\(\int \frac {c+d x}{x^3 \sqrt {a x^2+b x^3}} \, dx\) [230]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int \frac {c+d x}{x^3 \sqrt {a x^2+b x^3}} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{3 a x^4}+\frac {(5 b c-6 a d) \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {b (5 b c-6 a d) \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {b^2 (5 b c-6 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{8 a^{7/2}} \] Output:

-1/3*c*(b*x^3+a*x^2)^(1/2)/a/x^4+1/12*(-6*a*d+5*b*c)*(b*x^3+a*x^2)^(1/2)/a 
^2/x^3-1/8*b*(-6*a*d+5*b*c)*(b*x^3+a*x^2)^(1/2)/a^3/x^2+1/8*b^2*(-6*a*d+5* 
b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82 \[ \int \frac {c+d x}{x^3 \sqrt {a x^2+b x^3}} \, dx=\frac {-\sqrt {a} (a+b x) \left (15 b^2 c x^2+4 a^2 (2 c+3 d x)-2 a b x (5 c+9 d x)\right )+3 b^2 (5 b c-6 a d) x^3 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{7/2} x^2 \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(-(Sqrt[a]*(a + b*x)*(15*b^2*c*x^2 + 4*a^2*(2*c + 3*d*x) - 2*a*b*x*(5*c + 
9*d*x))) + 3*b^2*(5*b*c - 6*a*d)*x^3*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/S 
qrt[a]])/(24*a^(7/2)*x^2*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1944, 1931, 1931, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/3*(c*Sqrt[a*x^2 + b*x^3])/(a*x^4) - ((5*b*c - 6*a*d)*(-1/2*Sqrt[a*x^2 + 
 b*x^3]/(a*x^3) - (3*b*(-(Sqrt[a*x^2 + b*x^3]/(a*x^2)) + (b*ArcTanh[(Sqrt[ 
a]*x)/Sqrt[a*x^2 + b*x^3]])/a^(3/2)))/(4*a)))/(6*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 1931

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^(j - 1)*(c*x)^(m - j + 1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p 
+ 1))), x] - Simp[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1)))   I 
nt[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] 
 &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 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 = 0.50 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.45

method	result	size
pseudoelliptic	\(\frac {b \,x^{2} \left (a d -\frac {3 b c}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {3 \sqrt {b x +a}\, \left (\frac {2 \left (-2 d x -c \right ) a^{\frac {3}{2}}}{3}+\sqrt {a}\, b c x \right )}{4}}{a^{\frac {5}{2}} x^{2}}\)	\(64\)
risch	\(-\frac {\left (b x +a \right ) \left (-18 a b d \,x^{2}+15 b^{2} c \,x^{2}+12 a^{2} d x -10 a b c x +8 a^{2} c \right )}{24 a^{3} x^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {\left (6 a d -5 b c \right ) b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{8 a^{\frac {7}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(111\)
default	\(\frac {\sqrt {b x +a}\, \left (18 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {9}{2}} d -15 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {7}{2}} b c -48 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {11}{2}} d +40 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}} b c +30 \sqrt {b x +a}\, a^{\frac {13}{2}} d -33 \sqrt {b x +a}\, a^{\frac {11}{2}} b c -18 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{4} b^{3} d \,x^{3}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{3} b^{4} c \,x^{3}\right )}{24 x^{2} b \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {13}{2}}}\)	\(162\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.73 \[ \int \frac {c+d x}{x^3 \sqrt {a x^2+b x^3}} \, dx=\left [-\frac {3 \, {\left (5 \, b^{3} c - 6 \, a b^{2} d\right )} \sqrt {a} x^{4} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, a^{3} c + 3 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d\right )} x^{2} - 2 \, {\left (5 \, a^{2} b c - 6 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{4} x^{4}}, -\frac {3 \, {\left (5 \, b^{3} c - 6 \, a b^{2} d\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, a^{3} c + 3 \, {\left (5 \, a b^{2} c - 6 \, a^{2} b d\right )} x^{2} - 2 \, {\left (5 \, a^{2} b c - 6 \, a^{3} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{4} x^{4}}\right ] \] Input:

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

Output:

[-1/48*(3*(5*b^3*c - 6*a*b^2*d)*sqrt(a)*x^4*log((b*x^2 + 2*a*x - 2*sqrt(b* 
x^3 + a*x^2)*sqrt(a))/x^2) + 2*(8*a^3*c + 3*(5*a*b^2*c - 6*a^2*b*d)*x^2 - 
2*(5*a^2*b*c - 6*a^3*d)*x)*sqrt(b*x^3 + a*x^2))/(a^4*x^4), -1/24*(3*(5*b^3 
*c - 6*a*b^2*d)*sqrt(-a)*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + 
a*x)) + (8*a^3*c + 3*(5*a*b^2*c - 6*a^2*b*d)*x^2 - 2*(5*a^2*b*c - 6*a^3*d) 
*x)*sqrt(b*x^3 + a*x^2))/(a^4*x^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x}{x^3 \sqrt {a x^2+b x^3}} \, dx=-\frac {b^{3} {\left (\frac {3 \, {\left (5 \, b c - 6 \, a d\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3} b} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b c - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b c + 33 \, \sqrt {b x + a} a^{2} b c - 18 \, {\left (b x + a\right )}^{\frac {5}{2}} a d + 48 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} d - 30 \, \sqrt {b x + a} a^{3} d}{a^{3} b^{4} x^{3}}\right )}}{24 \, \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

-1/24*b^3*(3*(5*b*c - 6*a*d)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3* 
b) + (15*(b*x + a)^(5/2)*b*c - 40*(b*x + a)^(3/2)*a*b*c + 33*sqrt(b*x + a) 
*a^2*b*c - 18*(b*x + a)^(5/2)*a*d + 48*(b*x + a)^(3/2)*a^2*d - 30*sqrt(b*x 
 + a)*a^3*d)/(a^3*b^4*x^3))/sgn(x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.20 \[ \int \frac {c+d x}{x^3 \sqrt {a x^2+b x^3}} \, dx=\frac {-16 \sqrt {b x +a}\, a^{3} c -24 \sqrt {b x +a}\, a^{3} d x +20 \sqrt {b x +a}\, a^{2} b c x +36 \sqrt {b x +a}\, a^{2} b d \,x^{2}-30 \sqrt {b x +a}\, a \,b^{2} c \,x^{2}+18 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} d \,x^{3}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} c \,x^{3}-18 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} d \,x^{3}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} c \,x^{3}}{48 a^{4} x^{3}} \] Input:

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

Output:

( - 16*sqrt(a + b*x)*a**3*c - 24*sqrt(a + b*x)*a**3*d*x + 20*sqrt(a + b*x) 
*a**2*b*c*x + 36*sqrt(a + b*x)*a**2*b*d*x**2 - 30*sqrt(a + b*x)*a*b**2*c*x 
**2 + 18*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*d*x**3 - 15*sqrt(a)*l 
og(sqrt(a + b*x) - sqrt(a))*b**3*c*x**3 - 18*sqrt(a)*log(sqrt(a + b*x) + s 
qrt(a))*a*b**2*d*x**3 + 15*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**3*c*x** 
3)/(48*a**4*x**3)