\(\int \frac {c+d x}{x (a x^2+b x^3)^{3/2}} \, dx\) [289]

Optimal result

Integrand size = 24, antiderivative size = 174 \[ \int \frac {c+d x}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2 b^2 (b c-a d) x}{a^4 \sqrt {a x^2+b x^3}}-\frac {c \sqrt {a x^2+b x^3}}{3 a^2 x^4}+\frac {(11 b c-6 a d) \sqrt {a x^2+b x^3}}{12 a^3 x^3}-\frac {b (19 b c-14 a d) \sqrt {a x^2+b x^3}}{8 a^4 x^2}+\frac {5 b^2 (7 b c-6 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{8 a^{9/2}} \] Output:

-2*b^2*(-a*d+b*c)*x/a^4/(b*x^3+a*x^2)^(1/2)-1/3*c*(b*x^3+a*x^2)^(1/2)/a^2/ 
x^4+1/12*(-6*a*d+11*b*c)*(b*x^3+a*x^2)^(1/2)/a^3/x^3-1/8*b*(-14*a*d+19*b*c 
)*(b*x^3+a*x^2)^(1/2)/a^4/x^2+5/8*b^2*(-6*a*d+7*b*c)*arctanh((b*x^3+a*x^2) 
^(1/2)/a^(1/2)/x)/a^(9/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int \frac {c+d x}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-105 b^3 c x^3-4 a^3 (2 c+3 d x)+2 a^2 b x (7 c+15 d x)+5 a b^2 x^2 (-7 c+18 d x)\right )+15 b^2 (7 b c-6 a d) x^3 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{9/2} x^2 \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(Sqrt[a]*(-105*b^3*c*x^3 - 4*a^3*(2*c + 3*d*x) + 2*a^2*b*x*(7*c + 15*d*x) 
+ 5*a*b^2*x^2*(-7*c + 18*d*x)) + 15*b^2*(7*b*c - 6*a*d)*x^3*Sqrt[a + b*x]* 
ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(24*a^(9/2)*x^2*Sqrt[x^2*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1944, 1912, 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.

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

\(\Big \downarrow \) 1944

\(\displaystyle -\frac {(7 b c-6 a d) \int \frac {1}{\left (b x^3+a x^2\right )^{3/2}}dx}{6 a}-\frac {c}{3 a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1912

\(\displaystyle -\frac {(7 b c-6 a d) \left (\frac {5 \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\right )}{6 a}-\frac {c}{3 a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {(7 b c-6 a d) \left (\frac {5 \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\right )}{6 a}-\frac {c}{3 a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {(7 b c-6 a d) \left (\frac {5 \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\right )}{6 a}-\frac {c}{3 a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 1914

\(\displaystyle -\frac {(7 b c-6 a d) \left (\frac {5 \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\right )}{6 a}-\frac {c}{3 a x^2 \sqrt {a x^2+b x^3}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\left (\frac {5 \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{a}+\frac {2}{a x \sqrt {a x^2+b x^3}}\right ) (7 b c-6 a d)}{6 a}-\frac {c}{3 a x^2 \sqrt {a x^2+b x^3}}\)

Input:

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

Output:

-1/3*c/(a*x^2*Sqrt[a*x^2 + b*x^3]) - ((7*b*c - 6*a*d)*(2/(a*x*Sqrt[a*x^2 + 
 b*x^3]) + (5*(-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 
)))/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 1912

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

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 = 0.47 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.26

method	result	size
pseudoelliptic	\(\frac {-\frac {2 b c \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}-\frac {2 \left (a d -b c \right )}{a \sqrt {b x +a}}}{b}\)	\(46\)
risch	\(-\frac {\left (b x +a \right ) \left (-42 a b d \,x^{2}+57 b^{2} c \,x^{2}+12 a^{2} d x -22 a b c x +8 a^{2} c \right )}{24 a^{4} x^{2} \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{2} \left (-\frac {2 \left (30 a d -35 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \left (-16 a d +16 b c \right )}{\sqrt {b x +a}}\right ) \sqrt {b x +a}\, x}{16 a^{4} \sqrt {x^{2} \left (b x +a \right )}}\)	\(135\)
default	\(-\frac {\left (b x +a \right ) \left (90 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} d \,x^{3}-105 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} c \,x^{3}+12 a^{\frac {7}{2}} d x -30 a^{\frac {5}{2}} b d \,x^{2}-90 a^{\frac {3}{2}} b^{2} d \,x^{3}+8 a^{\frac {7}{2}} c -14 a^{\frac {5}{2}} b c x +35 a^{\frac {3}{2}} b^{2} c \,x^{2}+105 \sqrt {a}\, b^{3} c \,x^{3}\right )}{24 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {9}{2}}}\)	\(149\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.09 \[ \int \frac {c+d x}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left ({\left (7 \, b^{4} c - 6 \, a b^{3} d\right )} x^{5} + {\left (7 \, a b^{3} c - 6 \, a^{2} b^{2} d\right )} x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, a^{4} c + 15 \, {\left (7 \, a b^{3} c - 6 \, a^{2} b^{2} d\right )} x^{3} + 5 \, {\left (7 \, a^{2} b^{2} c - 6 \, a^{3} b d\right )} x^{2} - 2 \, {\left (7 \, a^{3} b c - 6 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}, -\frac {15 \, {\left ({\left (7 \, b^{4} c - 6 \, a b^{3} d\right )} x^{5} + {\left (7 \, a b^{3} c - 6 \, a^{2} b^{2} d\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, a^{4} c + 15 \, {\left (7 \, a b^{3} c - 6 \, a^{2} b^{2} d\right )} x^{3} + 5 \, {\left (7 \, a^{2} b^{2} c - 6 \, a^{3} b d\right )} x^{2} - 2 \, {\left (7 \, a^{3} b c - 6 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}}\right ] \] Input:

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

Output:

[-1/48*(15*((7*b^4*c - 6*a*b^3*d)*x^5 + (7*a*b^3*c - 6*a^2*b^2*d)*x^4)*sqr 
t(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(8*a^4*c 
 + 15*(7*a*b^3*c - 6*a^2*b^2*d)*x^3 + 5*(7*a^2*b^2*c - 6*a^3*b*d)*x^2 - 2* 
(7*a^3*b*c - 6*a^4*d)*x)*sqrt(b*x^3 + a*x^2))/(a^5*b*x^5 + a^6*x^4), -1/24 
*(15*((7*b^4*c - 6*a*b^3*d)*x^5 + (7*a*b^3*c - 6*a^2*b^2*d)*x^4)*sqrt(-a)* 
arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (8*a^4*c + 15*(7*a*b^ 
3*c - 6*a^2*b^2*d)*x^3 + 5*(7*a^2*b^2*c - 6*a^3*b*d)*x^2 - 2*(7*a^3*b*c - 
6*a^4*d)*x)*sqrt(b*x^3 + a*x^2))/(a^5*b*x^5 + a^6*x^4)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {5 \, {\left (7 \, b^{3} c - 6 \, a b^{2} d\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{4} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (b^{3} c - a b^{2} d\right )}}{\sqrt {b x + a} a^{4} \mathrm {sgn}\left (x\right )} - \frac {57 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} c - 136 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} c + 87 \, \sqrt {b x + a} a^{2} b^{3} c - 42 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{2} d + 96 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{2} d - 54 \, \sqrt {b x + a} a^{3} b^{2} d}{24 \, a^{4} b^{3} x^{3} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

-5/8*(7*b^3*c - 6*a*b^2*d)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4*sg 
n(x)) - 2*(b^3*c - a*b^2*d)/(sqrt(b*x + a)*a^4*sgn(x)) - 1/24*(57*(b*x + a 
)^(5/2)*b^3*c - 136*(b*x + a)^(3/2)*a*b^3*c + 87*sqrt(b*x + a)*a^2*b^3*c - 
 42*(b*x + a)^(5/2)*a*b^2*d + 96*(b*x + a)^(3/2)*a^2*b^2*d - 54*sqrt(b*x + 
 a)*a^3*b^2*d)/(a^4*b^3*x^3*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.13 \[ \int \frac {c+d x}{x \left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {90 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} d \,x^{3}-105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} c \,x^{3}-90 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} d \,x^{3}+105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} c \,x^{3}-16 a^{4} c -24 a^{4} d x +28 a^{3} b c x +60 a^{3} b d \,x^{2}-70 a^{2} b^{2} c \,x^{2}+180 a^{2} b^{2} d \,x^{3}-210 a \,b^{3} c \,x^{3}}{48 \sqrt {b x +a}\, a^{5} x^{3}} \] Input:

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

Output:

(90*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*d*x**3 - 105 
*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**3*c*x**3 - 90*sqrt( 
a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*d*x**3 + 105*sqrt(a)* 
sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**3*c*x**3 - 16*a**4*c - 24*a* 
*4*d*x + 28*a**3*b*c*x + 60*a**3*b*d*x**2 - 70*a**2*b**2*c*x**2 + 180*a**2 
*b**2*d*x**3 - 210*a*b**3*c*x**3)/(48*sqrt(a + b*x)*a**5*x**3)