\(\int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx\) [245]

Optimal result

Integrand size = 24, antiderivative size = 175 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=-\frac {c \sqrt {a x^2+b x^3}}{4 x^5}-\frac {(b c+8 a d) \sqrt {a x^2+b x^3}}{24 a x^4}+\frac {b (5 b c-8 a d) \sqrt {a x^2+b x^3}}{96 a^2 x^3}-\frac {b^2 (5 b c-8 a d) \sqrt {a x^2+b x^3}}{64 a^3 x^2}+\frac {b^3 (5 b c-8 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{64 a^{7/2}} \] Output:

-1/4*c*(b*x^3+a*x^2)^(1/2)/x^5-1/24*(8*a*d+b*c)*(b*x^3+a*x^2)^(1/2)/a/x^4+ 
1/96*b*(-8*a*d+5*b*c)*(b*x^3+a*x^2)^(1/2)/a^2/x^3-1/64*b^2*(-8*a*d+5*b*c)* 
(b*x^3+a*x^2)^(1/2)/a^3/x^2+1/64*b^3*(-8*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 = 0.24 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.78 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} \left (15 b^3 c x^3+8 a^2 b x (c+2 d x)+16 a^3 (3 c+4 d x)-2 a b^2 x^2 (5 c+12 d x)\right )+3 b^3 (5 b c-8 a d) x^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{192 a^{7/2} x^5 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*Sqrt[a*x^2 + b*x^3])/x^6,x]
 

Output:

(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a]*Sqrt[a + b*x]*(15*b^3*c*x^3 + 8*a^2*b*x*(c 
 + 2*d*x) + 16*a^3*(3*c + 4*d*x) - 2*a*b^2*x^2*(5*c + 12*d*x))) + 3*b^3*(5 
*b*c - 8*a*d)*x^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(192*a^(7/2)*x^5*Sqrt[a 
 + b*x])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1944, 1926, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

Int[((c + d*x)*Sqrt[a*x^2 + b*x^3])/x^6,x]
 

Output:

-1/4*(c*(a*x^2 + b*x^3)^(3/2))/(a*x^7) - ((5*b*c - 8*a*d)*(-1/3*Sqrt[a*x^2 
 + b*x^3]/x^4 + (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))/(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 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 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 = 0.61 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(\frac {\frac {5 b^{4} x^{5} \left (a d -\frac {7 b c}{10}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{64}+\frac {7 \left (-\frac {5 x^{3} \left (\frac {15 d x}{7}+c \right ) b^{3} a^{\frac {3}{2}}}{4}+b^{2} x^{2} \left (\frac {25 d x}{14}+c \right ) a^{\frac {5}{2}}-\frac {6 x b \left (\frac {5 d x}{3}+c \right ) a^{\frac {7}{2}}}{7}+\frac {12 \left (-5 d x -4 c \right ) a^{\frac {9}{2}}}{7}+\frac {15 \sqrt {a}\, b^{4} c \,x^{4}}{8}\right ) \sqrt {b x +a}}{240}}{a^{\frac {9}{2}} x^{5}}\)	\(118\)
risch	\(-\frac {\left (-24 a \,b^{2} d \,x^{3}+15 b^{3} c \,x^{3}+16 a^{2} b d \,x^{2}-10 a \,b^{2} c \,x^{2}+64 a^{3} d x +8 a^{2} b c x +48 c \,a^{3}\right ) \sqrt {x^{2} \left (b x +a \right )}}{192 x^{5} a^{3}}-\frac {\left (8 a d -5 b c \right ) b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 a^{\frac {7}{2}} x \sqrt {b x +a}}\)	\(132\)
default	\(\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (24 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {9}{2}} d -15 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {7}{2}} b c -88 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {11}{2}} d +55 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {9}{2}} b c +40 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {13}{2}} d -73 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {11}{2}} b c -24 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{4} b^{4} d \,x^{4}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{3} b^{5} c \,x^{4}+24 \sqrt {b x +a}\, a^{\frac {15}{2}} d -15 \sqrt {b x +a}\, a^{\frac {13}{2}} b c \right )}{192 b \,x^{5} \sqrt {b x +a}\, a^{\frac {13}{2}}}\)	\(189\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^6,x,method=_RETURNVERBOSE)
 

Output:

5/64*(b^4*x^5*(a*d-7/10*b*c)*arctanh((b*x+a)^(1/2)/a^(1/2))+28/75*(-5/4*x^ 
3*(15/7*d*x+c)*b^3*a^(3/2)+b^2*x^2*(25/14*d*x+c)*a^(5/2)-6/7*x*b*(5/3*d*x+ 
c)*a^(7/2)+12/7*(-5*d*x-4*c)*a^(9/2)+15/8*a^(1/2)*b^4*c*x^4)*(b*x+a)^(1/2) 
)/a^(9/2)/x^5
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.67 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=\left [-\frac {3 \, {\left (5 \, b^{4} c - 8 \, a b^{3} d\right )} \sqrt {a} x^{5} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (48 \, a^{4} c + 3 \, {\left (5 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{3} - 2 \, {\left (5 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x^{2} + 8 \, {\left (a^{3} b c + 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{384 \, a^{4} x^{5}}, -\frac {3 \, {\left (5 \, b^{4} c - 8 \, a b^{3} d\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (48 \, a^{4} c + 3 \, {\left (5 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{3} - 2 \, {\left (5 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x^{2} + 8 \, {\left (a^{3} b c + 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{192 \, a^{4} x^{5}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^6,x, algorithm="fricas")
 

Output:

[-1/384*(3*(5*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 + 3*(5*a*b^3*c - 8*a^2*b^2*d)*x^ 
3 - 2*(5*a^2*b^2*c - 8*a^3*b*d)*x^2 + 8*(a^3*b*c + 8*a^4*d)*x)*sqrt(b*x^3 
+ a*x^2))/(a^4*x^5), -1/192*(3*(5*b^4*c - 8*a*b^3*d)*sqrt(-a)*x^5*arctan(s 
qrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (48*a^4*c + 3*(5*a*b^3*c - 8* 
a^2*b^2*d)*x^3 - 2*(5*a^2*b^2*c - 8*a^3*b*d)*x^2 + 8*(a^3*b*c + 8*a^4*d)*x 
)*sqrt(b*x^3 + a*x^2))/(a^4*x^5)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x\right )} \left (c + d x\right )}{x^{6}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(1/2)/x**6,x)
 

Output:

Integral(sqrt(x**2*(a + b*x))*(c + d*x)/x**6, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=\int { \frac {\sqrt {b x^{3} + a x^{2}} {\left (d x + c\right )}}{x^{6}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^6,x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^3 + a*x^2)*(d*x + c)/x^6, x)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=-\frac {\frac {3 \, {\left (5 \, b^{5} c \mathrm {sgn}\left (x\right ) - 8 \, a b^{4} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} c \mathrm {sgn}\left (x\right ) - 55 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} c \mathrm {sgn}\left (x\right ) + 73 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} c \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {b x + a} a^{3} b^{5} c \mathrm {sgn}\left (x\right ) - 24 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{4} d \mathrm {sgn}\left (x\right ) + 88 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{4} d \mathrm {sgn}\left (x\right ) - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{4} d \mathrm {sgn}\left (x\right ) - 24 \, \sqrt {b x + a} a^{4} b^{4} d \mathrm {sgn}\left (x\right )}{a^{3} b^{4} x^{4}}}{192 \, b} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^6,x, algorithm="giac")
 

Output:

-1/192*(3*(5*b^5*c*sgn(x) - 8*a*b^4*d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a 
))/(sqrt(-a)*a^3) + (15*(b*x + a)^(7/2)*b^5*c*sgn(x) - 55*(b*x + a)^(5/2)* 
a*b^5*c*sgn(x) + 73*(b*x + a)^(3/2)*a^2*b^5*c*sgn(x) + 15*sqrt(b*x + a)*a^ 
3*b^5*c*sgn(x) - 24*(b*x + a)^(7/2)*a*b^4*d*sgn(x) + 88*(b*x + a)^(5/2)*a^ 
2*b^4*d*sgn(x) - 40*(b*x + a)^(3/2)*a^3*b^4*d*sgn(x) - 24*sqrt(b*x + a)*a^ 
4*b^4*d*sgn(x))/(a^3*b^4*x^4))/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=\int \frac {\sqrt {b\,x^3+a\,x^2}\,\left (c+d\,x\right )}{x^6} \,d x \] Input:

int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^6,x)
 

Output:

int(((a*x^2 + b*x^3)^(1/2)*(c + d*x))/x^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {(c+d x) \sqrt {a x^2+b x^3}}{x^6} \, dx=\frac {-96 \sqrt {b x +a}\, a^{4} c -128 \sqrt {b x +a}\, a^{4} d x -16 \sqrt {b x +a}\, a^{3} b c x -32 \sqrt {b x +a}\, a^{3} b d \,x^{2}+20 \sqrt {b x +a}\, a^{2} b^{2} c \,x^{2}+48 \sqrt {b x +a}\, a^{2} b^{2} d \,x^{3}-30 \sqrt {b x +a}\, a \,b^{3} c \,x^{3}+24 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} d \,x^{4}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} c \,x^{4}-24 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} d \,x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} c \,x^{4}}{384 a^{4} x^{4}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(1/2)/x^6,x)
 

Output:

( - 96*sqrt(a + b*x)*a**4*c - 128*sqrt(a + b*x)*a**4*d*x - 16*sqrt(a + b*x 
)*a**3*b*c*x - 32*sqrt(a + b*x)*a**3*b*d*x**2 + 20*sqrt(a + b*x)*a**2*b**2 
*c*x**2 + 48*sqrt(a + b*x)*a**2*b**2*d*x**3 - 30*sqrt(a + b*x)*a*b**3*c*x* 
*3 + 24*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b**3*d*x**4 - 15*sqrt(a)*lo 
g(sqrt(a + b*x) - sqrt(a))*b**4*c*x**4 - 24*sqrt(a)*log(sqrt(a + b*x) + sq 
rt(a))*a*b**3*d*x**4 + 15*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**4*c*x**4 
)/(384*a**4*x**4)