\(\int \frac {c+d x}{x^4 \sqrt {a x^2+b x^3}} \, dx\) [281]

Optimal result

Integrand size = 24, antiderivative size = 179 \[ \int \frac {c+d x}{x^4 \sqrt {a x^2+b x^3}} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}+\frac {(7 b c-8 a d) \sqrt {a x^2+b x^3}}{24 a^2 x^4}-\frac {5 b (7 b c-8 a d) \sqrt {a x^2+b x^3}}{96 a^3 x^3}+\frac {5 b^2 (7 b c-8 a d) \sqrt {a x^2+b x^3}}{64 a^4 x^2}-\frac {5 b^3 (7 b c-8 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{64 a^{9/2}} \] Output:

-1/4*c*(b*x^3+a*x^2)^(1/2)/a/x^5+1/24*(-8*a*d+7*b*c)*(b*x^3+a*x^2)^(1/2)/a 
^2/x^4-5/96*b*(-8*a*d+7*b*c)*(b*x^3+a*x^2)^(1/2)/a^3/x^3+5/64*b^2*(-8*a*d+ 
7*b*c)*(b*x^3+a*x^2)^(1/2)/a^4/x^2-5/64*b^3*(-8*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 = 135, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x}{x^4 \sqrt {a x^2+b x^3}} \, dx=\frac {-\sqrt {a} (a+b x) \left (-105 b^3 c x^3+16 a^3 (3 c+4 d x)-8 a^2 b x (7 c+10 d x)+10 a b^2 x^2 (7 c+12 d x)\right )-15 b^3 (7 b c-8 a d) x^4 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{192 a^{9/2} x^3 \sqrt {x^2 (a+b x)}} \] Input:

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

Output:

(-(Sqrt[a]*(a + b*x)*(-105*b^3*c*x^3 + 16*a^3*(3*c + 4*d*x) - 8*a^2*b*x*(7 
*c + 10*d*x) + 10*a*b^2*x^2*(7*c + 12*d*x))) - 15*b^3*(7*b*c - 8*a*d)*x^4* 
Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(192*a^(9/2)*x^3*Sqrt[x^2*(a 
 + b*x)])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 165, 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, 1931, 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.

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

\(\Big \downarrow \) 1944

\(\displaystyle -\frac {(7 b c-8 a d) \int \frac {1}{x^3 \sqrt {b x^3+a x^2}}dx}{8 a}-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {(7 b c-8 a d) \left (-\frac {5 b \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {(7 b c-8 a d) \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {(7 b c-8 a d) \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}\)

\(\Big \downarrow \) 1914

\(\displaystyle -\frac {(7 b c-8 a d) \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right ) (7 b c-8 a d)}{8 a}-\frac {c \sqrt {a x^2+b x^3}}{4 a x^5}\)

Input:

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

Output:

-1/4*(c*Sqrt[a*x^2 + b*x^3])/(a*x^5) - ((7*b*c - 8*a*d)*(-1/3*Sqrt[a*x^2 + 
 b*x^3]/(a*x^4) - (5*b*(-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)))/(8*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 = 82, normalized size of antiderivative = 0.46

method	result	size
pseudoelliptic	\(-\frac {3 \left (b^{2} x^{3} \left (a d -\frac {5 b c}{6}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {4 \left (-\frac {5 x b \left (\frac {9 d x}{5}+c \right ) a^{\frac {3}{2}}}{4}+\left (\frac {3 d x}{2}+c \right ) a^{\frac {5}{2}}+\frac {15 \sqrt {a}\, b^{2} c \,x^{2}}{8}\right ) \sqrt {b x +a}}{9}\right )}{4 a^{\frac {7}{2}} x^{3}}\)	\(82\)
risch	\(-\frac {\left (b x +a \right ) \left (120 a \,b^{2} d \,x^{3}-105 b^{3} c \,x^{3}-80 a^{2} b d \,x^{2}+70 a \,b^{2} c \,x^{2}+64 a^{3} d x -56 a^{2} b c x +48 c \,a^{3}\right )}{192 a^{4} x^{3} \sqrt {x^{2} \left (b x +a \right )}}+\frac {5 \left (8 a d -7 b c \right ) b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{64 a^{\frac {9}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(135\)
default	\(-\frac {\sqrt {b x +a}\, \left (120 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {11}{2}} d -105 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {9}{2}} b c -440 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {13}{2}} d +385 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {11}{2}} b c +584 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {15}{2}} d -511 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {13}{2}} b c -264 \sqrt {b x +a}\, a^{\frac {17}{2}} d +279 \sqrt {b x +a}\, a^{\frac {15}{2}} b c -120 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{5} b^{4} d \,x^{4}+105 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{4} b^{5} c \,x^{4}\right )}{192 x^{3} b \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {17}{2}}}\)	\(189\)

Input:

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

Output:

-3/4*(b^2*x^3*(a*d-5/6*b*c)*arctanh((b*x+a)^(1/2)/a^(1/2))+4/9*(-5/4*x*b*( 
9/5*d*x+c)*a^(3/2)+(3/2*d*x+c)*a^(5/2)+15/8*a^(1/2)*b^2*c*x^2)*(b*x+a)^(1/ 
2))/a^(7/2)/x^3
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.65 \[ \int \frac {c+d x}{x^4 \sqrt {a x^2+b x^3}} \, dx=\left [-\frac {15 \, {\left (7 \, b^{4} c - 8 \, a b^{3} d\right )} \sqrt {a} x^{5} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (48 \, a^{4} c - 15 \, {\left (7 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{3} + 10 \, {\left (7 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x^{2} - 8 \, {\left (7 \, a^{3} b c - 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{384 \, a^{5} x^{5}}, \frac {15 \, {\left (7 \, b^{4} c - 8 \, a b^{3} d\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) - {\left (48 \, a^{4} c - 15 \, {\left (7 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{3} + 10 \, {\left (7 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x^{2} - 8 \, {\left (7 \, a^{3} b c - 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{192 \, a^{5} x^{5}}\right ] \] Input:

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

Output:

[-1/384*(15*(7*b^4*c - 8*a*b^3*d)*sqrt(a)*x^5*log((b*x^2 + 2*a*x + 2*sqrt( 
b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(48*a^4*c - 15*(7*a*b^3*c - 8*a^2*b^2*d)* 
x^3 + 10*(7*a^2*b^2*c - 8*a^3*b*d)*x^2 - 8*(7*a^3*b*c - 8*a^4*d)*x)*sqrt(b 
*x^3 + a*x^2))/(a^5*x^5), 1/192*(15*(7*b^4*c - 8*a*b^3*d)*sqrt(-a)*x^5*arc 
tan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) - (48*a^4*c - 15*(7*a*b^3* 
c - 8*a^2*b^2*d)*x^3 + 10*(7*a^2*b^2*c - 8*a^3*b*d)*x^2 - 8*(7*a^3*b*c - 8 
*a^4*d)*x)*sqrt(b*x^3 + a*x^2))/(a^5*x^5)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x}{x^4 \sqrt {a x^2+b x^3}} \, dx=\frac {\frac {15 \, {\left (7 \, b^{5} c - 8 \, a b^{4} d\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {105 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} c - 385 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} c + 511 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} c - 279 \, \sqrt {b x + a} a^{3} b^{5} c - 120 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{4} d + 440 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{4} d - 584 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{4} d + 264 \, \sqrt {b x + a} a^{4} b^{4} d}{a^{4} b^{4} x^{4}}}{192 \, b \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

1/192*(15*(7*b^5*c - 8*a*b^4*d)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a 
^4) + (105*(b*x + a)^(7/2)*b^5*c - 385*(b*x + a)^(5/2)*a*b^5*c + 511*(b*x 
+ a)^(3/2)*a^2*b^5*c - 279*sqrt(b*x + a)*a^3*b^5*c - 120*(b*x + a)^(7/2)*a 
*b^4*d + 440*(b*x + a)^(5/2)*a^2*b^4*d - 584*(b*x + a)^(3/2)*a^3*b^4*d + 2 
64*sqrt(b*x + a)*a^4*b^4*d)/(a^4*b^4*x^4))/(b*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.15 \[ \int \frac {c+d x}{x^4 \sqrt {a x^2+b x^3}} \, dx=\frac {-96 \sqrt {b x +a}\, a^{4} c -128 \sqrt {b x +a}\, a^{4} d x +112 \sqrt {b x +a}\, a^{3} b c x +160 \sqrt {b x +a}\, a^{3} b d \,x^{2}-140 \sqrt {b x +a}\, a^{2} b^{2} c \,x^{2}-240 \sqrt {b x +a}\, a^{2} b^{2} d \,x^{3}+210 \sqrt {b x +a}\, a \,b^{3} c \,x^{3}-120 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} d \,x^{4}+105 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} c \,x^{4}+120 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} d \,x^{4}-105 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} c \,x^{4}}{384 a^{5} x^{4}} \] Input:

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

Output:

( - 96*sqrt(a + b*x)*a**4*c - 128*sqrt(a + b*x)*a**4*d*x + 112*sqrt(a + b* 
x)*a**3*b*c*x + 160*sqrt(a + b*x)*a**3*b*d*x**2 - 140*sqrt(a + b*x)*a**2*b 
**2*c*x**2 - 240*sqrt(a + b*x)*a**2*b**2*d*x**3 + 210*sqrt(a + b*x)*a*b**3 
*c*x**3 - 120*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b**3*d*x**4 + 105*sqr 
t(a)*log(sqrt(a + b*x) - sqrt(a))*b**4*c*x**4 + 120*sqrt(a)*log(sqrt(a + b 
*x) + sqrt(a))*a*b**3*d*x**4 - 105*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b* 
*4*c*x**4)/(384*a**5*x**4)