\(\int \frac {1}{(a x^2+b x^3)^{5/2}} \, dx\) [105]

Optimal result

Integrand size = 15, antiderivative size = 194 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {11 b}{24 a^2 \left (a x^2+b x^3\right )^{3/2}}-\frac {1}{4 a x \left (a x^2+b x^3\right )^{3/2}}-\frac {33 b^2 x}{32 a^3 \left (a x^2+b x^3\right )^{3/2}}+\frac {231 b^3 x^2}{64 a^4 \left (a x^2+b x^3\right )^{3/2}}+\frac {385 b^4 x^3}{64 a^5 \left (a x^2+b x^3\right )^{3/2}}+\frac {1155 b^4 x}{64 a^6 \sqrt {a x^2+b x^3}}-\frac {1155 b^4 \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{64 a^{13/2}} \] Output:

11/24*b/a^2/(b*x^3+a*x^2)^(3/2)-1/4/a/x/(b*x^3+a*x^2)^(3/2)-33/32*b^2*x/a^ 
3/(b*x^3+a*x^2)^(3/2)+231/64*b^3*x^2/a^4/(b*x^3+a*x^2)^(3/2)+385/64*b^4*x^ 
3/a^5/(b*x^3+a*x^2)^(3/2)+1155/64*b^4*x/a^6/(b*x^3+a*x^2)^(1/2)-1155/64*b^ 
4*arctanh((b*x^3+a*x^2)^(1/2)/a^(1/2)/x)/a^(13/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {a} \left (-48 a^5+88 a^4 b x-198 a^3 b^2 x^2+693 a^2 b^3 x^3+4620 a b^4 x^4+3465 b^5 x^5\right )-3465 b^4 x^4 (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{192 a^{13/2} x \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(Sqrt[a]*(-48*a^5 + 88*a^4*b*x - 198*a^3*b^2*x^2 + 693*a^2*b^3*x^3 + 4620* 
a*b^4*x^4 + 3465*b^5*x^5) - 3465*b^4*x^4*(a + b*x)^(3/2)*ArcTanh[Sqrt[a + 
b*x]/Sqrt[a]])/(192*a^(13/2)*x*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {1912, 1929, 1931, 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.

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

\(\Big \downarrow \) 1912

\(\displaystyle \frac {11 \int \frac {1}{x^2 \left (b x^3+a x^2\right )^{3/2}}dx}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1929

\(\displaystyle \frac {11 \left (\frac {9 \int \frac {1}{x^4 \sqrt {b x^3+a x^2}}dx}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {11 \left (\frac {9 \left (-\frac {7 b \int \frac {1}{x^3 \sqrt {b x^3+a x^2}}dx}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {11 \left (\frac {9 \left (-\frac {7 b \left (-\frac {5 b \int \frac {1}{x^2 \sqrt {b x^3+a x^2}}dx}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {11 \left (\frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x \sqrt {b x^3+a x^2}}dx}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1931

\(\displaystyle \frac {11 \left (\frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {b x^3+a x^2}}dx}{2 a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {11 \left (\frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}}{a}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {11 \left (\frac {9 \left (-\frac {7 b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}-\frac {\sqrt {a x^2+b x^3}}{a x^2}\right )}{4 a}-\frac {\sqrt {a x^2+b x^3}}{2 a x^3}\right )}{6 a}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}\right )}{8 a}-\frac {\sqrt {a x^2+b x^3}}{4 a x^5}\right )}{a}+\frac {2}{a x^3 \sqrt {a x^2+b x^3}}\right )}{3 a}+\frac {2}{3 a x \left (a x^2+b x^3\right )^{3/2}}\)

Input:

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

Output:

2/(3*a*x*(a*x^2 + b*x^3)^(3/2)) + (11*(2/(a*x^3*Sqrt[a*x^2 + b*x^3]) + (9* 
(-1/4*Sqrt[a*x^2 + b*x^3]/(a*x^5) - (7*b*(-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)))/a))/(3*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 1912

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[-(a*x^j + 
 b*x^n)^(p + 1)/(a*(n - j)*(p + 1)*x^(j - 1)), x] + Simp[(n*p + n - j + 1)/ 
(a*(n - j)*(p + 1))   Int[(a*x^j + b*x^n)^(p + 1)/x^j, x], x] /; FreeQ[{a, 
b}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && LtQ[p, -1]
 

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 1929

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*(n - j 
)*(p + 1))), x] + Simp[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))   In 
t[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] & 
&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, 
 -1]
 

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 = 0.40 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.07

method	result	size
pseudoelliptic	\(-\frac {2}{3 \left (b x +a \right )^{\frac {3}{2}} b}\)	\(13\)
default	\(-\frac {x \left (b x +a \right ) \left (3465 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{4} x^{4}+198 a^{\frac {7}{2}} b^{2} x^{2}-693 a^{\frac {5}{2}} b^{3} x^{3}-4620 a^{\frac {3}{2}} b^{4} x^{4}-3465 \sqrt {a}\, b^{5} x^{5}-88 a^{\frac {9}{2}} b x +48 a^{\frac {11}{2}}\right )}{192 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{\frac {13}{2}}}\)	\(109\)
risch	\(-\frac {\left (b x +a \right ) \left (-1545 b^{3} x^{3}+518 a \,b^{2} x^{2}-184 a^{2} b x +48 a^{3}\right )}{192 a^{6} x^{3} \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{4} \left (-\frac {2310 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {1280}{\sqrt {b x +a}}+\frac {256 a}{3 \left (b x +a \right )^{\frac {3}{2}}}\right ) \sqrt {b x +a}\, x}{128 a^{6} \sqrt {x^{2} \left (b x +a \right )}}\)	\(120\)

Input:

int(1/(b*x^3+a*x^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3/(b*x+a)^(3/2)/b
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.72 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [\frac {3465 \, {\left (b^{6} x^{7} + 2 \, a b^{5} x^{6} + a^{2} b^{4} x^{5}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (3465 \, a b^{5} x^{5} + 4620 \, a^{2} b^{4} x^{4} + 693 \, a^{3} b^{3} x^{3} - 198 \, a^{4} b^{2} x^{2} + 88 \, a^{5} b x - 48 \, a^{6}\right )} \sqrt {b x^{3} + a x^{2}}}{384 \, {\left (a^{7} b^{2} x^{7} + 2 \, a^{8} b x^{6} + a^{9} x^{5}\right )}}, \frac {3465 \, {\left (b^{6} x^{7} + 2 \, a b^{5} x^{6} + a^{2} b^{4} x^{5}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (3465 \, a b^{5} x^{5} + 4620 \, a^{2} b^{4} x^{4} + 693 \, a^{3} b^{3} x^{3} - 198 \, a^{4} b^{2} x^{2} + 88 \, a^{5} b x - 48 \, a^{6}\right )} \sqrt {b x^{3} + a x^{2}}}{192 \, {\left (a^{7} b^{2} x^{7} + 2 \, a^{8} b x^{6} + a^{9} x^{5}\right )}}\right ] \] Input:

integrate(1/(b*x^3+a*x^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/384*(3465*(b^6*x^7 + 2*a*b^5*x^6 + a^2*b^4*x^5)*sqrt(a)*log((b*x^2 + 2* 
a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(3465*a*b^5*x^5 + 4620*a^2*b 
^4*x^4 + 693*a^3*b^3*x^3 - 198*a^4*b^2*x^2 + 88*a^5*b*x - 48*a^6)*sqrt(b*x 
^3 + a*x^2))/(a^7*b^2*x^7 + 2*a^8*b*x^6 + a^9*x^5), 1/192*(3465*(b^6*x^7 + 
 2*a*b^5*x^6 + a^2*b^4*x^5)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/( 
b*x^2 + a*x)) + (3465*a*b^5*x^5 + 4620*a^2*b^4*x^4 + 693*a^3*b^3*x^3 - 198 
*a^4*b^2*x^2 + 88*a^5*b*x - 48*a^6)*sqrt(b*x^3 + a*x^2))/(a^7*b^2*x^7 + 2* 
a^8*b*x^6 + a^9*x^5)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(b*x**3+a*x**2)**(5/2),x)
 

Output:

Integral((a*x**2 + b*x**3)**(-5/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(b*x^3+a*x^2)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {1155 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{64 \, \sqrt {-a} a^{6} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (15 \, {\left (b x + a\right )} b^{4} + a b^{4}\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} \mathrm {sgn}\left (x\right )} + \frac {1545 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{4} - 5153 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{4} + 5855 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{4} - 2295 \, \sqrt {b x + a} a^{3} b^{4}}{192 \, a^{6} b^{4} x^{4} \mathrm {sgn}\left (x\right )} \] Input:

integrate(1/(b*x^3+a*x^2)^(5/2),x, algorithm="giac")
 

Output:

1155/64*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^6*sgn(x)) + 2/3*(15 
*(b*x + a)*b^4 + a*b^4)/((b*x + a)^(3/2)*a^6*sgn(x)) + 1/192*(1545*(b*x + 
a)^(7/2)*b^4 - 5153*(b*x + a)^(5/2)*a*b^4 + 5855*(b*x + a)^(3/2)*a^2*b^4 - 
 2295*sqrt(b*x + a)*a^3*b^4)/(a^6*b^4*x^4*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=-\frac {2\,x\,{\left (\frac {a}{b\,x}+1\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {13}{2};\ \frac {15}{2};\ -\frac {a}{b\,x}\right )}{13\,{\left (b\,x^3+a\,x^2\right )}^{5/2}} \] Input:

int(1/(a*x^2 + b*x^3)^(5/2),x)
 

Output:

-(2*x*(a/(b*x) + 1)^(5/2)*hypergeom([5/2, 13/2], 15/2, -a/(b*x)))/(13*(a*x 
^2 + b*x^3)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {3465 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{4} x^{4}+3465 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{5} x^{5}-3465 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{4} x^{4}-3465 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{5} x^{5}-96 a^{6}+176 a^{5} b x -396 a^{4} b^{2} x^{2}+1386 a^{3} b^{3} x^{3}+9240 a^{2} b^{4} x^{4}+6930 a \,b^{5} x^{5}}{384 \sqrt {b x +a}\, a^{7} x^{4} \left (b x +a \right )} \] Input:

int(1/(b*x^3+a*x^2)^(5/2),x)
 

Output:

(3465*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**4*x**4 + 346 
5*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**5*x**5 - 3465*sqrt 
(a)*sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**4*x**4 - 3465*sqrt(a)* 
sqrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*b**5*x**5 - 96*a**6 + 176*a**5* 
b*x - 396*a**4*b**2*x**2 + 1386*a**3*b**3*x**3 + 9240*a**2*b**4*x**4 + 693 
0*a*b**5*x**5)/(384*sqrt(a + b*x)*a**7*x**4*(a + b*x))