\(\int \frac {(c+d x) (a x^2+b x^3)^{5/2}}{x^{10}} \, dx\) [270]

Optimal result

Integrand size = 24, antiderivative size = 169 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=-\frac {b (5 b c+24 a d) \sqrt {a x^2+b x^3}}{32 x^3}-\frac {b^2 (5 b c+88 a d) \sqrt {a x^2+b x^3}}{64 a x^2}-\frac {(5 b c+8 a d) \left (a x^2+b x^3\right )^{3/2}}{24 x^6}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{4 x^9}+\frac {5 b^3 (b c-8 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{64 a^{3/2}} \] Output:

-1/32*b*(24*a*d+5*b*c)*(b*x^3+a*x^2)^(1/2)/x^3-1/64*b^2*(88*a*d+5*b*c)*(b* 
x^3+a*x^2)^(1/2)/a/x^2-1/24*(8*a*d+5*b*c)*(b*x^3+a*x^2)^(3/2)/x^6-1/4*c*(b 
*x^3+a*x^2)^(5/2)/x^9+5/64*b^3*(-8*a*d+b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a^ 
(1/2)/x)/a^(3/2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=\frac {\sqrt {x^2 (a+b x)} \left (-\sqrt {a} \sqrt {a+b x} \left (15 b^3 c x^3+16 a^3 (3 c+4 d x)+8 a^2 b x (17 c+26 d x)+2 a b^2 x^2 (59 c+132 d x)\right )+15 b^3 (b c-8 a d) x^4 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{192 a^{3/2} x^5 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^10,x]
 

Output:

(Sqrt[x^2*(a + b*x)]*(-(Sqrt[a]*Sqrt[a + b*x]*(15*b^3*c*x^3 + 16*a^3*(3*c 
+ 4*d*x) + 8*a^2*b*x*(17*c + 26*d*x) + 2*a*b^2*x^2*(59*c + 132*d*x))) + 15 
*b^3*(b*c - 8*a*d)*x^4*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(192*a^(3/2)*x^5*S 
qrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.89, 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, 1926, 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.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2} (c+d x)}{x^{10}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle -\frac {(b c-8 a d) \int \frac {\left (b x^3+a x^2\right )^{5/2}}{x^9}dx}{8 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{4 a x^{11}}\)

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(b c-8 a d) \left (\frac {5}{6} b \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^6}dx-\frac {\left (a x^2+b x^3\right )^{5/2}}{3 x^8}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{4 a x^{11}}\)

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(b c-8 a d) \left (\frac {5}{6} b \left (\frac {3}{4} b \int \frac {\sqrt {b x^3+a x^2}}{x^3}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{3 x^8}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{4 a x^{11}}\)

\(\Big \downarrow \) 1926

\(\displaystyle -\frac {(b c-8 a d) \left (\frac {5}{6} b \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{\sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{3 x^8}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{4 a x^{11}}\)

\(\Big \downarrow \) 1914

\(\displaystyle -\frac {(b c-8 a d) \left (\frac {5}{6} b \left (\frac {3}{4} b \left (-b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{3 x^8}\right )}{8 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{4 a x^{11}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\left (\frac {5}{6} b \left (\frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}}-\frac {\sqrt {a x^2+b x^3}}{x^2}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{3 x^8}\right ) (b c-8 a d)}{8 a}-\frac {c \left (a x^2+b x^3\right )^{7/2}}{4 a x^{11}}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(5/2))/x^10,x]
 

Output:

-1/4*(c*(a*x^2 + b*x^3)^(7/2))/(a*x^11) - ((b*c - 8*a*d)*(-1/3*(a*x^2 + b* 
x^3)^(5/2)/x^8 + (5*b*(-1/2*(a*x^2 + b*x^3)^(3/2)/x^5 + (3*b*(-(Sqrt[a*x^2 
 + b*x^3]/x^2) - (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/Sqrt[a]))/4) 
)/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 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 = 0.75 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {\left (264 a \,b^{2} d \,x^{3}+15 b^{3} c \,x^{3}+208 a^{2} b d \,x^{2}+118 a \,b^{2} c \,x^{2}+64 a^{3} d x +136 a^{2} b c x +48 c \,a^{3}\right ) \sqrt {x^{2} \left (b x +a \right )}}{192 x^{5} a}-\frac {5 \left (8 a d -b c \right ) b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{64 a^{\frac {3}{2}} x \sqrt {b x +a}}\)	\(132\)
pseudoelliptic	\(\frac {\frac {45 b^{8} x^{9} \left (a d -\frac {11 b c}{18}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16384}-\frac {11 \sqrt {b x +a}\, \left (\frac {80 x^{3} b^{3} \left (\frac {9 d x}{5}+c \right ) a^{\frac {11}{2}}}{99}+\frac {6592 x^{2} \left (\frac {243 d x}{206}+c \right ) b^{2} a^{\frac {13}{2}}}{33}+\frac {33152 x \left (\frac {297 d x}{259}+c \right ) b \,a^{\frac {15}{2}}}{99}+\left (\frac {1792 d x}{11}+\frac {14336 c}{99}\right ) a^{\frac {17}{2}}+\left (\frac {35 x^{3} \left (\frac {27 d x}{11}+c \right ) b^{3} a^{\frac {3}{2}}}{24}-\frac {7 \left (\frac {45 d x}{22}+c \right ) x^{2} b^{2} a^{\frac {5}{2}}}{6}+b x \left (\frac {21 d x}{11}+c \right ) a^{\frac {7}{2}}+\left (-\frac {18 d x}{11}-\frac {8 c}{9}\right ) a^{\frac {9}{2}}-\frac {35 \sqrt {a}\, b^{4} c \,x^{4}}{16}\right ) x^{4} b^{4}\right )}{14336}}{a^{\frac {13}{2}} x^{9}}\)	\(184\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (264 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {5}{2}} d +15 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {3}{2}} b c +120 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{4} d \,x^{4}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{5} c \,x^{4}-584 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {7}{2}} d +73 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {5}{2}} b c +440 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {9}{2}} d -55 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {7}{2}} b c -120 \sqrt {b x +a}\, a^{\frac {11}{2}} d +15 \sqrt {b x +a}\, a^{\frac {9}{2}} b c \right )}{192 b \,x^{9} \left (b x +a \right )^{\frac {5}{2}} a^{\frac {5}{2}}}\)	\(187\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^10,x,method=_RETURNVERBOSE)
 

Output:

-1/192*(264*a*b^2*d*x^3+15*b^3*c*x^3+208*a^2*b*d*x^2+118*a*b^2*c*x^2+64*a^ 
3*d*x+136*a^2*b*c*x+48*a^3*c)/x^5/a*(x^2*(b*x+a))^(1/2)-5/64*(8*a*d-b*c)*b 
^3/a^(3/2)*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 = 0.17 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.73 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=\left [-\frac {15 \, {\left (b^{4} c - 8 \, a b^{3} d\right )} \sqrt {a} x^{5} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (48 \, a^{4} c + 3 \, {\left (5 \, a b^{3} c + 88 \, a^{2} b^{2} d\right )} x^{3} + 2 \, {\left (59 \, a^{2} b^{2} c + 104 \, a^{3} b d\right )} x^{2} + 8 \, {\left (17 \, a^{3} b c + 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{384 \, a^{2} x^{5}}, -\frac {15 \, {\left (b^{4} c - 8 \, a b^{3} d\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (48 \, a^{4} c + 3 \, {\left (5 \, a b^{3} c + 88 \, a^{2} b^{2} d\right )} x^{3} + 2 \, {\left (59 \, a^{2} b^{2} c + 104 \, a^{3} b d\right )} x^{2} + 8 \, {\left (17 \, a^{3} b c + 8 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{192 \, a^{2} x^{5}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^10,x, algorithm="fricas")
 

Output:

[-1/384*(15*(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 + 88*a^2*b^2*d)*x^ 
3 + 2*(59*a^2*b^2*c + 104*a^3*b*d)*x^2 + 8*(17*a^3*b*c + 8*a^4*d)*x)*sqrt( 
b*x^3 + a*x^2))/(a^2*x^5), -1/192*(15*(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 + 3*(5*a*b^3*c 
 + 88*a^2*b^2*d)*x^3 + 2*(59*a^2*b^2*c + 104*a^3*b*d)*x^2 + 8*(17*a^3*b*c 
+ 8*a^4*d)*x)*sqrt(b*x^3 + a*x^2))/(a^2*x^5)]
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}{x^{10}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(5/2)/x**10,x)
 

Output:

Integral((x**2*(a + b*x))**(5/2)*(c + d*x)/x**10, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}}{x^{10}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^10,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(5/2)*(d*x + c)/x^10, x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=-\frac {\frac {15 \, {\left (b^{5} c \mathrm {sgn}\left (x\right ) - 8 \, a b^{4} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {15 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{5} c \mathrm {sgn}\left (x\right ) + 73 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{5} c \mathrm {sgn}\left (x\right ) - 55 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{5} c \mathrm {sgn}\left (x\right ) + 15 \, \sqrt {b x + a} a^{3} b^{5} c \mathrm {sgn}\left (x\right ) + 264 \, {\left (b x + a\right )}^{\frac {7}{2}} a b^{4} d \mathrm {sgn}\left (x\right ) - 584 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} b^{4} d \mathrm {sgn}\left (x\right ) + 440 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} b^{4} d \mathrm {sgn}\left (x\right ) - 120 \, \sqrt {b x + a} a^{4} b^{4} d \mathrm {sgn}\left (x\right )}{a b^{4} x^{4}}}{192 \, b} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^10,x, algorithm="giac")
 

Output:

-1/192*(15*(b^5*c*sgn(x) - 8*a*b^4*d*sgn(x))*arctan(sqrt(b*x + a)/sqrt(-a) 
)/(sqrt(-a)*a) + (15*(b*x + a)^(7/2)*b^5*c*sgn(x) + 73*(b*x + a)^(5/2)*a*b 
^5*c*sgn(x) - 55*(b*x + a)^(3/2)*a^2*b^5*c*sgn(x) + 15*sqrt(b*x + a)*a^3*b 
^5*c*sgn(x) + 264*(b*x + a)^(7/2)*a*b^4*d*sgn(x) - 584*(b*x + a)^(5/2)*a^2 
*b^4*d*sgn(x) + 440*(b*x + a)^(3/2)*a^3*b^4*d*sgn(x) - 120*sqrt(b*x + a)*a 
^4*b^4*d*sgn(x))/(a*b^4*x^4))/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}\,\left (c+d\,x\right )}{x^{10}} \,d x \] Input:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^10,x)
 

Output:

int(((a*x^2 + b*x^3)^(5/2)*(c + d*x))/x^10, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{5/2}}{x^{10}} \, dx=\frac {-96 \sqrt {b x +a}\, a^{4} c -128 \sqrt {b x +a}\, a^{4} d x -272 \sqrt {b x +a}\, a^{3} b c x -416 \sqrt {b x +a}\, a^{3} b d \,x^{2}-236 \sqrt {b x +a}\, a^{2} b^{2} c \,x^{2}-528 \sqrt {b x +a}\, a^{2} b^{2} d \,x^{3}-30 \sqrt {b x +a}\, a \,b^{3} c \,x^{3}+120 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{3} d \,x^{4}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} c \,x^{4}-120 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{3} d \,x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} c \,x^{4}}{384 a^{2} x^{4}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(5/2)/x^10,x)
 

Output:

( - 96*sqrt(a + b*x)*a**4*c - 128*sqrt(a + b*x)*a**4*d*x - 272*sqrt(a + b* 
x)*a**3*b*c*x - 416*sqrt(a + b*x)*a**3*b*d*x**2 - 236*sqrt(a + b*x)*a**2*b 
**2*c*x**2 - 528*sqrt(a + b*x)*a**2*b**2*d*x**3 - 30*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 - 15*sqrt( 
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 + 15*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**4* 
c*x**4)/(384*a**2*x**4)