\(\int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx\) [273]

Optimal result

Integrand size = 24, antiderivative size = 165 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 a^3 (b c-a d) \sqrt {a x^2+b x^3}}{b^5 x}+\frac {2 a^2 (3 b c-4 a d) \left (a x^2+b x^3\right )^{3/2}}{3 b^5 x^3}-\frac {6 a (b c-2 a d) \left (a x^2+b x^3\right )^{5/2}}{5 b^5 x^5}+\frac {2 (b c-4 a d) \left (a x^2+b x^3\right )^{7/2}}{7 b^5 x^7}+\frac {2 d \left (a x^2+b x^3\right )^{9/2}}{9 b^5 x^9} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.57 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {x^2 (a+b x)} \left (128 a^4 d+24 a^2 b^2 x (3 c+2 d x)-16 a^3 b (9 c+4 d x)+5 b^4 x^3 (9 c+7 d x)-2 a b^3 x^2 (27 c+20 d x)\right )}{315 b^5 x} \] Input:

Integrate[(x^4*(c + d*x))/Sqrt[a*x^2 + b*x^3],x]
 

Output:

(2*Sqrt[x^2*(a + b*x)]*(128*a^4*d + 24*a^2*b^2*x*(3*c + 2*d*x) - 16*a^3*b* 
(9*c + 4*d*x) + 5*b^4*x^3*(9*c + 7*d*x) - 2*a*b^3*x^2*(27*c + 20*d*x)))/(3 
15*b^5*x)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1945, 1922, 1922, 1922, 1920}

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.

\(\displaystyle \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx\)

\(\Big \downarrow \) 1945

\(\displaystyle \frac {(9 b c-8 a d) \int \frac {x^4}{\sqrt {b x^3+a x^2}}dx}{9 b}+\frac {2 d x^3 \sqrt {a x^2+b x^3}}{9 b}\)

\(\Big \downarrow \) 1922

\(\displaystyle \frac {(9 b c-8 a d) \left (\frac {2 x^2 \sqrt {a x^2+b x^3}}{7 b}-\frac {6 a \int \frac {x^3}{\sqrt {b x^3+a x^2}}dx}{7 b}\right )}{9 b}+\frac {2 d x^3 \sqrt {a x^2+b x^3}}{9 b}\)

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1922

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

\(\Big \downarrow \) 1920

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

Input:

Int[(x^4*(c + d*x))/Sqrt[a*x^2 + b*x^3],x]
 

Output:

(2*d*x^3*Sqrt[a*x^2 + b*x^3])/(9*b) + ((9*b*c - 8*a*d)*((2*x^2*Sqrt[a*x^2 
+ b*x^3])/(7*b) - (6*a*((2*x*Sqrt[a*x^2 + b*x^3])/(5*b) - (4*a*((2*Sqrt[a* 
x^2 + b*x^3])/(3*b) - (4*a*Sqrt[a*x^2 + b*x^3])/(3*b^2*x)))/(5*b)))/(7*b)) 
)/(9*b)
 

Defintions of rubi rules used

rule 1920

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] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[ 
n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])
 

rule 1922

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, j, m, n, 
p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(m + n*p + n - j + 1) 
/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c, 0])
 

rule 1945

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

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.56

method	result	size
pseudoelliptic	\(-\frac {512 \sqrt {b x +a}\, \left (-\frac {77 x^{4} \left (\frac {9 d x}{11}+c \right ) b^{5}}{256}+\frac {11 x^{3} \left (\frac {35 d x}{44}+c \right ) a \,b^{4}}{32}-\frac {33 \left (\frac {25 d x}{33}+c \right ) x^{2} a^{2} b^{3}}{80}+\frac {11 x \left (\frac {15 d x}{22}+c \right ) a^{3} b^{2}}{20}-\frac {11 \left (\frac {5 d x}{11}+c \right ) a^{4} b}{10}+a^{5} d \right )}{693 b^{6}}\)	\(92\)
trager	\(\frac {2 \left (35 d \,x^{4} b^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +128 a^{4} d -144 a^{3} b c \right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{5} x}\)	\(104\)
risch	\(\frac {2 \left (b x +a \right ) x \left (35 d \,x^{4} b^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +128 a^{4} d -144 a^{3} b c \right )}{315 \sqrt {x^{2} \left (b x +a \right )}\, b^{5}}\)	\(105\)
gosper	\(\frac {2 \left (b x +a \right ) \left (35 d \,x^{4} b^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +128 a^{4} d -144 a^{3} b c \right ) x}{315 b^{5} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(107\)
default	\(\frac {2 \left (b x +a \right ) \left (35 d \,x^{4} b^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +128 a^{4} d -144 a^{3} b c \right ) x}{315 b^{5} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(107\)
orering	\(\frac {2 \left (b x +a \right ) \left (35 d \,x^{4} b^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +128 a^{4} d -144 a^{3} b c \right ) x}{315 b^{5} \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(107\)

Input:

int(x^4*(d*x+c)/(b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-512/693*(b*x+a)^(1/2)*(-77/256*x^4*(9/11*d*x+c)*b^5+11/32*x^3*(35/44*d*x+ 
c)*a*b^4-33/80*(25/33*d*x+c)*x^2*a^2*b^3+11/20*x*(15/22*d*x+c)*a^3*b^2-11/ 
10*(5/11*d*x+c)*a^4*b+a^5*d)/b^6
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.64 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (35 \, b^{4} d x^{4} - 144 \, a^{3} b c + 128 \, a^{4} d + 5 \, {\left (9 \, b^{4} c - 8 \, a b^{3} d\right )} x^{3} - 6 \, {\left (9 \, a b^{3} c - 8 \, a^{2} b^{2} d\right )} x^{2} + 8 \, {\left (9 \, a^{2} b^{2} c - 8 \, a^{3} b d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{315 \, b^{5} x} \] Input:

integrate(x^4*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
 

Output:

2/315*(35*b^4*d*x^4 - 144*a^3*b*c + 128*a^4*d + 5*(9*b^4*c - 8*a*b^3*d)*x^ 
3 - 6*(9*a*b^3*c - 8*a^2*b^2*d)*x^2 + 8*(9*a^2*b^2*c - 8*a^3*b*d)*x)*sqrt( 
b*x^3 + a*x^2)/(b^5*x)
 

Sympy [F]

\[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {x^{4} \left (c + d x\right )}{\sqrt {x^{2} \left (a + b x\right )}}\, dx \] Input:

integrate(x**4*(d*x+c)/(b*x**3+a*x**2)**(1/2),x)
 

Output:

Integral(x**4*(c + d*x)/sqrt(x**2*(a + b*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.73 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (5 \, b^{4} x^{4} - a b^{3} x^{3} + 2 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x - 16 \, a^{4}\right )} c}{35 \, \sqrt {b x + a} b^{4}} + \frac {2 \, {\left (35 \, b^{5} x^{5} - 5 \, a b^{4} x^{4} + 8 \, a^{2} b^{3} x^{3} - 16 \, a^{3} b^{2} x^{2} + 64 \, a^{4} b x + 128 \, a^{5}\right )} d}{315 \, \sqrt {b x + a} b^{5}} \] Input:

integrate(x^4*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
 

Output:

2/35*(5*b^4*x^4 - a*b^3*x^3 + 2*a^2*b^2*x^2 - 8*a^3*b*x - 16*a^4)*c/(sqrt( 
b*x + a)*b^4) + 2/315*(35*b^5*x^5 - 5*a*b^4*x^4 + 8*a^2*b^3*x^3 - 16*a^3*b 
^2*x^2 + 64*a^4*b*x + 128*a^5)*d/(sqrt(b*x + a)*b^5)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.87 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, {\left (\frac {9 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} c}{b^{3}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} d}{b^{4}}\right )}}{315 \, b \mathrm {sgn}\left (x\right )} + \frac {32 \, {\left (9 \, a^{\frac {7}{2}} b c - 8 \, a^{\frac {9}{2}} d\right )} \mathrm {sgn}\left (x\right )}{315 \, b^{5}} \] Input:

integrate(x^4*(d*x+c)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")
 

Output:

2/315*(9*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^ 
2 - 35*sqrt(b*x + a)*a^3)*c/b^3 + (35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2 
)*a + 378*(b*x + a)^(5/2)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a 
)*a^4)*d/b^4)/(b*sgn(x)) + 32/315*(9*a^(7/2)*b*c - 8*a^(9/2)*d)*sgn(x)/b^5
 

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.63 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {\sqrt {b\,x^3+a\,x^2}\,\left (\frac {256\,a^4\,d-288\,a^3\,b\,c}{315\,b^5}+\frac {2\,d\,x^4}{9\,b}+\frac {x^3\,\left (90\,b^4\,c-80\,a\,b^3\,d\right )}{315\,b^5}+\frac {4\,a\,x^2\,\left (8\,a\,d-9\,b\,c\right )}{105\,b^3}-\frac {16\,a^2\,x\,\left (8\,a\,d-9\,b\,c\right )}{315\,b^4}\right )}{x} \] Input:

int((x^4*(c + d*x))/(a*x^2 + b*x^3)^(1/2),x)
 

Output:

((a*x^2 + b*x^3)^(1/2)*((256*a^4*d - 288*a^3*b*c)/(315*b^5) + (2*d*x^4)/(9 
*b) + (x^3*(90*b^4*c - 80*a*b^3*d))/(315*b^5) + (4*a*x^2*(8*a*d - 9*b*c))/ 
(105*b^3) - (16*a^2*x*(8*a*d - 9*b*c))/(315*b^4)))/x
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.56 \[ \int \frac {x^4 (c+d x)}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {b x +a}\, \left (35 b^{4} d \,x^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +128 a^{4} d -144 a^{3} b c \right )}{315 b^{5}} \] Input:

int(x^4*(d*x+c)/(b*x^3+a*x^2)^(1/2),x)
 

Output:

(2*sqrt(a + b*x)*(128*a**4*d - 144*a**3*b*c - 64*a**3*b*d*x + 72*a**2*b**2 
*c*x + 48*a**2*b**2*d*x**2 - 54*a*b**3*c*x**2 - 40*a*b**3*d*x**3 + 45*b**4 
*c*x**3 + 35*b**4*d*x**4))/(315*b**5)