\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{x^8} \, dx\) [256]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {(3 b c+8 a d) \sqrt {a x^2+b x^3}}{24 x^4}-\frac {b (3 b c+56 a d) \sqrt {a x^2+b x^3}}{96 a x^3}+\frac {b^2 (3 b c-8 a d) \sqrt {a x^2+b x^3}}{64 a^2 x^2}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{4 x^7}-\frac {b^3 (3 b c-8 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{64 a^{5/2}} \] Output:

-1/24*(8*a*d+3*b*c)*(b*x^3+a*x^2)^(1/2)/x^4-1/96*b*(56*a*d+3*b*c)*(b*x^3+a 
*x^2)^(1/2)/a/x^3+1/64*b^2*(-8*a*d+3*b*c)*(b*x^3+a*x^2)^(1/2)/a^2/x^2-1/4* 
c*(b*x^3+a*x^2)^(3/2)/x^7-1/64*b^3*(-8*a*d+3*b*c)*arctanh((b*x^3+a*x^2)^(1 
/2)/a^(1/2)/x)/a^(5/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.79 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (-9 b^3 c x^3+6 a b^2 x^2 (c+4 d x)+16 a^3 (3 c+4 d x)+8 a^2 b x (9 c+14 d x)\right )+3 b^3 (3 b c-8 a d) x^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{192 a^{5/2} x^5 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^8,x]
 

Output:

-1/192*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(-9*b^3*c*x^3 + 6*a*b^2 
*x^2*(c + 4*d*x) + 16*a^3*(3*c + 4*d*x) + 8*a^2*b*x*(9*c + 14*d*x)) + 3*b^ 
3*(3*b*c - 8*a*d)*x^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(5/2)*x^5*Sqrt[a 
 + b*x])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1944, 1926, 1926, 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.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{x^8} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle -\frac {(3 b c-8 a d) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^7}dx}{8 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 a x^9}\)

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(3 b c-8 a d) \left (\frac {1}{2} b \int \frac {\sqrt {b x^3+a x^2}}{x^4}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 a x^9}\)

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(3 b c-8 a d) \left (\frac {1}{2} b \left (\frac {1}{4} b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 a x^9}\)

\(\Big \downarrow \) 1931

\(\displaystyle -\frac {(3 b c-8 a d) \left (\frac {1}{2} b \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 a x^9}\)

\(\Big \downarrow \) 1914

\(\displaystyle -\frac {(3 b c-8 a d) \left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 a x^9}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\left (\frac {1}{2} b \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3}}{2 x^3}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{3 x^6}\right ) (3 b c-8 a d)}{8 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 a x^9}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^8,x]
 

Output:

-1/4*(c*(a*x^2 + b*x^3)^(5/2))/(a*x^9) - ((3*b*c - 8*a*d)*(-1/3*(a*x^2 + b 
*x^3)^(3/2)/x^6 + (b*(-1/2*Sqrt[a*x^2 + b*x^3]/x^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)) 
/2))/(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 = 0.62 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\left (24 a \,b^{2} d \,x^{3}-9 b^{3} c \,x^{3}+112 a^{2} b d \,x^{2}+6 a \,b^{2} c \,x^{2}+64 a^{3} d x +72 a^{2} b c x +48 c \,a^{3}\right ) \sqrt {x^{2} \left (b x +a \right )}}{192 x^{5} a^{2}}+\frac {\left (8 a d -3 b c \right ) b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 a^{\frac {5}{2}} x \sqrt {b x +a}}\)	\(132\)
pseudoelliptic	\(-\frac {7 \left (b^{6} x^{7} \left (a d -\frac {9 b c}{14}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-\frac {72 \left (-\frac {400 x b \left (\frac {91 d x}{75}+c \right ) a^{\frac {11}{2}}}{9}+\frac {160 \left (-\frac {7 d x}{3}-2 c \right ) a^{\frac {13}{2}}}{9}+x^{2} b^{2} \left (\frac {35 x^{3} b^{3} \left (\frac {7 d x}{3}+c \right ) a^{\frac {3}{2}}}{24}-\frac {7 x^{2} b^{2} \left (\frac {35 d x}{18}+c \right ) a^{\frac {5}{2}}}{6}+b x \left (\frac {49 d x}{27}+c \right ) a^{\frac {7}{2}}+\frac {2 \left (-7 d x -4 c \right ) a^{\frac {9}{2}}}{9}-\frac {35 \sqrt {a}\, b^{4} c \,x^{4}}{16}\right )\right ) \sqrt {b x +a}}{245}\right )}{512 a^{\frac {11}{2}} x^{7}}\)	\(152\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (24 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {7}{2}} d -9 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {5}{2}} b c +40 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {9}{2}} d +33 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {7}{2}} b c -24 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{3} b^{4} d \,x^{4}+9 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{5} c \,x^{4}-88 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {11}{2}} d +33 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}} b c +24 \sqrt {b x +a}\, a^{\frac {13}{2}} d -9 \sqrt {b x +a}\, a^{\frac {11}{2}} b c \right )}{192 b \,x^{7} \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}}}\)	\(189\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)
 

Output:

-1/192*(24*a*b^2*d*x^3-9*b^3*c*x^3+112*a^2*b*d*x^2+6*a*b^2*c*x^2+64*a^3*d* 
x+72*a^2*b*c*x+48*a^3*c)/x^5/a^2*(x^2*(b*x+a))^(1/2)+1/64*(8*a*d-3*b*c)*b^ 
3/a^(5/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*(x^2*(b*x+a))^(1/2)/x/(b*x+a)^(1/ 
2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.71 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\left [-\frac {3 \, {\left (3 \, b^{4} c - 8 \, a b^{3} d\right )} \sqrt {a} x^{5} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (48 \, a^{4} c - 3 \, {\left (3 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{3} + 2 \, {\left (3 \, a^{2} b^{2} c + 56 \, a^{3} b d\right )} x^{2} + 8 \, {\left (9 \, a^{3} b c + 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{384 \, a^{3} x^{5}}, \frac {3 \, {\left (3 \, b^{4} c - 8 \, a b^{3} d\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) - {\left (48 \, a^{4} c - 3 \, {\left (3 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{3} + 2 \, {\left (3 \, a^{2} b^{2} c + 56 \, a^{3} b d\right )} x^{2} + 8 \, {\left (9 \, a^{3} b c + 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{192 \, a^{3} x^{5}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="fricas")
 

Output:

[-1/384*(3*(3*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*(3*a*b^3*c - 8*a^2*b^2*d)*x^ 
3 + 2*(3*a^2*b^2*c + 56*a^3*b*d)*x^2 + 8*(9*a^3*b*c + 8*a^4*d)*x)*sqrt(b*x 
^3 + a*x^2))/(a^3*x^5), 1/192*(3*(3*b^4*c - 8*a*b^3*d)*sqrt(-a)*x^5*arctan 
(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) - (48*a^4*c - 3*(3*a*b^3*c - 
8*a^2*b^2*d)*x^3 + 2*(3*a^2*b^2*c + 56*a^3*b*d)*x^2 + 8*(9*a^3*b*c + 8*a^4 
*d)*x)*sqrt(b*x^3 + a*x^2))/(a^3*x^5)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{8}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/x**8,x)
 

Output:

Integral((x**2*(a + b*x))**(3/2)*(c + d*x)/x**8, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{x^{8}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(3/2)*(d*x + c)/x^8, x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\frac {\frac {3 \, {\left (3 \, b^{5} c \mathrm {sgn}\left (x\right ) - 8 \, a b^{4} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {9 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} c \mathrm {sgn}\left (x\right ) - 33 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} c \mathrm {sgn}\left (x\right ) - 33 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} c \mathrm {sgn}\left (x\right ) + 9 \, \sqrt {b x + a} a^{3} b^{5} c \mathrm {sgn}\left (x\right ) - 24 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{4} d \mathrm {sgn}\left (x\right ) - 40 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{4} d \mathrm {sgn}\left (x\right ) + 88 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{4} d \mathrm {sgn}\left (x\right ) - 24 \, \sqrt {b x + a} a^{4} b^{4} d \mathrm {sgn}\left (x\right )}{a^{2} b^{4} x^{4}}}{192 \, b} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="giac")
 

Output:

1/192*(3*(3*b^5*c*sgn(x) - 8*a*b^4*d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a) 
)/(sqrt(-a)*a^2) + (9*(b*x + a)^(7/2)*b^5*c*sgn(x) - 33*(b*x + a)^(5/2)*a* 
b^5*c*sgn(x) - 33*(b*x + a)^(3/2)*a^2*b^5*c*sgn(x) + 9*sqrt(b*x + a)*a^3*b 
^5*c*sgn(x) - 24*(b*x + a)^(7/2)*a*b^4*d*sgn(x) - 40*(b*x + a)^(5/2)*a^2*b 
^4*d*sgn(x) + 88*(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^2*b^4*x^4))/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}\,\left (c+d\,x\right )}{x^8} \,d x \] Input:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^8,x)
 

Output:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^8, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.19 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^8} \, dx=\frac {-96 \sqrt {b x +a}\, a^{4} c -128 \sqrt {b x +a}\, a^{4} d x -144 \sqrt {b x +a}\, a^{3} b c x -224 \sqrt {b x +a}\, a^{3} b d \,x^{2}-12 \sqrt {b x +a}\, a^{2} b^{2} c \,x^{2}-48 \sqrt {b x +a}\, a^{2} b^{2} d \,x^{3}+18 \sqrt {b x +a}\, a \,b^{3} c \,x^{3}-24 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} d \,x^{4}+9 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} c \,x^{4}+24 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} d \,x^{4}-9 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} c \,x^{4}}{384 a^{3} x^{4}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^8,x)
 

Output:

( - 96*sqrt(a + b*x)*a**4*c - 128*sqrt(a + b*x)*a**4*d*x - 144*sqrt(a + b* 
x)*a**3*b*c*x - 224*sqrt(a + b*x)*a**3*b*d*x**2 - 12*sqrt(a + b*x)*a**2*b* 
*2*c*x**2 - 48*sqrt(a + b*x)*a**2*b**2*d*x**3 + 18*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 + 9*sqrt(a)*l 
og(sqrt(a + b*x) - sqrt(a))*b**4*c*x**4 + 24*sqrt(a)*log(sqrt(a + b*x) + s 
qrt(a))*a*b**3*d*x**4 - 9*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**4*c*x**4 
)/(384*a**3*x**4)