\(\int \frac {(a x^2+b x^3)^{5/2}}{x^{11}} \, dx\) [298]

Optimal result

Integrand size = 19, antiderivative size = 168 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=-\frac {a^2 \sqrt {a x^2+b x^3}}{5 x^6}-\frac {21 a b \sqrt {a x^2+b x^3}}{40 x^5}-\frac {31 b^2 \sqrt {a x^2+b x^3}}{80 x^4}-\frac {b^3 \sqrt {a x^2+b x^3}}{64 a x^3}+\frac {3 b^4 \sqrt {a x^2+b x^3}}{128 a^2 x^2}-\frac {3 b^5 \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{128 a^{5/2}} \] Output:

-1/5*a^2*(b*x^3+a*x^2)^(1/2)/x^6-21/40*a*b*(b*x^3+a*x^2)^(1/2)/x^5-31/80*b 
^2*(b*x^3+a*x^2)^(1/2)/x^4-1/64*b^3*(b*x^3+a*x^2)^(1/2)/a/x^3+3/128*b^4*(b 
*x^3+a*x^2)^(1/2)/a^2/x^2-3/128*b^5*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x) 
/a^(5/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (128 a^4+336 a^3 b x+248 a^2 b^2 x^2+10 a b^3 x^3-15 b^4 x^4\right )+15 b^5 x^5 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{640 a^{5/2} x^6 \sqrt {a+b x}} \] Input:

Integrate[(a*x^2 + b*x^3)^(5/2)/x^11,x]
 

Output:

-1/640*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(128*a^4 + 336*a^3*b*x 
+ 248*a^2*b^2*x^2 + 10*a*b^3*x^3 - 15*b^4*x^4) + 15*b^5*x^5*ArcTanh[Sqrt[a 
 + b*x]/Sqrt[a]]))/(a^(5/2)*x^6*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1926, 1926, 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.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {1}{2} b \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^8}dx-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \int \frac {\sqrt {b x^3+a x^2}}{x^5}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} b \left (\frac {3}{8} b \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )-\frac {\sqrt {a x^2+b x^3}}{3 x^4}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{4 x^7}\right )-\frac {\left (a x^2+b x^3\right )^{5/2}}{5 x^{10}}\)

Input:

Int[(a*x^2 + b*x^3)^(5/2)/x^11,x]
 

Output:

-1/5*(a*x^2 + b*x^3)^(5/2)/x^10 + (b*(-1/4*(a*x^2 + b*x^3)^(3/2)/x^7 + (3* 
b*(-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))/2
 

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]
 

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {\left (-15 b^{4} x^{4}+10 a \,b^{3} x^{3}+248 a^{2} b^{2} x^{2}+336 a^{3} b x +128 a^{4}\right ) \sqrt {x^{2} \left (b x +a \right )}}{640 x^{6} a^{2}}-\frac {3 b^{5} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{128 a^{\frac {5}{2}} x \sqrt {b x +a}}\)	\(103\)
default	\(\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} \left (15 \left (b x +a \right )^{\frac {9}{2}} a^{\frac {5}{2}}-70 \left (b x +a \right )^{\frac {7}{2}} a^{\frac {7}{2}}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} x^{5} b^{5}-128 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {9}{2}}+70 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {11}{2}}-15 \sqrt {b x +a}\, a^{\frac {13}{2}}\right )}{640 x^{10} \left (b x +a \right )^{\frac {5}{2}} a^{\frac {9}{2}}}\)	\(113\)
pseudoelliptic	\(\frac {\frac {143 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{10} x^{10}}{131072}+\frac {13 \sqrt {b x +a}\, \left (-\frac {693 \sqrt {a}\, b^{9} x^{9}}{256}+\frac {231 a^{\frac {3}{2}} b^{8} x^{8}}{128}-\frac {231 a^{\frac {5}{2}} b^{7} x^{7}}{160}+\frac {99 a^{\frac {7}{2}} b^{6} x^{6}}{80}-\frac {11 a^{\frac {9}{2}} b^{5} x^{5}}{10}+a^{\frac {11}{2}} b^{4} x^{4}-\frac {12 a^{\frac {13}{2}} b^{3} x^{3}}{13}-\frac {21448 a^{\frac {15}{2}} b^{2} x^{2}}{65}-\frac {36736 a^{\frac {17}{2}} b x}{65}-\frac {16128 a^{\frac {19}{2}}}{65}\right )}{32256}}{a^{\frac {15}{2}} x^{10}}\)	\(138\)

Input:

int((b*x^3+a*x^2)^(5/2)/x^11,x,method=_RETURNVERBOSE)
 

Output:

-1/640*(-15*b^4*x^4+10*a*b^3*x^3+248*a^2*b^2*x^2+336*a^3*b*x+128*a^4)/x^6/ 
a^2*(x^2*(b*x+a))^(1/2)-3/128*b^5/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 = 0.09 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=\left [\frac {15 \, \sqrt {a} b^{5} x^{6} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{1280 \, a^{3} x^{6}}, \frac {15 \, \sqrt {-a} b^{5} x^{6} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (15 \, a b^{4} x^{4} - 10 \, a^{2} b^{3} x^{3} - 248 \, a^{3} b^{2} x^{2} - 336 \, a^{4} b x - 128 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{640 \, a^{3} x^{6}}\right ] \] Input:

integrate((b*x^3+a*x^2)^(5/2)/x^11,x, algorithm="fricas")
 

Output:

[1/1280*(15*sqrt(a)*b^5*x^6*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqr 
t(a))/x^2) + 2*(15*a*b^4*x^4 - 10*a^2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^4* 
b*x - 128*a^5)*sqrt(b*x^3 + a*x^2))/(a^3*x^6), 1/640*(15*sqrt(-a)*b^5*x^6* 
arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (15*a*b^4*x^4 - 10*a^ 
2*b^3*x^3 - 248*a^3*b^2*x^2 - 336*a^4*b*x - 128*a^5)*sqrt(b*x^3 + a*x^2))/ 
(a^3*x^6)]
 

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {5}{2}}}{x^{11}}\, dx \] Input:

integrate((b*x**3+a*x**2)**(5/2)/x**11,x)
 

Output:

Integral((x**2*(a + b*x))**(5/2)/x**11, x)
 

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}}}{x^{11}} \,d x } \] Input:

integrate((b*x^3+a*x^2)^(5/2)/x^11,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(5/2)/x^11, x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=\frac {1}{640} \, b^{5} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{2}} + \frac {15 \, {\left (b x + a\right )}^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 70 \, {\left (b x + a\right )}^{\frac {7}{2}} a \mathrm {sgn}\left (x\right ) - 128 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} \mathrm {sgn}\left (x\right ) + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} \mathrm {sgn}\left (x\right ) - 15 \, \sqrt {b x + a} a^{4} \mathrm {sgn}\left (x\right )}{a^{2} b^{5} x^{5}}\right )} \] Input:

integrate((b*x^3+a*x^2)^(5/2)/x^11,x, algorithm="giac")
 

Output:

1/640*b^5*(15*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^2) + (15*( 
b*x + a)^(9/2)*sgn(x) - 70*(b*x + a)^(7/2)*a*sgn(x) - 128*(b*x + a)^(5/2)* 
a^2*sgn(x) + 70*(b*x + a)^(3/2)*a^3*sgn(x) - 15*sqrt(b*x + a)*a^4*sgn(x))/ 
(a^2*b^5*x^5))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{5/2}}{x^{11}} \,d x \] Input:

int((a*x^2 + b*x^3)^(5/2)/x^11,x)
 

Output:

int((a*x^2 + b*x^3)^(5/2)/x^11, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a x^2+b x^3\right )^{5/2}}{x^{11}} \, dx=\frac {-256 \sqrt {b x +a}\, a^{5}-672 \sqrt {b x +a}\, a^{4} b x -496 \sqrt {b x +a}\, a^{3} b^{2} x^{2}-20 \sqrt {b x +a}\, a^{2} b^{3} x^{3}+30 \sqrt {b x +a}\, a \,b^{4} x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{5} x^{5}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{5} x^{5}}{1280 a^{3} x^{5}} \] Input:

int((b*x^3+a*x^2)^(5/2)/x^11,x)
 

Output:

( - 256*sqrt(a + b*x)*a**5 - 672*sqrt(a + b*x)*a**4*b*x - 496*sqrt(a + b*x 
)*a**3*b**2*x**2 - 20*sqrt(a + b*x)*a**2*b**3*x**3 + 30*sqrt(a + b*x)*a*b* 
*4*x**4 + 15*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**5*x**5 - 15*sqrt(a)*l 
og(sqrt(a + b*x) + sqrt(a))*b**5*x**5)/(1280*a**3*x**5)