\(\int \frac {x^2 (c+d x)}{(a x^2+b x^3)^{5/2}} \, dx\) [297]

Optimal result

Integrand size = 24, antiderivative size = 170 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {2 b (b c-a d) x^3}{3 a^3 \left (a x^2+b x^3\right )^{3/2}}+\frac {2 b (3 b c-2 a d) x}{a^4 \sqrt {a x^2+b x^3}}-\frac {c \sqrt {a x^2+b x^3}}{2 a^3 x^3}+\frac {(11 b c-4 a d) \sqrt {a x^2+b x^3}}{4 a^4 x^2}-\frac {5 b (7 b c-4 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{4 a^{9/2}} \] Output:

2/3*b*(-a*d+b*c)*x^3/a^3/(b*x^3+a*x^2)^(3/2)+2*b*(-2*a*d+3*b*c)*x/a^4/(b*x 
^3+a*x^2)^(1/2)-1/2*c*(b*x^3+a*x^2)^(1/2)/a^3/x^3+1/4*(-4*a*d+11*b*c)*(b*x 
^3+a*x^2)^(1/2)/a^4/x^2-5/4*b*(-4*a*d+7*b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a 
^(1/2)/x)/a^(9/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {x \left (\sqrt {a} \left (105 b^3 c x^3+a^2 b x (21 c-80 d x)+20 a b^2 x^2 (7 c-3 d x)-6 a^3 (c+2 d x)\right )+15 b (-7 b c+4 a d) x^2 (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{12 a^{9/2} \left (x^2 (a+b x)\right )^{3/2}} \] Input:

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

Output:

(x*(Sqrt[a]*(105*b^3*c*x^3 + a^2*b*x*(21*c - 80*d*x) + 20*a*b^2*x^2*(7*c - 
 3*d*x) - 6*a^3*(c + 2*d*x)) + 15*b*(-7*b*c + 4*a*d)*x^2*(a + b*x)^(3/2)*A 
rcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(12*a^(9/2)*(x^2*(a + b*x))^(3/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1943, 1912, 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.

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

\(\Big \downarrow \) 1943

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

\(\Big \downarrow \) 1912

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1931

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

\(\Big \downarrow \) 1914

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*(b*c - a*d)*x)/(3*a*b*(a*x^2 + b*x^3)^(3/2)) + ((7*b*c - 4*a*d)*(2/(a*x 
*Sqrt[a*x^2 + b*x^3]) + (5*(-1/2*Sqrt[a*x^2 + b*x^3]/(a*x^3) - (3*b*(-(Sqr 
t[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)))/a))/(3*a*b)
 

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 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 1943

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[(-e^(j - 1))*(b*c - a*d)*(e*x)^(m - j 
+ 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*b*n*(p + 1))), x] - Simp[e^j*((a*d*( 
m + j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(a*b*n*(p + 1)))   Int[(e*x)^(m 
 - j)*(a*x^j + b*x^(j + n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, j, m, 
n}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && LtQ[p, -1 
] && GtQ[j, 0] && LeQ[j, m] && (GtQ[e, 0] || IntegerQ[j])
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.34

method	result	size
pseudoelliptic	\(-\frac {32 \left (-\frac {3 x^{2} \left (\frac {d x}{3}+c \right ) b^{3}}{16}-\frac {3 \left (-\frac {d x}{2}+c \right ) x a \,b^{2}}{4}-\frac {a^{2} \left (-3 d x +c \right ) b}{2}+a^{3} d \right )}{3 \left (b x +a \right )^{\frac {3}{2}} b^{4}}\)	\(58\)
risch	\(-\frac {\left (b x +a \right ) \left (4 a d x -11 c b x +2 a c \right )}{4 a^{4} x \sqrt {x^{2} \left (b x +a \right )}}-\frac {b \left (-\frac {2 \left (20 a d -35 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \left (-16 a d +24 b c \right )}{\sqrt {b x +a}}+\frac {16 a \left (a d -b c \right )}{3 \left (b x +a \right )^{\frac {3}{2}}}\right ) \sqrt {b x +a}\, x}{8 a^{4} \sqrt {x^{2} \left (b x +a \right )}}\)	\(129\)
default	\(\frac {x^{3} \left (b x +a \right ) \left (60 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a b d \,x^{2}-105 \left (b x +a \right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} c \,x^{2}-12 a^{\frac {7}{2}} d x -80 a^{\frac {5}{2}} b d \,x^{2}-60 a^{\frac {3}{2}} b^{2} d \,x^{3}-6 a^{\frac {7}{2}} c +21 a^{\frac {5}{2}} b c x +140 a^{\frac {3}{2}} b^{2} c \,x^{2}+105 \sqrt {a}\, b^{3} c \,x^{3}\right )}{12 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {5}{2}} a^{\frac {9}{2}}}\)	\(150\)

Input:

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

Output:

-32/3/(b*x+a)^(3/2)*(-3/16*x^2*(1/3*d*x+c)*b^3-3/4*(-1/2*d*x+c)*x*a*b^2-1/ 
2*a^2*(-3*d*x+c)*b+a^3*d)/b^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.54 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\left [-\frac {15 \, {\left ({\left (7 \, b^{4} c - 4 \, a b^{3} d\right )} x^{5} + 2 \, {\left (7 \, a b^{3} c - 4 \, a^{2} b^{2} d\right )} x^{4} + {\left (7 \, a^{2} b^{2} c - 4 \, a^{3} b d\right )} x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (6 \, a^{4} c - 15 \, {\left (7 \, a b^{3} c - 4 \, a^{2} b^{2} d\right )} x^{3} - 20 \, {\left (7 \, a^{2} b^{2} c - 4 \, a^{3} b d\right )} x^{2} - 3 \, {\left (7 \, a^{3} b c - 4 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}, \frac {15 \, {\left ({\left (7 \, b^{4} c - 4 \, a b^{3} d\right )} x^{5} + 2 \, {\left (7 \, a b^{3} c - 4 \, a^{2} b^{2} d\right )} x^{4} + {\left (7 \, a^{2} b^{2} c - 4 \, a^{3} b d\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) - {\left (6 \, a^{4} c - 15 \, {\left (7 \, a b^{3} c - 4 \, a^{2} b^{2} d\right )} x^{3} - 20 \, {\left (7 \, a^{2} b^{2} c - 4 \, a^{3} b d\right )} x^{2} - 3 \, {\left (7 \, a^{3} b c - 4 \, a^{4} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{12 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}}\right ] \] Input:

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

Output:

[-1/24*(15*((7*b^4*c - 4*a*b^3*d)*x^5 + 2*(7*a*b^3*c - 4*a^2*b^2*d)*x^4 + 
(7*a^2*b^2*c - 4*a^3*b*d)*x^3)*sqrt(a)*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + 
 a*x^2)*sqrt(a))/x^2) + 2*(6*a^4*c - 15*(7*a*b^3*c - 4*a^2*b^2*d)*x^3 - 20 
*(7*a^2*b^2*c - 4*a^3*b*d)*x^2 - 3*(7*a^3*b*c - 4*a^4*d)*x)*sqrt(b*x^3 + a 
*x^2))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3), 1/12*(15*((7*b^4*c - 4*a*b^3 
*d)*x^5 + 2*(7*a*b^3*c - 4*a^2*b^2*d)*x^4 + (7*a^2*b^2*c - 4*a^3*b*d)*x^3) 
*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) - (6*a^4*c - 
15*(7*a*b^3*c - 4*a^2*b^2*d)*x^3 - 20*(7*a^2*b^2*c - 4*a^3*b*d)*x^2 - 3*(7 
*a^3*b*c - 4*a^4*d)*x)*sqrt(b*x^3 + a*x^2))/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a 
^7*x^3)]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.95 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {5 \, {\left (7 \, b^{2} c - 4 \, a b d\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (9 \, {\left (b x + a\right )} b^{2} c + a b^{2} c - 6 \, {\left (b x + a\right )} a b d - a^{2} b d\right )}}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} \mathrm {sgn}\left (x\right )} + \frac {11 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} c - 13 \, \sqrt {b x + a} a b^{2} c - 4 \, {\left (b x + a\right )}^{\frac {3}{2}} a b d + 4 \, \sqrt {b x + a} a^{2} b d}{4 \, a^{4} b^{2} x^{2} \mathrm {sgn}\left (x\right )} \] Input:

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

Output:

5/4*(7*b^2*c - 4*a*b*d)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^4*sgn(x 
)) + 2/3*(9*(b*x + a)*b^2*c + a*b^2*c - 6*(b*x + a)*a*b*d - a^2*b*d)/((b*x 
 + a)^(3/2)*a^4*sgn(x)) + 1/4*(11*(b*x + a)^(3/2)*b^2*c - 13*sqrt(b*x + a) 
*a*b^2*c - 4*(b*x + a)^(3/2)*a*b*d + 4*sqrt(b*x + a)*a^2*b*d)/(a^4*b^2*x^2 
*sgn(x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.88 \[ \int \frac {x^2 (c+d x)}{\left (a x^2+b x^3\right )^{5/2}} \, dx=\frac {-60 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a^{2} b d \,x^{2}+105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} c \,x^{2}-60 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} d \,x^{3}+105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} c \,x^{3}+60 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a^{2} b d \,x^{2}-105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} c \,x^{2}+60 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} d \,x^{3}-105 \sqrt {a}\, \sqrt {b x +a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} c \,x^{3}-12 a^{4} c -24 a^{4} d x +42 a^{3} b c x -160 a^{3} b d \,x^{2}+280 a^{2} b^{2} c \,x^{2}-120 a^{2} b^{2} d \,x^{3}+210 a \,b^{3} c \,x^{3}}{24 \sqrt {b x +a}\, a^{5} x^{2} \left (b x +a \right )} \] Input:

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

Output:

( - 60*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a**2*b*d*x**2 + 
105*sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*c*x**2 - 60* 
sqrt(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*d*x**3 + 105*sqr 
t(a)*sqrt(a + b*x)*log(sqrt(a + b*x) - sqrt(a))*b**3*c*x**3 + 60*sqrt(a)*s 
qrt(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a**2*b*d*x**2 - 105*sqrt(a)*sqrt 
(a + b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*c*x**2 + 60*sqrt(a)*sqrt(a + 
 b*x)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*d*x**3 - 105*sqrt(a)*sqrt(a + b* 
x)*log(sqrt(a + b*x) + sqrt(a))*b**3*c*x**3 - 12*a**4*c - 24*a**4*d*x + 42 
*a**3*b*c*x - 160*a**3*b*d*x**2 + 280*a**2*b**2*c*x**2 - 120*a**2*b**2*d*x 
**3 + 210*a*b**3*c*x**3)/(24*sqrt(a + b*x)*a**5*x**2*(a + b*x))