2*c + (a*b*c + 4*a^2*d)*x 
))/(a^2*x^3), -1/4*((b^2*c - 4*a*b*d)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a*x 
^2)*sqrt(-a)/(b*x^2 + a*x)) + sqrt(b*x^3 + a*x^2)*(2*a^2*c + (a*b*c + 4*a^ 
2*d)*x))/(a^2*x^3)]
 

Sympy [F]

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

integrate((d*x+c)*(b*x**3+a*x**2)**(1/2)/x**4,x)
 

Output:

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

Maxima [F]

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

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^4} \, dx=-\frac {\frac {{\left (b^{3} c \mathrm {sgn}\left (x\right ) - 4 \, a b^{2} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (b x + a\right )}^{\frac {3}{2}} b^{3} c \mathrm {sgn}\left (x\right ) + \sqrt {b x + a} a b^{3} c \mathrm {sgn}\left (x\right ) + 4 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{2} d \mathrm {sgn}\left (x\right ) - 4 \, \sqrt {b x + a} a^{2} b^{2} d \mathrm {sgn}\left (x\right )}{a b^{2} x^{2}}}{4 \, b} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="giac")
 

Output:

-1/4*((b^3*c*sgn(x) - 4*a*b^2*d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a))/(sq 
rt(-a)*a) + ((b*x + a)^(3/2)*b^3*c*sgn(x) + sqrt(b*x + a)*a*b^3*c*sgn(x) + 
 4*(b*x + a)^(3/2)*a*b^2*d*sgn(x) - 4*sqrt(b*x + a)*a^2*b^2*d*sgn(x))/(a*b 
^2*x^2))/b
 

Mupad [F(-1)]

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

int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^4,x)
 

Output:

int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.31 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^4} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2} c -8 \sqrt {b x +a}\, a^{2} d x -2 \sqrt {b x +a}\, a b c x +4 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b d \,x^{2}-\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} c \,x^{2}-4 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b d \,x^{2}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} c \,x^{2}}{8 a^{2} x^{2}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^4,x)
 

Output:

( - 4*sqrt(a + b*x)*a**2*c - 8*sqrt(a + b*x)*a**2*d*x - 2*sqrt(a + b*x)*a* 
b*c*x + 4*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b*d*x**2 - sqrt(a)*log(sq 
rt(a + b*x) - sqrt(a))*b**2*c*x**2 - 4*sqrt(a)*log(sqrt(a + b*x) + sqrt(a) 
)*a*b*d*x**2 + sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**2*c*x**2)/(8*a**2*x 
**2)